\left.\begin{array}{l}{[\text { Properties of medians. Center of gravity of a triangle. }]} \\ {[\text { Pythagorean Theorem (direct and inverse). }}\end{array}\right]

In triangle $A B C$, medians $A E$ and $B D$, drawn to sides $B C$ and $A C$, intersect at a right angle. Side $B C$ is equal to $a$. Find the other sides of triangle $A B C$ if $A E^{2}+B D^{2}=d^{2}$.
The medians of a triangle are divided by the point of intersection in the ratio 2:1, counting from the vertex.

## Solution

Let $O$ be the point of intersection of the medians. Denote $O E=x, O D=y$. Then

$$
A O=2 x, B O=2 y, A E=3 x, B D=3 y
$$

and by the condition of the problem

$$
9 x^{2}+9 y^{2}=9\left(x^{2}+y^{2}\right)=d^{2}
$$

Therefore,

$$
A B^{2}=4 x^{2}+4 y^{2}=4\left(x^{2}+y^{2}\right)=\frac{4}{9} d^{2}
$$

Thus,

$$
A B=\frac{2}{3} d, A C=2 A D=2 \sqrt{A O^{2}+D O^{2}}=2 \sqrt{4 x^{2}+y^{2}}
$$

and $x$ and $y$ are found from the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=\frac{d^{2}}{9} \\
x^{2}+4 y^{2}=\frac{a^{2}}{4}
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-23.jpg?height=369&width=740&top_left_y=2105&top_left_x=662)

## Answer

$$
\frac{2 d}{3}, \sqrt{\frac{20 d^{2}}{9}-a^{2}}
$$

A right triangle $M N C$ is inscribed in an equilateral triangle $A B C$ such that the vertex of the right angle $N$ lies on $A C$, and the vertex $M$ lies on side $A B$. In what ratio should point $N$ divide side $A C$ so that the area of triangle $M N C$ is $\frac{4}{9}$ of the area of triangle $A B C$?

## Solution

Let $A B=A C=B C=a, C N=x$. From the condition of the problem, it follows that $x>\frac{a}{2}$.

From the right triangle $A N M$, we find that

$$
M N=A N \tan \angle M A N=(a-x) \tan 60^{\circ}=(a-x) \sqrt{3}
$$

Then

$$
S_{\triangle \mathrm{MNC}}=\frac{1}{2} C N \cdot M N=\frac{1}{2} x \cdot(a-x) \sqrt{3}
$$

By the condition of the problem,

$$
S_{\triangle \mathrm{MNC}}=\frac{4}{9} S_{\Delta \mathrm{ABC}}, \text { or } \frac{x(a-x) \sqrt{3}}{2}=\frac{4}{9} \cdot \frac{a^{2} \sqrt{3}}{4} .
$$

From this equation, we find that $x=\frac{a}{3}$ or $x=\frac{2 a}{3}$. The condition $x>\frac{a}{2}$ is satisfied only by $\frac{2}{3} a$. Therefore,

$$
\frac{A N}{N C}=\frac{a-x}{x}=\frac{\frac{a}{3}}{\frac{2 a}{3}}=\frac{1}{2}
$$

## Answer

$\frac{A N}{N C}=\frac{1}{2}$