1959 IMO Problems/Problem 4
Contents
Problem
Construct a right triangle with a given hypotenuse such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.
Solutions
We denote the catheti of the triangle as and . We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse. (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)
Solution 1
The conditions of the problem require that
However, we notice that twice the area of the triangle is , since and form a right angle. However, twice the area of the triangle is also the product of and the altitude to . Hence the altitude to must have length . Therefore if we construct a circle with diameter and a line parallel to and of distance from , either point of intersection between the line and the circle will provide a suitable third vertex for the triangle. Q.E.D.
Solution 2
We denote the angle between and as . The problem requires that
or, equivalently, that
However, since , we can rewrite the condition as
or, equivalently, as
From this it becomes apparent that or ; hence the other two angles in the triangle must be and , which are not difficult to construct. Q.E.D.
Note. It is not difficult to reconcile these two constructions. Indeed, we notice that the altitude of the triangle is of length , which both of the solutions set equal to .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1959 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |