A rectangle with a perimeter of 24 is inscribed in a triangle with sides 10, 17, and 21, such that one of its sides lies on the largest side of the triangle.

Find the sides of the rectangle

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One of the sides of the rectangle cuts off a triangle similar to the given triangle.

## Solution

Let the vertices $P$ and $Q$ of the rectangle $M P Q K$ lie on the sides $A B=10$ and $B C=17$ of the triangle $A B C$, respectively, and the vertices $M$ and $K$ lie on the side $A C=21$. By Heron's formula, $S_{A B C}=\sqrt{24 \cdot 3 \cdot 14 \cdot 7}=7 \cdot 3 \cdot 4=$

84. 

If $B D$ is the height of triangle $A B C$, then $B D={ }^{2 S_{A B C}}{ }_{A C}=8$.

Let $P M=x$. Then $P Q=12-x$. From the similarity of triangles $P B Q$ and $A B C$, it follows that their heights $B T$ and $B D$ are in the same ratio as their bases $P Q$ and $A C$, that is,

$8-x / 8=12-x / 21$, from which $x=72 / 13$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-29.jpg?height=334&width=872&top_left_y=1857&top_left_x=593)

## Answer

$72 / 13,84 / 13$.