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B4. 16 Name the points and areas as in the figure. The area of triangle $A D F$ is equal to $a+b+e$ and also to half the area of rectangle $A B C D$. The area of triangle $C D E$ is equal to $b+c+d$ and also to half the area of rectangle $A B C D$. If we add these two areas, we get the area of the entire rectangle. Therefore, we find that
$$
(a+b+e)+(b+c+d)=a+b+c+d+e+3+5+8 .
$$
By subtracting $a+b+c+d+e$ from both sides of the equation, we find $b=3+5+8=16$. | 16 | B4. A rectangle is divided into eight pieces by four line segments as shown in the figure. The areas of three pieces are given, namely 3, 5, and 8.
What is the area of the shaded quadrilateral? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. $D=4-4 a$.
$\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=\left(\frac{2}{a}\right)^{2}-4 \cdot \frac{1}{a}=\frac{4-4 a}{a^{2}}=\frac{D}{a^{2}}$.
We obtain the equation: $\frac{D}{a^{2}} \cdot 9=D$. The condition $D>0$ is satisfied only by the root $a=-3$.
Answer: $a \in\{-3\}$. | \in{-3} | # Task 2.
Find the set of values of the parameter $a$ for which the discriminant of the equation $a x^{2}+2 x+1=0$ is 9 times the square of the difference of its two distinct roots. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3 C. The 1991st positive odd number is
$$
\begin{array}{c}
2 \times 1991-1=3981 \\
3969=63^{2}<3981<65^{2}=4225 \\
3981+2 \times \frac{63+1}{2}=4045
\end{array}
$$ | 4045 | 3 After removing all the perfect squares from the sequence of positive odd numbers, the 1991st number is
A. 4013
B. 4007
C. 4045
D. 4225 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
First, we will determine how many gingerbread cookies they decorated in total. There were 5 trays of twelve cookies each, which makes 60 cookies $(5 \cdot 12=60)$.
If all three of them took a cookie at the same time and started decorating it, then under the given conditions, all three would take another cookie at the same time only when the grandmother decorates the fifth, Marinka the third, and Jeník the second, and not before. Not even two of them would take a cookie at the same time before the specified time. We will call this time interval one "cycle." The number of cycles must be a whole number (the grandmother finished with Marinka at the same time).
First, let's imagine that Jeník only decorated and did not help with anything else. Then, in one cycle, all three together would decorate 10 cookies. To decorate sixty cookies, they would need exactly 6 cycles $(60: 10=6)$. However, since Jeník did not decorate for the entire 6 cycles (he also aligned the cookies on trays and carried them away), the grandmother and Marinka had to decorate for at least 7 cycles.
If they worked for 8 cycles, the grandmother would have decorated 40 cookies $(8 \cdot 5=40)$ and Marinka would have decorated 24 cookies $(8 \cdot 3=24)$. Together, they would have decorated 64 cookies, which is more than they had. Since they had to complete a cycle (otherwise, one would finish before the other), there could not have been 8 or more cycles.
This means that the grandmother and Marinka worked for exactly 7 cycles. The grandmother decorated 35 cookies $(7 \cdot 5=35)$ and Marinka decorated 21 cookies $(7 \cdot 3=21)$. Jeník decorated 4 cookies $(60-35-21=4)$.
If the grandmother decorates one cookie in 4 minutes, one cycle lasts 20 minutes $(4 \cdot 5=20)$. The entire work took them 140 minutes $(7 \cdot 20=140)$, which is 2 hours and 20 minutes.
Since Jeník decorated 4 cookies, he decorated for two full cycles $(4: 2=2)$, which is 40 minutes $(2 \cdot 20=40)$.
Another solution. The problem can also be solved by "ignoring" Jeník's decorating. The grandmother and Marinka together decorate 8 cookies in one cycle, so there can be at most 7 cycles $(60: 8=7$, remainder 4). The remaining 4 cookies will be decorated by Jeník. If there were only 6 cycles, Jeník would have been decorating the entire time and could not, for example, carry the trays. There could not have been fewer cycles. From then on, everything is the same. | 4,140,40 | Jeníček and Mařenka visit their grandmother, who has a confectionery and sells gingerbread. Both of them naturally help her, especially with decorating. In the time it takes for the grandmother to decorate five gingerbreads, Mařenka decorates three and Jeníček decorates two. During their last visit, all three of them decorated five full trays together. Mařenka and the grandmother decorated the whole time, while Jeníček, in addition to decorating, arranged the gingerbreads in groups of twelve on one tray and carried them to the pantry. All three started and finished at the same time.
1. How many gingerbreads did Jeníček decorate?
2. How long did the entire work take if the grandmother decorates one gingerbread in 4 minutes?
3. How long did Jeníček help with decorating?
(M. Petrová) | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.
Proof of the lemma. Introduce the origin at one of the grid nodes and denote by $\$ \mid v e c\{a\} \$$ and $\$ \mid v e c\{b\} \$$ the radius vectors to the two nearest nodes, as shown in the figure. Then the grid nodes are points of the form $\$ m \mid v e c\{a\}+ n \mid v e c\{b\} \$$ for integers \$m $\$$ and \$n $\$$. By the pigeonhole principle, among five points, there will be two points $\$m_1 \mid v e c\{a\}+ n_1 \mid v e c\{b\} \$$ and $\$m_2 \mid v e c\{a\}+ n_2 \mid v e c\{b\} \$$, for which the parity of \$m_1 and \$ \$m_2\$, and the parity of \$n_1 and \$n_2\$, coincide simultaneously. The midpoint of the segment connecting these two points is the point $\$ \backslash frac \left\{m \_1+m \_2\right\} 2 \backslash m k e r n 2 m u \operatorname{vec}\{a\}+\backslash f r a c\left\{n \_1+n \_2\right\} 2 \backslash m k e r n 2 m u \operatorname{vec}\{b\} \$$$. It is a grid node, since the numbers $\$ \backslash frac \left\{m \_1+m \_2\right\} 2 \$$ and $\$ \backslash frac \left\{n \_1+n \_2\right\} 2 \$$ are integers due to the same parity of \$ \mathrm{~m} \_1 \$$ and \$m_2\$, and \$n_1\$, and \$n_2\.
Solution. In the figure on the left, you can see an example of the placement of 8 grid nodes, among which there are no two such that the midpoint of the segment between them is a grid node. We will prove that nine nodes are sufficient. Note that the hexagonal grid is divided into the union of two triangular grids (see the figure on the right). By the pigeonhole principle, among any nine nodes, at least five will be in one of these two triangular grids. By the lemma, among these five nodes, there will be two of the required ones.
## Answer
9. | 9 | Kuuyggin A.K.
Among any five nodes of a regular square grid, there will always be two nodes such that the midpoint of the segment between them is also a node of the grid. What is the minimum number of nodes of a regular hexagonal grid that must be taken so that among them there will always be two nodes such that the midpoint of the segment between them is also a node of this grid? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The median line of a triangle cuts off a similar triangle from it, with a similarity coefficient of $\frac{1}{2}$. Therefore, the area of the cut-off triangle is one-fourth of the area of the original triangle, i.e., $\frac{1}{\mathbf{4}} 16=4$. Consequently, the area of the remaining trapezoid is $16-4=12$.
## Answer
12. | 12 | The area of the triangle is 16. Find the area of the trapezoid that is cut off from the triangle by its midline.
# | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
}
For this task, it is assumed that 0 is a natural number.
Then for each natural \(x \leq 199\), there are exactly \(1993-10 x\) different possibilities for \(y\) to choose (namely 0 to \(1993-10 x-1\)), such that the inequality is satisfied. Summing this over all \(x\), we thus obtain
\(\sum_{x=0}^{199}(1993-10 x)=200 \cdot 1993-10 \cdot \sum_{x=0}^{199} x=200 \cdot 1993-10 \cdot \frac{199 \cdot 200}{2}=200 \cdot 1993-1990 \cdot 100=199600\)
pairs of natural numbers that satisfy the inequality from the problem statement. | 199600 | \section*{Problem 2 - 320932}
How many pairs \((x, y)\) of natural numbers for which \(10 x+y<1993\) exist in total? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. We will show that the sought number is 5.
Let $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a cube. By coloring the 4 vertices of one face of the cube in red, for example $A, B, C, D$, there is no vertex that has all three adjacent vertices red, so $n \geq 5$.
Now, for $n=5$, no matter how we color $n$ vertices in red, one of the planes $(A B C D)$ and $\left(A^{\prime} B^{\prime} C^{\prime} D^{\prime}\right)$ will have at least three red vertices. We can assume that $A, B, C$ are red. If $D$ is also red, then there is only one red vertex in the plane $\left(A^{\prime} B^{\prime} C^{\prime} D^{\prime}\right)$. If $X^{\prime}$ is the red vertex, with $X \in\{A, B, C, D\}$, then vertex $X$ has all its neighbors red.
If $D$ is not red, then in the plane $\left(A^{\prime} B^{\prime} C^{\prime} D^{\prime}\right)$ there are two red vertices. If $B^{\prime}$ or $D^{\prime}$ is red, then $B$, respectively $D$, has all its neighbors red. In the contrary case, $A^{\prime}$ and $C^{\prime}$ are red and then $B^{\prime}$ has all its neighbors red.
In conclusion, no matter how we color 5 of the vertices in red, there will be at least one vertex that has all its neighbors red, so the minimum sought is $n=5$. | 5 | Problem 2. Determine the smallest natural number $n$ such that, no matter which $n$ vertices of a cube we color red, there exists a vertex of the cube that has all three adjacent vertices red. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Consider similar triangles.
## Solution
Let $P$ be the projection of vertex $C$ onto line $AK$. From the similarity of triangles $KRC$ and $KDA$, it follows that $CP: CK = AD: AK$. Therefore,
$CP = \frac{CK \cdot AD}{AK} = \frac{\frac{1}{3}}{\sqrt{1 + \left(\frac{2}{3}\right)^2}} = \frac{1}{\sqrt{13}}$.
Answer
$\frac{1}{\sqrt{13}}$.
Submit a comment | \frac{1}{\sqrt{13}} | Given a square $A B C D$ with side length 1. Point $K$ lies on side $C D$ and $C K: K D=1: 2$. Find the distance from vertex $C$ to the line $A K$.
# | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer. In the sixth entrance.
Solution. The shortest path from point $A$ to Vasya's entrance is the segment $A D$. The shortest path from point $B$ to Vasya's entrance is the path along segment $B C$, and then along segment $C D$. Since triangles $A E D$ and
$C E B$ are equal, $A D = B C$. Therefore, the paths from points $A$ and $B$ to Vasya's entrance differ by 4 cells.
Since the paths from Petya's entrance through the "upper" corner (i.e., through point $A$) and through the "lower" corner (i.e., through point $B$) are equal, the path from Petya's entrance to point $A$ must be 4 cells longer than the path to point $B$. Therefore, Petya lives in the sixth entrance. | 6 | Problem 1. Pete and Vasya live in neighboring houses (see plan). Vasya lives in the fourth entrance. It is known that for Pete to run to Vasya by the shortest path (not necessarily along the sides of the cells), it does not matter which side he runs around his house. Determine which entrance Pete lives in.
## [5 points] (A. V. Khachatryan) | 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】Solution: According to the problem, transporting four times, going four times and returning three times, eats up $5 \times(4+3)=35$ sticks, then the maximum number of sticks that can be transported to location $B$ is $200-35=165$, so the answer is: D. | 165 | 10. (10 points) A giant panda is transporting bamboo from location $A$ to location $B$. Each time, he can carry 50 bamboo sticks, but he eats 5 sticks both when walking from $A$ to $B$ and when returning from $B$ to $A$. There are currently 200 bamboo sticks at location $A$. How many bamboo sticks can the giant panda transport to location $B$ at most?
A. 150
B. 155
C. 160
D. 165 | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$\triangle A B C$ is a right angled isosceles triangle.
$$
\begin{array}{l}
\angle B A C=45^{\circ}(\angle \mathrm{s} \text { sum of } \triangle \text {, base } \angle \mathrm{s} \text { isos. } \triangle \text { ) } \\
\left.\angle A C D=45^{\circ} \text { (alt. } \angle \mathrm{s}, A B / / D C\right) \\
\angle B C D=135^{\circ}
\end{array}
$$
Apply sine law on $\triangle B C D$,
$$
\begin{array}{l}
\frac{B D}{\sin 135^{\circ}}=\frac{B C}{\sin D} \\
A B \sqrt{2}=\frac{A B \sin 45^{\circ}}{\sin D}, \text { given that } A B=B D \\
\sin D=\frac{1}{2} ; D=30^{\circ} \\
\angle C B D=180^{\circ}-135^{\circ}-30^{\circ}=15^{\circ}(\angle \mathrm{s} \text { sum of } \triangle B C D) \\
b=15
\end{array}
$$ | 15 | SG. 2 In figure $1, A B$ is parallel to $D C, \angle A C B$ is a right angle, $A C=C B$ and $A B=B D$. If $\angle C B D=b^{\circ}$, find the value of $b$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$D N$ and $E M$ are the altitudes of triangle $C D E$, which means that triangles $C M N$ and $C E D$ are similar with a coefficient of $\cos \angle C$.
Therefore, $M N=D E \cos \angle C=\frac{c}{2} \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b}$.
## Answer
$\frac{c\left(a^{2}+b^{2}-c^{2}\right)}{4 a b}$ | \frac{(^{2}+b^{2}-^{2})}{4} | In triangle $A B C$, on the midline $D E$, parallel to $A B$, a circle is constructed with $D E$ as its diameter, intersecting sides $A C$ and $B C$ at points $M$ and $N$.
Find $M N$, if $B C=a, A C=b, A B=c$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
7. E.
Notice,
$$
400\left(\frac{2}{5}\right)^{n}=2^{5} \times 5^{3}\left(\frac{2}{5}\right)^{n}=2^{n+5} \times 5^{3-n} \text {, }
$$
where, $n+5$ and $3-n$ are both non-negative integers. Therefore, $n=-5,-4, \cdots, 3$, a total of 9. | 9 | 7. There are $(\quad)$ integers $n$, such that $400\left(\frac{2}{5}\right)^{n}$ is an integer.
(A) 3
(B) 4
(C) 6
(D) 8
(E) 9 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 0 m/s or $0.8 \mathrm{~m} / \mathrm{s}$.
Solution. The mosquito flew a distance of $s=v t=0.5 \cdot 20=10$ m before landing.
That is, it moved along the trajectory $A B$.
The cosine of the angle between the trajectory and the water surface is 0.8.
Obviously, the speed of the shadow moving along the bottom coincides with the speed of the shadow moving on the surface of the pond.
Since the angle of incidence of the sunlight coincides with the angle between the trajectory and the water surface, there are two possible scenarios.
The first - the mosquito is moving along the sunbeam.
The speed of the shadow on the bottom of the pond $v_{\text {shadow }}=0 \mathrm{~m} / \mathrm{s}$.
The second - the mosquito is flying towards the sunbeam. 2
The speed of the shadow $v_{\text {shadow }}=2 v \cos \beta=2 \cdot 0.5 \cdot 0.8=0.8 \mathrm{~m} / \mathrm{s}$. | 0 | 8. (15 points) A mosquito was moving over the water in a straight line at a constant speed of \( v = 0.5 \) m/s and at the end of its movement, it landed on the water surface. 20 seconds before landing, it was at a height of \( h = 6 \) m above the water surface. The cosine of the angle of incidence of the sunlight on the water surface is 0.6. The incident sunlight, which creates the shadow of the mosquito, and its trajectory lie in the same vertical plane. Determine the speed at which the shadow of the mosquito moved along the bottom of the water body. | Other | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t b(b+t)+t^{3}
$$
the condition can be rewritten as
$$
(11-3 t) b(b+t) \leq t^{3} \leq(12-3 t) b(b+t)
$$
We can not have $t \geq 4$ since in that case $t^{3} \leq(12-3 t) b(b+t)$ is not satisfied as the right hand side is not positive. Therefore it remains to check the cases when $t \in\{1,2,3\}$. If $t=3$, the above condition becomes
$$
2 b(b+3) \leq 27 \leq 3 b(b+3)
$$
If $b \geq 3$, the left hand side is greater than 27 and if $b=1$ the right hand side is smaller than 27 so there are no solutions in these cases. If $b=2$, we get a solution $(a, b)=(5,2)$.
If $t \leq 2$, we have
$$
(11-3 t) b(b+t) \geq(11-6) \cdot 1 \cdot(1+1)=10>t^{3}
$$
so there are no solutions in this case.
In summary, the only solution is $(a, b)=(5,2)$.
| (5,2) |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
| Inequalities | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 1.
Solution. Let's introduce the functions $f(x)=\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}$ and $g(x)=\frac{2}{x^{2}}$.
For $x<0$, both functions $f$ and $g$ are positive, but $f(x)$ increases from $+\infty$ to 2 (not inclusive), since $f^{\prime}(x)=\frac{-2}{(x-1)^{3}}-\frac{2}{(x-2)^{3}}<0 \quad \text { for } \quad x \in(-\infty ; 0)$, while $g(x)$ decreases from $+\infty$ to 2 (not inclusive), since $g^{\prime}(x)=\frac{-4}{x^{3}}<0 \quad \text { for } \quad x \in(-\infty ; 0)$. Therefore, for $x<0$, the equation $f(x)=g(x)$ has no solutions.
