In a right triangle $ABC$, a rectangle ЕКМР is placed such that side $ЕК$ lies on the hypotenuse $BC$, and vertices $M$ and $P$ are on the legs $AC$ and $AB$ respectively. The leg $AC$ is equal to 3, and the leg $AB$ is equal to 4. Find the sides of the rectangle ЕКМР, if its area is $\frac{3}{5}$, and the perimeter is less than 9.
Denote $P E=M K=x$ and express segments $E B$ and $C K$ in terms of $x$.

## Solution

From the problem statement, it follows that

$$
B C=\sqrt{A C^{2}+A B^{2}}=5, \operatorname{tg} \angle B=
$$

$$
\operatorname{tg} \angle C M K=\frac{3}{4} .
$$

Let $M K=P E=x$. Since $S_{\text {MPEK }}=\frac{5}{3}$, then $M P=K E=\frac{5}{3 x}$.

From the right triangles $P E B$ and $С K M$, we find that

$$
B E=\frac{P E}{\operatorname{tg} \angle B}=\frac{4 x}{3}, C K=M K \operatorname{tg} \angle
$$

$$
C M K=\frac{3 x}{4}
$$

Since $C K+K E+E B=C B$, then

$$
\frac{3 x}{4}+\frac{5}{3 x}+\frac{4 x}{3}=5
$$

From this equation, we find that $x$ $=2$ or $x=\frac{2}{5}$. The second solution does not satisfy the condition $P_{\text {ЕКМР }}<9$. Therefore,

$$
P E=M K=2, M P=K E=\frac{5}{6} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-07.jpg?height=329&width=686&top_left_y=2437&top_left_x=0)

Answer

## Problem