[ Median perpendicular to a segment (GMT). ]

[ Equilateral (equiangular) triangle ]

In triangle $P Q R$, angle $Q P R$ is $60^{\circ}$. Perpendiculars are drawn from vertices $P$ and $R$ to sides $Q R$ and $P Q$ respectively. The point of intersection of these perpendiculars is at a distance of 1 from vertices $P$ and $Q$. Find the sides of triangle $P Q R$.
Use the property of the perpendicular bisector of a segment.

## Solution

Let $M$ be the point of intersection of the specified perpendiculars, and $A$ and $B$ be their points of intersection with the lines $Q R$ and $P Q$. Since the altitudes of a triangle intersect at one point, the line $Q M$ is perpendicular to the side $P Q$, and since point $M$ is equidistant from points $P$ and $Q$, $Q M$ is the perpendicular bisector of the side $P Q$.

Therefore, point $Q$ is equidistant from points $P$ and $R$. Hence, triangle $P Q R$ is isosceles. Since one of the angles of this triangle is $60^{\circ}$, triangle $P Q R$ is equilateral. Then

$$
P A = R B = \frac{3}{2} P M = \frac{3}{2}, \quad P Q = Q R = P R = \frac{P A}{\sin 60^{\circ}} = \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-46.jpg?height=323&width=520&top_left_y=-1&top_left_x=766)

## Answer

$P Q = Q R = P R = \sqrt{3}$.