Point $M$ is located on side $C D$ of square $A B C D$ with center $O$, such that $C M: M D=1: 2$.

Find the sides of triangle $A O M$, if the side of the square is 6.
Drop perpendiculars from the center of the square to the sides $C D$ and $A D$.

## Solution

Let $K$ be the projection of point $O$ onto side $C D$ of the square $A B C D$. Then $K$ is the midpoint of $C D$. From the right triangles $OMK$ and $AMD$, we find that

$O M^{2}=O K^{2}+K M^{2}=10, A M^{2}=A D^{2}+D M^{2}=52$.

## Answer

$3 \sqrt{2}, 2 \sqrt{13}, \sqrt{10}$.