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Answer: 6106
Solution: The equality $\frac{1}{x}+y+z=x+\frac{1}{y}+z$ implies $\frac{1}{x}+y=x+\frac{1}{y}$, so $x y=-1$ or $x=y$. Similarly, $y z=-1$ or $y=z$, and $z x=-1$ or $z=x$.
If no two elements multiply to -1 , then $x=y=z$. which implies $2 x+\frac{1}{x}=3$ and so $(x, y, z) \in$ $\left\{(1,1,1),\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\right\}$. Otherwise, we may assume $x y=-1$, which implies $z=\stackrel{x}{3}$ and $x+y=\frac{8}{3}$, whence $\{x, y, z\}=\left\{-\frac{1}{3}, 3,3\right\}$.
The final answer is $(1+1+1)+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+3+3\right)=\frac{61}{6}$. | \boxed{6106} | 18. [10] Let $x, y, z$ be real numbers satisfying
$$
\frac{1}{x}+y+z=x+\frac{1}{y}+z=x+y+\frac{1}{z}=3 .
$$
The sum of all possible values of $x+y+z$ can be written as $\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$. Find $100 m+n$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
4. Let $m=n-1, a_{i}=b_{i}-1$, then $1 \leqslant a_{0} \leqslant 2 m$, and
$$a_{i+1}=\left\{\begin{array}{ll}
2 a_{i}, & a_{i} \leqslant m, \\
2 a_{i}-(2 m+1), & a_{i}>m .
\end{array}\right.$$
This indicates: $a_{i+1} \equiv 2 a_{i}(\bmod 2 m+1)$, and for $i \in \mathbf{N}$, we have $1 \leqslant a_{i} \leqslant 2 m$.
(1) The required values are equivalent to finding $p\left(1,2^{k}-1\right)$ and $p\left(1,2^{k}\right)$ for $\left\{a_{i}\right\}$. The former is equivalent to finding the smallest $l \in \mathbf{N}^{*}$ such that
$$2^{l} \equiv 1\left(\bmod 2\left(2^{k}-1\right)+1\right) ;$$
The latter is equivalent to finding the smallest $t \in \mathbf{N}^{*}$ such that
$$2^{t} \equiv 1\left(\bmod 2^{k+1}+1\right)$$
Since $2\left(2^{k}-1\right)+1=2^{k+1}-1$, and for $1 \leqslant l \leqslant k$, it is clear that $2^{l} \neq 1\left(\bmod 2^{k+1}-1\right)$, hence $p\left(1,2^{k}-1\right)=k+1$. Also, $2^{2(k+1)} \equiv 1\left(\bmod 2^{k+1}+1\right)$, so $\delta_{2^{k+1}+1}(2) \mid 2(k+1)$. But for $1 \leqslant t \leqslant k+1$, we have $2^{t} \neq 1\left(\bmod 2^{k+1}+1\right)$, thus $p\left(1,2^{k}\right)=\delta_{2^{k+1}+1}(2)=2(k+1)$.
Therefore, for $\left\{b_{i}\right\}$, we have $p\left(2,2^{k}\right)=k+1, p\left(2,2^{k}+1\right)=2(k+1)$.
(2) We still discuss in terms of $\left\{a_{i}\right\}$, and need to prove: $p\left(a_{0}, m\right) \mid p(1, m)$.
First, let $p(1, m)=t$, then $2^{t} \equiv 1(\bmod 2 m+1)$, and thus $2^{t} a_{0} \equiv a_{0}(\bmod 2 m+1)$, so $p\left(a_{0}, m\right) \leqslant p(1, m)$.
Second, if $p\left(a_{0}, m\right) \times p(1, m)$, then we can set
$$p(1, m)=p\left(a_{0}, m\right) q+r, 0<r<p\left(a_{0}, m\right)$$
Let $s=p\left(a_{0}, m\right), t=p(1, m)$, then combining $2^{s} a_{0} \equiv a_{0}(\bmod 2 m+1)$, we know
$$a_{0} \equiv 2^{t} a_{0}=2^{n+r} a_{0} \equiv 2^{s(q-1)+r} a_{0} \equiv \cdots \equiv 2^{r} a_{0}(\bmod 2 m+1),$$
which contradicts the minimality of $s$.
Therefore, $p\left(a_{0}, m\right) \mid p(1, m)$, and the proposition is proved. | \boxed{p(2, 2^k) = k + 1 \text{ and } p(2, 2^k + 1) = 2(k + 1)} | 4. Let $n, b_{0} \in \mathbf{N}^{*}, n \geqslant 2, 2 \leqslant b_{0} \leqslant 2 n-1$. The sequence $\left\{b_{i}\right\}$ is defined as follows:
$$b_{i+1}=\left\{\begin{array}{ll}
2 b_{i}-1, & b_{i} \leqslant n, \\
2 b_{i}-2 n, & b_{i}>n,
\end{array} \quad i=0,1,2, \cdots\right.$$
Let $p\left(b_{0}, n\right)$ denote the smallest index $p$ such that $b_{p}=b_{0}$.
(1) For $k \in \mathbf{N}^{*}$, find the values of $p\left(2,2^{k}\right)$ and $p\left(2,2^{k}+1\right)$;
(2) Prove that for any $n$ and $b_{0}$, $p\left(b_{0}, n\right) \mid p(2, n)$. | Number Theory | AI-MO/NuminaMath-1.5/number_theory | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
27.7. Let's find the number of sequences ( $i_{1} ; \ldots ; i_{n}$ ), for which $i_{k} \geqslant k-3$ for $k=1,2, \ldots, n$. The number $i_{n}$ can take four values: $n, n-1, n-2, n-3$. The number $i_{n-1}$ can take five values: $f, n-1, n-2, n-3, n-4$, except for the value already occupied by the number $i_{n}$. Thus, the number $i_{n-1}$ can also take 4 values. Similarly, each of the numbers $i_{n-2}, \ldots, i_{4}$ can take 4 values. The numbers $i_{1}, i_{2}, i_{3}$ can be chosen arbitrarily from the three values remaining after selecting the numbers $i_{n}, \ldots, i_{4}$. Thus, among all $n$! possible sequences, there are exactly $4^{n-3} \cdot 3$! sequences that satisfy the required condition. Therefore, the desired probability is " 4 " $n$ - $3.3!/ n!$. | \boxed{\dfrac{4^{n-3} \cdot 6}{n!}} | 27.7. (New York, 76). The sequence ( $i_{1} ; i_{2} ; \ldots ; i_{n}$ ) is formed from the first $n>3$ natural numbers, arranged in a random order. What is the probability that for all $k=1,2, \ldots, n$ the inequality $i_{k} \geqslant k-3$ holds? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Extend the segment $L K$ beyond point $K$ to point $B$ such that $K B=K N$. Then $L B=b$ and triangles $B K N$ and $A M N$ are isosceles.
Let the angles of triangle $K L M$ be $2 \alpha, 2 \beta$, and $2 \gamma$ respectively. Then $\angle K L N=\angle N L M=\beta, \angle K B N=$ $\angle K N B=\alpha$,
$\angle L N A=\angle K N A-\angle K N L=90^{\circ}+\gamma-(\beta+2 \gamma)=90^{\circ}-\beta-\gamma=\alpha$. From this, the similarity of triangles $A L N$ and $N L B$ follows. Therefore, $A L: L N=L N: L B$, from which
$A L=L N^{2} / L B=a^{2} / b$.
## Answer
$a^{2} / b$. | \boxed{\dfrac{a^2}{b}} | [ The sum of the angles of a triangle. The theorem about the exterior angle. ]
In triangle $K L M$, all sides of which are different, the bisector of angle $K L M$ intersects side $K M$ at point $N$. A line through point $N$ intersects side $L M$ at point $A$, for which $M N=A M$. It is known that $L N=a, K L+K N=b$. Find $A L$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\begin{tabular}{c}
19 \\
\hline 6
\end{tabular} | \boxed{6} | 19. Given the lengths of the three sides of a triangle are $n+8, 2n+5, 5n-4$, if the three sides of the triangle are all unequal, and $n$ is a positive integer, then the possible values of $n$ are $\qquad$ in number. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Prove that $ABCD$ is a cyclic quadrilateral with mutually perpendicular diagonals.
## Solution
Let $C_{1}$ be the point symmetric to vertex $C$ with respect to the perpendicular bisector of diagonal $BD$. Then
$$
\begin{gathered}
S_{\triangle \mathrm{ABCD}}=S_{\Delta \mathrm{ABC}_{1} \mathrm{D}}=S_{\Delta \mathrm{ABC}_{1}}+S_{\Delta \mathrm{AC}_{1} \mathrm{D}}= \\
=\frac{1}{2} \cdot A B \cdot B C_{1} \sin \angle A B C_{1}+\frac{1}{2} \cdot A D \cdot D C_{1} \sin \angle A D C_{1} \leqslant \\
\leqslant \frac{1}{2}\left(A B \cdot B C_{1}+A D \cdot D C_{1}\right)=\frac{1}{2}(A B \cdot D C+A D \cdot B C)
\end{gathered}
$$
with equality only when
$$
\angle A B C_{1}=\angle A D C_{1}=90^{\circ}
$$
Therefore, quadrilateral $A B C_{1} D$ is cyclic and $A C_{1}$ is the diameter of its circumscribed circle.
The perpendicular bisector of diagonal $BD$ is the axis of symmetry of this circle. Therefore, vertex $C$ also lies on the circle. Hence, quadrilateral $A B C D$ is cyclic. Since $A C_{1}$ is the diameter of its circumscribed circle, $\angle A C C_{1}=90^{\circ}$. Therefore, diagonal $A C$ is parallel to the perpendicular bisector of diagonal $B D$. Consequently, diagonals $A C$ and $B D$ of quadrilateral $A B C D$ are mutually perpendicular. Then in triangle $COD$:
$$
O D=4, \angle C O D=90^{\circ}, \angle O C D=\angle A C D=\angle A B D=45^{\circ}
$$
Therefore, $O C=O D=4$. By the Pythagorean theorem in right triangle $COB$, we find that
$$
O B=\sqrt{B C^{2}-O C^{2}}=\sqrt{5^{2}-4^{2}}=3
$$
## Answer
3. | 3 | In quadrilateral $A B C D$, it is known that $D O=4, B C=5, \angle A B D=45^{\circ}$, where $O$ is the point of intersection of the diagonals. Find $B O$, if the area of quadrilateral $A B C D$ is equal to $\frac{1}{2}(A B \cdot C D+B C \cdot A D)$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. Let's choose the amount in rubles that will correspond to 1 euro (i.e., 100 European cents) at the end of the year. At the beginning of the year, the same amount corresponded to 108 European cents or $108 \cdot 100 / 80=135$ American cents. Consequently, at the end of the year, it will correspond to $135 \cdot 100 / 90=\mathbf{1 5 0}$ American cents.
Criteria. Arithmetic errors with the correct approach - 2 points are deducted.
If there is at least one "logical" error (the wrong percentage is taken, the wrong ratio is considered, etc.), the problem is considered unsolved. If the solution is completed and everything is correct except for one such error - 2 points can be given; in all other cases, 0. | \boxed{150} | 4. At the beginning of the year, the US dollar was worth 80 European cents. An expert predicted that over the course of the year, the euro's exchange rate against the ruble would increase by 8% (i.e., 1 euro would be able to buy 8% more rubles than at the beginning of the year), and the US dollar's exchange rate against the ruble would fall by 10%. If the forecast comes true, how many US cents will the euro be worth at the end of the year? | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$\angle A B D=\angle C B E=\beta$
## Solution
Since
$$
\angle A B D=\angle A B C-\angle D B C, \angle C B E=\angle D B E-\angle D B C, \angle A B C=\angle D B E \text {, }
$$
then $\angle A B D=\angle C B E$. Points $A, D, C, E$ lie on a circle with center at point $B$ and radius equal to $A B$. Therefore, $\angle D A K=\angle C E K$.
Segment $B K$ is seen from points $A$ and $D$ at the same angle ( $\angle B A K=\angle B D K$ ). Therefore, points $A, D, K$ and $B$ lie on the same circle. Hence, $\angle D B K=\angle D A K$. Similarly, we can prove that $\angle C B K=\angle C E K$. Therefore,
$$
\angle A B K=\angle A B D+\angle D B K=\beta+\frac{\alpha-\beta}{2}=\frac{\alpha+\beta}{2}, \angle K B C=\frac{1}{2} \angle D B C=\frac{\alpha-\beta}{2}
$$
Thus,
$$
\frac{\angle A B K}{\angle K B C}=\frac{\alpha+\beta}{\alpha-\beta}
$$
Answer
$\frac{\alpha+\beta}{\alpha-\beta}$. | \boxed{\dfrac{\alpha + \beta}{\alpha - \beta}} | [ Segment visible from two points at the same angle ]
Two equal isosceles triangles $A B C$ and $D B E$ ( $A B=B C=D B=B E$ ) share a common vertex $B$ and lie in the same plane, with points $A$ and $C$ located on opposite sides of line $B D$, and segments $A C$ and $D E$ intersecting at point $K$. It is known that $\angle A B C=\angle D B E=\alpha<\frac{\pi}{2}, \angle A K D=\beta<\alpha$. In what ratio does the line $B K$ divide the angle $A B C$? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution 1
Draw line segment $Q R$ parallel to $D C$, as in the following diagram. This segment divides square $A B C D$ into two halves. Since triangles $A B Q$ and $R Q B$ are congruent, each is half of rectangle $A B R Q$ and therefore one quarter of square $A B C D$. Draw line segment $P S$ parallel to $D A$, and draw line segment $P R$. Triangles $P D Q, P S Q, P S R$ and $P C R$ are congruent. Therefore each is one quarter of rectangle $D C R Q$ and therefore one eighth of square $A B C D$.
Quadrilateral $Q B C P$ therefore represents $\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{5}{8}$ of square $A B C D$. Its area is therefore $\frac{5}{8}$ of the area of the square.
Therefore, $\frac{5}{8}$ of the area of the square is equal to 15 . Therefore, $\frac{1}{8}$ of the area of the square is equal to 3 . Therefore the square has an area of 24 .
## Solution 2
Draw a line segment from $Q$ to $R$, the midpoint of $B C$.
Draw a line segment from $P$ to $S$, the midpoint of $Q R$.
Let the area of $\triangle Q S P$ equal $x$. Thus, the area of $\triangle Q D P$ is also $x$ and $Q D P S$ has area $2 x$.
Square $S P C R$ is congruent to square $Q D P S$ and thus has area $2 x$.
Rectangle $Q D C R$ has area $4 x$, as does the congruent rectangle $A Q R B$.
Also, $\triangle A Q B$ and $\triangle B R Q$ have equal areas and thus, each area is $2 x$.
Quadrilateral $Q B C P$ is made up of $\triangle B R Q, \triangle Q S P$ and square $S P C R$, and thus has area $2 x+x+2 x=5 x$.
Since quadrilateral $Q B C P$ has area 15 , then $5 x=15$ or $x=3$.
Therefore, the area of square $A B C D$, which is made up of quadrilateral $Q B C P, \triangle A Q B$ and $\triangle Q D P$, is $5 x+2 x+x=8 x=8(3)=24$.
ANSWER: (E) | \boxed{24} | In square $A B C D, P$ is the midpoint of $D C$ and $Q$ is the midpoint of $A D$. If the area of the quadrilateral $Q B C P$ is 15 , what is the area of square $A B C D$ ?
(A) 27.5
(B) 25
(C) 30
(D) 20
(E) 24
 | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
(1) From $2 S_{n}-n a_{n}=n$, we get
$$
2 S_{n+1}-(n+1) a_{n+1}=n+1 \text {. }
$$
Subtracting the two equations yields
$$
2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1 \text {. }
$$
Thus, $n a_{n}-(n-1) a_{n+1}=1$,
$$
(n+1) a_{n+1}-n a_{n+2}=1 \text {. }
$$
Subtracting (2) from (1) and rearranging gives
$$
a_{n}+a_{n+2}=2 a_{n+1} \text {. }
$$
Hence, $\left\{a_{n}\right\}$ is an arithmetic sequence.
From $2 S_{1}-a_{1}=1$ and $a_{2}=3$, we get $a_{1}=1$.
Thus, the common difference of the sequence $\left\{a_{n}\right\}$ is $d=2$.
Therefore, $a_{n}=1+2(n-1)=2 n-1$.
(2) From (1), we know
$$
\begin{aligned}
b_{n} & =\frac{1}{(2 n-1) \sqrt{2 n+1}+(2 n+1) \sqrt{2 n-1}} \\
& =\frac{1}{2} \times \frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{\sqrt{2 n-1} \sqrt{2 n+1}} \\
& =\frac{1}{2}\left(\frac{1}{\sqrt{2 n-1}}-\frac{1}{\sqrt{2 n+1}}\right) .
\end{aligned}
$$
Thus, $T_{n}=\frac{1}{2} \sum_{k=1}^{n}\left(\frac{1}{\sqrt{2 k-1}}-\frac{1}{\sqrt{2 k+1}}\right)$
$$
\begin{array}{l}
=\frac{1}{2}\left(1-\frac{1}{\sqrt{2 n+1}}\right)>\frac{9}{20} \\
\Rightarrow n \geqslant \frac{99}{2} .
\end{array}
$$
Therefore, the smallest positive integer $n$ that satisfies the condition is $n=50$. | \boxed{50} | 11. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $2 S_{n}-n a_{n}=n\left(n \in \mathbf{Z}_{+}\right)$, and $a_{2}=3$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$; for the sequence $\left\{b_{n}\right\}$, find the sum of the first $n$ terms, and determine the smallest positive integer $n$ such that $T_{n}>\frac{9}{20}$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
# Solution:
Let $O$ be the center of the circumscribed circle. Clearly, $A O = D O$ and $A B = D B$, so points $O$ and $B$ lie on the perpendicular bisector of segment $A D$. Since $A C$ is the diameter, $\angle A D C = 90^{\circ}$. Therefore, we get that $O B \| D C$. Consequently, triangle $O B P$ is similar to triangle $C D P$. Thus, we have
$$
\frac{C D}{O B} = \frac{C P}{O P} = \frac{C P}{C O - C P}
$$
From this, we can find the length of segment $C D$
$$
C D = \frac{O B \cdot C P}{C O - C P} = \frac{r \alpha}{r - \alpha}
$$
Answers:
| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 1.68 | 2.08 | 2.52 | 0.375 | 0.96 | 1.9125 | 0.36 | 2.6125 | 0.5 | 1.05 | 2.99 | | \boxed{\dfrac{r \alpha}{r - \alpha}} | 6. (8 points) Quadrilateral $ABCD$ is inscribed in a circle of radius $r$ such that diagonal $AC$ is the diameter of the circle. The diagonals of the quadrilateral intersect at point $P$. It is known that $BD = AB$ and $PC = \alpha < r$. Find the length of side $CD$.
# | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Since we would like the product to have 303 digits, then we would like this product to be greater than $10^{302}$ but less than $10^{303}$.
We will start by trying $k=300$. In this case, $\left(2^{k}\right)\left(5^{300}\right)=\left(2^{300}\right)\left(5^{300}\right)=10^{300}$, so we want $k$ to be bigger than 300 .
Each time we increase $k$ by 1 , the existing product is multiplied by 2 . For the final product to have 303 digits, we need to multiply $10^{300}$ by a power of 2 between 100 and 1000 . The smallest power of 2 that satisfies this is $2^{7}=128$.
Therefore, we would like $k=307$. In this case,
$$
\left(2^{307}\right)\left(5^{300}\right)=\left(2^{7}\right)\left(2^{300}\right)\left(5^{300}\right)=\left(2^{7}\right)\left(10^{300}\right)=128 \times 10^{300}
$$
When this number is expanded, the digits are 128 followed by 300 zeros. Therefore, the sum of the digits is 11 . | \boxed{11} | If $k$ is the smallest positive integer such that $\left(2^{k}\right)\left(5^{300}\right)$ has 303 digits when expanded, then the sum of the digits of the expanded number is
(A) 11
(B) 10
(C) 8
(D) 7
(E) 5 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 29.
Solution. The largest possible number that could have been named is 29 (in the case where there is a girl who danced with all the boys), and the smallest is 0. Thus, we have that the number of different numbers named is no more than 30.
We will prove that it cannot be exactly 30. Suppose the opposite, that all numbers from 0 to 29 were named. Then the numbers from 16 to 29 were named only by girls (this is already 14 numbers). The number 0 was definitely not named by a boy, since there is a girl who danced with all the boys. This means that the girls must have named exactly fifteen numbers: 0, 16, 17, 18, ..., 29.
At the same time, the number 15 was not named by any of the girls, since it is already known which numbers they named. If the number 15 was named by a boy, it would mean that he danced with all the girls. But there is a girl who named the number 0, who did not dance with anyone. Contradiction.
Now we will show how 29 different numbers could have been named. Let's number the girls from 1 to 15 and the boys from 1 to 29. Let boy $i$ dance with girl $j$ if and only if $i \geqslant j$.
We can illustrate this example with a $15 \times 29$ table: we will assign a column of the table to each boy and a row to each girl; if a boy danced with a girl, we will color the cell at the intersection of the corresponding row and column. In Fig. 4, above each column is written how many cells are colored in it; to the left of each row is written how many cells are colored in it.
Fig. 4: example for the solution to problem 9.6 | \boxed{29} | Problem 9.6. At a ball, 29 boys and 15 girls arrived. Some boys danced with some girls (no more than once in each pair). After the ball, each person told their parents how many times they danced. What is the maximum number of different numbers the children could have mentioned? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
11. E.
Let the number of campers who can only sing be $x$, the number of campers who can only dance be $y$, the number of campers who can only perform be $z$, the number of campers who can sing and dance be $a$, the number of campers who can sing and perform be $b$, and the number of campers who can dance and perform be $c$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
x+y+z+a+b+c=100, \\
y+z+c=42, \\
z+x+b=65, \\
x+y+a=29
\end{array}\right. \\
\Rightarrow a+b+c \\
=2(x+y+z+a+b+c)-(y+z+c)- \\
\quad(z+x+b)-(x+y+a) \\
=64 .
\end{array}
$$ | \boxed{64} | 11. In a certain summer camp, 100 campers each have at least one of the following talents: singing, dancing, or performing. Some have more than one talent, but no one has all three talents. If 42 campers cannot sing, 65 campers cannot dance, and 29 campers cannot perform, then the number of campers who have at least two talents is ( ).
(A) 16
(B) 25
(C) 36
(D) 49
(E) 64 | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
4. Let's first consider four points $A, B, C, D$ such that one of them is inside the triangle formed by the other three (without loss of generality, let $D \in \operatorname{int} \triangle A B C$). Since among the angles $\measuredangle A D B, \measuredangle A D C, \measuredangle B D C$ there is at least one not less than $120^{\circ}$, in this case it follows that $f(A, B, C, D) \geqslant 120^{\circ}$.
Now let's consider the case when none of the points $A, B, C, D$ lies inside the triangle formed by the other three, i.e., when they form a convex quadrilateral (without loss of generality, the quadrilateral $A B C D$). Since $\measuredangle A B C + \measuredangle B C D + \measuredangle C D A + \measuredangle D A B = 360^{\circ}$, at least one of these four angles must be not less than $90^{\circ}$, so in this case it follows that $f(A, B, C, D) \geqslant 90^{\circ}$.
Therefore, we conclude $\min f(A, B, C, D) \geqslant 90^{\circ}$. If the points $A, B, C, D$ form the vertices of a rectangle, then this bound is achieved, so the solution to the problem is $90^{\circ}$. | \boxed{90^\circ} | 4. For a quadruple of points $A, B, C, D$ in the plane, no three of which are collinear, let $f(A, B, C, D)$ denote the measure of the largest angle formed by these points (out of a total of 12 such angles). Determine $\min f(A, B, C, D)$, where the minimum is taken over all such quadruples of points. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. Let $\alpha, \beta$ and $\gamma$ be the angles in the triangle. Then the exterior angles are $180^{\circ}-\alpha$,
$180^{\circ}-\beta$ and $180^{\circ}-\gamma$, and their sum is
$$
\left(180^{\circ}-\alpha\right)+\left(180^{\circ}-\beta\right)+\left(180^{\circ}-\gamma\right)=540^{\circ}-(\alpha+\beta+\gamma)=540^{\circ}-180^{\circ}=360^{\circ}
$$
However, the exterior angles of the triangle are $72^{\circ}, 126^{\circ}$ and $162^{\circ}$. Therefore, its interior angles are $108^{\circ}, 54^{\circ}$ and $18^{\circ}$.
Let $ABC$ be a triangle with the calculated angles. Without loss of generality, we can assume that $\measuredangle ABC=108^{\circ}$ and $\measuredangle ACB=18^{\circ}$. If $CD$ is the angle bisector of the smallest angle ( $D \in AB$ ), then $\measuredangle DCB=9^{\circ}$ and
$$
\measuredangle CDB=180^{\circ}-\left(9^{\circ}+108^{\circ}\right)=180^{\circ}-127^{\circ}=53^{\circ}
$$ | \boxed{53^\circ} | Problem 1. The external angles of a triangle are $20\%, 35\%$ and $45\%$ of the sum of its external angles. Determine the angle between the bisector of the smallest angle and the shortest side. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3. $\frac{7}{8}$.
Connect $B D$ and $E C$ intersecting at point $O$. It is easy to see that $B D \perp E C$. Therefore, the angle $\angle B O D$ is the plane angle of the dihedral angle we are looking for.
$$
\begin{array}{l}
\text { In } \triangle B D O, \\
B D=2 \sqrt{2}, B O=\frac{8}{\sqrt{5}}, D O=\frac{2}{\sqrt{5}} . \\
\text { Hence } \cos \angle B O D=\frac{7}{8} .
\end{array}
$$ | \boxed{\dfrac{7}{8}} | 3. As shown in Figure 1, in rectangle $A B C D$, $A B=2, A D=$ 4, point $E$ is on line segment $A D$, and $A E=3$. Now, fold $\triangle A B E$ and $\triangle D C E$ along $B E$ and $C E$ respectively, so that point $D$ lands on line segment $A E$. At this moment, the cosine of the dihedral angle $D-E C-B$ is | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. If $n=1$, then $N=5^{n}+12 n^{2}+12 n+3=32$, which is not divisible by 100. Therefore, we can assume that $n \geq 2$.
$N$ will be divisible by 100 if and only if it is divisible by 4 and 25, since $100=4 \cdot 25$ and $(4 ; 25)=1$. Given that $n \geq 2$, $5^{2}=25 \mid 5^{n}$, so it must be a divisor of $12 n^{2}+12 n+3=3(2 n+1)^{2}$. Since $(3 ; 25)=1$, 25 must be a divisor of $(2 n+1)^{2}$. This is true if and only if $2 n+1$ is divisible by 5, i.e., $n=5 k+2$. In this case, $2 n+1=2(5 k+2)+1=10 k+5$.
4 will be a divisor of $N$ if and only if it is a divisor of $\left(5^{n}+3\right)$. But $5^{n}=(4+1)^{n}$ leaves a remainder of 1 when divided by 4, so $5^{n}+3$ leaves a remainder of $1+3=4$, which is 0, when divided by 4. Therefore, it is divisible by 4 for all $n \geq 1$.
In summary: 100 is a divisor of $5^{n}+12 n^{2}+12 n+3$ if and only if $n=5 k+2$, where $k=0,1,2,3, \ldots$. | \boxed{n \equiv 2 \pmod{5}} | Which are those natural numbers $n$ for which
$$
5^{n}+12 n^{2}+12 n+3
$$
is divisible by $100$? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
We set the radius of the circle to 1, and we determine the chord in question by the central angle $2 \alpha (<\pi)$ associated with it. The diameter of the circle inscribed in the sector - which touches both the chord and the arc - cannot be larger than the segment of the perpendicular erected at the point of tangency on the chord that falls within the sector. This segment is the largest when it lies on the axis of symmetry of the sector, and in this position, the circle inscribed in the sector is the circle whose diameter is the segment itself; this is what we need to consider.
The area of the sector is $\pi - (\alpha - \sin \alpha \cos \alpha)$, and the area of the inscribed circle is $\pi (1 + \cos \alpha)^2 / 4$. From these, the expression for the area in question can be transformed as follows:
$$
t = \frac{3 \pi}{4} - \alpha - \frac{\pi}{2} \cos \alpha - \frac{\pi}{4} \cos^2 \alpha + \sin \alpha \cos \alpha
$$
We seek the desired maximum based on the vanishing of the derivative.
$$
\begin{aligned}
t' = & -1 + \frac{\pi}{2} \sin \alpha + \frac{\pi}{2} \sin \alpha \cos \alpha + \cos^2 \alpha - \sin^2 \alpha = \\
& = \frac{\pi}{2} \sin \alpha \left(1 + \cos \alpha - \frac{4}{\pi} \sin \alpha \right) = \\
& = \frac{\pi}{2} \sin \alpha \left(2 \cos^2 \frac{\alpha}{2} - \frac{8}{\pi} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \right) = \\
& = \pi \sin \alpha \cos \frac{\alpha}{2} \left(\cos \frac{\alpha}{2} - \frac{4}{\pi} \sin \frac{\alpha}{2} \right)
\end{aligned}
$$
The zeros of the first two (variable) factors are of no interest to us: in the case of $\sin \alpha = 0$, $\alpha = 0$, the length of the chord is 0, the sector is identical to the given circle, the inscribed circle completely fills it, and the area under consideration is $0$; the root of $\cos \alpha / 2 = 0$ does not fall within the considered domain. The factor in parentheses vanishes if and only if
$$
\tan \frac{\alpha}{2} = \frac{\pi}{4} \approx 0.5708
$$
At this point, the derivative transitions from decreasing to increasing, because $\cos \alpha / 2$ is decreasing, and the second term is also decreasing, since $\sin \alpha / 2$ is increasing, and the factors in front are positive. Therefore, the area under consideration has a maximum at the angle defined by (1).
At the maximum point, the length of the chord is
$$
2 \sin \alpha = \frac{4 \tan \alpha / 2}{1 + \tan^2 \alpha / 2} = \frac{16 \pi}{16 + \pi^2} \approx 1.943
$$
$\left(\alpha = 76^\circ 17.5'\right)$. | \boxed{\dfrac{16}{16 + \pi^2}} | In a given circle, we draw a chord and inscribe the largest possible circle in the larger of the resulting circular segments. For what choice of the chord's length will the area of the segment outside the inscribed circle be the largest? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The instantaneous value of the measurable voltage between points $A$ and $B$ is composed of the $U / 2$ voltage dropping across the $R_{1}$ resistor and the $(U / 4) + (U / 4) \sin \omega t$ voltage taken from the $R$ resistor, where $\omega$ is the angular frequency of the harmonic oscillation. Therefore, the voltage measurable between $A$ and $B$ as a function of time is:
$$
U_{p}=(U / 4)(3+\sin \omega t)
$$
The definition of the effective voltage:
$$
U_{\text {eff }}^{2}=\frac{1}{T} \int_{0}^{T} U_{p}^{2} \mathrm{~d} t
$$
where $T$ is the period.
Substituting $U_{p}$ and performing the integration using the substitution $\sin ^{2} x=\frac{1-\cos 2 x}{2}$, we get $U_{\text {eff }}=\sqrt{19 / 32} U=$ $77.06 \mathrm{~V}$. György Molnár (Pannonhalma, Bencés Gym., IV. grade)
Note. The value of the effective voltage is independent of the specific values of $R$ and $R_{1}$, provided that $R=R_{1}$. | \boxed{77.06 \, \text{V}} | In the circuit shown in the figure, the wiper of the potentiometer performs harmonic oscillatory motion between the two extreme positions. What is the effective value of the voltage measurable between points $A$ and $B$? The internal resistance of the power supply can be neglected. Data: $U=100 \mathrm{~V}, R=R_{1}=1 \mathrm{k} \Omega$.
 | Other | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The interquartile range is defined as $Q3 - Q1$, which is $43 - 33 = 10$. $1.5$ times this value is $15$, so all values more than $15$ below $Q1$ = $33 - 15 = 18$ is an outlier. The only one that fits this is $6$. All values more than $15$ above $Q3 = 43 + 15 = 58$ are also outliers, of which there are none so there is only $\boxed{\textbf{(B) 1}}$ outlier in total. | \boxed{B} | The data set $[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]$ has median $Q_2 = 40$, first quartile $Q_1 = 33$, and third quartile $Q_3 = 43$. An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartle ($Q_1$) or more than $1.5$ times the interquartile range above the third quartile ($Q_3$), where the interquartile range is defined as $Q_3 - Q_1$. How many outliers does this data set have?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Algebra | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3.2. Two identical cylindrical vessels are connected at the bottom by a small-section pipe with a valve. While the valve was closed, water was poured into the first vessel, and oil into the second, so that the level of the liquids was the same and equal to \( h = 40 \, \text{cm} \). At what level will the water stabilize in the first vessel if the valve is opened? The density of water is 1000 kg \(/ \text{m}^3\), and the density of oil is 700 kg \(/ \text{m}^3\). Neglect the volume of the connecting pipe. Give the answer in centimeters.
Answer. \{34\}. | \boxed{34} | 3.2. Two identical cylindrical vessels are connected at the bottom by a small-section pipe with a valve. While the valve was closed, water was poured into the first vessel, and oil into the second, so that the level of the liquids was the same and equal to \( h = 40 \, \text{cm} \). At what level will the water stabilize in the first vessel if the valve is opened? The density of water is 1000 kg \(/ \text{m}^3\), and the density of oil is 700 kg \(/ \text{m}^3\). Neglect the volume of the connecting pipe. Provide the answer in centimeters. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Answer】Solution: $16 \times 3-3+1-15 \times 2+15$,
$$
\begin{array}{r}
=48-2-15, \\
=31 \text { (pieces), }
\end{array}
$$
Answer: The number of small equilateral triangles seen in the shadow is 31; hence the answer is: 31. | \boxed{31} | 7. (5 points) In the figure, three equally sized equilateral triangular transparent sheets of glass paper, each divided into 49 smaller equilateral triangles of the same size, have 16 of the smaller triangles shaded on each sheet. If these three sheets are overlapped, the number of shaded smaller equilateral triangles seen is $\qquad$.
| Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
We choose the notation such that each point without a comma lies on one of the rays originating from $O$, and the points with a comma lie on the other.
The 8 planes determined by the 2nd index points of the three lines form a regular octahedron $T_{2}$, whose volume is $32 / 3$. By taking $Z_{1}$ instead of $Z_{2}$ and $Z_{1}^{\prime}$ instead of $Z_{2}^{\prime}$ in these planes, the 8 planes form a body $T_{2}$ that can also be derived from $T_{2}$ by a contraction along the $z$-axis, so its volume is $16 / 3$. Similarly, the bodies $T_{x}, T_{y}$, and $T_{z}$, formed by contraction along the $x$ and $y$ axes, respectively, together with the faces of $T_{z}$, bound the body $T$ under investigation, which has 24 faces (Figure 1).
Figure 1
Each of $T_{x}, T_{y}$, and $T_{z}$ is symmetric with respect to the planes determined by the three axes taken in pairs, so the same is true for their common part, $T$. Therefore, it is sufficient to examine the part of $T$ that falls into one of the 8 regions defined by the 3 planes. Let this be the part that falls into the tetrahedron $O X_{2} Y_{2} Z_{2}$.
The plane $X_{2} Y_{2} Z_{1}$ divides the tetrahedron $O X_{2} Y_{2} Z_{2}$ into two parts of equal volume (since the non-common vertices $O, Z_{2}$ of the parts are at equal distances from the intersecting plane), and it intersects the faces $O Z_{2} Y_{2}$ and $O Z_{2} X_{2}$ along their medians $Y_{2} Z_{1}$ and $X_{2} Z_{1}$, respectively. The planes $X_{2} Y_{2} Z_{1}$ and $Y_{2} Z_{2} X_{1}$ intersect at point $Y_{2}$, and they intersect the face $O X_{2} Z_{2}$ along its medians $X_{2} Z_{1}$ and $X_{1} Z_{2}$, respectively, so the centroid $S_{2}$ of this face is also on both planes. The intersection line of the planes $X_{2} Y_{2} Z_{1}$ and $Y_{2} Z_{2} X_{1}$ is therefore the median $Y_{2} S_{2}$ of the tetrahedron $O X_{2} Y_{2} Z_{2}$ (Figure 2).
