1959 IMO Problems/Problem 4

Problem

Construct a right triangle with a given hypotenuse $c$ 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 $a$ and $b$ . 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 $abc$ is $ab$ , since $a$ and $b$ form a right angle. However, twice the area of the triangle is also the product of $c$ and the altitude to $c$ . Hence the altitude to $c$ must have length $\frac{c}{4}$ . Therefore if we construct a circle with diameter $c$ and a line parallel to $c$ and of distance $\frac{c}{4}$ from $c$ , 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 $b$ and $c$ as $\alpha$ . 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 $c \sin{\alpha}\cos{\alpha}$ , which both of the solutions set equal to $\frac{c}{4}$ .

Solution 3

If we let the legs be $a$ and $b$ with $a < b$ , then $c^2 = a^2 + b^2$ . Because $c^2 = a b/4$ as well, we immediately deduce via some short computations through quadratic formula that $a = (2 - \sqrt{3})b$ and $a = (2 + \sqrt{3})b$ . Thus, $\frac{a}{b} = \tan 15^\circ$ and $\frac{a}{b} = \tan 75^\circ$ , and so one of the angles of the triangle must be $15^\circ$ and the other must be $15^\circ$ . A $15^\circ$ angle is easily constructed by bisecting a $30^\circ$ angle (which is formed by constructing the altitude of an equilateral triangle), and $15^\circ$ angle is constructed by constructing a 15 degree angle on top of the 60 degree point.

Solution 4

WLOG, let the length of the hypotenuse be 2. The radius of the circle with the hypotenuse as the diameter will be 1. Let the circle has an center of (1,0) on a Cartesian plane. Let $a$ be the horizontal length from the origin to the desired point on the circumference which satisfies the question. Since the equation of the circle is ${(x-1)^2}+{y^2}=1$ , $y = |\sqrt{2x - {x^2}}|$ . The desired point which have a horizontal length of $a$ from the origin would have a vertical length of $\sqrt{a(2 - a)}$ when substituted into the previous equation. By using Pythagoras Theorem, the length from the origin to this point is $\sqrt{2a}$ and the length from this point to the point (2,0) is $\sqrt{-2a + 4}$ . So, we need to find the $a$ which $\sqrt {2a (-2a + 4)}=1$ . By solving this quadratic equation, we get $a=\frac{2 +/- \sqrt{5}}{2}$ , the length of the two sides would be $\sqrt{2 + \sqrt{3}}$ and $\sqrt{2 - \sqrt{3}}$ . With this, we get a conclusion of this triangle should have a length ratio of $2 + \sqrt{3}:2 - \sqrt{3}:2$ .

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

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