Zaslavsky A.A.

Given a circle with center $O$ and a point $P$ not lying on it. Let $X$ be an arbitrary point on the circle, $Y$ the intersection point of the bisector of angle $P O X$ and the perpendicular bisector of segment $P X$. Find the geometric locus of points $Y$.
Let $K, L$ be the projections of point $Y$ onto $O P$ and $O X$. From the definition of point $Y$, it follows that $Y P = Y X$ and $Y K = Y L$. Therefore, triangles $Y K P$ and $Y L X$ are equal, which means

$X L = P K$. Moreover, $O L = O K$. Since the lengths of segments $O P$ and $O X$ are not equal, one of them is equal to the sum of the lengths of segments $O K$ and $K P$, and the other is their difference. Consequently, $O K = 1 / 2(O P + O X)$ (see figure), which means point $K$ is fixed. It is obvious that any point on the perpendicular dropped from point $K$ to line $O P$ belongs to the desired locus of points.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-39.jpg?height=1790&width=1652&top_left_y=-6&top_left_x=186)

## Answer

The line perpendicular to ray $O P$ and intersecting it at a point that is a distance of ½ ( $O P + O X$) from $O$. Send a comment

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