Akopyan A.V.

On the plane, a point $M$ is marked, which does not lie on the coordinate axes. A point $Q$ moves along the y-axis, and a point $P$ moves along the x-axis such that the angle $P M Q$ is always right. Find the geometric locus of points $N$, which are symmetric to $M$ with respect to $P Q$.
Points $P, Q, M$ and the origin $O$ lie on a circle with diameter $P Q$. Therefore, point $N$ also lies on this circle and $\angle P O N = \angle P O M$ (see the figure). Thus, $N$ lies on the line symmetric to $O M$ with respect to the coordinate axes.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3ff818e52eab4ca192bg-31.jpg?height=847&width=900&top_left_y=1606&top_left_x=585)

On the other hand, if $N$ is any point on this line, and $P, Q$ are the points of intersection of the coordinate axes with the circle $O M N$, then

$\angle P M N = \angle P O N = \angle P O M = \angle P N M$ and $\angle P M Q = \angle P O Q = \angle P N Q = 90^{\circ}$, so points $M$ and $N$ are symmetric with respect to $P Q$.

## Answer

The line symmetric to $O M$ with respect to the coordinate axes.

Author: Shvetsvo D.V.

In triangle $A B C$, the altitude $A H$ is drawn. Points $I_{b}$ and $I_{c}$ are the centers of the inscribed circles of triangles $A B H$ and $C A H$; $L$ is the point of tangency of the inscribed circle of triangle $A B C$ with side $B C$. Find the angle $L I_{b} I_{c}$.

## Solution

Let $L_{b}, L_{c}$ be the projections of points $I_{b}, I_{c}$ on $B C$, and $r_{b}, r_{c}$ be the radii of the inscribed circles of triangles $A H B, A H C$ (see the figure). Since these triangles are right-angled, $r_{b} = 1 / 2(A H + B H - A B), r_{c} = 1 / 2(A H + C H - A C), r_{b} - r_{c} = 1 / 2(B H - C H) - 1 / 2(A B - A C) = (B H - C H) - 1 / 2(B L - C L) = L H$. Therefore, $I_{b} L_{b} = r_{b} = L L_{c}, I_{c} L_{c} = r_{c} = L L_{b}$, which means triangles $L I_{b} L_{b}$ and $I_{c} L_{c} L$ are equal, $L I_{b} = L I_{c}$ and $\angle L I_{b} L I_{c} = 90^{\circ}$. Thus, triangle $L I_{b} I_{c}$ is an isosceles right triangle.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3ff818e52eab4ca192bg-32.jpg?height=861&width=995&top_left_y=747&top_left_x=523)

## Answer

$45^{\circ}$