Figure 2
Similarly, we find that the other intersection lines are the medians $Z_{2} S_{3}$ and $X_{2} S_{1}$, where $S_{3}$ is the centroid of the face $O X_{2} Y_{2}$, and $S_{1}$ is the centroid of the face $O Y_{2} Z_{2}$. Thus, the intersection point of the planes $X_{2} Y_{2} Z_{1}$, $Y_{2} Z_{2} X_{1}$, and $Z_{2} X_{2} Y_{1}$ is the centroid $S$ of the tetrahedron $O X_{2} Y_{2} Z_{2}$. The part of $T$ that falls into this tetrahedron consists of the union of pyramids with apex $O$ and bases
$$
S_{1} S S_{2} Z_{1}, \quad S_{2} S S_{3} X_{1}, \quad S_{3} S S_{1} Y_{1}
$$
Since the area of the triangle $X_{2} Z_{2} Y_{1}$ is 1/6 of the area of the quadrilateral $S_{3} S S_{1} Y_{1}$ (because the area of the triangle $Y_{1} S S_{3}$ is half the area of the triangle $X_{2} S S_{3}$, which is 1/4 of the area of the triangle $X_{2} Z_{2} S_{3}$, which is 2/3 of the area of the triangle $X_{2} Z_{2} Y_{1}$), the volume of the pyramid $O S_{3} S S_{1} Y_{1}$ is 1/6 of the volume of the tetrahedron $O X_{2} Z_{2} Y_{1}$, or 1/12 of the volume of the tetrahedron $O X_{2} Y_{2} Z_{2}$. Therefore, the combined volume of the above three pyramids is 1/4 of the volume of the tetrahedron $O X_{2} Y_{2} Z_{2}$, and thus the volume of $T$ is also one-fourth of the volume of $T_{2}$:
$$
V_{T}=\frac{1}{4} V_{2}=\frac{1}{4} \cdot \frac{32}{3}=\frac{8}{3} \quad \text { cubic units. }
$$
Remark. The body under investigation is a deltoidic tetrahedron (deltoid icositetrahedron) of the regular (cubic) crystal system. Its general notation is $h, k, k$, where $h$ and $k (h < k)$ are the different segments cut from the axes by its planes, in our case $1,2,2$. | \boxed{\dfrac{8}{3}} | Three, pairwise perpendicular lines intersect at a point $O$. The points on one line that are one unit away from $O$ are $X_{1}$ and $X_{1}^{\prime}$, and the points that are two units away are $X_{2}$ and $X_{2}^{\prime}$. Similarly, we denote the points on the other two lines as $Y_{1}, Y_{1}^{\prime}, Y_{2}, Y_{2}^{\prime}, Z_{1}, Z_{1}^{\prime}, Z_{2}, Z_{2}^{\prime}$. - We form planes through every possible combination of a 1-indexed point and a 2-indexed point on each of the other lines, for example, through $X_{1}^{\prime}$, $Y_{2}$, and $Z_{2}^{\prime}$.
Determine the volume of the solid enclosed by these planes. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Answer】 426
【Explanation】 Case analysis
1 $\triangle$: 36, area $1 \times 36=36$
4 $\triangle$s: 15 upright, 6 inverted, total 21, area $1 \times 4 \times 21=84$
9 $\triangle$s: 10 upright, 1 inverted, total 11, area $1 \times 9 \times 11=99$
16 $\triangle$s: 6 upright, area $1 \times 16 \times 6=96$
25 $\triangle$s: 3 upright, area $1 \times 25 \times 3=75$
36 $\triangle$s: 1 upright, area $1 \times 36=36$
Total area: $36+84+99+96+75+36=426$ | 426 | 4. As shown in the figure, given that the area of a small triangle is 1, what is the sum of the areas of all the triangles in the figure?
The text is translated while preserving the original line breaks and format. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer. For 4 rubles. Solution. We will show how to manage with four rubles. First, we find the products $a b$ and $b c$. Since the numbers $a$ and $c$ have no common prime divisors, the greatest common divisor of these products is $b$. Thus, we find the number $b$, and with it the numbers $a=a b / b$ and $c=b c / b$. Similarly, we find the numbers $d$, $e$, and $f$ for two rubles.
We will show that three rubles are not enough. Suppose we know only three products. Then they must include all six numbers, otherwise we would not know anything about one of them. But in this case, each number appears in exactly one product, and if, for example, one of the products is 6, we would not be able to distinguish a pair of 2 and 3 from a pair of 1 and 6. | \boxed{4} | 4. Vasya thought of six natural numbers: $a, b, c, d, e, f$. For one ruble, you can indicate any two of them and find out their product. Petya knows that any two of the thought-of numbers are coprime (that is, they do not have common divisors greater than 1). What is the smallest amount he can spend to find out all the thought-of numbers? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution
a) Observe that
$$
11^{11}=\underbrace{11 \times 11 \times 11 \times 11 \times \cdots \times 11}_{11 \text { times }}
$$
Each time we multiply two numbers that have the digit 1 as the rightmost digit, we get a number that also has the digit 1 as the rightmost digit. For example, $11 \times 11=121$. Therefore, repeating the process, we will find that $11^{11}$ has the digit 1 as the rightmost digit.
b) Observe that
$$
9^{9}=\underbrace{9 \times 9 \times \cdots \times 9}_{9 \text { times }}
$$
In this case, when we perform the first multiplication, we get $9 \times 9=81$, which ends in 1. When we perform the next multiplication, we get $81 \times 9=729$, which ends in 9. Performing the next multiplication, we get a number that ends in 1 again. Then another number, now ending in 9. Thus, we have a pattern! Since we start with 9 and perform eight multiplications, we will end up with a number that ends in 9.
For the number $9219^{9219}$, we perform the same analysis as before. Since we start with a number that ends in 9 and perform 9218 multiplications, we conclude that the final result will be a number that ends in 9.
c) In this case, let's observe the pattern generated again. Note that $4 \times 4=16$. Multiplying the last digit (which in this case is 6) by 4, we get $6 \times 4=24$. Thus, the last digit becomes 4 again. In other words, we get a pattern where the last digit alternates between 4 and 6. Since
$$
2014^{2014}=\underbrace{2014 \times 2014 \times \cdots \times 2014}_{2014 \text { times }}
$$
we start with a 2014 and perform 2013 multiplications. Therefore, the rightmost digit of the result will be the digit 6. | \boxed{1}, \boxed{9}, \boxed{9}, \boxed{6} | We call the "last digit of a number" as the digit farthest to the right. For example, the last digit of 2014 is the digit 4.
a) What is the last digit of $11^{11}$?
b) What is the last digit of $9^{9}$? And what is the last digit of $9219^{9219}$?
c) What is the last digit of $2014^{2014}$? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5.
Answer: 10
Example: Let's break down 2015 into 11 numbers of 155 and one number of 310. In this case, the sum of the addends can range from $155 \times 1$ to $155 \times 10$.
We will prove that it is impossible to get fewer than 10 different numbers: first, note that 2015 is not divisible by 12, so we will have at least 2 different numbers. Arrange all the numbers in ascending order. First, select only the first number, then the first and second, and so on—finally, select the first 9 numbers—this gives us 9 different sums. Now, take the last 9 numbers—the new sum exceeds all previous sums, so there are at least 10 different numbers on the board. | \boxed{10} | 5. The number 2015 was divided into 12 addends, after which all numbers that can be obtained by adding some of these addends (from one to nine) were written down. What is the minimum number of numbers that could have been written down? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
After the first fold, point $A$ coincided with point $C$. The line along which the fold was made is the axis of the segment $A C$, which in the equilateral triangle $A B C$ passes through point $B$. After this fold, Petra obtained the triangle $B E C$, which is exactly half of the triangle $A B C$ (point $E$ denotes the midpoint of the segment $A C$, see the figure).
After the second fold, point $B$ also coincided with point $C$. The line along which the fold was made is the axis of the segment $B C$, which in the original triangle passes through point $A$. After this fold, Petra obtained the quadrilateral $O E C D$ ( $D$ denotes the midpoint of the segment $B C$ and $O$ is the intersection of the axes).
The axes of the sides in the equilateral triangle $A B C$ are axes of symmetry of this triangle. It follows that the quadrilaterals $O E C D, O D B F$, and $O F A E$ are congruent to each other and each has an area of $12 \mathrm{~cm}^{2}$ ( $F$ denotes the midpoint of the segment $A B$ ). The area of the triangle $A B C$ is equal to the sum of the areas of these three quadrilaterals:
$$
S_{A B C}=3 \cdot 12=36\left(\mathrm{~cm}^{2}\right) .
$$
Evaluation. 2 points for the correct interpretation of the first fold (triangle $B E C$); 2 points for the correct interpretation of the second fold (quadrilateral $O E C D$); 2 points for expressing the area of the triangle $A B C$. | \boxed{36} | Petra has an equilateral triangle $ABC$. First, she folded the triangle so that point $A$ coincided with point $C$. Then she folded the resulting shape so that point $B$ coincided with point $C$. She then drew the outline of this shape on paper and found that its area is $12 \, \text{cm}^2$. Determine the area of the original triangle.
(E. Novotná) | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
47. Let (Fig. 9) $\breve{B C}=2 \alpha, \breve{B L}=2 \beta$. Then
$$
|A C|=2 R \cos \alpha, \quad|C L|=2 R \sin (\alpha+\beta)
$$
$|C M|=|C L| \cos \left(90^{\circ}-\beta\right)=2 R \sin (\alpha+\beta) \sin \beta$,
$$
\begin{aligned}
A M \mid & =|A C \cdot|-|C M|=2 R[\cos \alpha-\sin (\alpha+\beta) \sin \beta]= \\
& =2 R\left[\cos \alpha-\frac{1}{2} \cos \alpha+\frac{1}{2} \cos (\alpha+2 \beta)\right]=2 R \cos \beta \cos (\alpha+\beta)
\end{aligned}
$$
and, $\quad$ finally, $\quad|A N|=a=|A M| \cos \alpha=2 R \cos \alpha \cos \beta \cos (\alpha+\beta)$. On the other hand, if $K, P$ and $Q$ are the midpoints of $A O, C O$ and $C L$ respectively, then $|K P|=\frac{1}{2}|A C|=R \cos \alpha,|P Q|=\frac{R}{2}, \widehat{K P Q}=$ $=\widehat{K P O}+\widehat{O P Q}=\alpha+180^{\circ}-\widehat{C O L}=\alpha+180^{\circ}-2 \alpha-2 \beta=180^{\circ}-\alpha-$ $-2 \beta$ and, by the cosine rule, $|K Q|^{2}=\frac{R^{2}}{4}+R^{2} \cos ^{2} \alpha+R^{2} \cos \alpha \times$ $x \cos (\alpha+2 \beta)=\frac{R^{2}}{4}+2 R^{2} \cos \alpha \cos \beta \cos (\alpha+\beta)=\frac{R^{2}}{4}+R a$.
Answer: $\sqrt{\frac{R^{2}}{4}+R a}$. | \boxed{\sqrt{\dfrac{R^2}{4} + R a}} | 47. Let $A B$ be the diameter of a circle, $O$ its center, $|A B|=2 R, C$ a point on the circle, $M$ a point on $A C$. From $M$, a perpendicular $M N$ is dropped onto $A B$ and a perpendicular to $A C$ is erected, intersecting the circle at point $L$ (segment $C L$ intersects $A B$). Find the distance between the midpoint of $A O$ and the midpoint of $C L$, if $|A N|=a$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
[Solution] . If $f(a)=f(b)=f(c)=f(d)=2$, it is clearly a mapping that fits the problem's requirements.
If $f(a), f(b), f(c), f(d)$ are not all 2, then at least one of them must be 1, and at most one of them can be 1 (note that $2^{3}=820$).
We discuss the following two scenarios:
(1) Among $f(a), f(b), f(c), f(d)$, there is one 1 and one 3 (two 2s), in this case, there are $P_{4}^{2}=4 \cdot 3=12$ mappings;
(2) Among $f(a), f(b), f(c), f(d)$, there is one 1 and two 3s (one 2), in this case, there are $P_{4}^{2}=4 \cdot 3=12$ mappings.
In total, there are $1+12+12=25$ mappings. Therefore, the answer is (C). | \boxed{25} | 15$\cdot$18 Mapping $f:\{a, b, c, d\} \rightarrow\{1,2,3\}$. If $10<f(a) \cdot f(b)$. $f(c) \cdot f(d)<20$, the number of such mappings is
(A) 23.
(B) 24.
(C) 25.
(D) 26.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991) | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The correct answer is $1+2+3+4+\ldots+11=66$, since there is no qualifying minute value between $0-1$ o'clock, there is 1 such minute between 1-2 o'clock, which is 1:01, and so on. We also accepted 67 as the correct answer if the solver noted that they included the boundaries. Indeed, according to the problem statement, only the number of minutes occurring within the interval is asked for. | \boxed{66} | In a 12-hour interval (from $0^{\mathrm{h}}$ to $12^{\mathrm{h}}$), how many minutes are there when the value of the hours is greater than the value of the minutes? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
# Answer: 16.
## Conditions and answers to the problems of the final stage of the 2012-13 academic year
# | \boxed{36} | Problem 6.
It is known that three numbers $a_{1}, a_{2}, a_{3}$ were obtained as follows. First, a natural number $A$ was chosen and the numbers $A_{1}=[A]_{16}, A_{2}=$ $[A / 2]_{16}, A_{3}=[A / 4]_{16}$ were found, where $[X]_{16}$ is the remainder of the integer part of the number $X$ divided by 16 (for example, $[53 / 2]_{16}=10$). Then an integer $B$ was chosen such that 0 $\leq \mathrm{B} \leq 15$. The numbers $A_{1}, A_{2}, A_{3}$ and $B$ are written in binary, i.e., each is represented as a string of 0s and 1s of length 4, adding the necessary number of zeros on the left. We agree to add such strings symbol by symbol "in a column" without carrying to the next digit according to the rule: $1+1=0+0=0$ and $0+1=1+0=0$, and denote the operation of symbol-by-symbol addition by the symbol $\diamond$. For example, $3 \diamond 14=(0011) \diamond$ $(1110)=(1101)=13$. Let $a_{1}=A_{1} \diamond B, a_{2}=A_{2} \diamond B, a_{3}=A_{3} \diamond B$. Find the sum of all possible values of the number $a_{3}$, given that $a_{1}=2, a_{2}=$ 9. Choose the correct answer:
# | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
12. 2004
From $|a-b| \leqslant \max \{a, b\}$ (where $a, b$ are non-negative numbers), it follows that $||a-b|-c| \leqslant \max \{|a-b|, c\} \leqslant \max \{a, b, c\}$, which leads to $M \leqslant 2004$.
On the other hand, $||2003-2002|-2001|-2000|=0$, $|||1999-1998|-1997|-1996|=0$, ..., $||7-6|-5|-4|=0$, $||3-2|-1|=0$. Let $x_{2004}=2004$, then $M$ can be equal to 2004. | \boxed{2004} | 12. $\left(x_{1}, x_{2}, \cdots, x_{2004}\right)$ is a permutation of $1,2,3, \cdots \cdots, 2004$, let $M=\|\cdots\|$ $\left|x_{1}-x_{2}\right|-x_{3}\left|-x_{4}\right|-\cdots\left|-x_{2004}\right|$, then the maximum value of $M$ is $\qquad$ | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
10. Find the sum of the roots of the equation param1 lying in the interval param2. Write the answer in degrees.
| param 1 | param2 | Answer |
| :---: | :---: | :---: |
| $\sin x+\sin ^{2} x+\cos ^{3} x=0$ | $\left[360^{\circ} ; 720^{\circ}\right]$ | 1800 |
| $\cos x-\cos ^{2} x-\sin ^{3} x=0$ | $\left[180^{\circ} ; 540^{\circ}\right]$ | 990 |
| $\sin x+\sin ^{2} x+\cos ^{3} x=0$ | $\left[-360^{\circ} ; 0^{\circ}\right]$ | -360 |
| $\cos x-\cos ^{2} x-\sin ^{3} x=0$ | $\left[-540^{\circ} ;-180^{\circ}\right]$ | -1170 | | 1800 | 10. Find the sum of the roots of the equation param1 that lie in the interval param2. Write the answer in degrees.
| param 1 | param2 | |
| :---: | :---: | :---: |
| $\sin x+\sin ^{2} x+\cos ^{3} x=0$ | $\left[360^{\circ} ; 720^{\circ}\right]$ | |
| $\cos x-\cos ^{2} x-\sin ^{3} x=0$ | $\left[180^{\circ} ; 540^{\circ}\right]$ | |
| $\sin x+\sin ^{2} x+\cos ^{3} x=0$ | $\left[-360^{\circ} ; 0^{\circ}\right]$ | |
| $\cos x-\cos ^{2} x-\sin ^{3} x=0$ | $\left[-540^{\circ} ;-180^{\circ}\right]$ | | | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
15. (1) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0), F(c, 0)$.
Then the line $l_{A B}: y=\frac{1}{2}(x-c)$.
Substituting into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, simplifying gives
$\left(a^{2}+4 b^{2}\right) x^{2}-2 a^{2} c x+a^{2} c^{2}-4 a^{2} b^{2}=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $x_{1}+x_{2}=\frac{2 a^{2} c}{a^{2}+4 b^{2}}, x_{1} x_{2}=\frac{a^{2} c^{2}-4 a^{2} b^{2}}{a^{2}+4 b^{2}}$.
Given $\overrightarrow{O A}+\overrightarrow{O B}=\left(x_{1}+x_{2}, y_{1}+y_{2}\right)$ is collinear with $\boldsymbol{a}=$ $(-3,1)$, we have
$3\left(y_{1}+y_{2}\right)+\left(x_{1}+x_{2}\right)=0$.
Also, $y_{1}=\frac{1}{2}\left(x_{1}-c\right), y_{2}=\frac{1}{2}\left(x_{2}-c\right)$, then
$3\left(\frac{1}{2}\left(x_{1}+x_{2}\right)-c\right)+\left(x_{1}+x_{2}\right)=0$
$\Rightarrow x_{1}+x_{2}=\frac{6}{5} c$
$\Rightarrow \frac{2 a^{2} c}{a^{2}+4 b^{2}}=\frac{6}{5} c$
$\Rightarrow a^{2}=6 b^{2}$
$\Rightarrow c=\sqrt{a^{2}-b^{2}}=\frac{\sqrt{30}}{6} a$
$\Rightarrow e=\frac{c}{a}=\frac{\sqrt{30}}{6}$.
(2) From (1), we know $a^{2}=6 b^{2}$.
Then the equation of the ellipse can be written as $x^{2}+6 y^{2}=6 b^{2}$.
Let $M(x, y)$.
From the given information,
$$
\begin{array}{l}
(x, y)=\lambda\left(x_{1}, y_{1}\right)+\mu\left(x_{2}, y_{2}\right) \\
\Rightarrow x=\lambda x_{1}+\mu x_{2}, y=\lambda y_{1}+\mu y_{2} .
\end{array}
$$
Since point $M(x, y)$ is on the ellipse, we have $\left(\lambda x_{1}+\mu x_{2}\right)^{2}+6\left(\lambda y_{1}+\mu y_{2}\right)^{2}=6 b^{2}$,
which simplifies to $\lambda^{2}\left(x_{1}^{2}+6 y_{1}^{2}\right)+\mu^{2}\left(x_{2}^{2}+6 y_{2}^{2}\right)+$
$2 \lambda \mu\left(x_{1} x_{2}+6 y_{1} y_{2}\right)=6 b^{2}$.
From (1), we know $a^{2}=\frac{6}{5} c^{2}, b^{2}=\frac{1}{5} c^{2}$, so $x_{1}+x_{2}=\frac{6}{5} c, x_{1} x_{2}=\frac{a^{2} c^{2}-4 a^{2} b^{2}}{a^{2}+4 b^{2}}=\frac{3}{25} c^{2}$.
