On the sides $A B, B C$ and $A C$ of triangle $A B C$, points $M, N$ and $K$ are taken respectively such that $A M: M B=2: 3, A K: K C=2: 1, B N: N C=1: 2$. In what ratio does the line $M K$ divide the segment $A N$?
Let $P$ be the intersection point of the line $M K$ with the segment $A N$. Express the areas of triangles $A M P, A K P$ and $A B C$ in terms of $S_{A B C}$ and $^{A P} /_{A N}$.

## Solution

Let $P$ be the intersection point of the line $M$ with the segment $A N$. Denote ${ }^{A P} /_{A N}=x$ and $S_{A B C}=S$. Then $S_{A M P}=$ $2 / 5 \cdot x \cdot S_{A B N}=2 x / 5 \cdot S_{/ 3}=2 x S / 15, S_{A K P}=2 / 3 \cdot x \cdot S_{A C N}=4 x S / 9, S_{A M K}=2 / 5 \cdot 2 / 3 S=4 S / 15 \cdot$ Since $S_{A M K}=S_{A M P}+S_{A K P}$, then ${ }^{2 x} / 15+{ }^{4 x} / 9=4 / 15$, from which $x=6 / 13$. Therefore, $A P: P N=6: 7$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-53.jpg?height=317&width=397&top_left_y=2334&top_left_x=836)

## Answer

$6: 7$, measured from point $A$.

In the rhombus $A B C D$, where $\angle B A D=60^{\circ}$, the perpendicular to side $A D$, erected from the midpoint of $A D$, intersects the diagonal $A C$ at point $M$, and the perpendicular to side $C D$, erected from the midpoint of $C D$, intersects the diagonal $A C$ at point $N$. Find the ratio of the area of triangle $M N D$ to the area of the rhombus $A B C D$.

## Hint

$M N=\frac{1}{3} A C$

## Solution

First method.

Let $a$ be the side of the rhombus. Then

$$
\begin{gathered}
C N=A M=\frac{A D}{2 \cos 30^{\circ}}=\frac{a}{\sqrt{3}}, \\
M N=A C-2 A M=a \sqrt{3}-\frac{2 a}{\sqrt{3}}=\frac{a}{\sqrt{3}}, \\
\frac{S_{\triangle D M N}}{S_{\triangle A D C}}=\frac{M N}{A C}=\frac{\frac{a}{\sqrt{3}}}{a \sqrt{3}}=\frac{1}{3} .
\end{gathered}
$$

Therefore, $\frac{S_{\triangle M N D}}{S_{A B C D}}=\frac{1}{6}$.

Second method.

Let $K$ be the intersection point of the diagonals of the rhombus. Since triangles $A B D$ and $B C D$ are equilateral, $M$ and $N$ are the points of intersection of their medians. Therefore,

$$
S_{\triangle \mathrm{DMK}}=S_{\triangle \mathrm{DNK}}=\frac{1}{6} S_{\Delta \mathrm{ABD}} .
$$

Therefore, $S_{\triangle M N D}=\frac{1}{6} S_{\mathrm{ABCD}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-54.jpg?height=331&width=732&top_left_y=1761&top_left_x=663)

## Answer