# 1959 IMO Problems/Problem 5

## Contents

## Problem

An arbitrary point is selected in the interior of the segment . The squares and are constructed on the same side of , with the segments and as their respective bases. The circles about these squares, with respective centers and , intersect at and also at another point . Let denote the point of intersection of the straight lines and .

(a) Prove that the points and coincide.

(b) Prove that the straight lines pass through a fixed point independent of the choice of .

(c) Find the locus of the midpoints of the segments as varies between and .

## Solution

### Part A

Since the triangles are congruent, the angles are congruent; hence is a right angle. Therefore must lie on the circumcircles of both quadrilaterals; hence it is the same point as .

### Part B

We observe that since the triangles are similar. Then bisects .

We now consider the circle with diameter . Since is a right angle, lies on the circle, and since bisects , the arcs it intercepts are congruent, i.e., it passes through the bisector of arc (going counterclockwise), which is a constant point.

### Part C

Denote the midpoint of as . It is clear that 's distance from is the average of the distances of and from , i.e., half the length of , which is a constant. Therefore the locus in question is a line segment.

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*

## See Also

Quadrados e Circulos circunscritos / IMO 1959-#5

###### ====================================

Link do vídeo: https://youtu.be/UNcHD5JI6wU