IMO 1959 Problem 4

IMO 1959 Problem 4

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 7m25s

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.

Exploration

The previous approach failed because it tried to manufacture several derived lengths from $c$ without a coherent straightedge-and-compass dependency chain and without a well-defined geometric configuration. A correct construction must begin with $c$ and directly define a point $A$ such that $AB$ and $AC$ are simultaneously constrained by a rigid geometric relation.

From the algebraic solution, any valid triangle must satisfy

$a^2+b^2=c^2,\quad ab=\frac{c^2}{4}.$

Testing small normalizations clarifies the structure. For $c=1$, the system gives two positive numbers whose product is $1/4$ and whose squares sum to $1$. Numerically this yields a fixed ratio between $a$ and $b$, so the geometry reduces to constructing a right triangle with a prescribed acute angle rather than independently constructing two lengths.

This suggests reparameterizing by a ratio $a/b$. Substituting $a=tb$ gives

$t^2b^2+b^2=c^2,\quad tb^2=\frac{c^2}{4}.$

Eliminating $b$ yields a quadratic equation for $t$ whose positive solution is a constant independent of $c$. This shows the shape is fixed and only scaling varies with $c$, so the construction should be angle based.

Dead-end attempts based on directly constructing $c\sqrt{6}/2$ or $c/\sqrt{2}$ are unnecessary and reintroduce the earlier inconsistency because they rebuild the triangle from algebra rather than defining it geometrically at the outset.

Problem Understanding

A right triangle is required with hypotenuse $c$ such that the median to the hypotenuse equals the geometric mean of the legs. In any right triangle, the midpoint of the hypotenuse is the circumcenter, so the median to the hypotenuse has length $c/2$. The condition becomes

$\frac{c}{2}=\sqrt{ab},$

so

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

The task is to construct a right triangle with hypotenuse $c$ and legs satisfying this product condition, using only straightedge and compass.

Key Observations

From

$a^2+b^2=c^2,\quad ab=\frac{c^2}{4},$

one computes

$\left(\frac{a}{b}+\frac{b}{a}\right)=\frac{a^2+b^2}{ab}=4.$

Let $t=\frac{a}{b}$. Then

$t+\frac{1}{t}=4,$

so

$t^2-4t+1=0,$

and the positive solution is

$t=2+\sqrt{3}.$

Thus the triangle is characterized by a fixed ratio of legs, independent of $c$. This converts the construction into building a right triangle with a fixed acute angle whose tangent equals $2+\sqrt{3}$.

Since $\tan 75^\circ=2+\sqrt{3}$, the required triangle is a right triangle with one acute angle equal to $75^\circ$.

Solution

Let $BC$ be a given segment of length $c$. Construct a circle with diameter $BC$. This ensures that any point $A$ on the circle satisfies $\angle BAC=90^\circ$.

At point $B$, construct a ray $BX$ such that $\angle CBX=75^\circ$. This angle is constructed by first constructing a right angle, bisecting it to obtain $45^\circ$, and combining it with a $30^\circ$ angle obtained from an equilateral triangle, producing $75^\circ$ as a sum of constructible angles.

Let $A$ be the second intersection of the ray $BX$ with the circle of diameter $BC$. Then $A$ lies on the circle, so $\angle BAC=90^\circ$, and the triangle $ABC$ is right-angled at $A$ with hypotenuse $BC=c$.

In triangle $ABC$, the angle at $B$ is $75^\circ$, so

$\frac{AC}{AB}=\tan 75^\circ=2+\sqrt{3}.$

Write $AB=b$ and $AC=a$, so $a=(2+\sqrt{3})b$.

Using $a^2+b^2=c^2$ gives

$(2+\sqrt{3})^2b^2+b^2=c^2.$

Since $(2+\sqrt{3})^2=7+4\sqrt{3}$, this becomes

$(8+4\sqrt{3})b^2=c^2,$

hence

$b^2=\frac{c^2}{8+4\sqrt{3}}.$

Then

$ab=(2+\sqrt{3})b^2=\frac{(2+\sqrt{3})c^2}{8+4\sqrt{3}}=\frac{c^2}{4}.$

Since the midpoint of the hypotenuse in a right triangle has distance $c/2$ from each vertex, the median to the hypotenuse has length $c/2$, which equals $\sqrt{ab}$. The constructed triangle therefore satisfies the required condition.

Verification of Key Steps

The construction begins with a fixed segment $BC$ and defines the triangle by placing a point $A$ on the circle with diameter $BC$, which guarantees the right angle at $A$ without assuming any side lengths. The additional constraint is imposed independently by fixing $\angle CBX=75^\circ$, so $A$ is determined as an intersection of two independently constructible loci, a circle and a ray.

The angle identity $\tan 75^\circ=2+\sqrt{3}$ ensures that the ratio of legs is constant across all values of $c$, so scaling is governed solely by the circle of diameter $BC$. Substituting this ratio into the Pythagorean relation produces a consistent value of $ab$, and the computation yields exactly $ab=c^2/4$ with no dependence on additional choices.

The median condition follows from the standard property of right triangles that the midpoint of the hypotenuse is the circumcenter, so the median equals $c/2$, which matches $\sqrt{ab}$ computed from the construction.

Alternative Approaches

A coordinate approach places $B=(0,0)$ and $C=(c,0)$ and describes $A$ as the intersection of the circle $x^2+y^2=cx$ with the line through $B$ forming angle $75^\circ$ with $BC$. Solving the resulting system reproduces the same ratio $a/b=2+\sqrt{3}$ and yields the same algebraic condition $ab=c^2/4$.

A similarity-based construction replaces angle $75^\circ$ by constructing a triangle similar to a fixed $15^\circ$–$75^\circ$–$90^\circ$ template, then scaling it so that the hypotenuse equals $c$, avoiding explicit trigonometric identification while preserving the same constraint structure.