Then $x_{1} x_{2}+6 y_{1} y_{2}$
$$
\begin{array}{l}
=x_{1} x_{2}+6 \times \frac{1}{4}\left(x_{1}-c\right)\left(x_{2}-c\right) \\
=\frac{5}{2} x_{1} x_{2}-\frac{3}{2}\left(x_{1}+x_{2}\right) c+\frac{3}{2} c^{2} \\
=\frac{3}{10} c^{2}-\frac{9}{5} c^{2}+\frac{3}{2} c^{2}=0 .
\end{array}
$$
Also, $x_{1}^{2}+6 y_{1}^{2}=6 b^{2}, x_{2}^{2}+6 y_{2}^{2}=6 b^{2}$, substituting into the equation
(1) gives $\lambda^{2}+\mu^{2}=1$.
Thus, $\lambda^{2}+\mu^{2}$ is a constant value of 1. | \boxed{\dfrac{\sqrt{30}}{6}} | 15. Given that the center of the ellipse is the origin $O$, the foci are on the $x$-axis, a line with a slope of $\frac{1}{2}$ passing through the right focus $F$ intersects the ellipse at points $A$ and $B$, and $\overrightarrow{O A}+\overrightarrow{O B}$ is collinear with $a=$ $(-3,1)$.
(1) Find the eccentricity of the ellipse;
(2) Let $M$ be a point on the ellipse, and
$$
\overrightarrow{O M}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R}),
$$
Prove: $\lambda^{2}+\mu^{2}$ is a constant. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
A5. To fence one side, which is $1500 \mathrm{~m}$, with $0.6 \mathrm{~m}$ long fence panels, we need $\frac{1500}{0.6}=\frac{15000}{6}=$ 2500 fence panels, so for both sides we need 5000 fence panels. | 5000 | A5. How many posts of length $60 \mathrm{~cm}$ are needed to line a $1.5 \mathrm{~km}$ long street on both sides?
(A) 25
(B) 90
(C) 5000
(D) 250
(E) 2.5 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5. D. From $y=2 \sin \left(x+\frac{2}{3} \pi\right)$, shifting left by $m$ units gives $y=2 \sin \left(x+m+\frac{2}{3} \pi\right)$, so $m+\frac{2}{3} \pi=k \pi+\frac{\pi}{2}$, therefore $m=k \pi-\frac{\pi}{6}$. Let $k=1$, then $m=\frac{5}{6} \pi$. | \boxed{\dfrac{5\pi}{6}} | 5. The graph of the function $y=\sqrt{3} \cos x-\sin x$ is shifted to the left by $m$ units, and the resulting graph is symmetric about the $y$-axis. Then the smallest positive value of $m$ is ( ).
A. $\frac{\pi}{6}$
B. $\frac{\pi}{3}$
C. $\frac{2 \pi}{3}$
D. $\frac{5 \pi}{6}$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The team names are abbreviated to A, B, C, and D. Each team played three matches against the remaining teams, and could earn 0, 1, or 3 points per match. The distribution of points for teams A, B, and D is therefore uniquely determined, while for team C there are two possibilities:
$$
7=3+3+1, \quad 4=3+1+0, \quad 3=3+0+0=1+1+1, \quad 2=1+1+0 .
$$
One point is always awarded to two teams, so the total number of ones in the above decompositions must be even. In the decompositions of 7, 3, and 2 (i.e., the points of teams A, B, and D), there are a total of four ones. Therefore, team C must have earned its 3 points as follows:
$$
3=3+0+0 \text{. }
$$
From this, it also follows that two matches ended in a draw.
Team D drew twice, and teams A and B each drew once. Therefore, team D drew with teams A and B.
Team A drew once and won twice. Since it drew with team D, it must have won against teams B and C.
Team B lost once, drew once, and won once. Since it lost to team A and drew with team D, it must have won against team C.
Note. From the above, we can construct a table of all matches:
| | A | B | C | D | total |
| :---: | :---: | :---: | :---: | :---: | :---: |
| A | - | 3 | 3 | 1 | 7 |
| B | 0 | - | 3 | 1 | 4 |
| C | 0 | 0 | - | 3 | 3 |
| D | 1 | 1 | 0 | - | 2 |
The answer to question a) can also be derived as follows: In total, 16 points were awarded in the tournament: \(7 + 4 + 3 + 2 = 16\). If no match had ended in a draw, a total of \(6 \times 3 = 18\) points would have been awarded. Each draw contributes two points to the total (i.e., one point less than a win), so two matches ended in a draw.
Evaluation. 2 points for the answer to each of the questions a) and b); 2 points for the quality of the commentary. | b) \boxed{B} | At the tournament, teams Akouska, Bovenska, Colska, and Demecka met. Each team played against each other exactly once. The winning team received three points, the losing team received no points, and in the event of a draw, each of the drawing teams received one point. After all six matches were played, Akouska had 7 points, Bovenska had 4 points, Colska had 3 points, and Demecka had 2 points.
a) How many matches ended in a draw?
b) What was the result of the match between Bovenska and Colska?
(J. Tkadlec) | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution.
Let in $\triangle ABC$, $AB=BC$, $\angle BAC=\angle BCA=72^{\circ}$, $AD$ is the bisector of $\angle BAC$, $AD=m$ (Fig. 10.43). $\angle BAD=\angle DAC=36^{\circ}$, $\angle ABC=180^{\circ}-2\angle BAC=36^{\circ}$, $\angle ADC=180^{\circ}-(\angle DAC+\angle BCA)=72^{\circ}$. Therefore, $\triangle ADB$ and $\triangle CAD$ are isosceles, $BD=AD=AC=m$. Let $AB=BC=x$. Then $DC=x-m$. Since $AD$ is the bisector of $\triangle ABC$, $\frac{BD}{DC}=\frac{AB}{AC}$; $\frac{m}{x-m}=\frac{x}{m}$; $x^2 - xm - m^2 = 0$, and since $x>0$, $x=\frac{m(1+\sqrt{5})}{2}$.
Answer: $m ; \frac{m(1+\sqrt{5})}{2}$. | \boxed{\frac{m(1+\sqrt{5})}{2}} | 10.235. In an isosceles triangle, the angle at the base is $72^{\circ}$, and the bisector of this angle has a length of $m$. Find the lengths of the sides of the triangle. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
153846 Solution Put n = 10 m a+b, where 1 ≤ a ≤ 9. Then 7n/2 = 10b + a, so 13b = a(7·10 m -2). Obviously 13 cannot divide a, so it must divide 7·10 m -2. Hence 10 m = 4 mod 13. We have 10 1 = 10, 10 2 = 9, 10 3 = -1, 10 4 = 3, 10 5 = 4 mod 13. So the smallest possible m is 5. Obviously the smallest possible a is 1, and we get b = 53846. Check: (7/2)153846 = 538461. 25th Swedish 1985 © John Scholes [email protected] 16 February 2004 Last corrected/updated 16 Feb 04 | \boxed{153846} | 25th Swedish 1985 Problem 2 Find the smallest positive integer n such that if the first digit is moved to become the last digit, then the new number is 7n/2. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
## Solution:
a) From the Pythagorean theorem, we obtain: $A M=4 \sqrt{2}, A C=8 \sqrt{2}$ and $M C=4 \sqrt{6}$, and from the converse of the Pythagorean theorem, we obtain that triangle $A M C$ is right-angled at $M$
b) Let $(A M C) \cap(E F H)=d$. We get $d\|A C\| E G$ and since $E M \| G N$, it follows that
MNGE is a parallelogram. $\mathrm{FN}=12 \mathrm{~cm}$ ..... 2p
c) Let $M P \| E A$ and since $E A \perp(A B D)$, we obtain $M P \perp(A B D)$ and $M P=4 \mathrm{~cm}$ ..... $1 p$
If $P Q \perp A C$, it follows that the angle corresponding to the dihedral angle is $M Q P$.
The tangent of the angle between the planes $(A B D)$ and $(A C M)$ has the value $\sqrt{2}$ ..... $1 p$ | c) \(\boxed{\sqrt{2}}\) | ## SUBJECT IV
$A B C D E F G H$ is a regular quadrilateral prism with $A B=8 \mathrm{~cm}$ and $A E=4 \mathrm{~cm}$, and $M$ is the midpoint of the edge $(E H)$.
a) Prove that the line $A M$ is perpendicular to the line $M C$.
b) Find the length of the segment $(F N)$ knowing that $(A C M) \cap F G=\{N\}$.
c) Find the tangent of the angle between the planes $(A B D)$ and $(A C M)$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let $K(A B D)$ denote the volume of the solid formed by the rotation of triangle $A B D$ around the axis $O_{1} O_{2}$, and similarly, $K\left(A E_{1} E_{2} B\right)$ denote the volume of the solid formed by the rotation of $A E_{1} E_{2} B$, etc.
$$
K(A B D)=K\left(A E_{1} E_{2} B\right)-K\left(E_{1} A D\right)-K\left(E_{2} B D\right)
$$
and
$$
\begin{gathered}
K(\widehat{A D B})=K\left(A E_{1} E_{2} B\right)-K\left(E_{1} \widehat{A D}\right)-K\left(E_{2} \widehat{B D}\right) \\
O_{1} E_{1}=R \cos \alpha, O_{2} R_{2}=r \cos \alpha, A E_{1}=R \sin \alpha \text {, and } B E_{2}=r \sin \alpha
\end{gathered}
$$
But
$$
\cos \alpha=\frac{R-r}{R+r} \text { and } \sin \alpha=\frac{2 \sqrt{R r}}{R+r}
$$
so
$$
\begin{gathered}
E_{1} E_{2}=R+r-O_{1} E_{1}+O_{2} E_{2}=R+r-R \frac{R-r}{R+r}+r \frac{R-r}{R+r}=\frac{4 R r}{R+r}, \\
A E_{1}=\frac{2 R}{R+r} \sqrt{R r} \text { and } B E_{2}=\frac{2 R}{R+r} \sqrt{R r} \\
K\left(A E_{1} E_{2} B\right)=\frac{E_{1} E_{2} \pi}{3}\left[A E_{1}^{2}+A E_{1} \cdot B E_{2}+B E_{2}^{2}\right]= \\
=\frac{16 \pi}{3} \frac{R^{2} r^{2}}{(R+r)^{3}}\left(R^{2}+R r+r^{2}\right) \\
K\left(E_{1} A D\right)=\frac{A E_{1}^{2} \cdot E_{1} D}{3} \pi=\frac{A E_{1}^{2}\left(R-O_{1} E_{1}\right)}{3} \pi=\frac{8}{3} \pi \frac{R^{4} r^{2}}{(R+r)^{3}} \\
K\left(E_{2} B D\right)=\frac{B E_{2}^{2} \cdot E_{2} D}{3} \pi=\frac{B E_{2}^{2}\left(r+O_{2} E_{2}\right)}{3} \pi=\frac{8}{3} \pi \frac{R^{2} r^{4}}{(R+r)^{3}} \\
K\left(E_{1}\right)=\frac{E_{1} D^{2} \cdot \pi}{3}\left(3 R-E_{1} D\right)=\frac{\pi}{3}\left(R-O_{1} E_{1}\right)^{2}\left(2 R+O_{1} E_{1}\right)= \\
=\frac{4}{3} \pi \frac{R^{3} r^{2}}{(R+r)^{3}}(3 R+r),
\end{gathered}
$$
and
$$
\begin{aligned}
K\left(E_{2} \widehat{B D}\right)=\frac{E_{2} D^{2} \cdot \pi}{3} & =\left(3 r-E_{2} D\right)=\frac{\pi}{3}\left(r+O_{2} E_{2}\right)^{2}\left(2 r-O_{2} E_{2}\right)= \\
& =\frac{4}{3} \pi \frac{R^{2} r^{3}}{(R+r)^{3}}(R+3 r)
\end{aligned}
$$
Therefore,
$$
\begin{gathered}
K(A B D)=\frac{16 \pi}{3} \frac{R^{2} r^{2}}{(R+r)^{3}}\left(R^{2}+R r+r^{2}\right)-\frac{8}{3} \pi \frac{R^{4} r^{2}}{(R+r)^{3}}- \\
-\frac{8}{3} \pi \frac{R^{2} r^{4}}{(R+r)^{3}}=\frac{8}{3} \pi \frac{R^{2} r^{2}}{R+r}
\end{gathered}
$$
and
$$
\begin{gathered}
K(\widehat{A D B})=\frac{16 \pi}{3} \frac{R^{2} r^{2}}{(R+r)^{3}}\left(R^{2}+R r+r^{2}\right)-\frac{4 \pi}{3} \frac{R^{3} r^{2}}{(R+r)^{3}}(3 R+r)- \\
-\frac{4 \pi}{3} \frac{R^{2} r^{3}}{(R+r)^{2}}(R+3 r)=\frac{4}{3} \pi \frac{R^{2} r^{2}}{(R+r)}
\end{gathered}
$$
thus the ratio of the volumes of the two solids of revolution is
$$
K(A B D): K(\widehat{A D B})=\frac{8}{3} \pi \frac{R^{2} r^{2}}{(R+r)}: \frac{4}{3} \pi \frac{R^{2} r^{2}}{(R+r)}=2: 1
$$
(Imre Dömény, Budapest.)
The problem was also solved by: Fodor H., Földes R., Fuchs I., Heimlich P., Jánosy Gy., Krampera Gy., Rosenberg J., Ruwald S., Székely J., Tandlich E. | \boxed{2:1} | Two circles touch each other externally at point $D$. We draw one of the external tangents $AB$ to these circles. We rotate the entire figure around the central line of the two circles. What is the ratio of the volumes of the solids formed by the rotation of the region defined by the lines $AB$, $AD$, and $BD$ (forming triangle $ABD$) and the region defined by the arcs $AD$ and $BD$ and the line $AB$? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
6. $\frac{5 \sqrt{2}}{6}$.
By symmetry, without loss of generality, assume point $M$ is inside $\triangle E F G$ (including the boundary).
Unfold the square $A B F E$ and the square $E F G H$ into a plane, as shown in Figure 3. Then point $M$ lies on the perpendicular bisector of $A G$.
When $M$ is the intersection of the perpendicular bisector of $A G$ and $E G$, $l(M, G)$ is maximized.
Since $M A = M G$, we have,
$$
\begin{aligned}
& \left(2-\frac{\sqrt{2}}{2} M G\right)^{2}+\left(1-\frac{\sqrt{2}}{2} M G\right)^{2}=M G^{2} \\
\Rightarrow & M G=\frac{5 \sqrt{2}}{6} .
\end{aligned}
$$
Therefore, the maximum value of $l(M, G)$ is $\frac{5 \sqrt{2}}{6}$. | \boxed{\dfrac{5\sqrt{2}}{6}} | 6. As shown in Figure 1, the cube $A B C D-E F G H$ has an edge length of 1, and point $M$ is on the surface $E F G H$ of the cube. Define a path as a route where every point lies on the surface of the cube. It is known that the shortest path length from point $M$ to $A$, $l(M, A)$, is equal to the shortest path length from point $M$ to $G$, $l(M, G)$. Then the maximum value of $l(M, G)$ is . $\qquad$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
For the higher permutation group, in addition to $p_{1}, p_{2}, p_{3}, p_{4}$ as in the previous example, it also includes:
(5) A $180^{\circ}$ flip along the $x^{\prime} x$ axis, corresponding to the permutation:
$$
\begin{array}{l}
p_{5}=\left(c_{1}\right)\left(c_{2}\right)\left(c_{3} c_{6}\right)\left(c_{4} c_{5}\right)\left(c_{7} c_{10}\right)\left(c_{8} c_{9}\right)\left(c_{11} c_{12}\right)\left(c_{13}\right)\left(c_{15}\right)\left(c_{14} c_{16}\right) \\
c_{1}\left(p_{5}\right)=4
\end{array}
$$
(6) A $180^{\circ}$ flip along the $y^{\prime} y$ axis, corresponding to the permutation:
$$
\begin{array}{l}
p_{6}=\left(c_{1}\right)\left(c_{2}\right)\left(c_{3} c_{4}\right)\left(c_{5} c_{6}\right)\left(c_{7} c_{8}\right)\left(c_{9} c_{10}\right)\left(c_{11} c_{12}\right)\left(c_{13} c_{15}\right)\left(c_{14}\right)\left(c_{16}\right) \\
c_{1}\left(p_{6}\right)=4
\end{array}
$$
(7) A $180^{\circ}$ flip along the diagonal 13, corresponding to the permutation:
$$
\begin{array}{l}
p_{7}=\left(c_{1}\right)\left(c_{2}\right)\left(c_{3}\right)\left(c_{4} c_{6}\right)\left(c_{5}\right)\left(c_{7}\right)\left(c_{8} c_{10}\right)\left(c_{9}\right)\left(c_{11}\right)\left(c_{12}\right)\left(c_{13} c_{14}\right)\left(c_{15}\right)\left(c_{16}\right), \\
c_{1}\left(p_{7}\right)=8
\end{array}
$$
(8) A $180^{\circ}$ flip along the diagonal 24, corresponding to the permutation:
$$
\begin{array}{l}
p_{8}=\left(c_{1}\right)\left(c_{2}\right)\left(c_{3} c_{5}\right)\left(c_{4}\right)\left(c_{6}\right)\left(c_{7} c_{9}\right)\left(c_{8}\right)\left(c_{10}\right)\left(c_{11}\right)\left(c_{12}\right)\left(c_{13} c_{16}\right)\left(c_{14} c_{15}\right) \\
c_{1}\left(p_{8}\right)=8
\end{array}
$$
According to Burnside's formula, the number of different equivalence classes is:
$$
l=\frac{1}{8} \times[16+2+4+2+4+4+8+8]=\frac{1}{8} \times[48]=6
$$
The 6 different schemes are consistent with Figure 4-7, but their meanings are not entirely the same, including various types of flips. Please understand them on your own. | \boxed{6} | Example 4-7 A circular ring, with beads of red or blue installed at the positions of $0^{\circ}, 90^{\circ}, 180^{\circ}, 270^{\circ}$ in a clockwise direction, how many different equivalence classes are there? That is, how many different schemes are there? Schemes that can be made to coincide by rigid body motion are considered the same.
This problem can also be viewed as the case of the four squares in the previous example, as shown in Figure 4-8, but the square is a transparent glass plate, colored with red and blue. How many coloring schemes are there? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: $36^{\circ}$.