For $0<x<1$, the function $f$ decreases from $+\infty$ to 2 (not inclusive), since $f^{\prime}(x)=\frac{-2}{(x-1)^{3}}-\frac{2}{(x-2)^{3}}<0 \quad \text { for } \quad x \in(0 ; 1)$, while the function $g$ decreases from $+\infty$ to 2 (not inclusive), since $g^{\prime}(x)=\frac{-4}{x^{3}}<0 \quad \text { for } \quad x \in(0 ; 1)$. However, for $0<x<1$, the inequality $f(x)>g(x)$ holds, because $\frac{1}{(x-1)^{2}}>\frac{1}{x^{2}}$ and $\frac{1}{(x-2)^{2}}>\frac{1}{x^{2}}$, so $f(x)>g(x)$. Therefore, for $0<x<1$, the equation has no solutions.
For $x>2$, the function $f$ increases from 2 (not inclusive) to $+\infty$, since $f^{\prime}(x)=\frac{-2}{(x-1)^{3}}-\frac{2}{(x-2)^{3}}>0 \quad \text { for } \quad x \in(2 ; +\infty)$, while the function $g$ decreases from 2 (not inclusive) to 0, since $g^{\prime}(x)=\frac{-4}{x^{3}}<0 \quad \text { for } \quad x \in(2 ; +\infty)$. For $x>2$, the inequalities $0<\frac{1}{(x-1)^{2}}<\frac{1}{x^{2}}$ and $\frac{1}{(x-2)^{2}}<\frac{1}{x^{2}}$ hold, so $f(x)<g(x)$. Therefore, for $x>2$, the equation has no solutions.
## Second Solution.
The original equation, under the conditions $x \neq 0, x \neq 1, x \neq 2$, is equivalent to the equation $6 x^{3}-21 x^{2}+24 x-8=0$. Consider the function $f(x)=6 x^{3}-21 x^{2}+24 x-8$. Since $f^{\prime}(x)=18 x^{2}-42 x+24$, $x=1$ is a point of maximum, and $x=\frac{4}{3}$ is a point of minimum. The function $f$ is increasing on the domain $(-\infty, 1) \cup\left(\frac{4}{3} ;+\infty\right)$ and decreasing on the interval $\left(1 ; \frac{4}{3}\right)$. Since $f(0)=-8, f(1)=1, f\left(\frac{4}{3}\right)=\frac{8}{9}$, the equation $f(x)=0$ has a unique root, which lies in the interval $(0 ; 1)$. | 1 | 7. How many solutions does the equation
\[
\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}=\frac{2}{x^{2}} ?
\]
 | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
60. Answer: $600=25 \cdot 24$ candies.
Let's show that a smaller number of candies might not be enough. If all participants solved the same number of problems, the number of candies must be a multiple of 25. Let $N=25 k$. Imagine that 24 people solved the same number of problems, while the 25th solved fewer. If each of the first 24 participants receives $k$ candies or even fewer, there will be extra candies left. Therefore, each of them must receive at least $k+1$ candies, which means $25 k \geqslant 24(k+1)$, or $k \geqslant 24$.
Now we will prove by induction that it is possible to distribute $m$ participants $m(m-1)$ candies such that each receives fewer than $2 s$ candies. The base case for $m=1$ is obvious: give the only participant 0 candies.
The transition from $m-1$ to $m$. Suppose there are $s \geqslant 1$ participants with the highest score, and besides them, there are $t$ participants, $s+t=m$. To the less successful participants, we will distribute $t(t-1)$ candies (each receiving fewer than $2 t$) by the induction hypothesis. After this, $(s+t)(s+t-1)-t(t-1)=s(s+2 t-1)$ candies will remain, and we will distribute them equally among the $s$ "top" participants. Each of them will receive $s+2 t-1$ candies. This number is not less than $2 t$, so it is more than what we distributed to each of the others. Additionally, $s+2 t-1 < 2 s+2 t=2 m$, which completes the transition. | 600 | 60. In a class, there are 25 students. The teacher wants to stock $N$ candies, conduct an olympiad, and distribute all $N$ candies for success in it (students who solve the same number of problems should receive the same number of candies, those who solve fewer should receive fewer, including possibly zero candies). What is the smallest $N$ for which this will be possible regardless of the number of problems in the olympiad and the success of the students? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
［Solution］The last digit of a perfect square can only be $1,4,5,6,9,0$.
This allows us to eliminate $(B),(C)$.
Furthermore, for a perfect square ending in 5, the last two digits should be 25.
By examining the last two digits of the squares in groups $(A),(D)$, we find that $(A)$ is the correct choice. Therefore, the answer is $(A)$. | A | $33 \cdot 17$ Given that among the following four sets of values for $x$ and $y$, there is exactly one set for which $\sqrt{x^{2}+y^{2}}$ is an integer. This set of $x$ and $y$ values is
(A) $x=88209, y=90288$.
(B) $x=82098, y=89028$.
(C) $x=28098, y=89082$.
(D) $x=90882, y=28809$.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1991) | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$$
-20^{\circ}
$$
7. [Analysis and Solution] According to the problem, we have $2000^{\circ}=180^{\circ} \times 11+20^{\circ}$,
$$
\begin{array}{l}
\therefore \sin 2000^{\circ}=\sin \left(180^{\circ} \times 11+20^{\circ}\right)=\sin \left(-20^{\circ}\right), \\
\therefore \arcsin \left(2000^{\circ}\right)=-20^{\circ}
\end{array}
$$ | -20 | 7. $\arcsin \left(\sin 2000^{\circ}\right)=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7. $\arcsin \left(\sin 2000^{\circ}\right)=$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Yihana is walking uphill exactly when the graph is increasing (that is, when the slope of the segment of the graph is positive).
This is between 0 and 3 minutes and between 8 and 10 minutes, which correspond to lengths of time of 3 minutes and 2 minutes in these two cases, for a total of 5 minutes.
ANSWER: (A)
# | 5 | Yihana walks for 10 minutes. A graph of her elevation in metres versus time in minutes is shown. The length of time for which she was walking uphill is
(A) 5 minutes
(B) 6 minutes
(C) 4 minutes
(D) 7 minutes
(E) 8 minutes
 | Other | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5. $4 \sqrt{3}+\frac{8 \pi}{3}$.
When $t=2$, the set $A=\{(2,0)\}$;
When $t=-2$, the set
$$
A=\left\{(x, y) \mid(x+2)^{2}+y^{2} \leqslant 4\right\} \text {. }
$$
Thus, the required region is the closed area formed by drawing two tangents from the point $(2,0)$ to the circle $(x+2)^{2}+$ $y^{2}=4$ (as shown in Figure 2), and its area is $4 \sqrt{3}+\frac{8 \pi}{3}$. | 4\sqrt{3}+\frac{8\pi}{3} | 5. The set
$$
A=\left\{\left.(x, y)\left|(x-t)^{2}+y^{2} \leqslant\left(1-\frac{t}{2}\right)^{2},\right| t \right\rvert\, \leqslant 2\right\}
$$
represents a plane region with an area of $\qquad$ . | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{2}{5}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{2}{5} * 15=12$ | 12 | 3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $15$, and $\sin \alpha = \frac{\sqrt{21}}{5}$? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
7. $\frac{18}{5}$.
First, prove that the height $D H$ of tetrahedron $A B C D$ is the diameter of the small sphere.
Draw $H E \perp A B$ at point $E$, and $H F \perp A C$ at point $F$. Then
$$
\begin{array}{l}
A E=A F=A D \cos \angle D A B=\frac{3}{\sqrt{2}}, \\
\cos \angle H A E=\sqrt{\frac{1+\cos \angle B A C}{2}}=\frac{3}{\sqrt{10}} \\
\Rightarrow A H=\frac{A E}{\cos \angle H A E}=\sqrt{5} \\
\Rightarrow D H=\sqrt{A D^{2}-A H^{2}}=2 .
\end{array}
$$
Thus, the center of the large sphere is also on $D H$.
Therefore, $A D=B D=C D, A B=A C=3 \sqrt{2}$.
So, $S_{\triangle A B C}=\frac{27}{5}, V_{\text {tetrahedron } A B C D}=\frac{18}{5}$. | \frac{18}{5} | 7. A sphere is circumscribed around the tetrahedron $ABCD$, and another sphere with radius 1 is tangent to the plane $ABC$ and the two spheres are internally tangent at point $D$. If
$$
\begin{array}{l}
AD=3, \cos \angle BAC=\frac{4}{5}, \\
\cos \angle BAD=\cos \angle CAD=\frac{1}{\sqrt{2}},
\end{array}
$$
then the volume of the tetrahedron $ABCD$ is | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The required minimum is 6 m and is achieved by a diagonal string of $m 4 \times 4$ blocks of the form below (bullets mark centres of blue cells):
In particular, this configuration shows that the required minimum does not exceed 6 m.
We now show that any configuration of blue cells satisfying the condition in the statement has cardinality at least 6 m.
Fix such a configuration and let $m_{1}^{r}$ be the number of blue cells in rows containing exactly one such, let $m_{2}^{r}$ be the number of blue cells in rows containing exactly two such, and let $m_{3}^{r}$ be the number of blue cells in rows containing at least three such; the numbers $m_{1}^{c}, m_{2}^{c}$ and $m_{3}^{c}$ are defined similarly.
Begin by noticing that $m_{3}^{c} \geq m_{1}^{r}$ and, similarly, $m_{3}^{r} \geq m_{1}^{c}$. Indeed, if a blue cell is alone in its row, respectively column, then there are at least two other blue cells in its column, respectively row, and the claim follows.
Suppose now, if possible, the total number of blue cells is less than 6 m. We will show that $m_{1}^{r}>m_{3}^{r}$ and $m_{1}^{c}>m_{3}^{c}$, and reach a contradiction by the preceding: $m_{1}^{r}>m_{3}^{r} \geq m_{1}^{c}>m_{3}^{c} \geq m_{1}^{r}$.
We prove the first inequality; the other one is dealt with similarly. To this end, notice that there are no empty rows - otherwise, each column would contain at least two blue cells, whence a total of at least $8 m>6 m$ blue cells, which is a contradiction. Next, count rows to get $m_{1}^{r}+m_{2}^{r} / 2+m_{3}^{r} / 3 \geq 4 m$,
and count blue cells to get $m_{1}^{r}+m_{2}^{r}+m_{3}^{r}m_{2}^{r} / 2 \geq 0$, and the conclusion follows. | 6m | Let $m$ be a positive integer. Consider a $4 m \times 4 m$ array of square unit cells. Two different cells are related to each other if they are in either the same row or in the same column. No cell is related to itself. Some cells are coloured blue, such that every cell is related to at least two blue cells. Determine the minimum number of blue cells. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads_ref | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
25. Ans: 13
Let $O$ be the centre of the circle. First we show that $P, B, O, M, E$ lie on a common circle. Clearly $P, B, O, E$ are concyclic. As $\angle B P M=\angle H B C=\angle B E M=30^{\circ}$, points $P, B, M, E$ are also concyclic. Thus $P, B, O, M, E$ all lie on the circumcircle of $\triangle P B E$.
Since $\angle P B O=90^{\circ}, P O$ is a diameter of this circle and $P O=\sqrt{24^{2}+10^{2}}=26$.
This circle is also the circumcircle of $\triangle P B M$. Therefore, $B M=26 \sin 30^{\circ}=13$. | 13 | 25. A pentagon $A B C D E$ is inscribed in a circle of radius 10 such that $B C$ is parallel to $A D$ and $A D$ intersects $C E$ at $M$. The tangents to this circle at $B$ and $E$ meet the extension of $D A$ at a common point $P$. Suppose $P B=P E=24$ and $\angle B P D=30^{\circ}$. Find $B M$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
# Answer: 664.
Solution. Consider the remainders of the numbers when divided by 3. Divisibility by 3 means that in each pair of numbers standing one apart, either both numbers are divisible by 3, or one has a remainder of 1 and the other has a remainder of 2 when divided by 3. Among the numbers from 1 to 1000, 333 are divisible by 3, 334 give a remainder of 1, and 333 give a remainder of 2.
1) Suppose \( \mathrm{N} \) is odd. Then the remainders 1 and 2 cannot alternate throughout the circle, so all numbers in the circle will be divisible by 3, and there will be no more than 333 of them.
2) Suppose \( \mathrm{N} \) is even. The numbers are divided into two cycles of equal length \( \mathrm{N}/2 \): those in even positions and those in odd positions. In each of them, either all numbers are divisible by 3, or they alternate with remainders of 1 and 2.
A cycle of numbers divisible by 3 has a length of no more than 333. A cycle with alternating remainders has an even length, not exceeding 666.
If there are numbers divisible by 3 among the listed ones, then there is a cycle of such numbers with a length of no more than 333. The second cycle would then consist of no more than 333 even numbers with remainders of 1 and 2, i.e., no more than 332 numbers. Then the total number of numbers would be no more than 664. This number is achievable: one cycle contains numbers divisible by 3 from 3 to 996, and the second: all numbers from 1 to 498 that are not divisible by 3.
If there are no numbers divisible by 3, then the lengths of the cycles are equal and even, so the total number of numbers written should be no more than 667 and divisible by 4, i.e., no more than 664.
Criteria. Consider the case 1) \( \mathrm{N} \) is odd: 1 point. Consider the case 2) \( \mathrm{N} \) is even - the idea of two cycles: 1 point, estimating the length of each cycle: 1 and 2 points respectively. Constructing an example: 2 points. | 664 | 9.4. $N$ different natural numbers, not exceeding 1000, are written in a circle such that the sum of any two of them, standing one apart, is divisible by 3. Find the maximum possible value of $N$.
# | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Rectangle $A B C D$ is divided into five congruent rectangles as shown. The ratio $A B: B C$ is
(A) $3: 2$
(B) 2:1
(C) $5: 2$
(D) $5: 3$
(E) $4: 3$
## Solution
If we let the width of each rectangle be $x$ units then the length of each rectangle is $3 x$ units. (This is illustrated in the diagram.) The length, $A B$, is now $3 x+x+x$ or $5 x$ units and $B C=3 x$. Thus $A B: B C=5 x: 3 x$
$$
=5: 3, x \neq 0
$$
ANSWER: (D) | 5:3 | Rectangle $A B C D$ is divided into five congruent rectangles as shown. The ratio $A B: B C$ is
(A) 3:2
(D) $5: 3$
(B) $2: 1$
(E) $4: 3$
(C) $5: 2$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
To solve for $a$, $b$, and $c$ to form the sides of a triangle, it is clear that they cannot be 0, i.e., $a, b, c \in \{1, 2, \cdots, 9\}$.
(1) If they form an equilateral triangle, let the number of such three-digit numbers be $n_{1}$. Since all three digits in the three-digit number are the same, we have:
$$
n_{1} = \mathrm{C}_{9}^{1} = 9.
$$
(2) If they form an isosceles (non-equilateral) triangle, let the number of such three-digit numbers be $n_{2}$. Since the three-digit number contains only 2 different digits, let them be $a$ and $b$. Noting that the legs and base of the triangle can be interchanged, the number of digit pairs $(a, b)$ is $2 \mathrm{C}_{9}^{2}$. However, when the larger number is the base, if $a > b$, it must satisfy $b < a < 2b$. In this case, the digit pairs that cannot form a triangle are as follows, totaling 20 cases:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$a$ & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\hline$b$ & 4,3 & 4,3 & 3,2 & 3,2 & 2,1 & 2,1 & 1 & 1 & \\
\hline
\end{tabular}
At the same time, for each digit pair $(a, b)$, the two digits can be placed in three positions in $\mathrm{C}_{3}^{2}$ ways. Therefore,
$$
n_{2} = \mathrm{C}_{3}^{2} \left(2 \mathrm{C}_{9}^{2} - 20\right) = 6 \left(\mathrm{C}_{9}^{2} - 10\right) = 156.
$$
In summary, $n = n_{1} + n_{2} = 165$. Therefore, the answer is C. | 165 | (5) Let the three-digit number $n=\overline{a b c}$, if the lengths of the sides of a triangle can be formed by $a, b, c$ to constitute an isosceles (including equilateral) triangle, then the number of such three-digit numbers $n$ is ( ).
(A) 45
(B) 81
(C) 165
(D) 216 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The pattern is quite simple to see after listing a couple of terms.
\[\begin{tabular}{|r|r|r|} \hline \#&\text{Removed}&\text{Left}\\ \hline 1&10&90\\ 2&9&81\\ 3&9&72\\ 4&8&64\\ 5&8&56\\ 6&7&49\\ 7&7&42\\ 8&6&36\\ 9&6&30\\ 10&5&25\\ 11&5&20\\ 12&4&16\\ 13&4&12\\ 14&3&9\\ 15&3&6\\ 16&2&4\\ 17&2&2\\ 18&1&1\\ \hline \end{tabular}\]
Thus, the answer is $\boxed{\text{(C) } 18}$. | 18 | A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a [perfect square](https://artofproblemsolving.com/wiki/index.php/Perfect_square), and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$ | Number Theory | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$$
2 b=a+c \ldots \ldots
$$
(1), $3 a=b+c$
(1) - (2): $2 b-3 a=a-b \Rightarrow 3 b=4 a \Rightarrow a: b=3: 4$
Let $a=3 k, b=4 k$, sub. into (1): $2(4 k)=3 k+c \Rightarrow c=5 k$
$$
Q=\frac{a+b+c}{a}=\frac{3 k+4 k+5 k}{3 k}=4
$$ | 4 | I2.2 Let $a, b$ and $c$ be real numbers with ratios $b:(a+c)=1: 2$ and $a:(b+c)=1: P$.
If $Q=\frac{a+b+c}{a}$, find the value of $Q$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
6. $\frac{3+2 \sqrt{2}}{5}$.
$$
\begin{array}{l}
\text { Given } 1=\frac{1}{x+3 y}+\frac{1}{2 x+y}=\frac{1}{x+3 y}+\frac{2}{4 x+2 y} \\
\geqslant \frac{(1+\sqrt{2})^{2}}{5(x+y)} \\
\Rightarrow x+y \geqslant \frac{(1+\sqrt{2})^{2}}{5}=\frac{3+2 \sqrt{2}}{5} .
