5. In triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=5, \mathrm{BC}=\sqrt{17}$, and $\mathrm{AC}=4$, a point $\mathrm{M}$ is taken on side $\mathrm{AC}$ such that $\mathrm{CM}=1$. Find the distance between the centers of the circumcircles of triangles $\mathrm{ABM}$ and $\mathrm{BCM}$.
Answer: 2.

Solution: Draw the height BH to side AC. Let $\mathrm{CH}=x$, then $\mathrm{BH}=$ 4 - $x$. By the Pythagorean theorem from two triangles, we have $B H^{2}=B C^{2}-C H^{2}=17-x^{2}$ and $B H^{2}=A B^{2}-A H^{2}=25-(4-x)^{2}$. Equating the right sides of both equations, we get $x=1$, which means points M and H coincide. Then triangles ABM and BCM are right triangles, the centers of the circumscribed circles lie on the midpoints of the hypotenuses AB and BC. The distance between the centers is equal to the length of the midline parallel to side AC, i.e., it is equal to 2.

## MATHEMATICS

10th GRADE

Grading criteria: Correct and justified solution - 7 points, incorrect solution or only answer $-\mathbf{0}$ points.