Solution. Inscribed angles $D A F$ and $D E F$ subtend the arc $D F$ and are therefore equal. Given that $\angle F A B = \angle A B C$, it follows that $B C \parallel A F$. Then the line $\ell$, passing through point $M$ and perpendicular to the tangent $B C$, contains the diameter of the circle $\omega$ and is perpendicular to its chord $A F$. Thus, points $A$ and $F$ are symmetric with respect to $\ell$, meaning that $\ell$ is the axis of symmetry of the trapezoid $B C A F$. This implies that $A C = B F$ and $A M = F M$. Therefore, triangles $B F M$ and $C A M$ are congruent by the three sides, so $\angle B F M = \angle E A M$. Additionally, $\angle E F M = \angle E A M$ as inscribed angles subtending the same chord. Thus, $F M$ is the angle bisector of $\angle B F E$, and
$$
\angle E A M = \angle E F M = \frac{1}{2} \angle B F E = 36^{\circ}.
$$
Notice now that the angle between the chord $E M$ and the tangent $B C$ is equal to the inscribed angle subtending $E M$. Therefore, $\angle C M E = \angle E A M = 36^{\circ}$. | \boxed{36^\circ} | 3. In triangle $ABC$, the median $AM$ is drawn. Circle $\alpha$ passes through point $A$, touches line $BC$ at point $M$, and intersects sides $AB$ and $AC$ at points $D$ and $E$ respectively. On the arc $AD$ that does not contain point $E$, a point $F$ is chosen such that $\angle BFE = 72^{\circ}$. It turns out that $\angle DEF = \angle ABC$. Find the angle $\angle CME$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
2. The answer is (B). There are three possible cases to analyze:
(a) 150 is a multiple of 2 and 3, but not of 5;
(b) 150 is a multiple of 2 and 5, but not of 3;
(c) 150 is a multiple of 3 and 5, but not of 2.
First, we notice that 1500 is a multiple of $2 \cdot 3 \cdot 5=30$, and there are no other multiples of 10 or 15 between 1500 and 1509. The last two cases are therefore excluded. The only multiple of 6 that is not also a multiple of 5 among the 10 possible values is 1506, so only the first of the three cases can be correct, and there is only one possible choice for the missing digit. | \boxed{B} | 2. An ancient text states that Methuselah lived 150 years, where the symbol replaces the unit digit, which scholars cannot read. Fortunately, we possess three other manuscripts about the life of Methuselah; the first claims that he lived an even number of years, the second that he lived a number of years that is a multiple of 3, the third that he lived a number of years that is a multiple of 5. Knowing that exactly one of these three manuscripts contains false information, how many different digits could be behind the symbol?
(A) None
(B) One
(C) Two
(D) Three
(E) Four | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Analysis】The surface area is increased by two rectangles compared to the original cube.
As shown in the figure, $M J=12-4-3=5, M I=12$, by the Pythagorean theorem, $I J=13$ so the area of the new rectangle is $12 \times 13=156$
The sum of the two parts of the surface area is $6 \times 12^{2}+2 \times 156=1176$ square centimeters. | \boxed{1176} | 13. As shown in the figure, a cube with an edge length of 12 cm is cut with one slice. The cut is made along $I J$ and exits at $L K$, such that $A I=D L=4$ cm, $J F=K G=3$ cm, and the cross-section $I J K L$ is a rectangle. The cube is cut into two parts, and the sum of the surface areas of these two parts is $\qquad$ square centimeters. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
4. Ans: 2844
If only one digit appears, then there are 9 such numbers. If the two digits that appear are both nonzero, then the number of such numbers is
$$
2 \times\binom{ 7}{3}\binom{9}{2}=2520 .
$$
If one of two digits that appear is 0 , then the number of such numbers is
$$
\left(\binom{6}{4}+\binom{6}{3}\right) \times\binom{ 9}{1}=315
$$
Hence the answer is $9+2520+315=2844$. | \boxed{2844} | 4. In each of the following 7-digit natural numbers:
1001011,5550000,3838383,7777777,
every digit in the number appears at least 3 times. Find the number of such 7 -digit natural numbers. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let's denote the centers of the circles by $A, B$, their intersection points by $C, D$, and the angle $ACB$ by $x$.
The sought distance
$$
AB = 2 AC \sin \frac{x}{2} = 4 \sin \frac{x}{2}
$$
The area of the common part of the two circles is
$$
T = AC^2 (\pi - x) - 2 T_{ABC} = 4 (\pi - x) - 4 \sin x
$$
This is how much smaller the area covered together is compared to twice the area of one circle:
$$
\begin{gathered}
4 (\pi - x) - 4 \sin x = 8 \pi - 6 \pi \\
x = \frac{\pi}{2} - \sin x
\end{gathered}
$$
It is easy to see that $ABx_8 \quad \text{and} \quad 1.5708 - 0.7446 = 0.8262 < x_9
$$
Thus, the root lies between these two values, somewhere where $\sin x_{10} = 0.74$. Here (based on the tables of cyclometric functions) $x_{10} = 0.8331$, which is smaller at $\left(\pi / 2 - \sin x_{10}\right)$ than $x_{9}$ at $\left(\pi / 2 - \sin x_{9}\right)$. Therefore,
$$
x_8 = 0.83 < x < x_{10} = 0.8331
$$
It is not advisable to directly read the appropriate bounds for $\sin x / 2$ from the table; a more accurate estimate is provided by the
$$
\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}
$$
relationship. From this, we get that
$$
\begin{gathered}
2.600 = 8(1 - 0.675) < 16 \sin^2 \frac{x}{2} < 8 \left(1 - \sqrt{1 - 0.74^2}\right) < 8(1 - 0.672) = 2.624 \\
1.612 < AB < 1.620
\end{gathered}
$$
Thus, the two sought fractions are 161/100 and $162 / 100$. | \boxed{\frac{161}{100} < AB < \frac{162}{100}} | Two circular disks, each with a radius of 2 units, lie on top of each other in such a way that together they cover an area of $6 \pi$ square units. Let's confine the distance between the centers of the two disks between two fractions, where the denominators are 100, the numerators are integers, and the difference between the numerators is 1. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 20 We claim that $O$ should be the orthocenter of the triangle $A B C$. If $O$ is not on an altitude of $\triangle A B C$, suppose (without loss of generality) that $\overline{A O}$ is not perpendicular to $\overline{B C}$. We can
rotate $A$ around $O$, leaving $B$ and $C$ fixed, to make $\overline{A O}$ perpendicular to $\overline{B C}$, which strictly increases the area. Therefore, if $[A B C]$ is maximal then $\triangle A B C$ is an isosceles triangle with orthocenter $O$ and base $\overline{A B}$.
Let $F$ be the foot of the perpendicular from $C$ to $\overline{A B}$. Since $\angle F O A$ and $\angle C O E$ are vertical, $\angle F A O=$ $\angle O C E$. Then $\triangle F A O$ is similar to $\triangle F C B$, so we have $\frac{A F}{O F}=\frac{C F}{B F}=\frac{O F+7}{A F}$, so $A F^{2}=O F^{2}+7 \cdot O F$. Since $A F^{2}=225-O F^{2}, 2 \cdot O F^{2}+7 \cdot O F-225=0$, so $O F \stackrel{=}{=}$. Then $A F=12$, so $A B=24$ and $B C=20$. Thus, the length of the shortest side of $\triangle A B C$ is 20 . | \boxed{20} | 8. [6] Let $O$ be the point $(0,0)$. Let $A, B, C$ be three points in the plane such that $A O=15, B O=15$, and $C O=7$, and such that the area of triangle $A B C$ is maximal. What is the length of the shortest side of $A B C$ ? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Jean writes five tests and achieves the marks shown on the graph. What is her average mark on these five tests?
(A) 74
(B) 76
(C) 70
(D) 64
(E) 79
Solution
Jean's average is $\frac{80+70+60+90+80}{5}=\frac{380}{5}=76$.
ANSWER: (B) | $\boxed{76}$ | Jean writes five tests and achieves the marks shown on the graph. What is her average mark on these five tests?
(A) 74
(D) 64
(B) 76
(E) 79
(C) 70
 | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
4. 49894
4. Let the answer be $\overline{a b c b a}$. Note that
$$
\overline{a b c b a}=10001 a+1010 b+100 c=101(99 a+10 b+c)+2 a-c
$$
For the number to be divisible by 101 , we must have $2 a-c=0$. For the number to be largest, we may take $a=4, c=8$ and $b=9$. This gives the answer is 49894 . | $\boxed{49894}$ | 4. A palindrome is a positive integer which is the same when reading from the right hand side or from the left hand side, e.g. 2002. Find the largest five-digit palindrome which is divisible by 101 .
(1 mark)
若某正整數不論從左面或右面讀起皆相同（例如：2002），則該數稱為「回文數」。求可被 101 整除的最大五位回文數。 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
1. The answer is $c=\frac{1000}{91}=11-\frac{11}{1001}$. Clearly, if $c^{\prime}$ works, so does any $c>c^{\prime}$. First we prove that $c=11-\frac{11}{1001}$ is good.
We start with 100 empty boxes. First, we consider only the coins that individually value more than $\frac{1000}{1001}$. As their sum cannot overpass 1000, we deduce that there are at most 1000 such coins. Thus we are able to put (at most) 10 such coins in each of the 100 boxes. Everything so far is all right: $10 \cdot \frac{1000}{1001} < 11 - \frac{11}{1001}$.
for all $i=1,2, \ldots, 100$. Then
$$
x_{1}+x_{2}+\cdots+x_{100}+100 x > 100 \cdot\left(11-\frac{11}{1001}\right) .
$$
But since $1000 \geqslant x_{1}+x_{2}+\cdots+x_{100}+x$ and $\frac{1000}{1001} \geqslant x$ we obtain the contradiction
$$
1000+99 \cdot \frac{1000}{1001} > 100 \cdot\left(11-\frac{11}{1001}\right) \Longleftrightarrow 1000 \cdot \frac{1100}{1001} > 100 \cdot 11 \cdot \frac{1000}{1001} .
$$
Thus the algorithm does not fail and since we have finitely many coins, we will eventually reach to a happy end.
Now we show that $c=11-11 \alpha$, with $1>\alpha>\frac{1}{1001}$ does not work.
Take $r \in\left[\frac{1}{1001}, \alpha\right)$ and let $n=\left\lfloor\frac{1000}{1-r}\right\rfloor$. Since $r \geqslant \frac{1}{1001}$, then $\frac{1000}{1-r} \geqslant 1001$, therefore $n \geqslant 1001$.
Now take $n$ coins each of value $1-r$. Their sum is $n(1-r) \leqslant \frac{1000}{1-r} \cdot(1-r)=1000$. Now, no matter how we place them in 100 boxes, as $n \geqslant 1001$, there exist 11 coins in the same box. But $11(1-r)=11-11 r>11-11 \alpha$, so the constant $c=11-11 \alpha$ indeed does not work. | \boxed{\dfrac{1000}{91}} | In an exotic country, the National Bank issues coins that can take any value in the interval $[0,1]$. Find the smallest constant $c>0$ such that the following holds, no matter the situation in that country:
Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000, can split those coins into 100 boxes, such that the total value inside each box is at most $c$.
## Proposed by Romania | Combinatorics | AI-MO/NuminaMath-1.5/olympiads_ref | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
If t is the side of the triangle, and s is the side of the square, then
\[3t-4s=1989\]
\[t-s=d\]
Solving the first equation for t gives
\[t = \frac{4s+1989}{3}\]
Substituting into the second equation,
\[\frac{4s+1989}{3} - s = d\]
\[\frac{s+1989}{3} = d\]
\[s+1989 = 3d\]
If s = 0, d = 663. But s has to be greater than 0, so the first 663 positive integers aren't possible $\to\boxed{\textbf{(D)}}$ | \boxed{D} | The perimeter of an equilateral triangle exceeds the perimeter of a square by $1989 \ \text{cm}$. The length of each side of the triangle exceeds the length of each side of the square by $d \ \text{cm}$. The square has perimeter greater than 0. How many positive integers are NOT possible value for $d$?
$\text{(A)} \ 0 \qquad \text{(B)} \ 9 \qquad \text{(C)} \ 221 \qquad \text{(D)} \ 663 \qquad \text{(E)} \ \infty$ | Algebra | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
21. B
Each 'triple' consisting of a cell and the two cells immediately below can have at most two odds (for if the bottom two are both odd, the one above is even, so they cannot be all odd). The whole diagram can be dissected into six of these (shaded) triples as shown in the top diagram, with three other (white) cells left over. These six triples have at most $6 \times 2=12$ odds between them. Moreover, the three remaining white cells cannot all be odd; if we assign the values $A$ and $C$ to the lowest of these white cells, and $B$ to the cell between them, then the cells above have values $A+B$ and $B+C$. The top white cell then contains $A+2 B+C$, which is even when $A$ and $C$ are both odd. Hence the three white cells have at most two odds, giving the whole diagram at most $12+2=14$ odds.
The second diagram shows one possible way of achieving this maximum of 14 odds. | \boxed{14} | 21. Carlos wants to put numbers in the number pyramid shown in such a way that each number above the bottom row is the sum of the two numbers immediately below it. What is the largest number of odd numbers that Carlos could put in the pyramid?
A 13
B 14
C 15
D 16
E 17 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. Let's consider the initial seating arrangement. From the statement of each Liar, it follows that none of those sitting in front of or behind him is a Liar, i.e., each Liar is surrounded by two Knights. Knights and Liars cannot alternate, as in front of or behind each Knight, there must be at least one Knight. Therefore, Liars and Knights in the seating arrangement form sequences ...KKLKKLKK... Between two Knights, whose neighbors are Liars, any number of Knights can be added in the initial seating, for example, ...KKLKKKKKKLKK... Therefore, the minimum number of Knights will only be in the case where the seating is in triplets ...KKLKK.L... Initially, 39 seats are occupied, this number is divisible by 3, so the maximum number of Liars is 13, and the minimum number of Knights is 26. Note that if one Liar is replaced by a Knight, then we get 12 Liars, for the formation of triplets, 24 Knights are needed, and 3 more Knights can be placed arbitrarily between other Knights.
The problem states that the seating is redone when another Liar appears. Let's see how 40 people, including the additional Liar, can be seated. From his statement, it follows that it is impossible to seat them so that at least one of those sitting in front of or behind each belongs to his team. If we had an extra triplet of Knights, which we placed freely during the first seating, then the additional Liar together with two of them could form a triplet ..KKL.. Therefore, there could not have been more than 26 Knights in the initial seating.
Answer. The team of Knights consisted of 26 people. | \boxed{26} | 4. Once, a team of Knights and a team of Liars met in the park and decided to ride the circular carousel, which can accommodate 40 people (the "Chain" carousel, where everyone sits one behind the other). When they sat down, each person saw two others, one in front of them and one behind them, and said: "At least one of the people sitting in front of me or behind me belongs to my team." One seat remained empty, and they invited another Liar. This Liar said: "Together with me, we can arrange ourselves on the carousel so that this rule is satisfied again." How many people were in the team of Knights? (A Knight always tells the truth, a Liar always lies.) | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$$
\begin{array}{l}
2=2 \times 1 \\
6=2(1+2) \\
12=2(1+2+3) \\
20=2(1+2+3+4) \\
30=2(1+2+3+4+5)
\end{array}
$$
The last number of the $50^{\text {th }}$ group
$$
\begin{array}{l}
=2(1+2+\ldots+50) \\
=2 \cdot \frac{1}{2} \cdot 50 \cdot(1+50)=2550
\end{array}
$$
There are 50 numbers in the $50^{\text {th }}$ group.
The first number of the $50^{\text {th }}$ group $=2550-2(50-1)=2452$
$$
\begin{array}{l}
2452+2454+\ldots+2550=50 P \\
\frac{1}{2} \cdot 50 \cdot(2452+2550)=50 P \\
P=2501
\end{array}
$$ | \boxed{2550} | SG. 1 Find the last number of the $50^{\text {th }}$ group.
SG. 2 Find the first number of the $50^{\text {th }}$ group.
SG. 3 Find $P$ if the sum of the numbers in the $50^{\text {th }}$ group is $50 P$. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
15. Consider the plane passing through the axes of symmetry of the given cylinders. The cross-section of the common part of the cylinders by this plane is a square. If we draw a plane parallel to this one, it is clear from the figure that the cross-section will also be a square, and the circle inscribed in this square will
represent the cross-section of the plane with the sphere inscribed in the common part of the cylinders. Therefore, the volume of the common part of our cylinders is related to the volume of the inscribed sphere as the area of the square is to the area of the inscribed circle.
$$
V=\frac{4 \pi r^{3}}{3} \cdot \frac{4}{\pi}=\frac{16 r^{3}}{3}=\frac{16}{3} \mathrm{~cm}^{3}
$$
[L. Moser, M. M., 25, 290 (May 1952).]
From this, it follows that the common part of the two given cylinders can be divided into infinitely small pyramids, the vertices of which lie at the point of intersection of the axes of the cylinders, and their bases are elements of the cylinders. All such pyramids have a height of one unit. Therefore, the surface area of the common part of the cylinders is $16 \mathrm{~cm}^{2}$.
[J. H. Butchart, M. M., 26, 54 (September 1952).] | \boxed{\dfrac{16}{3}} | 15. Intersecting Cylinders. The axes of symmetry of two straight circular cylinders, each with a diameter of 2 cm, intersect at a right angle. What is the volume of the common part of these cylinders? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
6. 27 .
Let the three vertices of $\triangle A B C$ have coordinates
$$
A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right), C\left(x_{C}, y_{C}\right) \text {. }
$$
The focus of the parabola $y^{2}=4 \sqrt{2} x$ is $F(\sqrt{2}, 0)$.
From the given conditions, we have
$$
\left\{\begin{array}{l}
x_{A}+x_{B}+x_{C}=3 \sqrt{2}, \\
y_{A}+y_{B}+y_{C}=0 .
\end{array}\right.
$$
Then $y_{A}^{2}+y_{B}^{2}+y_{C}^{2}$
$$
\begin{array}{l}
=4 \sqrt{2}\left(x_{A}+x_{B}+x_{C}\right)=24, \\
y_{B} y_{C}+y_{C} y_{A}+y_{A} y_{B} \\
=\frac{1}{2}\left(y_{A}+y_{B}+y_{C}\right)^{2}-\frac{1}{2}\left(y_{A}^{2}+y_{B}^{2}+y_{C}^{2}\right) \\
=-12 .
\end{array}
$$
Therefore, $y_{A}^{4}+y_{B}^{4}+y_{C}^{4}$
$$
\begin{aligned}
= & \left(y_{A}^{2}+y_{B}^{2}+y_{C}^{2}\right)^{2}-2\left(y_{B} y_{C}+y_{C} y_{A}+y_{A} y_{B}\right)^{2}+ \\
& 4 y_{A} y_{B} y_{C}\left(y_{A}+y_{B}+y_{C}\right) \\
= & 24^{2}-2 \times 12^{2}=288 .
\end{aligned}
$$
Accordingly,
$$
\begin{array}{l}
|F A|^{2}+|F B|^{2}+|F C|^{2} \\
=\left(x_{A}-\sqrt{2}\right)^{2}+y_{A}^{2}+\left(x_{B}-\sqrt{2}\right)^{2}+ \\
y_{B}^{2}+\left(x_{C}-\sqrt{2}\right)^{2}+y_{C}^{2} \\
= \frac{y_{A}^{4}+y_{B}^{4}+y_{C}^{4}}{(4 \sqrt{2})^{2}}-2 \sqrt{2}\left(x_{A}+x_{B}+x_{C}\right)+ \\
3 \times 2+y_{A}^{2}+y_{B}^{2}+y_{C}^{2} \\
= 9-12+6+24=27 .