\end{array}
$$
When $x=\frac{4+\sqrt{2}}{10}, y=\frac{2+3 \sqrt{2}}{10}$, the equality holds.
Thus, the minimum value of $x+y$ is $\frac{3+2 \sqrt{2}}{5}$. | \frac{3+2\sqrt{2}}{5} | 6. Given positive real numbers $x, y$ satisfy
$$
\frac{1}{x+3 y}+\frac{1}{2 x+y}=1 \text {. }
$$
Then the minimum value of $x+y$ is | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
For each knob, the 8 different positions are denoted as $1,2, \cdots, 8$, and each combination is denoted as $(i, j, k)$, where $1 \leqslant i, j, k \leqslant 8$. Below, using the dictionary order, we write down 16 triples $(i, j, k)$ for $1 \leqslant i, j, k \leqslant 4$ such that any two triples have at most one number in common. By adding 4 to each component of these triples, we obtain another 16 triples, making a total of 32 triples as follows:
$$
\begin{array}{l}
(1,1,1),(2,1,2),(3,1,3),(4,1,4), \\
(1,2,2),(2,2,1),(3,2,4),(4,2,3), \\
(1,3,3),(2,3,4),(3,3,1),(4,3,2), \\
(1,4,4),(2,4,3),(3,4,2),(4,4,1), \\
(5,5,5),(6,5,6),(7,5,7),(8,5,8), \\
(5,6,6),(6,6,5),(7,6,8),(8,6,7), \\
(5,7,7),(6,7,8),(7,7,5),(8,7,6), \\
(5,8,8),(6,8,7),(7,8,6),(8,8,5) .
\end{array}
$$
For any combination $(i, j, k)$ of the 3 knob positions, by symmetry, we can assume $1 \leqslant i, j \leqslant 4$. It is easy to see that $(i, j)$ must appear in the first two components of one of the first 16 triples, so testing these 32 combinations will definitely open the cabinet.
On the other hand, testing 31 times does not guarantee that the cabinet will be opened.
Let $K$ be the set of these 31 triples. If two triples have at least two components in common, we say one covers the other. Testing these 31 times to ensure the cabinet opens is equivalent to the 31 triples in $K$ covering all possible triples.
We consider each triple as an integer point in three-dimensional space, then all possible $8^{3}$ triples correspond to all integer points $(i, j, k)$, $1 \leqslant i, j, k \leqslant 8$ within and on the surface of a cube with side length 7. Clearly, each integer point covers 21 other integer points, which are the 7 points on each of the 3 lines parallel to the 3 coordinate axes passing through the point $(a, b, c)$.
We color the points in $K$ red, then the 31 red points are distributed on 8 planes: $z=1,2, \cdots, 8$. By the pigeonhole principle, there must be a plane with at most 3 red points. Without loss of generality, assume this plane is $z=1$ and the 3 red points are $(1,1,1),(2,2,1)$, and $(3,3,1)$ (other cases cover fewer points than this one). Thus, there are 25 points
$$
(i, j, k), 4 \leqslant i, j \leqslant 8, k=1
$$
not covered by these 3 red points. Therefore, these 25 points can only be covered by red points on other planes. Since one red point can only cover one of these 25 points, 25 red points are needed to cover them all. In addition to these 25 red points, $K$ has 6 more red points, so the red points $(a, b, c)$ satisfying $1 \leqslant a, b \leqslant 3$ are at most 6. Thus, among the 8 sets
$$
P_{k}=\{(i, j, k) \mid 1 \leqslant i, j \leqslant 3\}, k=1,2, \cdots, 8
$$
there must be one set without red points, and thus the 9 integer points in this set need 9 red points to cover, which is impossible.
In summary, at least 32 tests are required to ensure the cabinet is opened. | 32 | 7.109 Given that the lock on a safe is composed of 3 dials, each with 8 different positions. Due to disrepair, the safe can now be opened as long as two out of the three dials are in the correct position. How many combinations must be tried at a minimum to ensure the safe can be opened? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
［Solution］By Vieta's formulas, we get $a^{2}+b^{2}=c, a^{2} b^{2}=c$, hence $\quad a^{2}+b^{2}=a^{2} b^{2}, \quad \therefore \quad b^{2}-1=\frac{b^{2}}{a^{2}}, \quad a^{2}-1=\frac{a^{2}}{b^{2}}$. If $a>0, b>0$, then
$$
\begin{aligned}
& a \sqrt{1-\frac{1}{b^{2}}}+b \sqrt{1-\frac{1}{a^{2}}} \\
= & \sqrt{a^{2}-\frac{a^{2}}{b^{2}}}+\sqrt{b^{2}-\frac{b^{2}}{a^{2}}} \\
= & \sqrt{a^{2}-\left(a^{2}-1\right)}+\sqrt{b^{2}-\left(b^{2}-1\right)}=2 .
\end{aligned}
$$
If $a<0, b<0$, the original expression $=0$. Therefore, the answer is $(D)$. | D | 2.71 $a, b$ are two non-zero real numbers, $a^{2}, b^{2}$ are the roots of the equation $x^{2}-c x+c=0$, then the value of the algebraic expression $a \sqrt{1-\frac{1}{b^{2}}}+b \sqrt{1-\frac{1}{a^{2}}}$ is
(A) 0.
(B) 2.
(C) Does not exist.
(D) None of the above answers is correct.
(China Jiangxi Province Nanchang City Junior High School Mathematics Competition, 1990) | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
8. 8 .
Let point $P\left(2 t^{2}, 2 t\right)$.
Since it is an incircle, we know $|t|>1$.
Let the slope of the line passing through point $P$ be $k$. Then the equation of the line is
$$
y-2 t=k\left(x-2 t^{2}\right) \text {. }
$$
Since this line is tangent to the circle, we have
$$
d=\frac{\left|2 t^{2} k-2 t-k\right|}{\sqrt{1+k^{2}}}=1,
$$
which simplifies to $\left(4 t^{4}-4 t^{2}\right) k^{2}-4\left(2 t^{2}-1\right) t k+4 t^{2}-1=0$.
By Vieta's formulas, we can express $k_{1}+k_{2}$ and $k_{1} k_{2}$ in terms of $\iota$, and calculate
$$
\begin{array}{l}
\left|y_{B}-y_{C}\right|=\frac{2 t^{2}}{t^{2}-1} . \\
\text { Hence } S_{\triangle P B C}=2 \frac{t^{4}}{t^{2}-1}=2\left(t^{2}-1+\frac{1}{t^{2}-1}+2\right) \\
\geqslant 8,
\end{array}
$$
and when $t^{2}=2$, $S_{\triangle P B C}$ reaches its minimum value of 8. | 8 | 8. Given that $P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and $(x-1)^{2}+y^{2}=1$ is the incircle of $\triangle P B C$. Then the minimum value of $S_{\triangle P B C}$ is $\qquad$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Answer】Solution: 5 can only be in the first or last position, $52314, 54132, 41325, 23145$, so the answer is: $B$. | 4 | 14. (12 points) Arrange the natural numbers from 1 to 5 in a row from left to right, such that starting from the third number, each number is the sum or difference of the previous two numbers. How many ways are there to arrange them?
A. 2
B. 4
C. 6
D. 8 | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let $x_{n+1}=x_{1}$ so that the system can be written in the form $x_{i}^{2}+x_{i+1}^{2}+$ $50=16 x_{i}+x_{i+1}$ for all $i \in\{1, \ldots, n\}$. This latter equation is equivalent to the equality $\left(x_{i}-8\right)^{2}+$ $\left(x_{i+1}-6\right)^{2}=50$, which is much more manageable for this exercise, since by examining how 50 can be written as the sum of two squares, it implies that all pairs $\left(x_{i}, x_{i+1}\right)$ must be one of the following:
$$
\begin{aligned}
& (7,-1) \quad ; \quad(7,13) \quad ; \quad(9,-1) \quad ; \quad(9,13) \quad ; \quad(3,1) \text {; } \\
& (13,1) \text {; (13,11) ; (1,5) ; (1,7) ; (15,5) ; }(15,7)
\end{aligned}
$$
Furthermore, one of the previous pairs $(x, y)$ can only appear if $x$ (resp. $y$) is the second (resp. the first) of another pair. This eliminates the pairs $(7,-1),(9,-1),(9,13),(3,1)$, $(3,11),(13,11),(1,5),(15,5)$ and $(15,7)$, leaving only $(7,13),(13,1)$ and $(1,7)$ as candidates. Thus, if there is a solution, the $x_{i}$ must form a periodic sequence ..., $7,13,1,7,13,1, \ldots$ and this is obviously possible only if $n$ is a multiple of 3. Conversely, if $n$ is a multiple of 3, this periodic sequence is indeed a solution. | n | . Find all integers $n \geqslant 2$ for which the system:
$$
\left\{\begin{aligned}
x_{1}^{2}+x_{2}^{2}+50 & =16 x_{1}+12 x_{2} \\
x_{2}^{2}+x_{3}^{2}+50 & =16 x_{2}+12 x_{3} \\
& \vdots \\
x_{n-1}^{2}+x_{n}^{2}+50 & =16 x_{n-1}+12 x_{n} \\
x_{n}^{2}+x_{1}^{2}+50 & =16 x_{n}+12 x_{1}
\end{aligned}\right.
$$
admits at least one solution in integers. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let's solve the problem more generally for $n$ teams, where $n$ is an odd number. We will represent each team with a point. After a match between two teams, an arrow should point from the losing team to the winning team. (In this context, we can assume that there has never been a tie.)
If the teams $A$, $B$, and $C$ have beaten each other in a cycle, which can happen in two ways, then the triangle $ABC$ is called directed. Otherwise, the triangle is not directed.
Provide an estimate for the number of non-directed triangles.
In every non-directed triangle, there is exactly one vertex such that the team represented by this vertex has beaten the other two. We will count the non-directed triangles according to the "winning" vertex.
Let the teams be represented by the points $A_{1}, A_{2}, \ldots, A_{n}$, and let $d_{i}$ denote the number of wins for $A_{i}$. Since there were a total of $\binom{n}{2}$ matches, $\sum_{i=1}^{n} d_{i}=\frac{n(n-1)}{2}$. Then, $\binom{d_{i}}{2}$ non-directed triangles are associated with the vertex $A_{i}$, since we need to choose two from the teams it has beaten.
Thus, the total number of non-directed triangles is: $\sum_{i=1}^{n}\binom{d_{i}}{2}$.
By the inequality between the arithmetic mean and the quadratic mean:
$$
\begin{gathered}
\sum_{i=1}^{n}\binom{d_{i}}{2}=\frac{1}{2} \sum_{i=1}^{n} d_{i}^{2}-\frac{1}{2} \sum_{i=1}^{n} d_{i}=\frac{1}{2} n \frac{\sum_{i=1}^{n} d_{i}^{2}}{n}-\frac{1}{2} \sum_{i=1}^{n} d_{i} \geq \frac{1}{2} n\left(\frac{\sum_{i=1}^{n} d_{i}}{n}\right)^{2}-\frac{1}{2} \sum_{i=1}^{n} d_{i}= \\
=\frac{1}{2} n\left(\frac{n-1}{2}\right)^{2}-\frac{1}{2} \frac{n(n-1)}{2}=\frac{n^{3}-4 n^{2}+3 n}{8}
\end{gathered}
$$
Can equality hold here? Yes, if we can provide a directed complete graph for which
$$
d_{1}=d_{2}=\cdots=d_{n}=\frac{n-1}{2}
$$
Let $n=2 k+1$. Then one possible construction is as follows:
If $1 \leq i<j \leq 2 k+1$ and $j-i$ is greater than $k$, then $A_{j}$ beat $A_{i}$, otherwise $A_{i}$ beat $A_{j}$. It is easy to verify that in this case,
$$
d_{1}=d_{2}=\cdots=d_{n}=\frac{n-1}{2}
$$
Since there are $\binom{n}{3}$ triangles in total, the number of directed triangles in our case is
$$
\binom{n}{3}-\frac{n^{3}-4 n^{2}+3 n}{8}=\frac{n(n-1)(n-2)}{6}-\frac{n^{3}-4 n^{2}+3 n}{8}=\frac{n^{3}-n}{24}
$$
For the case $n=23$, this is 506, so at most 506 cycles of wins could have occurred during the tournament.
Based on the work of András Juhász (Fazekas M. Főv. Gyak. Gimn., 10th grade) | 506 | In a round-robin tournament, 23 teams participated. Each team played exactly once against all the others. We say that 3 teams have cycled victories if, considering only their games against each other, each of them won exactly once. What is the maximum number of cycled victories that could have occurred during the tournament? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3. The answer is $\mathbf{( B )}$.
It is necessary to calculate the radius of the inscribed circle and the radius of the circumscribed circle of the hexagon. The radius $R$ of the circumscribed circle is equal to the side of the hexagon, which measures $2 \mathrm{~m}$. The radius $r$ of the inscribed circle is given by the height of an equilateral triangle with a side length of $2 \mathrm{~m}$ (see figure) and thus measures $\sqrt{3} \mathrm{~m}$. Therefore, the area of the circumscribed circle is $4 \pi \mathrm{m}^{2}$ and the area of the inscribed circle is $3 \pi \mathrm{m}^{2}$, and the difference between these, which is the required area, is $\pi \mathrm{m}^{2}$.
[Problem proposed by U. Bindini]
 | \pi\mathrm{~}^{2} | 3) A regular hexagon with side length $2 \mathrm{~m}$ is given. Calculate the area of the circular ring delimited by the inscribed circle and the circumscribed circle of the hexagon.
(A) $\frac{\pi}{2} \mathrm{~m}^{2}$
(B) $\pi \mathrm{~m}^{2}$
(C) $\frac{4 \pi}{3} \mathrm{~m}^{2}$
(D) $2 \pi \mathrm{~m}^{2}$
(E) $\frac{\pi}{9} \mathrm{~m}^{2}$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
6. C.
Connect $M C, M N$, and draw $M E \perp A C$ at point $E$.
$$
\begin{array}{l}
\text { Then } \angle A M E=\frac{1}{2} \angle A M C=\angle A B D=60^{\circ} \\
\Rightarrow \frac{A M}{A C}=\frac{A M}{2 A E}=\frac{1}{\sqrt{3}} .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
\triangle A M N \cong \triangle B M N, \triangle A M N \backsim \triangle A C D . \\
\text { Therefore } \frac{S_{\text {quadrilateral } A M B N}}{S_{\triangle A C D}}=\frac{2 S_{\triangle A M N}}{S_{\triangle A C D}}=2\left(\frac{A M}{A C}\right)^{2}=\frac{2}{3} .
\end{array}
$$ | \frac{2}{3} | 6. As shown in Figure $3, \odot M$ and
$\odot N$ intersect at points $A$ and $B$, a line $CD$ through point $B$ intersects $\odot M$ and $\odot N$ at points $C$ and $D$ respectively, and $\angle ABD = 60^{\circ}$. Then the ratio of the area of quadrilateral $AMBN$ to the area of $\triangle ACD$ is ( ).
(A) $1: 2$
(B) $1: 3$
(C) $2: 3$
(D) $3: 4$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
## Solution
$$
\begin{aligned}
& \int_{0}^{\pi} 2^{4} \sin ^{8} x d x=\int_{0}^{\pi} 2^{4}\left(\sin ^{2} x\right)^{4} d x=\int_{0}^{\pi} 2^{4}\left(\frac{1-\cos 2 x}{2}\right)^{4} d x=\int_{0}^{\pi}(1-\cos 2 x)^{4} d x= \\
& =\int_{0}^{\pi}\left(1-4 \cos 2 x+6 \cos ^{2} 2 x-4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\int_{0}^{\pi}\left(1-4 \cos 2 x+6 \frac{1+\cos 4 x}{2}-4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\int_{0}^{\pi}\left(4-4 \cos 2 x+3 \cos 4 x-4 \cos ^{3} 2 x+\cos ^{4} 2 x\right) d x= \\
& =\left.\left(4 x-2 \sin 2 x+\frac{3 \sin 4 x}{4}\right)\right|_{0} ^{\pi}-2 \int_{0}^{\pi}\left(1-\sin ^{2} 2 x\right) d(\sin 2 x)+\int_{0}^{\pi} \cos ^{4} 2 x d x= \\
& =4 \pi-\left.2\left(\sin 2 x-\frac{\sin ^{3} 2 x}{3}\right)\right|_{0} ^{\pi}+\int_{0}^{\pi}\left(\frac{1+\cos 4 x}{2}\right)^{2} d x= \\
& =4 \pi+\frac{1}{4} \cdot \int_{0}^{\pi}\left(1+2 \cos 4 x+\cos ^{2} 4 x\right) d x=4 \pi+\frac{1}{4} \cdot \int_{0}^{\pi}\left(1+2 \cos 4 x+\frac{1+\cos 8 x}{2}\right) d x= \\
& =4 \pi+\frac{1}{8} \cdot \int_{0}^{\pi}(3+4 \cos 4 x+\cos 8 x) d x=4 \pi+\left.\frac{1}{8}\left(3 x+\sin 4 x+\frac{\sin 8 x}{8}\right)\right|_{0} ^{\pi}= \\
& =4 \pi+\frac{3 \pi}{8}=\frac{35 \pi}{8}
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_10-26" Categories: Kuznetsov's Problem Book Integrals Problem 10 | Integrals
## Problem Kuznetsov Integrals 10-27
## Material from PlusPi | \frac{35\pi}{8} | ## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi} 2^{4} \cdot \sin ^{8} x d x
$$ | Calculus | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
# Solution.
Let $f(x)=k=x-\frac{2}{x}, f(f(x))=f(k)=k-\frac{2}{k}, f(f(f(x)))=f(f(k))=f(k)-\frac{2}{f(k)}$. The equation $f(f(f(x)))=f(k)-\frac{2}{f(k)}=1 \Leftrightarrow f^{2}(k)-f(k)-2=0$ has two solutions $f_{1}(k)=-1$ and $f_{2}(k)=2$.
