The centers of three pairwise externally tangent circles are located at points $A, B, C$,

$\angle A B C=90^{\circ}$. The points of tangency are $K, P$, and $M$; point $P$ lies on side $A C$. Find the angle $K P M$.
Express the desired angle in terms of the acute angles of triangle $ABC$.

## Solution

Let $\angle B A C=\alpha, \angle A C B=\gamma\left(\alpha+\gamma=90^{\circ}\right)$.

Suppose point K lies on segment $A B$. From the isosceles triangles KAP and MCP, we find that $\angle A P K=90^{\circ} - \alpha / 2, \angle M P C=90^{\circ} - \gamma / 2$.

Thus, $\angle K P M=180^{\circ}-(\angle A P K+\angle M P C)=1 / 2(\alpha+\gamma)=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-29.jpg?height=672&width=874&top_left_y=1173&top_left_x=592)

Answer

$45^{\circ}$.