SOLUTION. The problem is about a well-known right-angled triangle, specifically the one with legs of lengths 3 and 4, and the hypotenuse (denoted as $BC$) of length 5, whose area is $\frac{1}{2} \cdot 3 \cdot 4 = 6$. Let's also denote $P$ as the foot of the perpendicular from point $N$ to line $BC$ (Fig. 1). Since point $N$ lies on the angle bisector of $\angle ABC$, it follows that $|NA| = |NP|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8a9c2ddf9e5bc17afa4eg-02.jpg?height=465&width=600&top_left_y=967&top_left_x=728)

Fig. 1

Throughout the solution, we will be looking for relationships between the areas of different parts of triangle $ABC$. Since $M$ is the midpoint of side $BC$, the following equalities hold:

$$
S_{ABM} = S_{ACM} = \frac{1}{2} S_{ABC} = 3 \quad \text{and} \quad S_{BMK} = S_{CMK},
$$

because in both cases, we have pairs of triangles with the same height to the same bases. Comparing the bases and heights of two triangles leads to the proportions

$$
S_{ABN} : S_{CBN} = S_{AKN} : S_{CKN} = |AN| : |CN| = 4 : 5,
$$

since in both pairs, the triangles have the same heights to their respective bases $AN$ and $CN$, and the triangles in the first pair $ABN, CBN$ also have the same heights $NA$, resp. $NP$ from the common vertex $N$, so the ratio of their areas is indeed equal to the ratio $4:5$ of the lengths of their bases $AB$ and $CB$. From the known value $S_{ABN} + S_{CBN} = S_{ABC} = 6$, it follows that

$$
S_{ABN} = \frac{8}{3} \quad \text{and} \quad S_{CBN} = \frac{10}{3}.
$$

Our solution will be based on calculating the unknown area of triangle $AKN$, which we will denote as $S$ as shown in the figure, and for which we will derive an equation from the previously derived relationships.

To do this, we will sequentially express

$$
\begin{aligned}
& S_{ABK} = S_{ABN} - S_{AKN} = \frac{8}{3} - S, \\
& S_{BMK} = S_{ABM} - S_{ABK} = 3 - \left(\frac{8}{3} - S\right) = S + \frac{1}{3}, \\
& S_{BCK} = 2 S_{BMK} = 2S + \frac{2}{3}, \\
& S_{CKN} = S_{CBN} - S_{BCK} = \frac{10}{3} - \left(2S + \frac{2}{3}\right) = \frac{8}{3} - 2S.
\end{aligned}
$$

Substituting into the proportion $S_{AKN} : S_{CKN} = 4 : 5$ gives us the equation

$$
\frac{S}{\frac{8}{3} - 2S} = \frac{4}{5}
$$

from which we get $S = \frac{32}{39}$. The areas we need to compare thus have the values

$$
\begin{gathered}
S_{ABK} = S_{ABN} - S_{AKN} = \frac{8}{3} - S = \frac{72}{39}, \\
S_{CNKM} = S_{ACM} - S_{AKN} = 3 - S = \frac{85}{39}.
\end{gathered}
$$

Answer. The desired ratio is $72:85$.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Prove that any triangle is divided into two smaller triangles of equal area by any of its medians. [The relevant triangles have the same height to the same bases.]

N2. Prove that any triangle is divided into six smaller triangles of equal area by its three medians. [For triangle $ABC$ with medians $AA_1, BB_1, CC_1$ and centroid $T$, use equalities of the type $S_{ABA_1} = S_{ACA_1}$ and $S_{BTA_1} = S_{CTA_1}$, valid according to the result of problem N1.]

N3. Derive the rule for the ratio in which the bisector of an internal angle of a given triangle divides its opposite side: The bisector of angle $BAC$ intersects side $BC$ of triangle $ABC$ at point $P$ determined by the ratio $|PB| : |PC| = |AB| : |AC|$. [Show that both ratios have the same value as the ratio of the areas of triangles $ABP$ and $ACP$.]

D1. Inside the sides $AB, AC$ of a given triangle $ABC$, points $E, F$ are chosen in sequence, such that $EF \parallel BC$. Segment $EF$ is then divided by point $D$ such that $p = |ED| : |DF| = |BE| : |EA|$.

a) Show that the ratio of the areas of triangles $ABC$ and $ABD$ is the same for $p = 2:3$ as for $p = 3:2$.

b) Explain why the ratio of the areas of triangles $ABC$ and $ABD$ has a value of at least 4. $[65-C-I-4]$

D2. In a right-angled trapezoid $ABCD$ with a right angle at vertex $A$, the base $AB$ is intersected by the height $CP$ of the trapezoid with its diagonal $BD$ at point $K$. The area of quadrilateral $APCD$ is half the area of trapezoid $ABCD$. Determine what fraction of the area of triangle $ABC$ is occupied by triangle $BCK$. $[65-C-II-3]$

D3. A trapezoid $ABCD$ with bases $AB, CD$ is given, such that $2|AB| = 3|CD|$.

a) Find a point $P$ inside the trapezoid such that the areas of triangles $ABP$ and $CDP$ are in the ratio $3:1$ and the areas of triangles $BCP$ and $DAP$ are also in the ratio $3:1$.

b) For the found point $P$, determine the successive ratio of the areas of triangles $ABP, BCP, CDP$, and $DAP$. $[64-C-II-3]$
2. For the right-angled triangle $ABC$ with the right angle at vertex $A$, it is given that $|AB|=4$ and $|AC|=3$. Let $M$ be the midpoint of the hypotenuse $BC$, and $N$ the intersection of the angle bisector of the internal angle at vertex $B$ with the leg $AC$. The segments $AM$ and $BN$ intersect at a point, which we will denote as $K$. Calculate the ratio of the area of triangle $BAK$ to the area of quadrilateral $CNKM$.

(Patrik Bak)