We obtain $\left[\begin{array}{l}f(k)=k-\frac{2}{k}=-1 \\ f(k)=k-\frac{2}{k}=2\end{array} \Leftrightarrow\left[\begin{array}{l}k^{2}+k-2=0 \\ k^{2}-2 k-2=0\end{array}\right.\right.$.
The system has four distinct roots $k_{1}, k_{2}, k_{3}, k_{4}$.
Notice that the equations $x-\frac{2}{x}=k \Leftrightarrow x^{2}-k_{i} x-2=0$ have two distinct roots and these equations with different $k_{i}$ do not have common roots.
(Schoolchildren should strictly prove this or find the roots $x$ explicitly).
Thus, the original equation has 8 distinct roots.
Answer. 8.
| Content of the criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 12 |
| The answer is correct. All main logical steps of the solution are provided. There are arithmetic errors or typos that did not affect the overall course of the solution. | +. | 10 |
| The answer is correct. All main logical steps of the solution are provided. There is no strict justification for some conclusions. | $\pm$ | 9 |
| All main logical steps of the solution are provided. The incorrect reasoning led to an incorrect answer. | $+/ 2$ | 6 |
| The answer is correct. The main logical steps of the solution are provided. There is no proof that the equations $x^{2}-k_{i} x-2=0$ for different $k_{i}$ do not have common roots. | | |
| The answer is correct. The solution is missing or incorrect. | $\mp$ | 2 |
| :---: | :---: | :---: |
| Some steps reflecting the general idea of the solution are provided. The answer is missing or incorrect. | | |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score | 12 | | | 8 | # Task 11.3. (12 points)
How many distinct roots does the equation $f(f(f(x)))=1$ have, if $f(x)=x-\frac{2}{x}$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
[Solution] Consider the set $\{-3,0,3\}$ satisfies conditions (1) - (4), then (B), (C), (D) cannot be true.
Therefore, the answer is $(A)$. | A | 15.19 A set $P$ with certain integers as elements has the following properties:
(1) $P$ contains positive and negative numbers;
(2) $P$ contains odd and even numbers;
(3) $-1 \notin P$; (4) If $x, y \in P$, then $x+y \in P$. For this set $P$, it can be concluded that
(A) $0 \in P, 2 \notin P$.
(B) $0 \notin P, 2 \in P$.
(C) $0 \in P, 2 \in P$.
(D) $0 \notin P, 2 \notin P$.
（1st “Hope Cup” National Mathematics Invitational Competition, 1990） | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Answer】1011, 2022,
【Analysis】 $\frac{2022 m}{2022+m}$ is an integer, $\frac{2022 m+2022^{2}-2022^{2}}{2022+m}=2022-\frac{2022^{2}}{2022+m}$ is an integer. Therefore, $2022+m$ is a factor of $2022^{2}$. $2022^{2}=2^{2} \times 3^{2} \times 337^{2}$, then, $\mathrm{m}$ can be 1011, 2022. | 1011,2022 | 11. Find all possible values of $m$, given that $m$ is a positive integer greater than 2022, and $2022+m$ divides $2022m$. Find $m$ equals $\qquad$. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Since squares $A B C D$ and $D E F G$ have equal side lengths, then $D C=D E$, ie. $\triangle C D E$ is isosceles. Therefore,
$$
\angle D E C=\angle D C E=70^{\circ} \text { and so }
$$
$$
\angle C D E=180^{\circ}-70^{\circ}-70^{\circ}=40^{\circ}
$$
and
$$
\begin{aligned}
y^{\circ} & =360^{\circ}-\angle A D C-\angle C D E-\angle G D E \\
y^{\circ} & =360^{\circ}-90^{\circ}-40^{\circ}-90^{\circ} \\
y^{\circ} & =140^{\circ} \\
y & =140
\end{aligned}
$$
ANSWER: (E) | 140 | In the diagram, $A B C D$ and $D E F G$ are squares with equal side lengths, and $\angle D C E=70^{\circ}$. The value of $y$ is
(A) 120
(D) 110
(B) 160
(C) 130
(E) 140
 | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: $-\frac{29}{2}$ $\overrightarrow{A B}=-\frac{1}{2}|\overrightarrow{A B}|^{2}$ . Similarly, we get $\overrightarrow{O B} \cdot \overrightarrow{B C}=-\frac{1}{2}|\overrightarrow{B C}|^{2}, \overrightarrow{O C} \cdot \overrightarrow{C A}=-\frac{1}{2}|\overrightarrow{C A}|^{2}$, hence $\overrightarrow{O A} \cdot \overrightarrow{O B}+\overrightarrow{O B} \cdot \overrightarrow{O C}+$ $\overrightarrow{O C} \cdot \overrightarrow{O A}=-\frac{1}{2}\left(|\overrightarrow{A B}|^{2}+|\overrightarrow{B C}|^{2}+|\overrightarrow{C A}|^{2}\right)=-\frac{29}{2}$ | -\frac{29}{2} | 3. Suppose the side lengths of $\triangle A B C$ are $2,3,4$, and the circumcenter is 0, then $\overrightarrow{O A} \cdot \overrightarrow{O B}+\overrightarrow{O B} \cdot \overrightarrow{O C}+\overrightarrow{O C} \cdot \overrightarrow{O A}=$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
[Solution] Let $\alpha$ and $\beta$ be the solutions of $x^{2}+m x+n=0$. Then, according to the problem, we have
$$
\left\{\begin{array}{rl}
-m = \alpha + \beta, \\
n & = \alpha \beta;
\end{array} \quad \left\{\begin{array}{rl}
-p & =\alpha^{3}+\beta^{3}, \\
q & =\alpha^{3} \beta^{3}.
\end{array}\right.\right.
$$
Since
$$
\begin{array}{c}
\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta), \\
-p=-m^{3}+3 m n.
\end{array}
$$
$$
\therefore
$$
Therefore, the answer is $(B)$. | ^{3}-3n | 2・54 If the solutions of the equation $x^{2}+p x+q=0$ are the cubes of the solutions of the equation $x^{2}+m x+n=0$, then
(A) $p=m^{3}+3 m n$.
(B) $p=m^{3}-3 m n$.
(C) $p+q=m^{3}$.
(D) $\left(\frac{m}{n}\right)^{3}=\frac{p}{q}$.
(E) None of these.
(28th American High School Mathematics Examination, 1977) | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Analysis】Before folding, the degree of $\angle C$ can be obtained by using the sum of the interior angles of a triangle. After folding, the degree of $\angle CDA$ can be obtained by using the sum of the exterior angles of a triangle and the sum of the interior angles of a quadrilateral.
【Solution】According to the analysis, before folding, by the sum of the interior angles of a triangle, $\angle C=180^{\circ}-74^{\circ}-70^{\circ}=36^{\circ}$, after folding,
$$
\begin{array}{l}
\angle E O D=\angle C+\angle C E O=36^{\circ}+20^{\circ}=56^{\circ} ; \\
\angle B O D=180^{\circ}-\angle D O E=180^{\circ}-56^{\circ}=124^{\circ}, \\
\angle C D A=360^{\circ}-\angle A B E-\angle B A E-\angle B O D=360^{\circ}-70^{\circ}-74^{\circ}-124^{\circ}=92^{\circ} .
\end{array}
$$
Therefore, the answer is: $92^{\circ}$. | 92 | 4. (10 points) As shown in the figure, fold a triangular paper piece $A B C$ so that point $C$ lands on the plane of triangle $A B C$, with the fold line being $D E$. Given $\angle A B E=74^{\circ}, \angle D A B=70^{\circ}, \angle C E B=20^{\circ}$, then $\angle C D A$ equals $\qquad$ . | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution: (a) Let $P_{1}$ be the number of students who passed before the score increase, and $P_{2}$ be the number of students who passed after the score increase. From the information we have, we can write:
$$
66 N=71 P_{1}+56\left(N-P_{1}\right), \quad 71 N=75 P_{2}+59\left(N-P_{2}\right)
$$
From the first relation, solving the equations, we get $10 N=15 P_{1}$ or $2 N=3 P_{1}$, from which, since $N$ and $P_{1}$ are integers, we can conclude that $N$ is a multiple of 3. Similarly, from the second relation, we have $12 N=16 P_{2}$ or $3 N=4 P_{2}$, and thus $N$ is a multiple of 4. In conclusion, $N$ must be a multiple of 3 and 4, i.e., of 12, and can therefore be 12, 24, or 36. These three cases are indeed possible. Let's look for an example with $N=12$, trying to have scores as equal as possible. If 8 people, i.e., those who were immediately passing, had a score of 71 before the increase, one person got 62, and the remaining three got 54, we have verified all the assumptions of the problem. The cases $N=24$ and $N=36$ are analogous, respectively doubling and tripling the number of people in each score range.
(b) No $N$ can satisfy the conditions of this point. Let $M_{p}$ be the average score - before the increase - of those who were not passing before the increase but would pass after. Then we would have:
$$
76 P_{1}+\left(M_{p}+5\right)\left(P_{2}-P_{1}\right)=79 P_{2}
$$
But $M_{p}<65$, so $M_{p}+5<70<76$, from which
$$
79 P_{2}=76 P_{1}+\left(M_{p}+5\right)\left(P_{2}-P_{1}\right)<76 P_{1}+76\left(P_{2}-P_{1}\right)=76 P_{2}<79 P_{2}
$$
which is a contradiction. | 12,24,36 | 1. In a math test, $N<40$ people participate. The passing score is set at 65. The test results are as follows: the average score of all participants is 66, that of the promoted is 71, and that of the failed is 56. However, due to an error in the formulation of a question, all scores are increased by 5. At this point, the average score of the promoted becomes 75 and that of the non-promoted 59.
(a) Find all possible values of $N$.
(b) Find all possible values of $N$ in the case where, after the increase, the average score of the promoted became 79 and that of the non-promoted 47. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Compose a system of three equations with respect to the sides of a triangle. Use the equality $\sqrt{6}-\sqrt{2}=4 \cos 75^{\circ}$ when solving it.
## Solution
Let $C D$ be the height of the triangle. Denote $A B=c, B C=a, A C=b$. Then $S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot C D=\frac{1}{2} c$. On the other hand, $S_{\triangle \mathrm{ABC}}=\frac{1}{2} B C \cdot A C \cdot \sin \angle A C B=\frac{1}{2} a b \sin 75^{\circ}$, so $c=a b \sin 75^{\circ}$. By the cosine theorem,
$$
A B^{2}=B C^{2}+A C^{2}-2 B C \cdot A C \cdot \cos \angle A C B, \quad
$$
or $c^{2}=a^{2}+b^{2}-2 a b \cos 75^{\circ}$. Finally, by the problem's condition, $a+b+c=4+\sqrt{6}-\sqrt{2}$. Thus, we have the system
$$
\left\{\begin{array}{l}
c=a b \sin 75^{\circ} \\
A^{2}=a^{2}+b^{2}-2 a b \cos 75^{\circ} \\
a+b+c=4+\sqrt{6}-\sqrt{2}
\end{array}\right.
$$
Noting that $\sqrt{6}-\sqrt{2}=4 \cos 75^{\circ}$, we rewrite the system as
$$
\left\{\begin{array}{l}
a b=\frac{c}{\sin 75^{\circ}} \\
A^{2}=(a+b)^{2}-2 a b\left(1+\cos 75^{\circ}\right) \\
a+b=4+4 \cos 75^{\circ}-A
\end{array}\right.
$$
Substituting $a b$ from the first equation and $a+b$ from the third into the second, we get an equation in terms of $c$:
$$
c^{2}=\left(4\left(1+\cos 75^{\circ}\right)-c\right)^{2}-\frac{2 c\left(1+\cos 75^{\circ}\right)}{\sin 75^{\circ}}
$$
from which we find that
$$
c=\frac{8\left(1+\cos 75^{\circ}\right) \sin 75^{\circ}}{4 \sin 75^{\circ}+1}=\frac{8 \sin 75^{\circ}+4 \sin 150^{\circ}}{4 \sin 75^{\circ}+1}=\frac{8 \sin 75^{\circ}+2}{4 \sin 75^{\circ}+1}=2 .
$$
Let $R$ be the radius of the circumscribed circle of the given triangle. Then
$$
R=\frac{A B}{2 \sin \angle A C B}=\frac{c}{2 \sin 75^{\circ}}=2 \cdot \frac{\sqrt{6}+\sqrt{2}}{4}=\sqrt{6}-\sqrt{2} \text {. }
$$
## Answer
$\$|\operatorname{sqrt}\{6\}-| \operatorname{sqrt}\{2\} . \$-->\sqrt{6}-\sqrt{2}$. | \sqrt{6}-\sqrt{2} | [Law of Cosines] [Law of Sines]
In an acute triangle $ABC$, the angle $\angle ACB = 75^{\circ}$, and the altitude dropped from this vertex is 1.
Find the radius of the circumscribed circle if it is known that the perimeter of triangle $ABC$ is $4 + \sqrt{6} - \sqrt{2}$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Establish a Cartesian coordinate system, with $A$ as the origin, $AB$ as the $x$-axis, $AC$ as the $y$-axis, and $AA_1$ as the $z$-axis, then $E\left(0,1, \frac{1}{2}\right), G\left(\frac{1}{2}, 0,1\right)$. Let $F\left(t_{1}, 0,0\right)\left(0<t_{1}<1\right), D\left(0, t_{2}, 0\right)\left(0<t_{2}<1\right)$, then $\overrightarrow{E F}=\left(t_{1},-1,-\frac{1}{2}\right), \overrightarrow{G D}=\left(-\frac{1}{2}, t_{2},-1\right)$. Since $G D \perp E F$, we have $t_{1} + 2 t_{2} = 1$, which implies $0 < t_{2} < \frac{1}{2}$. Also, $\overrightarrow{D F}=\left(t_{1},-t_{2}, 0\right), |\overrightarrow{D F}|^{2} = \sqrt{t_{1}^{2} + t_{2}^{2}} = \sqrt{5 t_{2}^{2} - 4 t_{2} + 1} = \sqrt{5\left(t_{2} - \frac{2}{5}\right)^{2} + \frac{1}{5}}$, thus $\sqrt{\frac{1}{5}} \leqslant |\overrightarrow{D F}| < 1$. Therefore, the answer is A. | [\frac{1}{\sqrt{5}},1) | 12.2.5 * In the right triangular prism $A_{1} B_{1} C_{1}-A B C$, $\angle B A C=\frac{\pi}{2}, A B=A C=A A_{1}=$ 1. Given that $G$ and $E$ are the midpoints of $A_{1} B_{1}$ and $C C_{1}$, respectively, and $D$ and $F$ are moving points on segments $A C$ and $A B$ (excluding the endpoints). If $G D \perp E F$, then the range of the length of segment $D F$ is ( ).
(A) $\left[\frac{1}{\sqrt{5}}, 1\right)$
(B) $\left[\frac{1}{5}, 2\right)$
(C) $[1, \sqrt{2})$
(D) $\left[\frac{1}{\sqrt{5}}, \sqrt{2}\right)$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. Let $x$ be the fifth part of the reference books, $y$ be the seventh part of all fiction books. Then on the first shelf, there are $-x+y+150$ books, and on the second shelf, there are $-4x+6y$ books. From the condition of the problem, the number of books on the shelves is equal, therefore, $x+y+150=4x+6y$, from which $3x+5y=150$. But the number of reference books is a multiple of 2, and the number of fiction books is a multiple of 3, i.e., $x=2k, y=3n$. Then the equation takes the form: $6k+15n=150$. Solving in natural numbers: $n=10-\frac{6}{15}k$, therefore, $k=15, n=4$, from which $x=30, y=12$. Finally, we get that a total of 150 reference books were brought by 75 participants, 84 fiction books were brought by 42 participants. Thus, 267 people participated in the campaign, bringing 384 books.
Answer: 267 people, 384 books. | 267 | 3. Each participant in the school charity event brought either one encyclopedia, or three fiction books, or two reference books. In total, 150 encyclopedias were collected. After the event, two bookshelves in the library were filled, with an equal number of books on each. On the first shelf, there was one fifth of all the reference books, one seventh of all the fiction books, and all the encyclopedias. How many participants were there in the event, and how many books did they bring in total? | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 400.
Solution. Consider ten consecutive natural numbers. We will prove that no more than four of them are selected. If at least five numbers are selected, then three of them have the same parity, but then their pairwise differences cannot be only 2 and 8. Indeed, if $a<b<c$, then $b-a=2$ and $c-a=8$, but then $c-b=6$, which is impossible. Therefore, in each set of ten, no more than four numbers are selected, and in the first thousand numbers, there are no more than 400, since there are a hundred sets of ten in a thousand.
If we take all the numbers ending in $1, 2, 3$ or 4, then there will be exactly 400, but no difference will be equal to 4, 5 or 6. | 400 | 2. What is the maximum number of different numbers from 1 to 1000 that can be chosen so that the difference between any two chosen numbers is not equal to any of the numbers 4, 5, 6. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3. (C)
If the line $l$ passes through the point $(a, 0)$ and through two rational points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, then it must be that $x_{1} \neq x_{2}$.
Otherwise, the equation of $l$ would be $x=a$, and thus $x_{1}=x_{2}=a$, which contradicts the fact that $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$ are rational points.