\end{array}
$$ | \boxed{27} | 6. If the centroid of the inscribed $\triangle A B C$ of the curve $y^{2}=4 \sqrt{2} x$ is its focus $F$. Then
$$
|F A|^{2}+|F B|^{2}+|F C|^{2}=
$$
$\qquad$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$$
\begin{array}{l}
a=\frac{1}{9}+\frac{2}{90}+\frac{3}{900}+\cdots+\frac{9}{900000000}=\frac{100000000+20000000+3000000+\cdots+9}{900000000} \\
a=\frac{123456789}{900000000}=\frac{13717421}{100000000}=0.13717421 \\
\frac{\text { Area of shaded part }}{\text { Area of the circle }}=\frac{\pi-2}{\pi-2+3 \pi+2}=\frac{\pi-2}{4 \pi} \\
\frac{\pi(1)^{2} \cdot \frac{b}{360}-\frac{1}{2}(1)^{2} \sin b^{\circ}}{\pi(1)^{2}}=\frac{\pi-2}{4 \pi} \Rightarrow \frac{\pi b}{90}-2 \sin b^{\circ}=\pi-2 ; b=90 \\
\text { L }) \longrightarrow \mathrm{S}_{0} \\
0 \quad \frac{1}{3} \quad \frac{2}{3} \quad 1 \\
1=(\quad) \longrightarrow(\quad)-1 S_{1} \\
\begin{array}{llllllll}
0 & \frac{1}{9} & \frac{2}{9} & \frac{1}{3} & \frac{2}{3} & \frac{7}{9} & \frac{8}{9} & 1
\end{array} \\
\text { |-( )-( )-( )-( )-( )-( )-( )-| S S } \\
\end{array}
$$
The total length in $S_{0}=\frac{2}{3}$
The total length in $S_{1}=4 \times \frac{1}{9}=\frac{4}{9}$
The total length in $S_{2}=8 \times \frac{1}{27}=\frac{8}{27}$
Deductively, the total length in $S_{5}=2^{6} \times \frac{1}{3^{6}}=\frac{64}{729}$
The total length removed in $S_{5}=1-\frac{64}{729}=\frac{665}{729}$
Sum of integers code as $102,104, \ldots, 200$ is $51+52+\ldots+100$
Sum of integers code as $101,103, \ldots, 199$ is $-50-51-\ldots-99$
Sum of all integers $=1+1+\ldots+1(50$ times $)=50$ | G4.4: \boxed{50} | G4.1 If $0 . \dot{1}+0.0 \dot{2}+0.00 \dot{3}+\ldots+0.00000000 \dot{9}=a$, find the value of $a$ (Give your answer in decimal)
G4.2 The circle in the figure has centre $O$ and radius $1, A$ and $B$ are points on the circle. Given that $\frac{\text { Area of shaded part }}{\text { Area of unshaded part }}=\frac{\pi-2}{3 \pi+2}$ and $\angle A O B=b^{\circ}$, find the value of $b$.
G4.3 A sequence of figures $S_{0}, S_{1}, S_{2}, \ldots$ are constructed as follows. $S_{0}$ is obtained by removing the middle third of $[0,1]$ interval; $S_{1}$ by removing the middle third of each of the two intervals in $S_{0} ; S_{2}$ by removing the middle third of each of the four intervals in $S_{1} ; S_{3}, S_{4}, \ldots$ are obtained similarly. Find the total length $c$ of the intervals removed in the construction of $\mathrm{S}_{5}$ (Give your answer in fraction).
G4.4 All integers are coded as shown in the following table. If the sum of all integers coded from 101 to 200 is $d$, find the value of $d$.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline Integer & $\ldots$ & $\ldots$ & -3 & -2 & -1 & 0 & 1 & 2 & 3 & $\ldots$ & $\ldots$ \\
\hline Code & $\ldots$ & $\ldots$ & 7 & 5 & 3 & 1 & 2 & 4 & 6 & $\ldots$ & $\ldots$ \\
\hline
\end{tabular} | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
# Solution:
Convert all speeds to SI units:
$36 \kappa m / h = 10 m / s$
$54 \kappa m / h = 15 m / s$
$72 \kappa m / h = 20 m / s$
The acceleration of a car of this make:
$a=\frac{v^{2}}{2 S}=\frac{20^{2}}{2 \cdot 40}=5 \mathrm{M} / s^{2}$
Distance traveled by the first car before stopping:
$S_{1}=\frac{v_{2}^{2}}{2 a}=\frac{10^{2}}{2 \cdot 5}=10 \mathrm{M}$
Distance traveled by the second car before stopping:
$S_{2}=v_{2} \cdot 2+\frac{v_{2}^{2}}{2 a}=15 \cdot 2+\frac{15^{2}}{2 \cdot 5}=30+22.5=52.5 \mathrm{M}$
Therefore, the distance between the cars should be no less than:
$l=S_{2}-S_{1}=52.5-10=42.5$ meters
(3 points) | \boxed{42.5} | # Problem №1 (15 points)
Two identical cars are driving in the same direction. The speed of one is $36 \kappa \mu / h$, and the other is catching up at a speed of $54 \mathrm{km} / h$. It is known that the reaction time of the driver of the rear car to the activation of the brake lights of the front car is 2 seconds. What should be the distance between the cars so that they do not collide if the first driver decides to brake sharply? For a car of this make, the braking distance is 40 meters at a speed of $72 \kappa \mu / h$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$, $y$, and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also, $P$ lies on the extensions of segments $DM,$ $NQ,$ and $CB$ so the solid $DPCN = DMBCQN + MPBQ$ is a right triangular pyramid. Note also that pyramid $MPBQ$ is similar to $DPCN$ with scale factor $.5$ and thus the volume of solid $DMBCQN,$ which is one of the solids bounded by the cube and the plane, is $[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].$ But the volume of $DPCN$ is simply the volume of a pyramid with base $1$ and height $.5$ which is $\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.$ So $[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.$ Note, however, that this volume is less than $.5$ and thus this solid is the smaller of the two solids. The desired volume is then $[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}$
Another way to finish after using coordinates: Take the volume of DMBCNQ as the sum of the volumes of DMBQ and DBCNQ. Then, we have the sum of the volumes of a tetrahedron and a pyramid with a trapezoidal base. Their volumes, respectively, are $\tfrac{1}{48}$ and $\tfrac{1}{8}$, giving the same answer as above. $\blacksquare$ ~mathtiger6 | \boxed{89} | Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Geometry | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The scores of all ten runners must sum to $55$. So a winning score is anything between $1+2+3+4+5=15$ and $\lfloor\tfrac{55}{2}\rfloor=27$ inclusive. It is easy to check that this range is covered by considering $1+2+3+4+x$, $1+2+3+x+10$ and $1+2+x+9+10$, so the answer is $\boxed{13}$. | \boxed{13} | In a certain cross country meet between 2 teams of 5 runners each, a runner who finishes in the $n$th position contributes $n$ to his teams score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?
(A) 10 (B) 13 (C) 27 (D) 120 (E) 126 | Combinatorics | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 9800.
Solution. Examples of unlimited movement on the board of 9800 turtles.
Example 1. Place the reptiles in the cells of a rectangle consisting of cells at the intersection of the 98 lower horizontals and 100 leftmost verticals. They will all move in the same way: first all to the right, then all up, then all to the left, and then all down.
After 4 moves, the situation will coincide with the initial one, so the movement can continue in a permitted manner indefinitely.
Example 2. The initial placement of turtles is the same as in Example 1, but the movement is organized such that they are divided into 2x2 square cells, in each of which they simultaneously move clockwise.
We will prove that it is impossible to properly place more than 9800 turtles on a 101x99 board. Suppose the contrary, that the turtles are placed on the board in some way such that they have the possibility to move around the board indefinitely in the manner specified in the condition and not end up on the same cell in a quantity of two or more simultaneously. Color the cells of the board in a checkerboard pattern, with the lower left cell, as usual, being black. This will result in 5000 black and 4999 white cells. Among the black cells, we will call the cells with both vertical and horizontal coordinates being odd "odd" cells, and those with both coordinates being even "even" cells. Note that for white cells, one of these coordinates is even, and the other is odd. There will be $51 \cdot 50=2550$ odd black cells and $50 \cdot 49=2450$ even black cells. Clearly, at any moment, the number of turtles on even black cells should not exceed 2450.
Note that when two consecutive moves are performed, the parity of both coordinates of each turtle changes, so after two moves, all turtles from odd black cells will move to even black cells and vice versa, while turtles from white cells will move back to white cells. Consequently, at any time, the number of turtles on all black cells does not exceed $2 \cdot 2450=4900$.
Finally, when one move is performed, turtles from black cells move to white cells and vice versa, so the total number of turtles on the entire board cannot exceed $2 \cdot 4900=9800$.
Remark. In fact, we have proven that if there are more than 9800 turtles on the board, they will be able to make no more than two moves without ending up on the same cell in a quantity of two or more simultaneously.
Grading criteria. Proof of the maximality of the number 9800: 6 points. Construction of an example for 9800 turtles: 1 point. Attempt to prove any other estimate except 9800: 0 points. | \boxed{9800} | 11.5. On some cells of a rectangular board of size 101 by 99, there is one turtle each. Every minute, each of them simultaneously crawls to one of the cells adjacent to the one they are in, by side. In doing so, each subsequent move is made in a direction perpendicular to the previous one: if the previous move was horizontal - to the left or right, then the next one will be vertical - up or down, and vice versa. What is the maximum number of turtles that can move around the board indefinitely such that at each moment, no more than one turtle is in any cell? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let the angle at centre be $\theta$ radians.
$$
\begin{array}{l}
2+1 \times \theta=3 \\
\theta=1 \text { radian }=\frac{180^{\circ}}{\pi} \approx 57.3^{\circ}=57^{\circ} \text { (correct to the nearest degree) }
\end{array}
$$ | \boxed{57} | I4.4 A wire of $c \mathrm{~cm}$ is bent to form a sector of radius $1 \mathrm{~cm}$. What is the angle of the sector in degrees (correct to the nearest degree)? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution.
$$
\begin{aligned}
& \frac{\left(1-\frac{\sqrt{3}-1}{2}\right) \cdot\left(\frac{\sqrt{3}-1}{2}+2\right)}{\left(\frac{\sqrt{3}-1}{2}\right)^{2} \cdot\left(\frac{\sqrt{3}-1}{2}+1\right)^{2}}=\frac{-\left(\frac{\sqrt{3}-1}{2}-1\right) \cdot\left(\frac{\sqrt{3}-1}{2}+2\right)}{\left(\frac{\sqrt{3}-1}{2} \cdot\left(\frac{\sqrt{3}-1}{2}+1\right)\right)^{2}}= \\
& =-\frac{\left(\frac{\sqrt{3}-1}{2}\right)^{2}+\frac{\sqrt{3}-1}{2}-2}{\left(\left(\frac{\sqrt{3}-1}{2}\right)^{2}+\frac{\sqrt{3}-1}{2}\right)^{2}}=-\frac{\frac{4-2 \sqrt{3}}{4}+\frac{\sqrt{3}-1}{2}-2}{\left(\frac{4-2 \sqrt{3}}{4}+\frac{\sqrt{3}-1}{2}\right)^{2}}= \\
& =-\frac{\frac{2-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}-2}{\left(\frac{2-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}\right)^{2}}=-\frac{\frac{2-\sqrt{3}+\sqrt{3}-1}{2}-2}{\left(\frac{2-\sqrt{3}+\sqrt{3}-1}{2}\right)^{2}}=6 .
\end{aligned}
$$
Answer: 6. | \boxed{6} | 2.141. $\frac{(1-y)(y+2)}{y^{2}(y+1)^{2}} ; \quad y=\frac{\sqrt{3}-1}{2}$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
20. 8
20. Let $a, b, c$ (where $a \leq b \leq c$ ) be the lengths of the sides of such a triangle. By Heron's formula, we have
$$
\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)\left(\frac{c+a-b}{2}\right)\left(\frac{b+c-a}{2}\right)}=2(a+b+c),
$$
which simplifies to
$$
(a+b-c)(c+a-b)(b+c-a)=64(a+b+c) .
$$
Observe that $a+b-c, c+a-b, b+c-a, a+b+c$ have the same parity and hence must be even. Set $a+b-c=2 r, c+a-b=2 s, b+c-a=2 t$, where $r, s, t$ are positive integers with $3 \leq r \leq s \leq t$. The above equation is then reduced to
$$
r s t=16(r+s+t) \text {. }
$$
As $t<r+s+t \leq 3 t$, we have $16<r s \leq 48$. Also, the above equation implies $t=\frac{16(r+s)}{r s-16}$. Hence we need to find $r$ and $s$ so that $16<r s \leq 48$ and $r s-16$ divides $16(r+s)$. We can then list out all such pairs of $(r, s)$ and compute the corresponding $t$, getting 8 different solutions, namely, $(r, s, t)=(3,6,72),(3,7,32),(3,8,22),(3,12,12),(4,5,36),(4,6,20),(4,8,12)$ and $(6,7,8)$. Therefore there are 8 such triangles. | \boxed{8} | 20. How many different triangles with integer side lengths are there such that the sum of the lengths of any two sides exceeds the length of the third side by at least 5 units, and that the area is numerically twice the perimeter? (Two triangles are regarded to be the same if they are congruent.)
(2 marks)
有多少個不同的三角形各邊的長度均為整數, 任何兩邊的長度之和均比第三邊長 5 單位或以上, 且其面積在數值上是其周界的兩倍?（兩個全等的三角形視為相同。) | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. Let $MK$ and $PH$ be segments connecting the midpoints of opposite sides of a convex quadrilateral $ABCD$, with $MK = PH$, $AC = 18$, and $BD = 7$.
We have: $MP \| AC$, $MP = \frac{1}{2} AC$ (as the midline of $\triangle ABC$); $HK \| AC$, $HK = \frac{1}{2} AC$ (as the midline of $\triangle ADC$). $\Rightarrow MP \| HK$, $MP = HK \Rightarrow MPKH$ is a parallelogram.
Since $MK = PH$, the quadrilateral $MPKH$ is a rectangle, with sides parallel to the diagonals $AC$ and $BD$ of the given quadrilateral $ABCD$, therefore $AC \perp BD$. This means that $S_{ABCD} = \frac{1}{2} AC \cdot BD = \frac{1}{2} \cdot 18 \cdot 7 = 63$ (sq.units).
Answer: 63.
| Criterion Description | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 15 |
| The solution contains a computational error, possibly leading to an incorrect answer, but the solution has a correct sequence of all steps. | 10 |
| In the solution, one or two formulas are correctly written, which could be the beginning of a possible solution. | 5 |
| The solution does not meet any of the criteria described above. | 0 |
| Maximum score | 15 | | \boxed{63} | 2. (15 points) Find the area of a convex quadrilateral with equal diagonals, if the lengths of the segments connecting the midpoints of its opposite sides are 13 and 7. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5 Substitute $x-2$ for $x$ in
$$
f(x+2)+f(2-x)=4
$$
to get
$$
f(x)+f(4-x)=4 .
$$
If the point $(x, y)$ is on the graph of $y=f(x)$, then $4-y=f(4-x)$, meaning the point $(x, y)$'s symmetric point about $(2,2)$, which is $(4-x, 4-y)$, is also on the graph of $y=f(x)$. Conversely, this is also true, so (1) is a true statement. Substitute $x-2$ for $x$ in
$$
f(x+2)=f(2-x)
$$
to get $f(x)=f(4-x)$. If the point $(x, y)$ is on the graph of $y=f(x)$, then $y=f(4-x)$, meaning the point $(x, y)$'s symmetric point about the line $x=2$, which is $(4-x, y)$, is also on the graph of $y=f(x)$, so (2) is a true statement. Given that (2) is a true statement, it is not difficult to deduce that (3) is also a true statement. Therefore, the answer is D. | \boxed{D} | (5) Let the function $f(x)$ be defined on $\mathbf{R}$, and consider the following three statements:
(1) The graph of the function $f(x)$ that satisfies the condition $f(x+2)+f(2-x)=4$ is symmetric about the point $(2, 2)$;
(2) The graph of the function $f(x)$ that satisfies the condition $f(x+2)=f(2-x)$ is symmetric about the line $x=2$;
(3) The graphs of the functions $f(x-2)$ and $f(-x+2)$ in the same coordinate system are symmetric about the line $x=2$.
The number of true statements is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$7(1,2]$ Prompt: (1) When $F$ is on the line segment $C B$ ( $F$ does not coincide with $C$), from $|P D|=|P F|$, we know
$$
|| P F|-| P A||=|| P D|-| P A||=|A D|=6=2 a \text {. }
$$
Thus, the locus of point $P$ is a hyperbola with foci at $A$ and $F$, and $|A F|=2 c, 6<2 c \leqslant 12$, so $3<c \leqslant 6$, hence $e=\frac{2 c}{2 a}=\frac{c}{3} \in(1,2]$.
(2) When $F$ is on the line segment $A C$ (not coinciding with $C$), from $|P D|=|P F|$ we get
$$
|P F|+|P A|=|P D|+|P A|=|A D|=6=2 a .
$$
Thus, the locus of point $P$ is an ellipse with foci at $A$ and $F$, or a circle when $A$ and $F$ coincide. In summary, the range of the eccentricity of the hyperbola when the locus of $P$ is a hyperbola is (1, 2].
(Changqi No.2 High School Liu Anning provided the problem) | \boxed{(1,2]} | (7) Given that $C$ and $F$ are two points on line segment $AB$, $AB=12$, $AC=6$, $D$ is any point on the circle with center $A$ and radius $AC$, the perpendicular bisector of line segment $FD$ intersects line $AD$ at point $P$. If the locus of point $P$ is a hyperbola, then the range of the eccentricity of this hyperbola is $\qquad$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
An example of coloring a square grid in four colors, such that any two cells at a distance of 6 are colored differently, is shown in the figure on the left.
Another example. Introduce a coordinate system on the square grid. It is sufficient to color the cells with even coordinate sums (the coloring of the remaining cells can be obtained from this by shifting 1 to the right). Points with coordinate sums divisible by 4 will be colored in two colors: if both coordinates are even - blue, if both are odd - red. Points with even but not divisible by 4 coordinate sums will be similarly colored in the remaining two colors. To prove that fewer colors cannot suffice, it is enough to consider the four cells shown in the figure on the right. Any two of them are located at a distance of 6 from each other, and therefore, all of them must be colored differently.
## Answer
In 4 colors. | \boxed{4} | [ Chessboards and chess pieces $]$ [ Examples and counterexamples. Constructions ] [ Evenness and oddness
Authors: Pechkovsky A.N., Itenberg I.
Given an infinite grid paper with a cell side equal to one. The distance between two cells is defined as the length of the shortest path of a rook from one cell to another (the path of the center of the rook is considered). What is the minimum number of colors needed to color the board (each cell is colored with one color) so that two cells, located at a distance of 6, are always colored differently?
# | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 223332.
Sketch of the solution. If a number is divisible by 3, then the sum of its digits is divisible by 3, which means the number of twos is a multiple of three, and thus, there must be at least three twos. Therefore, the desired number is a six-digit number. Since it is divisible by 2, it ends in 2. The number is smaller the smaller the leading digits are, hence the answer.
Criteria. Correct answer without justification: 2 points. | \boxed{223332} | 1. Find the smallest number that contains only the digits 2 and 3 in equal quantities, and is divisible by 2 and 3. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Answer】 4
【Analysis】 Number puzzle, breakthrough, multiplication and division. There are 3 different numbers from 1 to 8 in the multiplication: $2 \times 3=6, 2 \times 4=8$. The numbers not involved are $1, 5, 7$. 2 can only be filled in the lower left corner or the upper right corner, with adjacent sides forming a multiplication equation and a division equation. In fact, there are 4 ways to fill, all of which are derived from the same filling method through symmetric exchange. | \boxed{4} | 3. Put the operation symbols " +", " -", " ×", and " ÷" into the four positions in the figure, and fill the numbers " $1, 2, 3, 4, 5, 6, 7, 8$ " into the eight blank spaces in the figure, so that the four equations formed by the quadrilateral all hold true. How many ways are there to do this? | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: $120,125$.
Solution. There are essentially two cases for painting the faces of the large cube:
- three painted faces form a "P" shape;
- three painted faces share a common vertex.
In the first case, if the painted $1 \times 1 \times 1$ cubes are "cut off," a parallelepiped $4 \times 5 \times 6$ remains. Then the number of small cubes without red faces is $4 \cdot 5 \cdot 6 = 120$.
In the second case, if the painted $1 \times 1 \times 1$ cubes are "cut off," a cube $5 \times 5 \times 5$ remains. Then the number of small cubes without red faces is $2 \cdot 5 \cdot 5 = 125$. | \boxed{120}, \boxed{125} | Problem 6.6. Zhenya painted three faces of a white cube $6 \times 6 \times 6$ red. Then he sawed it into 216 identical small cubes $1 \times 1 \times 1$. How many of the small cubes could have no red faces? List all possible options. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
We break the problem into stages, one for each card revealed, then further into cases based on the number of remaining unrevealed cards of each color. Since [expected value](https://artofproblemsolving.com/wiki/index.php/Expected_value) is linear, the expected value of the total number of correct card color guesses across all stages is the sum of the expected values of the number of correct card color guesses at each stage; that is, we add the probabilities of correctly guessing the color at each stage to get the final answer (See [https://brilliant.org/wiki/linearity-of-expectation/](https://artofproblemsolving.comhttps://brilliant.org/wiki/linearity-of-expectation/))
At any stage, if there are $a$ unrevealed cards of one color and $b$ of the other color, and $a \geq b$, then the optimal strategy is to guess the color with $a$ unrevealed cards, which succeeds with probability $\frac{a}{a+b}.$
Stage 1:
There are always $3$ unrevealed cards of each color, so the probability of guessing correctly is $\frac{1}{2}$.
Stage 2:
There is always a $3$-$2$ split ($3$ unrevealed cards of one color and $2$ of the other color), so the probability of guessing correctly is $\frac{3}{5}$.
Stage 3:
There are now $2$ cases:
The guess from Stage 2 was correct, so there is now a $2$-$2$ split of cards and a $\frac{1}{2}$ probability of guessing the color of the third card correctly.
The guess from Stage 2 was incorrect, so the split is $3$-$1$ and the probability of guessing correctly is $\frac{3}{4}$.
Thus, the overall probability of guessing correctly is $\frac{3}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{5}$.
Stage 4:
This stage has $2$ cases as well:
The guesses from both Stage 2 and Stage 3 were incorrect. This occurs with probability $\frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10}$ and results in a $3$-$0$ split and a certain correct guess at this stage.
Otherwise, there must be a $2$-$1$ split and a $\frac{2}{3}$ probability of guessing correctly.
The probability of guessing the fourth card correctly is therefore $\frac{1}{10} \cdot 1 + \frac{9}{10} \cdot \frac{2}{3} = \frac{7}{10}$.
Stage 5:
Yet again, there are $2$ cases:
In Stage 4, there was a $2$-$1$ split and the guess was correct. This occurs with probability $\frac{9}{10} \cdot \frac{2}{3} = \frac{3}{5}$ and results in a $1$-$1$ split with a $\frac{1}{2}$ chance of a correct guess here.
Otherwise, there must be a $2$-$0$ split, making a correct guess certain.
In total, the fifth card can be guessed correctly with probability $\frac{3}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot 1 = \frac{7}{10}$.
Stage 6:
At this point, only $1$ card remains, so the probability of guessing its color correctly is $1$.
In conclusion, the expected value of the number of cards guessed correctly is \[\frac{1}{2} + \frac{3}{5} + \frac{3}{5} + \frac{7}{10} + \frac{7}{10} + 1 = \frac{5+6+6+7+7+10}{10} = \frac{41}{10},\] so the answer is $41 + 10 = \boxed{051}.$
~OrangeQuail9 | \boxed{51} | Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Combinatorics | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Every time we sharpen the pencil, we carve a small cylindrical part of the pencil lead into a conical shape. The radius and height of the base circle of the cylinder and the cone are equal, so the volume of the cone is $\frac{1}{3}$ of the volume of the cylinder. Therefore, with each sharpening, $\frac{2}{3}$ of the pencil lead is wasted. If we continue this until the length of the pencil is reduced to $\frac{1}{10}$ of its original length, the total waste is $\frac{9}{10} \cdot \frac{2}{3}=\frac{3}{5}$ of the lead, which means $60\%$ of the lead is wasted. | \boxed{60\%} | We have a brand new pencil. We sharpen the pencil. When the tip wears down, we always sharpen it again, until the length of the pencil decreases to one-tenth of its original length. Estimate what percentage of the pencil lead is lost during the sharpening process. | Other | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
: 12. A real number may have infinitely many digits. The strategy therefore has to be to exhibit some rational number which requires N to get a 2, and then to show that for any real number some special digit, such as the first, is a 2 for some multiple <= N. A little experimentation reveals that for most integers fairly small multiples suffice. We notice that 15 requires 8, but most seem to need less. If x has first digit 7, 8 or 9, then 3x has first digit 2. If x has first digit 5 or 6, then 4x has first digit 2. If x has first digit 4, then 5x has first digit 2. If x has first digit 3, then 7x has first digit 2. Obviously, if x has first digit 2, then 1x has first digit 2. So we need only consider numbers with first digit 1. If x has second digit 2, we are home. If x has second digit 0 (including of course the case x has no second digit), 1, 3 or 4, then 2x has first digit 2. If x has second digit 5, then 8x has second digit 2. It is convenient to notice that x and 10x behave the same way, so we can restrict x to be between 1 and 10. The analysis above shows that we need only consider 1.6 ≤ x < 1.9. If 5/3 ≤ x < 2, then 20 ≤ 12x < 30. We might also look at 5/3. Its multiples are 1.66... , 3.33... , 5, 6.66... , 8.33... , 10, 11.66... , 13.33... , 15, 16.66... , 18.33... , 20. So it requires N = 12. So we are left with 1.6 ≤ x < 5/3. The first digit is not 2 for N up to 12. But we notice that 2x has a 2 for 1.6 ≤ x < 1.65, and for 1.66 ≤ x < 1.665 and for 1.666 ≤ x ≤ 1.6665 and so on. Also 5x has a 2 for 1.64 ≤ x < 1.66 and for 1.664 ≤ x < 1.666 and so on. Thus 2x or 5x has a 2 for any x in the range 1.6 ≤ x < 5/3. Note that there are other x which require N = 12, such as 1.6795 and 1.6835. 27th BMO 1991 © John Scholes [email protected] 15 Aug 2002 | \boxed{12} | 27th BMO 1991 Problem 4 Let N be the smallest positive integer such that at least one of the numbers x, 2x, 3x, ... , Nx has a digit 2 for every real number x. Find N. Failing that, find upper and lower bounds and show that the upper bound does not exceed 20. Solution | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. In both cases, the same amount of water passes through in the same time with the same flow rate, and the same amount of heat is transferred to it. Therefore, the outlet temperature is the same: $t_{3}=t_{2}=40^{\circ} \mathrm{C}$.
Answer: $40^{\circ} \mathrm{C}$.
Criteria: 20 points - correct solution, possibly through the heat balance equation; 10 points - it is shown that when the flow rate through the heater decreases, the temperature of the hot water increases, but the final temperature is found incorrectly; $\mathbf{0}$ points - everything else. | \boxed{40^{\circ} \mathrm{C}} | 3. In the cottage village where Gavrila spends his summers, there is a water supply with cold water. The boy's parents installed a water heater, which has a fixed power as long as the temperature of the water in it is below $100^{\circ} \mathrm{C}$. After the water pipe enters the house, a tee was installed so that part of the water goes through the heater to the hot tap, and the rest goes directly to the cold tap. Before exiting, the hot and cold water mix. Gavrila fully opened the cold tap and found that the water temperature was $20^{\circ} \mathrm{C}$. When he closed the cold tap and opened the hot one, water with the same flow rate came out at a temperature of $40^{\circ} \mathrm{C}$. Then Gavrila opened both taps equally so that the flow rate remained the same. What is the water temperature in this case? | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Prove that $\angle K C M=75^{\circ}$.
## Solution
Since $C M<M K$ and the angles of triangle $C M K$ are $60^{\circ}, 45^{\circ}$, and $75^{\circ}$, $C M$ is the smallest side of the triangle. Therefore,
$$
\angle M K C=45^{\circ}, \angle K M C=75^{\circ} .
$$
Using the Law of Sines in triangles $A B C$ and $M K C$, we find that
$$
A C=\frac{B C \sin 45^{\circ}}{\sin 75^{\circ}}, M K=\frac{K C \sin 60^{\circ}}{\sin 75^{\circ}}=\frac{A C \sin 60^{\circ}}{\sin 75^{\circ}}=\frac{B C \sin 45^{\circ} \sin 60^{\circ}}{\sin ^{2} 75^{\circ}}
$$
Therefore,
$$
\frac{B C}{M K}=\frac{\sin ^{2} 75^{\circ}}{\sin 45^{\circ} \sin 60^{\circ}}=\frac{1-\cos 150^{\circ}}{2 \sin 45^{\circ} \sin 60^{\circ}}=\frac{2\left(1+\frac{\sqrt{3}}{2}\right)}{\sqrt{2} \cdot \sqrt{3}}=\frac{2+\sqrt{3}}{\sqrt{6}}
$$
## Answer
$\frac{2+\sqrt{3}}{\sqrt{6}}$. | \boxed{\dfrac{2+\sqrt{3}}{\sqrt{6}}} | In triangle $A B C$, given: $\angle A C B=60^{\circ}, \angle A B C=45^{\circ}$. On the extension of $A C$ beyond vertex $C$, a point $K$ is taken such that $A C=C K$. On the extension of $B C$ beyond vertex $C$, a point $M$ is taken such that the triangle with vertices $C, M$, and $K$ is similar to the original triangle. Find $B C: M K$, given that $C M: M K<1$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The ball changes the direction of its trajectory each time it hits one of the table edges. Since the trajectory always makes an angle of $45^{\circ}$ with the edge, the trajectory of the ball, hit from a corner, will always follow the diagonals of the squares it crosses. By tracing this trajectory, we conclude that (b) the ball will hit the edges of the table five times before (a) falling into the upper left pocket.
Counting the squares crossed, we see that (c) it will cross 23 squares diagonally. | (c) \boxed{23} | In the figure below, we see a checkered billiard table and part of the trajectory of a ball, which is hit from one corner of the table, such that every time the ball hits one of the edges of the table, it continues its movement forming angles of $45^{\circ}$ with the edge.
(a) In which of the four pockets will the ball fall?
(b) How many times will the ball hit the edges of the table before falling into the pocket?
(c) The ball will follow the diagonal of how many of these squares during its trajectory?
 | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
10. Each of such five-digit numbers is divisible by 3, since $1+2+$ $+3+4+5=15$, and $15: 3$. These numbers do not have any common divisors greater than 3, as, for example, $12345=3 \cdot 5 \cdot 823$, but the number 12354 does not divide evenly by either 5 or 823. Therefore, the greatest common divisor of these numbers is 3. | \boxed{3} | 10. Find the greatest common divisor of all five-digit numbers written using the digits
$$
1,2,3,4,5
$$
without repetition. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
A4. The inscribed angle $x$ is subtended by an arc whose central angle measures $360^{\circ}-164^{\circ}$, so it measures half of the angle $196^{\circ}$, which is $98^{\circ}$. | \boxed{98^{\circ}} | A4. How many degrees is angle $x$?
(A) $16^{\circ}$
(B) $82^{\circ}$
(C) $98^{\circ}$
(D) $164^{\circ}$
(E) $328^{\circ}$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Problem 1. $u c\left(2^{x}\right) \in\{2,4,6,8\}, x \in \mathbb{N}^{*}$ ..... $.1 p$
$u c\left(2^{x}+2^{x+1}+2^{x+2}+2^{x+3}\right)=0, x \in \mathbb{N}^{*}$ ..... $.2 p$
$u c\left(2^{1}+2^{2}+2^{3}+2^{4}\right)=u c\left(2^{5}+2^{6}+2^{7}+2^{8}\right)=\ldots=u c\left(2^{2009}+2^{2010}+2^{2011}+2^{2012}\right)=0$ ..... $2 p$
$u c\left(2^{0}+2^{2013}+2^{2014}\right)=7$ ..... $1 p$
$u c(n)=7$ ..... $1 p$ | \boxed{7} | Problem 1. Find the last digit of the number $n=2^{0}+2^{1}+2^{2}+\ldots+2^{2014}$. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
12. (3 points) A ship sails from Port A to Port B, traveling 24 kilometers per hour against the current. After reaching Port B, it returns to Port A with the current. It is known that the downstream journey takes 5 hours less than the upstream journey, and the speed of the current is 3 kilometers per hour. The distance between Port A and Port B is __ $\frac{200}{3}$ kilometers.
【Solution】Solution: The downstream speed is:
$$
24+3+3=30 \text { (km/h); }
$$
The distance between Port A and Port B is:
$$
\begin{array}{l}
5 \div\left(\frac{1}{24}+\frac{1}{30}\right) \\
=5 \div \frac{3}{40}, \\
=\frac{200}{3} \text { (km); }
\end{array}
$$
Answer: The distance between Port A and Port B is $\frac{200}{3}$ kilometers.
Therefore, the answer is: $\frac{200}{3}$. | \boxed{1080000} | 12. (3 points) A ship sails from Port A to Port B, traveling 24 kilometers per hour against the current. After reaching Port B, it returns to Port A with the current. It is known that the downstream journey takes 5 hours less than the upstream journey, and the speed of the current is 3 kilometers per hour. The distance between Port A and Port B is $\qquad$ meters. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution
D
ComMENTARY
In the solution below we make use of the formula
$$
1+2+\cdots+n=\frac{1}{2} n(n+1)
$$
for the sum of the positive integers from 1 to $n$ inclusive.
If you are not familiar with this formula, see Problems 10.3 and 10.4 below.
Because the total that Milly gets is the same as the total that Billy gets, each of their totals is half the sum of the positive integers from 1 to 20 inclusive. We can obtain this total by putting $n=20$ in the formula $\frac{1}{2} n(n+1)$. This gives $\frac{1}{2} \times 20 \times 21=210$. Therefore the total of the numbers that Milly adds up is half of 210 , that is, 105 . So we need to find the positive integer $n$ such that $1+2+3+\cdots+n=105$
Meтrod 1
We find the value of $n$ by trying each of the given options in turn until we find the correct value. To test whether option $\mathrm{A}$ is correct, we need to see if the sum of the integers from 1 to 11 inclusive is equal to 105 . By putting $n=11$ in the formula $\frac{1}{2} n(n+1)$, we see that
$$
1+2+3+\cdots+11=\frac{1}{2} \times 11 \times 12=66 .
$$
So option A is not the correct answer.
We now test the other options in turn. We have
$$
\begin{array}{l}
1+2+\cdots+12=(1+2+\cdots+11)+12=66+12=78 . \\
1+2+\cdots+13=(1+2+\cdots+12)+13=78+13=91 . \\
1+2+\cdots+14=(1+2+\cdots+13)+14=91+14=105 .
\end{array}
$$
Therefore $n=14$. So option $\mathrm{D}$ is correct.
In the context of the SMC we can stop here.
However, a complete solution should explain why there is no other value of $n$ that works. See Problem 10.1, below
Meтноd 2
To find the smallest positive integer $n$ such that the sum of the integers from 1 to $n$, inclusive, is 105 , we solve the equation $\frac{1}{2} n(n+1)=105$. We have
$$
\begin{aligned}
\frac{1}{2} n(n+1)=105 & \Leftrightarrow n(n+1)=210 \\
& \Leftrightarrow n^{2}+n=210 \\
& \Leftrightarrow n^{2}+n-210=0 \\
& \Leftrightarrow(n+15)(n-14)=0 \\
& \Leftrightarrow n=-15 \text { or } n=14 .
\end{aligned}
$$
Because $n$ is a positive integer, we deduce that $n=14$.
Note that, as in the solution to Question 2, here the symbol $\Leftrightarrow$ stands for if, and only if.
For INVESTIGATION
10.1 Prove that there is exactly one positive integer $n$ that satisfies the equation
$$
\frac{1}{2} n(n+1)=105 \text {. }
$$
10.2 Show that for each positive integer $m$ the equation
$$
\frac{1}{2} n(n+1)=m
$$
has at most one positive integer solution.
10.3 The numbers obtained by adding up all the positive integers from 1 to $n$ inclusive are called the triangular numbers. This is because they correspond to the number of dots in a triangular array. For example, the figure on the right shows an array of dots corresponding to the sum $1+2+3+4+5+6$.
The notation $T_{n}$ is often used for the $n$-th triangular number. That is,
$$
T_{n}=1+2+3+\cdots+n .
$$
In the figure on the right we have put together two triangular arrays of dots corresponding to the sum $1+2+3+4+5+6$. The two triangular arrays together form a rectangle with 6 rows and 7 columns. The rectangle therefore contains $6 \times 7$ dots. So the figure illustrates the fact that $2 T_{6}=6 \times 7$, and hence that $T_{6}=\frac{1}{2}(6 \times 7)$.
Generalize the above to give a proof of the formula $T_{n}=\frac{1}{2} n(n+1)$. for all positive integers $n$.
