1959 IMO Problems/Problem 4

Problem

( Proposed by Hungary ) Construct a right triangle with a given hypotenuse $\displaystyle c$ 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 $\displaystyle a$ and $\displaystyle b$ . 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

$ab = \frac{c^2}{4}.$

However, we notice that twice the area of the triangle $\displaystyle abc$ is $\displaystyle ab$ , since $\displaystyle a$ and $\displaystyle b$ form a right angle. However, twice the area of the triangle is also the product of $\displaystyle c$ and the altitude to $\displaystyle c$ . Hence the altitude to $\displaystyle c$ must have length $\frac{c}{4}$ . Therefore if we construct a circle with diameter $\displaystyle c$ and a line parallel to $\displaystyle c$ and of distance $\frac{c}{4}$ from $\displaystyle c$ , 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 $\displaystyle b$ and $\displaystyle c$ as $\displaystyle \alpha$ . The problem requires that

$ab = \frac{c^2}{4},$

or, equivalently, that

$2 \frac{ab}{c^2} = \frac{1}{2}.$

However, since $\frac{a}{c} = \sin{\alpha};\; \frac{b}{c} = \cos{\alpha}$ , we can rewrite the condition as

$2\sin{\alpha}\cos{\alpha} = \frac{1}{2},$

or, equivalently, as

$\sin{2\alpha} = \frac{1}{2}.$

From this it becomes apparent that $2\alpha = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ ; hence the other two angles in the triangle must be $\frac{ \pi }{12}$ and $\frac{ 5 \pi }{12}$ , 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 $\displaystyle c \sin{\alpha}\cos{\alpha}$ , which both of the solutions set equal to $\frac{c}{4}$ .

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

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