The slope of the line $l$ is
$$
k=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y_{1}}{x_{1}-a},
$$
thus
$$
y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x_{1}-a\right) .
$$
Since $a$ is an irrational number, the above equation holds if and only if $y_{1}=y_{2}$, hence the line passing through the point $(a, 0)$ and through two rational points is unique, and its equation is $y=0$. | C | 3. In the Cartesian coordinate system, a point whose both coordinates are rational numbers is called a rational point. If $a$ is an irrational number, then among all lines passing through the point $(a, 0)$,
(A) there are infinitely many lines, each of which contains at least two rational points;
(B) there are exactly $n(2 \leqslant n<+\infty)$ lines, each of which contains at least two rational points;
(C) there is exactly one line that passes through at least two rational points;
(D) each line passes through at most one rational point. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution: We have
$$2 \sum_{c y c} \frac{a}{b}=\left(\sum_{c y c} \frac{a}{b}+\sum_{c y c} \frac{b}{a}\right)+\left(\sum_{c y c} \frac{a}{b}-\sum_{c y c} \frac{b}{a}\right)=\frac{\sum_{s y m} a^{2}(b+c)}{a b c}+\frac{(a-b)(b-c)(c-a)}{a b c}$$
Thus, the inequality equivalent to
$$\sum_{c y c} \frac{a}{b}+2 k \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geq 6+2 k \Leftrightarrow \frac{\sum_{s y m} a^{2}(b+c)}{a b c}+2 k \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geq 6+2 k+\frac{(a-b)(b-c)(a-c)}{a b c}$$
We only need prove inequality in case $(a-b)(b-c)(c-a) \geq 0$
$$\begin{array}{c}
\Leftrightarrow \frac{p q-3 r}{r}+\frac{2 k q}{p^{2}-2 q} \geq 6+2 k+\frac{\sqrt{p^{2} q^{2}+18 p q r-27 r^{2}-4 q^{3}-4 p^{3} r}}{r} \\
\left(p^{2} q^{2}+18 p q r-27 r^{2}-4 q^{3}-4 p^{3} q\right)\left(p^{2}-2 q\right)^{2} \leq\left[(p q-3 r)\left(p^{2}-2 q\right)+2 k q r-(6+2 k) r\left(p^{2}-2 q\right)\right]^{2}
\end{array}$$
Letting $f(r)=A r^{2}+B r+C \geq 0$ (Assume $a+b+c=p=3$ ).
$$\begin{array}{c}
A=324 k^{2}+36 k^{2} q^{2}+216 k q^{2}+2916 k-3888 q+432 q^{2}+8748-216 k^{2} q-1620 k q \\
B=8748-432 q^{3}+4320 q^{2}-126361-72 k q^{3}-972 k q+540 k q^{2}
\end{array}$$
And $C=16 q^{5}-144 q^{4}+324 q^{3}$
Case 1: $0 \leq q \leq \frac{3\left(k+11-\sqrt{k^{2}+10 k+49}\right)}{2(k+6)} \Rightarrow B \geq 0$. We have $C \geq 0, A \geq 0 \Rightarrow f(r) \geq 0$
Case 2: $\frac{3\left(k+11-\sqrt{k^{2}+10 k+49}\right)}{2(k+6)} \leq q \leq 3$. We have
$$\begin{array}{c}
\Delta=B^{2}-4 A C=-144(q-3)^{2}(2 q-9)^{2}\left(48 q^{3}+24 k q^{3}+4 k^{2} q^{3}-144 k q^{2}-468 q^{2}\right. \\
\left.-9 k^{2} q^{2}+162 k q+1296 q-719\right)
\end{array}$$
So we have $k_{\max }=3 \sqrt[3]{4}-2$ | 3 \sqrt[3]{4}-2 | Problem 24 Let $a, b, c$ be nonnegative real numbers. Find the best constant $k$ such that the inequality always holds
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geq 3+k$$ | Inequalities | AI-MO/NuminaMath-1.5/inequalities | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
# Solution.
Since it is necessary to find the largest number of three-digit numbers that are multiples of 4, the deviation between the members should be minimal. Note that the arithmetic
progression with a difference of $d=2$, defined by the formula $a_{k}=2 k$, satisfies the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$.
Indeed,
$$
a_{k}=2 k, a_{k+1}=2 k+2, a_{k+2}=2 k+4, \text { or by the formula }
$$
$$
a_{k+2}=3 a_{k+1}-2 a_{k}-2=3(2 k+2)-2 \cdot 2 k-2=2 k+4
$$
The sequence containing 2022: 4, 6,
This finite sequence can contain all even three-digit numbers from 100 to 999. Among them, the numbers divisible by 4 are: 100, 104, 108, 112, ..., 992, 925, 950, 996 - 25 in each of the nine hundreds, i.e., 225 numbers.
Answer. 225. | 225 | Task 4. (20 points) A finite increasing sequence of natural numbers $a_{1}, a_{2}, \ldots, a_{n}(n \geq 3)$ is given, and for all $\kappa \leq n-2$ the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$ holds. The sequence must contain $a_{k}=2022$. Determine the maximum number of three-digit numbers, divisible by 4, that this sequence can contain. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Analysis】Let's assume the four input numbers are $\mathrm{a}<\mathrm{b}<\mathrm{c}<\mathrm{d}$. Since the final output is the same, let's assume this result is $\mathrm{m}$. If $\mathrm{m}$ is an even number, since $4 \mathrm{k}+1$ is odd, odd $\neq$ even, so the final input result must also be an even number, which is $2 \mathrm{~m}$. By analogy, the four numbers input by the people would all be $8 \mathrm{~m}$, which contradicts the fact that the input numbers are all different positive integers. Therefore, $\mathrm{m}$ is an odd number.
Then, there are 2 choices for the previous step: $2 \mathrm{~m}$ or $\frac{m-1}{4}$.
If the previous step is $2 \mathrm{~m}$, since $2 \mathrm{~m}$ is an even number, the process must be:
$$
8 \mathrm{~m} \rightarrow 4 \mathrm{~m} \rightarrow 2 \mathrm{~m} \rightarrow \mathrm{m}
$$
Next, we examine the parity of $\frac{m-1}{4}$.
Similarly, if $\frac{m-1}{4}$ is even, then the process can only be:
$$
\mathrm{m}-1 \rightarrow \frac{m-1}{2} \rightarrow \frac{m-1}{4} \rightarrow \mathrm{m}
$$
This results in only 2 different input values, which contradicts the fact that the input numbers are all different positive integers. Therefore, $\frac{m-1}{4}$ is also odd.
Then, there are 2 choices for the previous step: $\frac{m-1}{2}$ or $\frac{\frac{m-1}{4}-1}{4}=\frac{m-5}{16}$. Next, we examine the parity of $\frac{m-5}{16}$.
Similarly, $\frac{m-5}{16}$ is odd.
Therefore, the four processes are:
$$
\begin{aligned}
8 \mathrm{~m} \rightarrow 4 \mathrm{~m} \rightarrow 2 \mathrm{~m} \rightarrow \mathrm{m} \\
\mathrm{m}-1 \rightarrow \frac{m-1}{2} \rightarrow \frac{m-1}{4} \rightarrow \mathrm{m} \\
\frac{m-5}{8} \rightarrow \frac{m-5}{16} \rightarrow \frac{m-1}{4} \rightarrow \mathrm{m} \\
\frac{m-21}{64} \rightarrow \frac{m-5}{16} \rightarrow \frac{m-1}{4} \rightarrow \mathrm{m}
\end{aligned}
$$
Since these numbers are all positive integers, $64 \mid \mathrm{m}-21$, and $\frac{m-21}{64}$ is odd. To make $\mathrm{m}$ the smallest, then $\frac{m-21}{64}=1 \Rightarrow m=85$.
At this point, the four input numbers are: $680, 84, 10, 1$.
Therefore, the value input by Dad Sheep is 680. | 680 | 13. The left figure below is a strange black box. This black box has one input port and one output port. When we input a number into the input port, a number result will be produced at the output port, following these rules:
(1) If the input is an odd number $\mathrm{k}$, the output is $4 \mathrm{k}+1$.
(2) If the input is an even number $\mathrm{k}$, the output is $\mathrm{k} \div 2$.
For example, if the input is the number 8, the output will be $8 \div 2=2$. If the input is the number 3, the output will be $3 \times 4+1=13$. Now, if 3 such black boxes are connected in series, as shown in the right figure below, the output of the first black box becomes the input of the second black box, and so on. For example, if the input number is 16, after the first black box, the result is 8, which becomes the input of the second black box. After the second black box, the result is 4, which becomes the input of the third black box. After the third black box, the result is 2, which is the final output. We can represent this process as $16 \rightarrow 8 \rightarrow 4 \rightarrow 2$.
Now, Meiyangyang, Xiyangyang, Lazy Sheep, and Father Sheep input different positive integers into this series of black boxes, with Father Sheep inputting the largest number. The 4 final output results are the same. When this output result is the smallest, what is the input value of Father Sheep? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 38 overtakes.
Solution. First, let's prove that no more than 38 overtakes occurred. Note that between the start and the first overtake, and between two consecutive overtakes, at least one of the teams must have changed the runner. There were 19 changes of runners in each team, meaning a total of 38 changes, so there could not have been more than 38 overtakes.
Let's prove that 38 overtakes were possible. We will illustrate the movement trajectories of the teams on a graph with the horizontal axis representing time and the vertical axis representing the segments, directed upwards and to the right. The lower-left end of each broken line coincides with the origin, and the upper-right ends of the broken lines have the same vertical coordinate. In this situation, an overtake is a point of intersection of the broken lines that is not a vertex.
So, let's provide an example. First, draw a broken line consisting of 40 consecutive sides of a regular 160-gon, such that the last side is horizontal. In other words, the first segment of the broken line starts from the point with coordinates \((0,0)\) and is rotated at an angle of \(9/4^{\circ}\), and each subsequent segment is rotated by \(9/4^{\circ}\) clockwise compared to the previous one. Thus, the last segment will be horizontal, i.e., rotated by \(40 \cdot 9/4^{\circ} = 90^{\circ}\) relative to the vertical axis. Connect the vertices of the auxiliary broken line with numbers \(1, 2, 4, 6, \ldots, 50\) sequentially with line segments; this will be the trajectory of the first team. The trajectory of the second team will be a broken line connecting the vertices with numbers \(1, 3, 5, \ldots, 51\). The segments \((i, i+2)\) and \((i+1, i+3)\) for \(i=1, 2, \ldots, 38\) will intersect, and their points of intersection will be the 38 overtakes we are looking for.
Let's illustrate the example for the case of teams consisting of two participants.
Fig. 1: to the solution of problem 2 | 38 | 2. In the relay race Moscow - Petushki, two teams of 20 people each participated. Each team divided the distance into 20 not necessarily equal segments and distributed them among the participants so that each ran exactly one segment (the speed of each participant is constant, but the speeds of different participants may vary). The first participants of both teams started simultaneously, and the baton handover happens instantaneously. What is the maximum number of overtakes that could occur in such a race? Overtaking at the segment boundaries is not counted as an overtake.
(2023-96, E. Neustroeva) | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
## Solution
Let ${ }^{B C} /{ }_{A D}=3 / 5$. The similarity coefficient of triangles $M B C$ and $M A D$ is $3 / 5$. Therefore, $S_{M B C}=9 / 25 S_{M A D}=18$. Hence, $S_{A B C D}=S_{M A D}-S_{M B C}=50-18=32$.
## Answer
32. | 32 | $3+$
Find the area of trapezoid $ABCD (AD \| BC)$, if its bases are in the ratio $5:3$, and the area of triangle $ADM$ is 50, where $M$ is the point of intersection of lines $AB$ and $CD$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5. 31.5
square meters
5.【Solution】The surface area of the object is exactly equal to the surface area of a large cylinder plus the lateral surface areas of the medium and small cylinders. That is,
$$
\begin{aligned}
& 2 \times \pi \times 1.5^{2}+2 \times \pi \times 1.5 \times 1+2 \times \pi \times 1 \times 1+2 \times \pi \times 0.5 \times 1 \\
= & 4.5 \pi+3 \pi+2 \pi+\pi \\
= & 10.5 \pi \text { (square meters) }
\end{aligned}
$$
Taking the value of $\pi$ as 3, the above expression equals 31.5 (square meters)
Answer: The surface area of this object is 31.5 square meters | 31.5 | 5. (As shown in the right figure) Three cylinders with heights of 1 meter and base radii of 1.5 meters, 1 meter, and 0.5 meters respectively are combined to form an object. Find the surface area of this object (take $\pi=3$). | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 16.
Solution. Divide the $5 \times 5$ square without the central cell into four $2 \times 3$ rectangles, and each of these into two $1 \times 3$ rectangles.
This results in 8 rectangles of $1 \times 3$, the sum of the numbers in each of which is 23. Since the sum of all the numbers is 200, we find the number in the central cell as $200 - 23 \cdot 8 = 16$. | 16 | Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
 | 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.
It is clear that the minimum value of the function is zero (achieved when $\left.x_{1}=x_{2}=x_{3}=0\right)$.
Let's find the maximum. We can assume that $x_{1} \geq x_{2} \geq x_{3} \geq 0$. We will prove two inequalities:
$$
\begin{gathered}
\sqrt{x_{1}^{2}+x_{2} x_{3}} \leq x_{1}+\frac{x_{3}}{2} \\
\sqrt{x_{2}^{2}+x_{3} x_{1}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq \frac{2 x_{1}+3 x_{2}+x_{3}}{2}
\end{gathered}
$$
The first inequality is equivalent to the inequality $4 x_{3}\left(x_{1}-x_{2}\right)+x_{3}^{2} \geq 0$.
To prove the second inequality, we use the fact that for $u \geq 0$ and $v \geq 0$, $\sqrt{u}+\sqrt{v} \leq \sqrt{2(u+v)}$ holds.
Thus,
$$
\sqrt{x_{2}^{2}+x_{3} x_{1}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq \sqrt{2\left(x_{2}^{2}+x_{3} x_{1}+x_{3}^{2}+x_{1} x_{2}\right)}
$$
It is not difficult to verify that
$2\left(x_{2}^{2}+x_{3} x_{1}+x_{3}^{2}+x_{1} x_{2}\right) \leq\left(\frac{x_{1}+3 x_{2}+2 x_{3}}{2}\right)^{2} \Leftrightarrow\left(x_{1}-x_{2}-2 x_{3}\right)^{2}+8 x_{3}\left(x_{2}-x_{3}\right) \geq 0$.
We have:
$$
\begin{aligned}
\sqrt{x_{1}^{2}+x_{2} x_{3}}+\sqrt{x_{2}^{2}+x_{1} x_{3}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} \leq & x_{1}+\frac{x_{3}}{2}+\frac{x_{1}+3 x_{2}+2 x_{3}}{2}= \\
& =\frac{3}{2}\left(x_{1}+x_{2}+x_{3}\right) \leq \frac{3}{2} \cdot 2=3 .
\end{aligned}
$$
Answer: $\quad \max f\left(x_{1}, x_{2}, x_{3}\right)=3, \min f\left(x_{1}, x_{2}, x_{3}\right)=0$
for $x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0$. | 3 | # Problem 1.
Three electric generators have powers $x_{1}, x_{2}, x_{3}$, the total power of all three does not exceed 2 MW. In the power system with such generators, a certain process is described by the function
$$
f\left(x_{1}, x_{2}, x_{3}\right)=\sqrt{x_{1}^{2}+x_{2} x_{3}}+\sqrt{x_{2}^{2}+x_{1} x_{3}}+\sqrt{x_{3}^{2}+x_{1} x_{2}} .
$$
Find the maximum and minimum values of this function. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let $p$ be a prime number. Then for $x=y=p$ the given condition gives us that the number $f^{2}(p)+3 p f(p)$ is a perfect square. Then, $f^{2}(p)+3 p f(p)=k^{2}$ for some positive integer $k$. Completing the square gives us that $(2 f(p)+3 p)^{2}-9 p^{2}=4 k^{2}$, or
$$
(2 f(p)+3 p-2 k)(2 f(p)+3 p+2 k)=9 p^{2} .
$$
Since $2 f(p)+3 p+3 k>3 p$, we have the following 4 cases.
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 2 f ( p ) + 3 p + 2 k = 9 p } \\
{ 2 f ( p ) + 3 p - 2 k = p }
\end{array} \text { or } \left\{\begin{array}{l}
2 f(p)+3 p+2 k=p^{2} \\
2 f(p)+3 p-2 k=9
\end{array}\right.\right. \text { or } \\
& \left\{\begin{array} { l }
{ 2 f ( p ) + 3 p + 2 k = 3 p ^ { 2 } } \\
{ 2 f ( p ) + 3 p - 2 k = 3 }
\end{array} \text { or } \left\{\begin{array}{l}
2 f(p)+3 p+2 k=9 p^{2} \\
2 f(p)+3 p-2 k=1
\end{array}\right.\right.
\end{aligned}
$$
Solving the systems, we have the following cases for $f(p)$.
$$
f(p)=p \text { or } f(p)=\left(\frac{p-3}{2}\right)^{2} \text { or } f(p)=\frac{3 p^{2}-6 p-3}{4} \text { or } f(p)=\left(\frac{3 p-1}{2}\right)^{2} .
$$
In all cases, we see that $f(p)$ can be arbitrary large whenever $p$ grows.
Now fix a positive integer $x$. From the given condition we have that
$$
(f(y)+x)^{2}+x f(x)-x^{2}
$$
is a perfect square. Since for $y$ being a prime, let $y=q, f(q)$ can be arbitrary large and $x f(x)-x^{2}$ is fixed, it means that $x f(x)-x^{2}$ should be zero, since the difference of $(f(q)+x+1)^{2}$ and $(f(q)+x)^{2}$ can be arbitrary large.