10.4 We can express the sum of the first $n$ positive integers more succinctly using the sigme notation. That is, we write $\sum_{k=1}^{n} k$ for the sum $1+2+3+\cdots+n$, and, more generally, $\sum_{k=1}^{n} s_{k}$ for the sum $s_{1}+s_{2}+s_{3}+\cdots+s_{n}$. We use this notation to give an algebraic proof of the formula for the triangular numbers.
We begin with the observation that
$$
(k+1)^{2}-k^{2}=\left(k^{2}+2 k+1\right)-k^{2}=2 k+1 .
$$
Therefore summing for $k$ going from 1 to $n$, we obtain
$$
\begin{aligned}
\sum_{k=1}^{n}\left((k+1)^{2}-k^{2}\right) & =\sum_{k=1}^{n}(2 k+1) \\
& =2 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1 .
\end{aligned}
$$
That is,
$$
\sum_{k=1}^{n}\left((k+1)^{2}-k^{2}\right)=2 T_{n}+n .
$$
Show how the left hand side of the above equation simplifies. Then rearrange the equation to obtain the formula for $T_{n}$.
10.5 The solution to Question 10 shows that the equation $T_{m}=\frac{1}{2} T_{n}$ has the positive integer solution $m=14, n=20$.
Show that if $m=a, n=b$ is a solution of the equation $T_{m}=\frac{1}{2} T_{n}$, then so also is $m=3 a+2 b+2$ and $n=4 a+3 b+3$.
Deduce that the equation $T_{m}=\frac{1}{2} T_{n}$, has infinitely many positive integer solutions. [Note that another way of putting this is to say that there are infinitely many positive integers $T$ such that both $T$ and $2 T$ are triangular numbers.] | \boxed{14} | 10. The positive integer $n$ is between 1 and 20 . Milly adds up all the integers from 1 to $n$ inclusive. Billy adds up all the integers from $n+1$ to 20 inclusive. Their totals are the same.
What is the value of $n$ ?
A 11
B 12
C 13
D 14
E 15 | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
a) We start with those number combinations in which the digits 1, 4, 6 appear in this order. In addition to them, there is another unknown digit $x$. Since $x$ is not identical to the digits 1, 4, 6, there are six possibilities for $x$.
Since $x$ can stand at the 1st, 2nd, 3rd, or 4th position in the setting sequence, there are a total of $4 \cdot 6 = 24$ possibilities for the sequence 1, 4, 6.
Analogously, we proceed with the other five possible sequences: 1, 6, 4; 4, 1, 6; 4, 6, 1; 6, 1, 4; 6, 4, 1. In the worst case, therefore, $6 \cdot 24 = 144$ settings have to be made.
b) We start with those number combinations in which the digits 1, 6 appear in this order. In addition to them and the digit 4, there is another unknown digit $x$.
Again, there are six possibilities for $x$. Since $x$ can stand exactly at those three positions in the setting sequence where the digit 4 does not stand, there are a total of $3 \cdot 6 = 18$ possibilities for the sequence 1, 6.
Analogously, we proceed with the other sequence 6, 1. Thus, under the conditions of part b), at most $2 \cdot 18 = 36$ settings would be necessary. | b) \boxed{36} | ## Task 3 - 250823
A security lock has four wheels, each marked with the digits 1 to 9. The lock can only be opened by setting exactly one number combination (a specific digit on each wheel). The owner of such a lock has forgotten the exact number combination at the time of purchase. He only remembers that the digits 1, 4, and 6 each appear exactly once in the number combination. The order of the individual digits has also slipped his mind.
a) How many different settings would have to be tried in the worst case to open the lock?
b) How many different settings would be necessary at most if the owner still remembers which wheel he needs to set the digit 4 on? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
49. The polyhedron $A B M D C N$ is a triangular prism with base $A B M$, lateral edges $A D, B C', M N$,
$$
\text { Answer: } \frac{b}{2 a} \sqrt{4 a^{2}-b^{2}}
$$ | \boxed{\dfrac{b}{2a} \sqrt{4a^2 - b^2}} | 49. At the base of the pyramid $A B C D M$ lies a square $A B C D$ with side $a$, the lateral edges $A M$ and $B M$ are also equal to $a$, the lateral edges $C M$ and $D M$ have length $b$. On the face $C D M$, as the base, a triangular pyramid $C D M N$ is constructed outward, with lateral edges of length $a$. Find the distance between the lines $A D$ and $M N$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution
a) From $\angle D A P=\angle P A F$ and $A P \perp D F$, it follows that triangle $A D F$ is isosceles and that $P$ is the midpoint of $D F$. Since $E Q$ and $D F$ are perpendicular to $A E$, we have $D F \| E Q$ and $\angle C E Q=\angle C D F=\angle D F A$. In the parallelogram $A B C D, \angle D A B=\angle D C B$ and thus triangles $A D F$ and $E C Q$ are similar. Therefore, $C E=C Q$ and
$$
\frac{C Q}{A D}=\frac{E Q}{D F}
$$
Since the opposite sides of quadrilateral $D E Q P$ are parallel, it follows that it is a parallelogram. Consequently, $D P=E Q$ and
$$
\frac{C Q}{A D}=\frac{D P}{2 \cdot D P}=\frac{1}{2}
$$
b) Let $x$ be the length of segment $A D$. From $\angle D A E=\angle E A B=\angle D E A$, we can conclude that triangle $A D E$ is isosceles and thus $A D=D E=x$. From the previous item, it follows that $C Q=x / 2$. Thus
$$
\begin{aligned}
C D & =D E+E C \\
20 & =x+x / 2 \\
& =3 x / 2
\end{aligned}
$$
Finally, $x=40 / 3 \mathrm{~cm}$. | b) \(\boxed{\dfrac{40}{3} \, \text{cm}}\) | In the parallelogram $A B C D$, the angle $B A D$ is acute and the side $A D$ is shorter than the side $A B$. The bisector of the angle $\angle B A D$ intersects the side $C D$ at $E$. Through $D$, a perpendicular to $A E$ is drawn, intersecting $A E$ at $P$ and $A B$ at $F$. Through $E$, a perpendicular to $A E$ is drawn, intersecting the side $B C$ at $Q$. Additionally, the segment $P Q$ is parallel to $A B$ and the length of $A B$ is $20 \mathrm{~cm}$.
a) Find the value of $\frac{C Q}{A D}$.
b) Find the measure of the length of the side $A D$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
7.2027091.
Let \( f(n) = \left[\frac{n}{2}\right] + \left[\frac{n}{3}\right] + \left[\frac{n}{6}\right] \). Then
\[
\begin{array}{l}
f(0)=0, f(1)=0, f(2)=1, \\
f(3)=2, f(4)=3, f(5)=3 .
\end{array}
\]
For a positive integer \( k \), we have
\[
\begin{array}{l}
f(6 k) = \left[\frac{6 k}{2}\right] + \left[\frac{6 k}{3}\right] + \left[\frac{6 k}{6}\right] \\
= 3 k + 2 k + k = 6 k, \\
f(6 k + 1) \\
= \left[\frac{6 k + 1}{2}\right] + \left[\frac{6 k + 1}{3}\right] + \left[\frac{6 k + 1}{6}\right] \\
= 3 k + 2 k + k = 6 k, \\
f(6 k + 2) \\
= \left[\frac{6 k + 2}{2}\right] + \left[\frac{6 k + 2}{3}\right] + \left[\frac{6 k + 2}{6}\right] \\
= (3 k + 1) + 2 k + k = 6 k + 1, \\
f(6 k + 3) \\
= \left[\frac{6 k + 3}{2}\right] + \left[\frac{6 k + 3}{3}\right] + \left[\frac{6 k + 3}{6}\right] \\
= (3 k + 1) + (2 k + 1) + k = 6 k + 2, \\
f(6 k + 4) \\
= \left[\frac{6 k + 4}{2}\right] + \left[\frac{6 k + 4}{3}\right] + \left[\frac{6 k + 4}{6}\right] \\
= (3 k + 2) + (2 k + 1) + k = 6 k + 3, \\
f(6 k + 5) \\
= \left[\frac{6 k + 5}{2}\right] + \left[\frac{6 k + 5}{3}\right] + \left[\frac{6 k + 5}{6}\right] \\
= (3 k + 2) + (2 k + 1) + k = 6 k + 3. \\
\text{ Hence } \sum_{n=1}^{2014} \left( \left[\frac{n}{2}\right] + \left[\frac{n}{3}\right] + \left[\frac{n}{6}\right] \right) \\
= \sum_{n=1}^{2014} f(n) = \sum_{n=0}^{2014} f(n) \\
= \sum_{k=0}^{335} \sum_{i=0}^{5} f(6 k + i) - f(2015) \\
= \sum_{k=0}^{335} (36 k + 9) - f(6 \times 335 + 5) \\
= \frac{(9 + 12069) \times 336}{2} - (6 \times 335 + 3) \\
= 2027091 \\
\end{array}
\] | \boxed{2027091} | 7. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $n$ is a positive integer, then
$$
\sum_{n=1}^{2014}\left(\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]\right)=
$$
$\qquad$ . | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution
## ALTERNATIVE E
One way to fill in the table according to the conditions of the statement is given below. In each step, we indicate with gray color the new cells filled; the reader can justify each of the illustrated steps. We note that the final table is unique, regardless of how it is filled.
Referring back to the statement of this question, we see that the sum of the numbers in the gray squares marked in the drawing of this statement is equal to $6+8+5+1=20$. | \boxed{20} | The small squares of the board in the figure must be filled in such a way that:
- in the small squares of each of the regions in the shape of $\hookleftarrow$ appear the numbers 1, 3, 5, and 7 or the numbers 2, 4, 6, and 8;
- in small squares with a common side, consecutive numbers do not appear.
What is the sum of the numbers that will appear in the gray squares?
A) 12
B) 14
C) 16
D) 18
E) 20
# | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let $n$ be the number of cups (the number of people in the family), and $x$ be the amount of milk consumed (in cups). Then the amount of coffee consumed is $n-x$. Katya drank one cup of coffee with milk, which consisted of one quarter of all the milk $(x / 4)$ and one sixth of all the coffee $((n-x) / 6)$. We get
$$
\begin{gathered}
\frac{x}{4}+\frac{(n-x)}{6}=1 \\
3 x+2(n-x)=12 \\
x+2 n=12
\end{gathered}
$$
Since $n$ is an integer, it follows from the last equation that $x$ is an integer, and even ($x=12 - 2 n$). Moreover, $x \leq n$, since the amount of milk consumed is certainly no more than the total amount of the drink.
By a small enumeration, we find that the last equation has three solutions:
$$
n=6, x=0 ; \quad n=5, x=2 ; \quad n=4, x=4 .
$$
The first and last solutions correspond to the case where everyone drank just milk or just coffee, while the second solution corresponds to the case where they drank coffee with milk.
## Answer
5 people. | $\boxed{5}$ | Booin D.A.
The whole family drank a full cup of coffee with milk, and Katya drank a quarter of all the milk and a sixth of all the coffee. How many people are in the family? | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
A2. In the graph, there is a rational function with a pole at $x=-1$, a double zero at $x=2$, and an initial value of 2, that is, the function $f(x)=\frac{(x-2)^{2}}{2 x+2}$. | \boxed{\frac{(x-2)^{2}}{2 x+2}} | A2. The graph of the function on the image is:
(A) $f(x)=\log _{3}(x+1)-1$
(B) $f(x)=\frac{(x-2)^{2}}{2 x+2}$
(C) $f(x)=2^{x+1}+3$
(D) $f(x)=\frac{2}{3} x-1$
(E) None of the above. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
1. C.
$$
1-70 \%-(1-70 \%) \times \frac{1}{3}=20 \% \text {. }
$$ | \boxed{C} | 1. 甲拿走了整张甜饼的 $70 \%$, 乙又拿走了剩下的三分之一. 则还剩下的为整张甜饼的 ( ).
(A) $10 \%$
(B) $15 \%$
(C) $20 \%$
(D) $30 \%$
(E) $35 \%$
1. A takes $70 \%$ of the whole cookie, and B takes one-third of the remainder. The part still left is ( ) of the whole cookie.
(A) $10 \%$
(B) $15 \%$
(C) $20 \%$
(D) $30 \%$
(E) $35 \%$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
# Solution.
Note that if $D$ and $E$ have different parity, their sum is odd, and if they have the same parity, their sum is even. The second number in the pair is always odd. Initially, we have a pair $О Н$ (Odd, Odd).
After one year
After two years $\quad \mathrm{H}+\mathrm{H}=$ О, Н. And so on.
That is, after an odd number of years, there will be a pair of odd numbers.
Since they do not coincide, the smallest difference can only be 2.
We will also prove that after an even number of years, a difference of 1 can be achieved. We will prove this fact by mathematical induction.
For $n=1$, i.e., after one year $(13,11)$, the difference is 2. For $n=2(24,13 \cdot 2$ - $1=25)$. The difference is 1. Suppose for $n=2 k$ the difference is 1, i.e., $(2 m ; 2 m-1) \Rightarrow(4 m-$ $1 ; 4 m-3)$, the difference is 2. If $n=2 k-1$, the difference, by the induction hypothesis, is 2, then for $n=2 k$ from two courses $(2 m ; 2 m+2) \Rightarrow(4 m+2 ; 4 m+3)$. The difference is 1. Therefore, after 101 years, the smallest difference can be 2.
Answer: 2.
## Recommendations for checking.
Proved only the fact of the alternation of the parity of the difference of the courses - 2 points
Answer without justification - 0 points | \boxed{2} | 3. At present, the exchange rates of the US dollar and the euro are as follows: $D=6$ yuan and $E=7$ yuan. The People's Bank of China determines the yuan exchange rate independently of market conditions and adheres to a policy of approximate equality of currencies. One bank employee proposed the following scheme for changing the exchange rate to the management. Over one year, the exchange rates $D$ and $E$ are allowed to change according to the following four rules. Either change $D$ and $E$ to the pair $(D+E, 2D \pm 1)$, or to the pair $(D+E, 2E \pm 1)$. Moreover, it is forbidden for the exchange rates of the dollar and euro to be equal at any time.
For example: From the pair $(6,7)$, after one year, the following pairs can be made: $(13,11)$, $(11,13)$, $(13,15)$, or $(15,13)$. What is the smallest value that the difference between the larger and smaller of the resulting exchange rates can take after 101 years? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
# Solution:
family income
$71000+11000+2600=84600$ rubles
average monthly expenses
$8400+18000+3200+2200+18000=49800$ rubles
expenses for forming a financial safety cushion
$(84600-49800) * 0.1=3480$ rubles
the amount the Petrovs can save monthly for the upcoming vacation
$84600-49800-3480=31320$ rubles
## Answer: 31320 | \boxed{31320} | Task 12. (16 points)
The Vasilievs' family budget consists of the following income items:
- parents' salary after income tax deduction - 71000 rubles;
- income from renting out property - 11000 rubles;
- daughter's scholarship - 2600 rubles
The average monthly expenses of the family include:
- utility payments - 8400 rubles;
- food - 18000 rubles;
- transportation expenses - 3200 rubles;
- tutor services - 2200 rubles;
- other expenses - 18000 rubles.
10 percent of the remaining amount is transferred to a deposit for the formation of a financial safety cushion. The father wants to buy a car on credit. Determine the maximum amount the Vasilievs family can pay monthly for the car loan.
In your answer, provide only the number without units of measurement! | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$\begin{array}{l}x^{2}-2 x-6=0 \\ a=1+\sqrt{7} \\ a^{2}-2 a-6=0 \Rightarrow a^{2}=2 a+6 \Rightarrow a=2+\frac{6}{a} \\ 2+\frac{T}{a}=2+\frac{6}{a}=a \Rightarrow 2+\frac{T}{2+\frac{T}{a}}=2+\frac{6}{a}=a \Rightarrow 2+\frac{T}{2+\frac{T}{2+\frac{T}{a}}}=2+\frac{6}{a}=a \\ P=3+\frac{6}{a}=1+2+\frac{6}{a}=1+a=1+1+\sqrt{7}=2+\sqrt{7}\end{array}$ | \boxed{2+\sqrt{7}} | I3.4 Let $a$ be the positive root of the equation $x^{2}-2 x-T=0$.
If $P=3+\frac{T}{2+\frac{T}{2+\frac{T}{T}}}$, find the value of $P$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Solution】Solution: Let the side length of the smallest equilateral triangle be 1. As shown in Figure 1, with $A$ as the vertex, equilateral triangles with side lengths of 4, 3, 2, and 1 can be formed, so point $A$ must be removed. Similarly, points $B$ and $C$ must also be removed.
As shown in Figure 2 (blanks indicate points that must be removed), four equilateral triangles with side lengths of 2 and several equilateral triangles with side lengths of 1 are formed, so points $O$, $D$, $E$, and $F$ must be removed.
Therefore, a total of 7 points have been removed.
Hence, the answer is: $B$. | \boxed{B} | 12. (12 points) The figure below shows a triangular array of 15 points. To ensure that no equilateral triangles remain with the points in the figure as vertices, at least ( ) points must be removed.
A. 6
B. 7
C. 8
D. 9 | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
13.2. Students who received threes and fours constitute
$$
\frac{1}{3}+\frac{5}{13}=\frac{28}{39} \text { of the entire class. }
$$
Therefore, the number of students in this class is divisible by 39. Since the class size does not exceed 50 students, there are 39 students in total in the class, with 28 of them receiving threes and fours on the test. Therefore, fives were received by $39-28-1=10$ students. | \boxed{10} | 13.2. A third of the sixth-grade students received threes for their math test. How many students received fives if only one student received a two, and fives were received by $\frac{5}{13}$ of the sixth-graders?
$$
(6-7 \text { grade) }
$$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
8. Answer: (C).
Let the side of the square be 2. Then the radius of the circle is $\sqrt{2}$. Let $\theta=\angle X O Y$. So
$$
\tan (\theta / 2)=\frac{M Y}{M O}=M Y=P M \tan 30^{\circ}=\frac{\sqrt{2}-1}{\sqrt{3}} .
$$
Then
$$
\tan \theta=\frac{2 \tan (\theta / 2)}{1-\tan ^{2}(\theta / 2)}=\frac{\sqrt{6}-\sqrt{3}}{\sqrt{2}} .
$$ | \boxed{\frac{\sqrt{6} - \sqrt{3}}{\sqrt{2}}} | 8. A square $A B C D$ and an equilateral triangle $P Q R$ are inscribed in a circle centred at $O$ in such a way that $A B \| Q R$. The sides $P Q$ and $P R$ of the triangle meet the side $A B$ of the square at $X$ and $Y$ respectively.
The value of $\tan \angle X O Y$ is
(A) $\frac{1}{\sqrt{3}}$;
(B) 1 ;
(C) $\frac{\sqrt{6}-\sqrt{3}}{\sqrt{2}}$;
(D) $\frac{2 \sqrt{2}-2}{\sqrt{3}}$;
(E) $\sqrt{3}$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
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