Given a right triangle $ABC$ with a right angle at vertex $C$ and the lengths of the legs in the ratio $1:3$. Points $K$, respectively $L$, are the centers of squares that have one side in common with the legs $AC$, respectively $BC$, and do not overlap with triangle $ABC$. Point $M$ is the midpoint of the hypotenuse $AB$.

a) Justify that point $C$ lies on the segment $KL$.

b) Calculate the ratio of the areas of triangles $ABC$ and $KLM$.

(J. Švrček)
Segments $K C$ and $C L$ are parts of the diagonals in squares, so angles $K C A$ and $L C B$ have a size of $45^{\circ}$. Angle $A C B$ is a right angle, thus angle $K C L$ has a size of

$$
45^{\circ}+90^{\circ}+45^{\circ}=180^{\circ} .
$$

From this, it follows that point $C$ is an internal point of segment $K L$.

![](https://cdn.mathpix.com/cropped/2024_04_17_76d9fb87d3e1f8305746g-2.jpg?height=739&width=926&top_left_y=1041&top_left_x=565)

In the following, we assume such a labeling of vertices that $|A C|:|C B|=1: 3$. The size of the leg $A C$ is denoted by $b$. With this notation, the area of the right triangle $A B C$ is expressed as

$$
\frac{1}{2}(b \cdot 3 b)=\frac{3}{2} b^{2}
$$

From the problem statement, it follows that segments $K M$ and $M L$ are (extended) midlines of triangle $A B C$. Thus, triangle $K L M$ is also a right triangle and, moreover, isosceles, with leg lengths $\frac{1}{2} b+\frac{3}{2} b=2 b$. The area of triangle $K L M$ is therefore

$$
\frac{1}{2}(2 b \cdot 2 b)=2 b^{2}
$$

The ratio of the areas of triangles $A B C$ and $K L M$ is

$$
\frac{3}{2} b^{2}: 2 b^{2}=3: 4
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_76d9fb87d3e1f8305746g-3.jpg?height=740&width=943&top_left_y=287&top_left_x=565)

Evaluation. 2 points for the answer to each of the questions a) and b); 2 points for the quality of the commentary.

Notes. Instead of the previous notation $|A C|=b,|A B|=3 b$ etc., one can use $s$ for additional division, or representation on a grid:

![](https://cdn.mathpix.com/cropped/2024_04_17_76d9fb87d3e1f8305746g-3.jpg?height=925&width=943&top_left_y=1348&top_left_x=565)

The collinearity of points $K, C, L$, as well as the right-angled and isosceles nature of triangle $K L M$, is valid for any right triangle $A B C$. The given ratio of the lengths of the legs speaks only to the ratio of the areas of triangles $A B C$ and $K L M$.