After all, we conclude that $x f(x)=x^{2}$, so $f(x)=x$, which clearly satisfies the given condition. | f(x)=x | Find all functions $f: \mathbf{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that the number $x f(x)+f^{2}(y)+2 x f(y)$ is a perfect square for all positive integers $x, y$. | Number Theory | AI-MO/NuminaMath-1.5/olympiads_ref | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let $C D$ be a diameter, $O$ be the midpoint of $C D$, and $D A, A B$ and $B C$ be chords. Then the required area is equal to the area of
sector $A O B$,
## Solution
Let $C D$ be a diameter, $O$ be the midpoint of $C D$, and $D A, A B$ and $B C$ be chords. Triangles $A O D, A O B$ and $B O C$ are equilateral, $\angle A O D = \angle A O B = \angle B O C = 60^{\circ}$, so $A B \parallel D C$. Therefore, $S_{\triangle \text{ ADB }} = S_{\triangle \text{ AOB }}$. The problem requires finding the area of the figure composed of triangle $A D B$ and the segment $A B$ bounded by the arc $A B$ not containing point $D$. Note that the sector $A O B$ consists of triangle $A O B$ and this same segment, hence the required area is equal to the area of sector $A O B$, i.e., one-sixth of the area of the corresponding circle. Thus, the required area is $\frac{\pi r^{2}}{6}$.
Answer
Problem | \frac{\pir^{2}}{6} | A semicircle of radius $r$ is divided into 3 equal parts by points, and the points of division are connected by chords to one end of the diameter that spans this semicircle. Find the area of the figure bounded by the two chords and the arc between them. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
[Solution] Let $y=-x$, then $y \in\left(0, \frac{1}{2}\right)$, and
$$
\begin{array}{l}
a_{1}=\cos (\sin y \pi), \quad a_{2}=\sin (\cos y \pi), \quad a_{3}=\cos (1-y) \pi<0 . \\
\because \quad \sin y \pi+\cos y \pi=\sqrt{2} \sin \left(y \pi+\frac{\pi}{4}\right) \leqslant \sqrt{2} \leqslant \frac{\pi}{2}, \\
\therefore \quad 0<\cos y \pi<\frac{\pi}{2}-\sin y \pi<\frac{\pi}{2} .
\end{array}
$$
Thus $0<\sin (\cos y \pi)<\cos (\sin y \pi)$. Therefore, the correct choice is $(A)$. | A | 9.26 Let $x \in\left(-\frac{1}{2}, 0\right)$, then $a_{1}=\cos (\sin x \pi), a_{2}=\sin (\cos x \pi)$, $a_{3}=\cos (x+1) \pi$ have the following size relationship:
(A) $a_{3}<a_{2}<a_{1}$.
(B) $a_{1}<a_{3}<a_{2}$.
(C) $a_{3}<a_{1}<a_{2}$.
(D) $a_{2}<a_{3}<a_{1}$.
(China High School Mathematics League, 1996) | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. The areas of figures $\Omega$ bounded by closed piecewise-smooth lines $L$ without self-intersections can be calculated using formula (2.65):
$$
\mu(\Omega)=\oint_{L}-y d x+x d y
$$
where the contour $L$ is traversed counterclockwise.
a) In the case of a cardioid $t \in [0, 2\pi]$, $d x = (-2 \sin t + 2 \sin 2t) d t$, $d y = (2 \cos t - 2 \cos 2t) d t$. The desired area is
$$
\begin{aligned}
& \oint_{L}-y d x + x d y = \int_{0}^{2\pi}[(2 \cos t - 2 \cos 2t)(2 \cos t - \cos 2t) + \\
& + (2 \sin t - 2 \sin 2t)(2 \sin t - \sin 2t)] d t = \int_{0}^{2\pi}(6 - 6 \cos t) d t = 12 \pi.
\end{aligned}
$$
b) In the case of an astroid $t \in [0, 2\pi]$, $d x = -3a \cos^2 t \sin t d t$, $d y = 3a \sin^2 t \cos t d t$. The area bounded by the astroid is
$$
\begin{aligned}
& \oint -y d x + x d y = \int_{0}^{2\pi} \left(3a^2 \cos^2 t \sin^4 t + 3a^2 \cos t \sin^2 t\right) d t = \\
& = 3a^2 \int_{0}^{2\pi} \cos^2 t \sin^2 t d t = \frac{3a^2}{4} \int_{0}^{2\pi} \sin^2 2t d t = \\
& = \frac{3a^2}{8} \int_{0}^{2\pi} (1 - \cos 4t) d t = \frac{3\pi a^2}{4}
\end{aligned}
$$
c) First, we find the parametric representation of the curve $y^2 = x^2 + x^3$. For this, let $y = tx$, where $t$ is a parameter. After substitution and simplification by $x^2$, we get:
$$
\begin{aligned}
& x = t^2 - 1 \\
& y = tx = t^3 - t
\end{aligned}
$$
The curve is shown in Fig. 2.8. The closed loop is traversed counterclockwise as the parameter $t$ increases from -1 to +1. We calculate the area of the region bounded by this loop.
Fig. 2.8
$$
\begin{aligned}
& \mu(\Omega) = \oint_{L} -y d x + x d y = \int_{-1}^{1} \left(2(t^3 - t)t + (t^2 - 1)(3t^2 - 1)\right) d t = \\
& = \int_{-1}^{1} (t^2 - 1)^2 d t = 2 \int_{0}^{1} (t^4 - 2t^2 + 1)^2 d t = 2 \left(\frac{1}{5} - \frac{2}{3} + 1\right) = \frac{16}{15}
\end{aligned}
$$ | 12\pi,\frac{3\pi^2}{4},\frac{16}{15} | 2.6. Find the area of the figure bounded by the given line:
a) $x=2 \cos t-\cos 2 t, y=2 \sin t-\sin 2 t$ (cardioid);
b) $x=a \cos ^{3} t, y=a \sin ^{3} t, a>0$ (astroid);
c) $y^{2}=x^{2}+x^{3}$. | Calculus | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
A number that is divisible by 5 and less than $5^{x}$ is $5^{x-1}$. The
$$
5^{x}-5^{x-1}=4 \cdot 5^{x-1}
$$
difference gives the number of numbers that do not share a common divisor with $5^{x}$. Thus:
$$
\begin{gathered}
4 \cdot 5^{x-1}=7812000 \\
5^{x-1}=1953125=5^{9}
\end{gathered}
$$
From this, it follows that
$$
x-1=9
$$
that is,
$$
x=10
$$ | 10 | Determine $x$ from the condition that the integer of the form $5^{x}$ is preceded by the number 7812500, which has no common divisor with $5^{x}$. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
A $10 \mathrm{~cm}$ by $10 \mathrm{~cm}$ board has 9 rows of 9 holes, or $9 \times 9=81$ pegs in total.
Each hole on the 2 main diagonals has a peg in it.
There are 9 holes on each diagonal, with the centre hole on both diagonals, since there is an odd number of holes in each row.
Therefore, the total number of holes on the two diagonals is $9+9-1=17$.
This means that the number of empty holes is $81-17=64$.
ANSWER: 64 | 64 | A $5 \mathrm{~cm}$ by $5 \mathrm{~cm}$ pegboard and a $10 \mathrm{~cm}$ by $10 \mathrm{~cm}$ pegboard each have holes at the intersection of invisible horizontal and vertical lines that occur in $1 \mathrm{~cm}$ intervals from each edge. Pegs are placed into the holes on the two main diagonals of both pegboards. The $5 \mathrm{~cm}$ by $5 \mathrm{~cm}$ pegboard is shown; it has 16 holes. The 8 shaded holes have pegs, and the 8 unshaded holes do not. How many empty holes does the $10 \mathrm{~cm}$ by $10 \mathrm{~cm}$ pegboard have?
 | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
2. $\left[\frac{\sqrt{2}}{3}, 1\right)$.
As shown in Figure 3, take the midpoint $D$ of $B C$, and connect $O D$ and $A D$.
Figure 3
Since the circumcenter of $\triangle A B C$ is $O$, we have
$O D \perp B C$.
$$
\begin{array}{l}
\text { By } \overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C} \\
=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{D O} \cdot \overrightarrow{B C}=\overrightarrow{A D} \cdot \overrightarrow{B C}, \\
\overrightarrow{A D} \cdot \overrightarrow{B C}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \\
=\frac{1}{2}\left(|\overrightarrow{A C}|^{2}-|\overrightarrow{A B}|^{2}\right)=\frac{1}{2}\left(b^{2}-c^{2}\right)
\end{array}
$$
Thus, $\overrightarrow{A O} \cdot \overrightarrow{B C}=\frac{1}{2}\left(b^{2}-c^{2}\right)$.
Similarly, $\overrightarrow{B O} \cdot \overrightarrow{A C}=\frac{1}{2}\left(a^{2}-c^{2}\right)$,
$$
\begin{array}{l}
\overrightarrow{C O} \cdot \overrightarrow{B A}=\frac{1}{2}\left(b^{2}-a^{2}\right) . \\
\text { Therefore, } \overrightarrow{A O} \cdot \overrightarrow{B C}=3 \overrightarrow{B O} \cdot \overrightarrow{A C}+4 \overrightarrow{C O} \cdot \overrightarrow{B A} \\
\Rightarrow \frac{1}{2}\left(b^{2}-c^{2}\right) \\
\quad=3 \times \frac{1}{2}\left(a^{2}-c^{2}\right)+4 \times \frac{1}{2}\left(b^{2}-a^{2}\right) \\
\Rightarrow a^{2}+2 c^{2}=3 b^{2} .
\end{array}
$$
By the cosine rule,
$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$
$$
\begin{array}{l}
=\frac{a^{2}+c^{2}-\frac{1}{3}\left(a^{2}+2 c^{2}\right)}{2 a c} \\
=\frac{1}{3} \times \frac{2 a^{2}+c^{2}}{2 a c} .
\end{array}
$$
Notice that,
$$
\frac{2 a^{2}+c^{2}}{2 a c} \geqslant \frac{2 \sqrt{2} a c}{2 a c}=\sqrt{2},
$$
with equality holding if and only if $\sqrt{2} a=c$.
Combining with $0<\angle B<\pi$, we get
$$
\frac{\sqrt{2}}{3} \leqslant \cos B<1 \text {. }
$$ | [\frac{\sqrt{2}}{3},1) | 2. Given that the circumcenter of $\triangle A B C$ is $O$, and
$$
\overrightarrow{A O} \cdot \overrightarrow{B C}=3 \overrightarrow{B O} \cdot \overrightarrow{A C}+4 \overrightarrow{C O} \cdot \overrightarrow{B A} \text {. }
$$
Then the range of $\cos B$ is $\qquad$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$A B=4 \mathrm{~cm}, B C=5 \mathrm{~cm}$ and $D E=3 \mathrm{~cm} . E F=c \mathrm{~cm}$
Draw $B G \perp D F, C H \perp D F$
$D G=A B=4 \mathrm{~cm}, G H=B C=5 \mathrm{~cm}$
$E G=D G-D E=4 \mathrm{~cm}-3 \mathrm{~cm}=1 \mathrm{~cm}$
Let $O$ be the centre.
Let $M$ be the foot of perpendicular of $O$ on $E F$ and produce $O M$ to $N$ on $B C$.
$O N \perp B C$ (corr. $\angle \mathrm{s} A C / / D F$ )
$B N=N C=2.5 \mathrm{~cm}(\perp$ from centre bisect chord $)$
$M F=E M(\perp$ from centre bisect chords $)$
$=E G+G M=1 \mathrm{~cm}+B N=1 \mathrm{~cm}+2.5 \mathrm{~cm}=3.5 \mathrm{~cm}$
$E F=2 E M=7 \mathrm{~cm}$ | 7\mathrm{~} | G3.3 In Figure 2, a rectangle intersects a circle at points $B, C, E$ and $F$. Given that $A B=4 \mathrm{~cm}, B C=5 \mathrm{~cm}$ and $D E=3 \mathrm{~cm}$. If $E F=c \mathrm{~cm}$, find the value of $c$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
. Only one of the triangles in the tiling is acute. Indeed, the fundamental remark is that acute triangles are recognized as those that have the center of their circumscribed circle as an interior point. Here, all the triangles in the tiling have the same circumscribed circle, namely the circumscribed circle of the regular polygon. It is then obvious that the center of this common circle is inside a single triangle. (It might seem a priori that the center could be located on a side that is common to two triangles, but in reality this cannot happen since the sides of the triangles are the diagonals of the regular polygon, and these do not pass through the center of the circle because the polygon has an odd number of sides.) | 1 | . A convex regular polygon with 1997 vertices has been decomposed into triangles using diagonals that do not intersect internally. How many of these triangles are acute? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
25. E Each floor has 35 rooms. On every floor except floor 2 , the digit 2 will be used for rooms ' $n 02$ ', ' $n 12$ ', ' $n 20$ ' to ' $n 29$ ' (including ' $n 22$ ') and ' $n 32$ '. Hence the digit 2 will be used 14 times on each floor except floor 2 . On floor 2, the digit 2 will be used an extra 35 times as the first digit of the room number. Therefore the total number of times the digit 2 will be used is $5 \times 14+35=105$. | 105 | 25. The room numbers of a hotel are all three-digit numbers. The first digit represents the floor and the last two digits represent the room number. The hotel has rooms on five floors, numbered 1 to 5 . It has 35 rooms on each floor, numbered $\mathrm{n} 01$ to $\mathrm{n} 35$ where $\mathrm{n}$ is the number of the floor. In numbering all the rooms, how many times will the digit 2 be used?
A 60
В 65
C 95
D 100
E 105 | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3. Answer: 5 trucks. Indeed, 5 trucks will be always enough. First four trucks can carry at least 8 tons of stones and the fifth will be able to carry all the rest. If there were 13 stones weighing $10 / 13$ tons each, then each truck would be able to carry only 3 of them, hence five cars might be needed in this case. | 5 | 3. Several pounamu stones weigh altogether 10 tons and none of them weigh more than 1 tonne. A truck can carry a load which weight is at most 3 tons. What is the smallest number of trucks such that bringing all stones from the quarry will be guaranteed? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let $\alpha$ be a parameter such that
$$\begin{aligned}
f^{2}(x) & =\frac{1}{\alpha^{2}} \sin ^{2} x(a \alpha+\alpha \cos x)^{2} \leqslant \frac{1}{\alpha^{2}} \sin ^{2} x\left(\alpha^{2}+\cos ^{2} x\right)\left(a^{2}+\alpha^{2}\right) \\
& \leqslant \frac{1}{\alpha^{2}}\left(\frac{\sin ^{2} x+\alpha^{2}+\cos ^{2} x}{2}\right)^{2}\left(a^{2}+\alpha^{2}\right)=\frac{1}{\alpha^{2}}\left(\frac{\alpha^{2}+1}{2}\right)^{2}\left(a^{2}+\alpha^{2}\right),
\end{aligned}$$
with equality holding if and only if $\alpha^{2}=a \cos x, \sin ^{2} x=\alpha^{2}+\cos ^{2} x$.
Eliminating $x$, we get $2 \alpha^{4}+a^{2} \alpha^{2}-a^{2}=0$.
Solving the equation, we get $\alpha^{2}=\frac{1}{4}\left(\sqrt{a^{4}+8 a^{2}}-a^{2}\right)$, thus $\cos x=\frac{1}{4}\left(\sqrt{a^{2}+8}-a\right)$.
Therefore, when $x=2 k \pi \pm \arccos \left[\frac{1}{4}\left(\sqrt{a^{2}+8}-a\right)\right](k \in \mathbf{Z})$,
$$f(x)_{\max }=\frac{\sqrt{a^{4}+8 a^{2}}-a^{2}+4}{8} \cdot \sqrt{\frac{\sqrt{a^{4}+8 a^{2}}+a^{2}+2}{2}} .$$ | \frac{\sqrt{a^{4}+8 a^{2}}-a^{2}+4}{8} \cdot \sqrt{\frac{\sqrt{a^{4}+8 a^{2}}+a^{2}+2}{2}} | Example 4 Let $a$ be a real number, find the maximum value of the function $f(x)=|\sin x(a+\cos x)|(x \in \mathbf{R})$. | Algebra | AI-MO/NuminaMath-1.5/inequalities | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution. Multiply by $\sin \pi x$. All $x=n, n \in \mathbb{Z}$ need to be removed from the answer, as they are not roots of the original equation. Next, using the equality $8 \sin \pi x \cos \pi x \cos 2 \pi x \cos 4 \pi x=\sin 8 \pi x$, we find
$$
\begin{aligned}
\sin 8 \pi x \cdot \cos 2 \pi x & =\sin \pi x \cdot \cos 9 \pi x \Longleftrightarrow \\
\sin 10 \pi x+\sin 6 \pi x & =\sin 10 \pi x-\sin 8 \pi x \Longleftrightarrow \\
\sin 6 \pi x & =\sin (-8 \pi x)
\end{aligned}
$$
From which $x=k / 7$ or $x=1 / 2+l, k, l \in \mathbb{Z}$.
The roots of the equation in the interval $[0 ; 1]$ are: $1 / 7, 2 / 7, 3 / 7, 4 / 7, 5 / 7, 6 / 7$ and $1 / 2$.
