Problem 5. Given an equilateral triangle $ABC$. On side $AB$, point $K$ is marked, and on side $BC$ - points $L$ and $M$ ( $L$ lies on segment $BM$ ) such that $KL=KM, BL=2, AK=3$. Find $CM$.

[7 points]

![](https://cdn.mathpix.com/cropped/2024_05_06_fbe01430ce51a0699f98g-3.jpg?height=390&width=387&top_left_y=655&top_left_x=533)

(E. V. Bakayev)
Answer: 5.

Solution. Mark a point $T$ on the extension of segment $L M$ beyond point $M$ such that $M T=2$. Angles $B L K$ and $T M K$ are equal because they are adjacent to the equal angles of the isosceles triangle $K L M$. Therefore, triangles $B L K$ and $T M K$ are congruent by two sides and the included angle. Then their corresponding angles are equal: $\angle K T M=\angle K B L=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_fbe01430ce51a0699f98g-3.jpg?height=394&width=393&top_left_y=1426&top_left_x=530)

In triangle $K B T$, two angles are $60^{\circ}$, so it is equilateral, and $B K=B T$. Since triangle $A B C$ is also equilateral and $B A=B C$, then $C T=B C-B T=B A-B K=A K=3$ (and point $T$ lies on side $B C$, not on its extension). Then $C M=C T+M T=3+2=5$.

Second solution. Draw the height $K H$ of the isosceles triangle $K L M$. It is also its median, so $L H=H M$. Let $L H=H M=x$. Triangle $K B H$ is a right triangle with angle $B$ equal to $60^{\circ}$, so its hypotenuse $K B$ is twice its leg $B H$. Since $B H=2+x$, then $K B=2 B H=4+2 x$, and thus $B A=B K+K A=4+2 x+3=7+2 x$. Triangle $A B C$ is equilateral, so $B C=B A=7+2 x$. Therefore, $M C=B C-B M=(7+2 x)-(2+2 x)=5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_fbe01430ce51a0699f98g-4.jpg?height=458&width=410&top_left_y=780&top_left_x=526)