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 $c^2 = 4 a b$ 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 $75^\circ$ . 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 $75^\circ$ angle is constructed by constructing a 15 degree angle on top of the 60 degree point.

Solution 4

We draw perpendicular $h$ on $c$ and denote the angle between the median and $c$ as $\alpha$ . The area of the triangle is $\frac{ab}{2}$ or $\frac{ch}{2}$

We can write $c$ as $2\sqrt{ab}$ So we have,

$\frac{ab}{2} = \frac{2\sqrt{ab}h}{2}$

Which simplifies to be

$\frac{\sqrt{ab}}{h} = \frac{1}{2}$ Or

$\sin\alpha = \sin\frac{\pi}{6}$

So, we draw $c$ and from it's midpoint, we draw a segment $\frac{c}{2}$ forming an angle $\frac{\pi}{6}$ Join the vertices to get the required triangle.

Solution 4

This is a geometric solution. We start with some observations in the first paragraph and describe the construction in the next paragraph. Consider any right triangle with hypotenuse c. Construct square with side length c on the opposite side of c from the triangle. Construct 3 more triangles congruent to the original triangle on the outside of each of the three other sides of the square in such a way that a larger square with side length a+b is formed. The area of this larger square is c^2+2ab = 1.5c^2 so a+b = (3/2)^(1/2) *c

Now go back to the given c. Construct a square with side length c first. Let two adjacent corners of it be M and N. Construct the square's diagonals and let center of the square be O. Construct a circle around O with radius (3/2)^(1/2) * OM. To do so, take OM, first draw a 2:1 right triangle with OM being the short side to obtain a hypothenuse with length 3^(1/2) OM, then draw a 1:1 right triangle from that side to obtain a hypothenuse with length 6^(1/2)*OM. then bisect it for desired length. Next draw the circular locus of points X such that MXO is 45 degrees. To accomplish this, simply find the point on the perpendicular bisector of OM that is 1/(2OM) from OM. There are two such points, pick the one closer to N. Draw circle around this new point going through O and M. The intersection of the two circles is the desired third vertex of the triangle with the given hypotenuse c.

Video Solutions

https://youtu.be/ih2wDhRKzwI

- little fermat

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
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1959 IMO Problems/Problem 4

Contents

Problem

Solutions

Solution 1

Solution 2

Solution 3

Solution 4

Solution 4

Video Solutions

See Also