Answer: 3.5. | 3.5 | 8-1. Find all common points of the graphs
$$
y=8 \cos \pi x \cdot \cos ^{2} 2 \pi x \cdot \cos 4 \pi x \quad \text { and } \quad y=\cos 9 \pi x
$$
with abscissas belonging to the segment $x \in[0 ; 1]$. In your answer, specify the sum of the abscissas of the found points. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Solution】Solution: The total number of matches played, $4 \times 6 \div 2=12$ (matches), among which 4 matches were draws, the number of matches with a winner is: $12-4=8$ (matches), the total points scored by the six teams: $3 \times 8+2 \times 4=32$ (points), because, the top three teams scored at least: $7+8+9=24$ (points), so, the bottom three teams scored at most: $32-24=8$ (points), also, because, the fourth team scored more points than the fifth team, so, the fifth team scored at most 3 points, because, the sixth team might score 0 points, so, the fifth team scored at least 1 point, hence the answer is: 3,1. | 3,1 | 13. (8 points) Six football teams are playing a round-robin tournament, where each team plays against every other team. If a match ends in a draw, each team gets 1 point; otherwise, the winning team gets 3 points and the losing team gets 0 points. Now, after four rounds (each team has played 4 matches), the total points of each team are all different. It is known that the team in third place has a total of 7 points, and there have been 4 matches that ended in a draw. Therefore, the team in fifth place can have a maximum of $\qquad$ points and a minimum of $\qquad$ points. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution.
Let $x$ be the speed of the oncoming train.
According to the condition, $(40+x) \cdot \frac{3}{3600}=75 \cdot 10^{-3}$, from which $x=50($ km $/ h)$.
Answer: $50 \mathrm{km} /$ h. | 50\mathrm{}/\mathrm{} | 13.278. A passenger on a train knows that the speed of this train on this section of the track is 40 km/h. As soon as a passing train started to go by the window, the passenger started a stopwatch and noticed that the passing train took $3 \mathrm{s}$ to pass by the window. Determine the speed of the passing train, given that its length is 75 m. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$2020$ | 2020 | 4. Given that when $x=1$, $4 a x^{3}+3 b x^{2}-2 c x=8$, and $3 a x^{3}+2 b x^{2}-c x=-6$, then, when $x=-1$, the value of $10 a x^{3}-7 b x^{2}-4 c x+2016$ is $\qquad$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, then no more than two numbers are divisible by $d$. Suppose there are three such numbers. Then they must be pairwise connected, and thus form a three-link cycle.
3) If $q$ is a prime divisor of $n$, then among four sequentially connected numbers, there exists a pair of adjacent numbers whose sum of squares is not divisible by $q$. Suppose there is a chain $(a, b, c, d)$ of sequentially connected numbers such that the sum of squares of any pair of adjacent numbers is divisible by $q$. Then
$$
a^{2}+d^{2}=\left(a^{2}+b^{2}\right)-\left(b^{2}+c^{2}\right)+\left(c^{2}+d^{2}\right) \vdots q
$$
This means that the numbers $a$ and $d$ are connected, so there is a cycle of length 4 in the picture, which does not exist.
From 1) and 3), it follows that the number $n$ has at least two different odd prime divisors. Let there be exactly two (say, $p$ and $q$). We will show that they are different from $3, 7, 11$. Suppose, for example, that $p \in \{3, 7, 11\}$. Note that for any natural number $a$
$$
a^{2} \bmod 3 \in \{0,1\}, \quad a^{2} \bmod 7 \in \{0,1,2,4\}, \quad a^{2} \bmod 11 \in \{0,1,3,4,5,9\}
$$
Therefore, $a^{2}+b^{2}$ is divisible by $p$ if and only if $a^{2}$ and $b^{2}$ are divisible by $p$, and thus $a$ and $b$ are divisible by $p$. By 2), this condition can be satisfied by only one pair. From the remaining four pairs of connected numbers, a three-link chain can be formed in which the sum of squares of each pair of adjacent numbers is divisible by $q$. But this contradicts 3).
Thus, if $n$ has exactly two different odd prime divisors, then $n \geqslant 5 \cdot 13=65$. If there are more than two such divisors, then $n \geqslant 3 \cdot 5 \cdot 7=105>65$. The arrangement for $n=65$ is shown in the figure.
 | 65 | 1. In the picture, several circles are drawn, connected by segments. Tanya chooses a natural number n and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers a and b are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with n; if connected, then the numbers $a^{2}+b^{2}$ and n must have a common natural divisor greater than 1. For what smallest n does such an arrangement exist?
 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: $\frac{7}{24}$ From the starting point of $(1,0)$, there is a $\frac{1}{4}$ chance we will go directly to $(1,1)$, a $\frac{1}{2}$ chance we will end at $(2,0)$ or $(1,-1)$, and a $\frac{1}{4}$ chance we will go to $(0,0)$. Thus, if $p$ is the probability that we will reach $(1,1)$ from $(0,0)$, then the desired probability is equal to $\frac{1}{4}+\frac{1}{4} p$, so we need only calculate $p$. Note that we can replace the condition $|x|+|y| \geq 2$ by $|x|+|y|=2$, since in each iteration the quantity $|x|+|y|$ can increase by at most 1 . Thus, we only have to consider the eight points $(2,0),(1,1),(0,2),(-1,1),(-2,0),(-1,-1),(0,-2),(1,-1)$. Let $p_{1}, p_{2}, \ldots, p_{8}$ be the probability of reaching each of these points from $(0,0)$, respectively. By symmetry, we see that $p_{1}=p_{3}=p_{5}=p_{7}$ and $p_{2}=p_{4}=p_{6}=p_{8}$. We also know that there are two paths from $(0,0)$ to $(1,1)$ and one path from $(0,0)$ to $(2,0)$, thus $p_{2}=2 p_{1}$. Because the sum of all probabilities is 1 , we have $p_{1}+p_{2}+\ldots+p_{8}=1$. Combining these equations, we see that $4 p_{1}+4 p_{2}=12 p_{1}=1$, so $p_{1}=\frac{1}{12}$ and $p_{2}=\frac{1}{6}$. Since $p=p_{2}=\frac{1}{6}$, then the final answer is $\frac{1}{4}+\frac{1}{4} \cdot \frac{1}{6}=\frac{7}{24}$ | \frac{7}{24} | 4. [4] An ant starts at the point $(1,0)$. Each minute, it walks from its current position to one of the four adjacent lattice points until it reaches a point $(x, y)$ with $|x|+|y| \geq 2$. What is the probability that the ant ends at the point $(1,1)$ ? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer. For 5 rubles.
Solution. Let's mark the numbers $17, 19, 23$, and 29, spending four rubles. Then mark the number 2, spending another ruble. After this, we can freely mark all even numbers (since they are divisible by 2), and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$), the even number $2n$ is marked, and we can mark $n$ as its divisor. It remains to mark 21, 25, and 27, and this is also done for free: 25 is divisible by the marked number 5, and 21 and 27 are divisible by the marked number 3. In any way of solving the problem, the prime numbers 17, 19, 23, and 29, which exceed 15, will have to be marked for money, as they are not divisors or multiples of any numbers on the board. So, 4 rubles will be spent only on them. To mark anything else, we will have to spend a fifth ruble. Therefore, it is impossible to fulfill the conditions of the problem for less than five rubles.
Comment. In fact, after marking the "large" prime numbers, we could have marked any of the remaining numbers on the board instead of two. Indeed, then we will freely mark its smallest prime divisor $p$. If $p=2$, we act according to the algorithm described above. If not, we mark $2p$ (this is possible since $p<15$), then mark two, and then everything else in the known way.
A similar solution is applicable for an arbitrarily long set $2, 3, 4, \ldots, N$ - we are forced to mark all "large" prime numbers (exceeding $N / 2$) for money, and then mark any of the remaining numbers for a ruble. Then we freely mark two in the way described above, then all even numbers, then all "small" prime numbers (not exceeding $N / 2$), because any "small" $p$ will be a divisor of $2p$. Now we can mark all other unmarked numbers: each of them will be divisible by its minimal prime divisor - a "small" prime number. | 5 | Problem 3. The numbers $2,3,4, \ldots, 29,30$ are written on the board. For one ruble, you can mark any number. If a number is already marked, you can freely mark its divisors and numbers that are multiples of it. What is the minimum number of rubles needed to mark all the numbers on the board? [6 points]
(I.V. Yashchenko) | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
4. A
Math, Chinese, English, Chemistry, the failure rates of these 4 subjects are: $1\%, 2\%, 4\%, 8\%$; the sum is $15\%$. Moreover, for those who fail at least two subjects, among the above 4 subjects, at least one subject is failed, which is no less than $15\%$, so the number of people who pass at least 4 subjects is no less than $85\%$. Furthermore, it is assumed that for those who fail at least two subjects, the combination of failures reaches the minimum value. Since $1-84\%=16\%>15\%$. | 85 | 4. A prestigious school has the following pass rates in a college entrance examination simulation:
Math: $99 \%$
Chinese: $98 \%$
English: $96 \%$
Chemistry: $92 \%$
Comprehensive: $84 \%$
Then the minimum possible value for the number of people who pass at least 4 subjects is ( ).
A. $85 \%$
B. $84.5 \%$
C. $86 \%$
D. $86.5 \%$ | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Analysis】Since $\frac{A E}{A B}=\frac{A F}{A C}=\frac{1}{3}$, therefore $\mathrm{EF} / / \mathrm{BC}$
Therefore $\mathrm{S}_{\mathrm{EBD}}=\mathrm{S}_{\mathrm{FBD}}=\mathrm{S}_{1} \rightarrow \mathrm{S}_{1}+\mathrm{S}_{2}=\mathrm{S}_{\mathrm{EBC}}=\frac{2}{3} \mathrm{~S}_{\mathrm{ABC}}=40$
When the sum is constant, the smaller the difference, the greater the product, so when $S_{1}=S_{2}$, that is, when $D$ is the midpoint, $S_{1} \times S_{2}$ is maximized at $20 \times 20=400$ | 400 | 5. As shown in the figure, the area of $\triangle \mathrm{ABC}$ is $60, \mathrm{E}$ and $\mathrm{F}$ are points on $\mathrm{AB}$ and $\mathrm{AC}$ respectively, satisfying $\mathrm{AB}=3 \mathrm{AE}, \mathrm{AC}=3 \mathrm{AF}$, point $D$ is a moving point on segment $B C$, let the area of $\triangle F B D$ be $S_{1}$, the area of $\triangle E D C$ be $S_{2}$, then the maximum value of $S_{1} \times S_{2}$ is $\qquad$ | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
6 $a=0$ Hint: Let $\cos x=-1, \sin x=0, a \sin ^{2} x+\cos x \geqslant$ $a^{2}-1$, we get $a^{2} \leqslant 0$. And when $a=0$, the original inequality obviously always holds. Therefore, $a=0$. | 0 | 6 If the inequality $a \sin ^{2} x+\cos x \geqslant a^{2}-1$ holds for any $x \in \mathbf{R}$, then the range of real number $a$ is $\qquad$. | Inequalities | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5. Vova has 19 math grades in his journal, all twos and threes, and the first four grades are twos. It turned out that among the quartets of consecutive grades, all 16 possible combinations of four twos and threes are present. What are Vova's last four grades?
ANSWER. 3222.
SOLUTION. Note that there are exactly 16 quartets of consecutive grades in a string of 19 grades, meaning each combination appears exactly once. After the first four twos, there must be a three, otherwise the combination of four twos would appear twice. Therefore, after the combination 3222, neither 2 nor 3 can follow, as in both cases, a combination that has already appeared would reappear. Thus, the combination 3222 must be at the end of the string. | 3222 | 5. Vova has 19 math grades in his journal, all twos and threes, and the first four grades are twos. It turned out that among the quartets of consecutive grades, all 16 possible combinations of four twos and threes are present. What are Vova's last four grades? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
}
For the lengths \(A B=c, D C=h, A D=q, D B=p\) it holds that
\[
h=\frac{2}{5} c \quad(1) \quad \text { and } \quad p+q=c \quad(2) \quad \text { and } \quad q<p
\]
Furthermore, according to the height theorem \(p g=h^{2} \quad\) (4). Substituting \(h\) from (1) into (4) yields
\[
p q=\frac{4}{25} c^{2}
\]
Substituting \(q\) from (2) into (5) yields
\[
\begin{aligned}
p(c-p) & =\frac{4}{25} c^{2} \\
p_{1,2} & =\frac{1}{2} c \pm \sqrt{\frac{1}{4} c^{2}-\frac{4}{25} c^{2}}=\frac{1}{2} c \pm \sqrt{\frac{1}{100} c^{2}(25-16)} \\
& =\frac{c}{10}(5 \pm 3)
\end{aligned}
\]
thus either \(p=\frac{4}{5} c\) and then further \(q=\frac{1}{5} c\) according to (2), or \(p=\frac{1}{5} c\) and then further \(q=\frac{4}{5} c\) according to (2). Of these two possibilities, only the first remains due to (3). Therefore, the sought ratio \(A D: D B=q: p=1: 4\). | 1:4 | \section*{Problem 1 - 211021}
In a right-angled triangle \(A B C\), the height \(D C\) perpendicular to the hypotenuse \(A B\) is exactly \(\frac{2}{5}\) times as long as the hypotenuse \(A B\). For the foot of the perpendicular \(D\), it holds that \(A D < D B\).
In what ratio \(A D: D B\) does it divide the hypotenuse? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution.
$$
\begin{aligned}
& \sin ^{2}\left(\alpha-\frac{3 \pi}{2}\right)\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{tg}\left(\frac{\pi}{4}+\alpha\right) \cos ^{-2}\left(\frac{\pi}{4}-\alpha\right)= \\
& =\left(\sin \left(\frac{3}{2} \pi-\alpha\right)\right)^{2}\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{tg}\left(\frac{\pi}{4}+\alpha\right) \cdot \frac{1}{\cos ^{2}\left(\frac{\pi}{4}-\alpha\right)}= \\
& =\cos ^{2} \alpha\left(1-\frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}\right) \cdot \frac{1-\cos \left(\frac{\pi}{2}+2 \alpha\right)}{\sin \left(\frac{\pi}{2}+2 \alpha\right)} \cdot \frac{1}{\frac{1+\cos \left(\frac{\pi}{2}-2 \alpha\right)}{2}}= \\
& =\frac{1+\cos 2 \alpha}{2} \cdot \frac{1+\cos 2 \alpha-1+\cos 2 \alpha}{1+\cos 2 \alpha} \cdot \frac{1+\sin 2 \alpha}{\cos 2 \alpha} \cdot \frac{2}{1+\sin 2 \alpha}=2
\end{aligned}
$$
Answer: 2. | 2 | 3.081. $\sin ^{2}\left(\alpha-\frac{3 \pi}{2}\right)\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{tg}\left(\frac{\pi}{4}+\alpha\right) \cos ^{-2}\left(\frac{\pi}{4}-\alpha\right)$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
7. $\pi f(x)=\cos 2 x-2 \sqrt{3} \sin x \cos x=\cos 2 x-\sqrt{3} \sin 2 x=2 \cos \left(2 x+\frac{\pi}{3}\right)$ Therefore, $T=\pi$ | \pi | 7. (2004 National College Entrance Examination - Beijing Paper) The smallest positive period of the function $f(x)=\cos 2 x-2 \sqrt{3} \sin x \cos x$ is | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 974
Solution: $A B^{2}=1+(b-a)^{2}, B C^{2}=1+(c-b)^{2}, A C^{2}=4+(c-a)^{2}$.
If triangle $A B C$ is a right triangle with hypotenuse $A C$, then by the Pythagorean theorem $A C^{2}=A B^{2}+B C^{2}, 1+(b-a)^{2}+1+(b-c)^{2}=4+(a-c)^{2}$, which simplifies to $(b-a)(b-c)=1$. Since both factors are integers, we have only the following cases: $b=a+1=c+1$ and $b=a-1=c-1$, for each of which there are 99 triples $(a, b, c)$, i.e., 198 ways.
If the hypotenuse is side $A B$, then similarly we get the relation $(c-a)(c-b)=-2$, which is possible only in the following cases:
$c=a+1=b-2$
$c=a-1=b+2$
$c=a+2=b-1$
$c=a-2=b+1$
for each of which there are 97 triples $(a, b, c)$, i.e., a total of $97 \cdot 4=388$ ways.
If the hypotenuse is side $B C$, then we get the relation $(a-b)(a-c)=-2$. Similarly, we find 388 ways.
In total, we get $198+388+388=974$ ways. | 974 | 8. In how many different ways can integers $a, b, c \in [1,100]$ be chosen so that the points with coordinates $A(-1, a), B(0, b)$, and $C(1, c)$ form a right triangle? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
II. (50 points)
Make a trigonometric substitution. Since $|a|>1$, we can set
$$
a=\frac{1}{\cos \theta}(\theta \in(0,2 \pi)) \text {. }
$$
Then $b=\frac{a^{2}}{2-a^{2}}=\frac{\frac{1}{\cos ^{2} \theta}}{2-\frac{1}{\cos ^{2} \theta}}=\frac{1}{\cos ^{2} \theta}$,
Similarly, $c=\frac{1}{\cos ^{4} \theta}, a=\frac{1}{\cos ^{8} \theta}$.
Thus, $\cos 8 \theta=\cos \theta, \cos \theta-\cos 8 \theta=0$. And $\cos \theta \neq 0, \cos \theta \neq \pm 1$,
$$
\begin{array}{l}
2 \sin \frac{7}{2} \theta \cdot \sin \frac{9}{2} \theta=0 . \\
\frac{7}{2} \theta=k \pi \quad k \in \mathbb{Z} . \\
\frac{9}{2} \theta=k \pi \quad k \in \mathbb{Z} .
\end{array}
$$
Since we only need to consider the value of $\cos \theta$,
$\cos \theta$ can only be $\cos \frac{4}{7} \pi, -\cos \frac{\pi}{7}, \cos \frac{2}{7} \pi, \cos \frac{4}{9} \pi, -\cos \frac{\pi}{9}, -\cos \frac{2}{9} \pi$.
1. When $\cos \theta=\cos \frac{4}{7} \pi$ or $-\cos \frac{\pi}{7}$ or $\cos \frac{2}{7} \pi$:
$$
\begin{array}{l}
a+b+c=\frac{1}{\cos \frac{4}{7} \pi}+\frac{1}{\cos \frac{8}{7} \pi}+\frac{1}{\cos \frac{16}{7} \pi} \\
=\frac{1}{\cos \frac{4}{7} \pi}+\frac{1}{-\cos \frac{\pi}{7}}+\frac{1}{\cos \frac{2}{7} \pi} \\
=\frac{2}{\varepsilon^{2}+\frac{1}{\varepsilon^{2}}}+\frac{2}{\varepsilon^{4}+\frac{1}{\varepsilon^{4}}}+\frac{2}{\varepsilon^{8}+\frac{1}{\varepsilon^{8}}} \\
\left(\text { Let } \varepsilon=\cos \frac{2}{7} \pi+i \sin \frac{2}{7} \pi\right) \\
=2 \cdot\left(\frac{\varepsilon^{2}}{\varepsilon^{4}+1}+\frac{\varepsilon^{4}}{\varepsilon+1}+\frac{\varepsilon}{\varepsilon^{2}+1}\right) \\
=2 \cdot \frac{-2 \varepsilon^{7}}{\varepsilon^{7}}=-4 .
\end{array}
$$
2. When $\cos \theta=\cos \frac{4}{9} \pi$ or $-\cos \frac{\pi}{9}$ or $\cos \frac{2}{9} \pi$:
$$
\begin{array}{l}
a+b+c=\frac{1}{\cos \frac{4}{9} \pi}+\frac{1}{\cos \frac{8}{9} \pi}+\frac{1}{\cos \frac{16}{9} \pi} \\
=\frac{2}{z^{2}+\frac{1}{z^{2}}}+\frac{2}{z^{4}+\frac{1}{z^{4}}}+\frac{2}{z^{8}+\frac{1}{z^{8}}}
\end{array}
$$
(Let $\left.z=\cos \frac{2}{9} \pi+i \sin \frac{2}{9} \pi\right)$
$$
\begin{array}{l}
=2\left(\frac{z^{2}}{z^{4}+1}+\frac{z^{4}}{z^{8}+1}+\frac{z^{8}}{z^{16}+1}\right) \\
=2 \cdot \frac{2 z^{2}+2 z^{8}-z^{5}}{-z^{5}} \\
=2 \cdot\left(\frac{2}{-z^{3}}+\frac{2}{-z^{-3}}+1\right)=6 .
\end{array}
$$
3. When $\cos \theta=-\cos \frac{\pi}{3}$:
$$
a+b+c=\frac{1}{\cos \frac{4}{3} \pi}+\frac{1}{\cos \frac{8}{3} \pi}+\frac{1}{\cos \frac{12}{3} \pi}=-6 .
$$
Therefore, the possible values of $a+b+c$ are $6, -4, -6$. | 6,-4,-6 | II. (50 points $\}$
$a, b, c \in \mathbf{R}$. Satisfy $|a|>1,|b|>1,|c|>1$ and $b=\frac{a^{2}}{2-a^{2}}, c=\frac{b^{2}}{2-b^{2}}, a=$ $\frac{c^{2}}{2-c^{2}}$. Find all possible values of $a+b+c$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
[Solution] From the given, we have $x y=2000+x+y$,
which means
$$
(x-1)(y-1)=2001=3 \cdot 23 \cdot 29 \text {. }
$$
Since $x$ and $y$ are both two-digit numbers, we have
$$
\begin{array}{l}
(x-1)(y-1)=29 \cdot 69, \\
(x-1)(y-1)=23 \cdot 87,
\end{array}
$$
which gives
$$
\left\{\begin{array} { l }
{ x - 1 = 2 9 , } \\
{ y - 1 = 6 9 , }
\end{array} \quad \left\{\begin{array}{l}
x-1=23, \\
y-1=87 .
\end{array}\right.\right.
$$
Solving these, we get
$$
x=30, y=70 \text {, or } x=24, y=88 \text {. }
$$ | 30,70\text{,or}24,88 | $10 \cdot 7$ Let $x$ and $y$ be two natural numbers with two digits, and $x<y$. The product $xy$ is a four-digit number. The first digit is 2, and if this first digit 2 is removed, the remaining number is exactly $x+y$, for example $x=30, y=70$, besides this pair, there is another pair of numbers with the above property, try to find these two numbers.
(Dutch Mathematics Competition, 1983) | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
【Analysis and Solution】
Number Theory, Remainder.
(1) If the 40 odd numbers taken out contain "odd numbers with a units digit of 5"; $5 \times$ odd number has a units digit of 5;
the units digit of the product is 5.
(2) If the 40 odd numbers taken out do not contain "odd numbers with a units digit of 5"; then we need to take all the odd numbers with units digits of $1, 3, 7, 9$ from $1 \sim 100$; $1 \times 3 \times 7 \times 9$ has a units digit of 9;
$9^{1}$ has a units digit of $9, 9^{2}$ has a units digit of $1, 10 \div 2=5, 9^{10}$ has a units digit of 1; the units digit of the product is 1.
In summary, the units digit of the product could be 1 or 5. | 1or5 | 【Question 17】
From $1,2, \cdots, 100$, if you arbitrarily take out 40 odd numbers, the possible unit digit of their product is $\qquad$ - | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Answer: 4.5
Solution. Let's choose any five people standing in a circle in a row, with the outermost being boys. Let their numbers be $x, x y, y, y z, z$. Then $y=x y+y z$. Since $y \neq 0$, we can divide by it, obtaining $x+z=1$. Thus, the sum of any two numbers said by boys standing three apart is 1.
Let the boys say the numbers $a_{1}, a_{2}, \ldots, a_{6}$ (in this order around the circle). Then
$$
1=a_{1}+a_{3}=a_{2}+a_{4}=\ldots=a_{5}+a_{1}=a_{6}+a_{2} .
$$
We can notice that
$$
6=\left(a_{1}+a_{3}\right)+\left(a_{2}+a_{4}\right)+\ldots+\left(a_{5}+a_{1}\right)+\left(a_{6}+a_{2}\right)=2\left(a_{1}+a_{2}+\ldots+a_{6}\right)
$$
from which the sum of all the boys' numbers is 3.
On the other hand, the sum of the boys' numbers is twice the sum of all the girls' numbers, since each girl's number is a summand for two boys. Therefore, the sum of the girls' numbers is 1.5. The total sum is then 4.5.
It remains to note that such a situation is indeed possible: all boys can write the number $\frac{1}{2}$, and all girls can write the number $\frac{1}{4}$.
## Criteria
The highest applicable criterion is used:
7 p. Any complete solution to the problem.
7 p. It is proven that the sum can only be 4.5, but no valid example of a permissible arrangement of numbers is provided.
1 p. It is stated that the sum of the girls' numbers is half the sum of the boys' numbers.
3 p. It is proven that the sum of the numbers of boys standing "three apart" is 1.
0 p. Correct answer.
1 p. Correct example of number arrangement. | 4.5 | Problem 5. Six boys and six girls stood in a circle, alternating. Each of them wrote a non-zero number in their notebook. It is known that each number written by a boy is equal to the sum of the numbers written by the adjacent girls, and each number written by a girl is equal to the product of the numbers written by the adjacent boys. What can the sum of all twelve numbers be? | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
## Solution
In the section of the given figure by the plane $z=$ const, there is an ellipse:
$$
\begin{aligned}
& \frac{x^{2}}{16}+\frac{y^{2}}{9}=\frac{64-z^{2}}{64} \\
& \frac{x^{2}}{16 \cdot \frac{64-z^{2}}{64}}+\frac{y^{2}}{9 \cdot \frac{64-z^{2}}{64}}=1 \rightarrow a=\frac{1}{2} \sqrt{64-z^{2}} ; b=\frac{3}{8} \sqrt{64-z^{2}} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{1}{2} \sqrt{64-z^{2}} \cdot \frac{3}{8} \sqrt{64-z^{2}}=\frac{3 \pi}{16} \cdot\left(64-z^{2}\right) \\
& V=\int_{0}^{4} S(z) d z=\frac{3 \pi}{16} \int_{0}^{4}\left(64-z^{2}\right) d z=\left.\frac{3 \pi}{16}\left(64 z-\frac{z^{3}}{3}\right)\right|_{0} ^{4}= \\
& =\frac{3 \pi}{16}\left(64 \cdot 4-\frac{4^{3}}{3}-64 \cdot 0+\frac{0^{3}}{3}\right)=\frac{\pi}{16}(3 \cdot 256-64)=\frac{704 \cdot \pi}{16}=44 \pi
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-20$ » Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals
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## Problem Kuznetsov Integrals 20-21
## Material from PlusPi | 44\pi | ## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{z^{2}}{64}=1, z=4, z=0
$$ | Calculus | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
If $a_{1}, a_{2}, a_{3}$ meet the conditions, then $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-2, a_{3}^{\prime}=a_{3}-4$ are distinct, so $a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}$ are selected in ascending order from the numbers $1,2, \cdots, 10$. Conversely, this is also true.
Therefore, the number of different ways to choose them is $C_{10}^{3}=120$. | 120 | 11. If numbers $a_{1}, a_{2}, a_{3}$ are taken in increasing order from the set $1,2, \cdots, 14$, such that both $a_{2}-$ $a_{1} \geqslant 3$ and $a_{3}-a_{2} \geqslant 3$ are satisfied, then the total number of different ways to choose such numbers is $\qquad$. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$$
\begin{array}{l}
\frac{1}{x+3}=\frac{2}{x^{2}-1}-\frac{1}{x-1} \Rightarrow \frac{1}{x+3}=\frac{2-(x+1)}{(x-1)(x+1)} \\
\frac{1}{x+3}=\frac{1-x}{(x-1)(x+1)} \Rightarrow \frac{1}{x+3}=-\frac{1}{x+1} \\
x+1=-x-3 \Rightarrow x=-2=a \\
\left|\frac{2 \log 2}{\log b}-1-\frac{\log 3}{\log b}\right|+\frac{2 \log 2}{\log b}+1+\frac{\log 3}{\log b}=3 \Rightarrow\left|\frac{\log 4 / 3 b}{\log b}\right|+\frac{\log 12}{\log b}=2 \Rightarrow\left|\log \frac{4}{3 b}\right|=\log \frac{b^{2}}{12} \\
\log \frac{4}{3 b}= \pm \log \frac{b^{2}}{12} \\
\log \frac{4}{3 b}=\log \frac{b^{2}}{12} \text { or } \log \frac{4}{3 b}=\log \frac{12}{b^{2}} \\
b^{3}=16 \text { or } b=9
\end{array}
$$
When $b^{3}=16,\left(b^{3}\right)^{2}=2560$ accepted.
Method 2
Remark Define the maximum function of $x, y$ as: $\operatorname{Max}(x, y)=\frac{x+y+|x-y|}{2}$
Similarly, the minimum function of $x, y$ is: $\operatorname{Min}(x, y)=\frac{x+y-|x-y|}{2}$
$$
f(b)=\frac{-a}{\log _{2} b}=\frac{2 \log 2}{\log b}=\frac{\log 4}{\log b}, g(b)=1+\frac{1}{\log _{3} b}=\frac{\log b}{\log b}+\frac{\log 3}{\log b}=\frac{\log 3 b}{\log b}
$$
The given equation is equivalent to $2 \operatorname{Max}(f(b), g(b))=3$
If $f(b)>g(b)$, i.e. $\frac{\log 4}{\log b}>\frac{\log 3 b}{\log b} \Rightarrow b<\frac{4}{3}$, then the equation is $2 f(b)=3$
$$
\frac{2 \log 4}{\log b}=3 \Rightarrow \log 16=\log b^{3} \Rightarrow b^{3}=16<\left(\frac{4}{3}\right)^{3} \Rightarrow 1<\frac{4}{27}
$$
$\Rightarrow 27<4$, which is a contradiction; $\therefore$ rejected
If $f(b) \leq g(b)$, the equation is equivalent to $2 g(b)=3$
$$
\begin{array}{l}
\text { i.e. } \frac{2 \log 3 b}{\log b}=3 \Rightarrow \log 9 b^{2}=\log b^{3} \Rightarrow 9 b^{2}=b^{3} \Rightarrow b=9 \\
x^{2}-5 x+1=0, x^{2}+1=5 x, x^{4}+2 x^{2}+1=25 x^{2}, x^{4}+x^{2}+1=24 x^{2} \\
c=\frac{x_{0}^{2}}{x_{0}^{4}+x_{0}^{2}+1}=\frac{x_{0}^{2}}{24 x_{0}^{2}}=\frac{1}{24}
\end{array}
$$ | -2,9,\frac{1}{24} | I3.1 Let $x \neq \pm 1$ and $x \neq-3$. If $a$ is the real root of the equation $\frac{1}{x-1}+\frac{1}{x+3}=\frac{2}{x^{2}-1}$, find the value of $a$
I3.2 If $b>1, f(b)=\frac{-a}{\log _{2} b}$ and $g(b)=1+\frac{1}{\log _{3} b}$. If $b$ satisfies the equation $|f(b)-g(b)|+f(b)+g(b)=3$, find the value of $b$.
I3.3 Given that $x_{0}$ satisfies the equation $x^{2}-5 x+(b-8)=0$. If $c=\frac{x_{0}^{2}}{x_{0}^{4}+x_{0}^{2}+1}$, find the value of $c$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
2. 1 Detailed Explanation: It is clear that $x \geqslant 1$. The function $f(x)=\left(\frac{3}{17}\right)^{x}+\left(\frac{5}{17}\right)^{x}+\left(\frac{9}{17}\right)^{x}$ is a decreasing function on $[1,+\infty)$, so when $x=1$, $f(x)_{\max }=1$. Also, $g(x)=\sqrt{x-1}$ is an increasing function on $[1,+\infty)$, so when $x=1$, $g(x)_{\min }=0$. From the graph, it is known that the original equation has only one solution. | 1 | 2. The number of real roots of the equation $\left(\frac{3}{17}\right)^{x}+\left(\frac{5}{17}\right)^{x}+\left(\frac{9}{17}\right)^{x}=\sqrt{x-1}$ is $\qquad$ . | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Ivan took three times two pears, in the end, he had 6 pears. To determine how many pears Mirka ended up with, we will trace back the changes in the number of pears step by step. $\mathrm{K}$ to do this, it is enough to realize that before each of Ivan's takings, there were two more pears on the plate, and before each of Mirka's takings, there were twice as many pears on the plate.
Ivan took the last 2 pears during his third taking.
Mirka took 2 pears during her second taking, before that, there were 4 pears on the plate. Ivan took 2 pears during his second taking, before that, there were 6 pears on the plate.
Mirka took 6 pears during her first taking, before that, there were 12 pears on the plate. Ivan took 2 pears during his first taking, originally, there were 14 pears on the plate. Mirka took a total of 8 pears, so in the end, she had two more pears than Ivan. | 2 | Ivan and Mirka were sharing pears from a plate. Ivan always took two pears, and Mirka took half of what was left on the plate. They proceeded in this manner: Ivan, Mirka, Ivan, Mirka, and finally Ivan, who took the last two pears.
Determine who ended up with more pears and by how many.
(M. Dillingerová)
Hint. How many pears did Mirka take the second time? | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Let $S$ be the area of triangle $A B F$. Express through $S$ the areas of triangles $C D F, A F D$ and $B F C$.
## Solution
Let $S$ be the area of triangle $A B F$. From the similarity of triangles $A B F$ and $C D F$, it follows that $S_{\triangle \mathrm{CDF}}=4 S$.
Since
$$
\frac{D F}{F B}=\frac{C F}{A F}=\frac{C D}{A B}=2 \text {, }
$$
then
$$
S_{\triangle \mathrm{AFD}}=2 S, S_{\triangle \mathrm{BFC}}=2 S .
$$
$$
\frac{S_{\triangle A B F}+S_{\triangle C D F}}{S_{\triangle A F D}+S_{\triangle B F C}}=\frac{S+4 S}{2 S+2 S}=\frac{5}{4}
$$
This statement is true for any trapezoid where one base is twice the length of the other.
## Answer
$\frac{5}{4}$. | \frac{5}{4} | [ [Ratio of areas of triangles with a common base or common height] Difficult
In a circle, a trapezoid $ABCD$ is inscribed, with its bases $AB=1$ and $DC=2$. Let the point of intersection of the diagonals of this trapezoid be denoted as $F$. Find the ratio of the sum of the areas of triangles $ABF$ and $CDF$ to the sum of the areas of triangles $AFD$ and $BCF$.
# | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
Solution.
$$
\begin{aligned}
& \frac{\left(\left(4,625-\frac{13}{18} \cdot \frac{9}{26}\right): \frac{9}{4}+2.5: 1.25: 6.75\right): 1 \frac{53}{68}}{\left(\frac{1}{2}-0.375\right): 0.125+\left(\frac{5}{6}-\frac{7}{12}\right):(0.358-1.4796: 13.7)}= \\
& =\frac{\left(\left(\frac{37}{8}-\frac{1}{4}\right) \cdot \frac{4}{9}+2: 6.75\right) \cdot \frac{68}{121}}{0.125: 0.125+0.25:(0.358-0.108)}=\frac{\left(\frac{35}{18}+2: \frac{27}{4}\right) \frac{68}{121}}{1+0.25: 0.25}= \\
& =\frac{121}{54} \cdot \frac{68}{121} \cdot \frac{1}{2}=\frac{17}{27}
\end{aligned}
$$
Answer: $\frac{17}{27}$. | \frac{17}{27} | 1.037. $\frac{\left(\left(4,625-\frac{13}{18} \cdot \frac{9}{26}\right): \frac{9}{4}+2.5: 1.25: 6.75\right): 1 \frac{53}{68}}{\left(\frac{1}{2}-0.375\right): 0.125+\left(\frac{5}{6}-\frac{7}{12}\right):(0.358-1.4796: 13.7)}$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
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