IMO 1959 Problem 4
We are given a positive length $c$, and we must construct a right triangle whose hypotenuse has length $c$.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m05s
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.
Problem Understanding
We are given a positive length $c$, and we must construct a right triangle whose hypotenuse has length $c$. If the legs are denoted by $a$ and $b$, and the midpoint of the hypotenuse is $M$, then the condition is
$CM=\sqrt{ab},$
where $CM$ denotes the median drawn to the hypotenuse.
In every right triangle, the midpoint of the hypotenuse is equidistant from the three vertices. Hence the median to the hypotenuse equals half the hypotenuse. Therefore the required condition becomes
$\frac c2=\sqrt{ab}.$
Since the triangle is right angled, the legs also satisfy
$a^2+b^2=c^2.$
The problem reduces to finding positive numbers $a,b$ satisfying
$ab=\frac{c^2}{4}, \qquad a^2+b^2=c^2,$
and then constructing a right triangle with those side lengths.
Key Observations
Let
$s=a+b, \qquad p=ab.$
From
$a^2+b^2=c^2,$
we obtain
$s^2-2p=c^2.$
Using
$p=\frac{c^2}{4},$
this becomes
$s^2-\frac{c^2}{2}=c^2,$
hence
$s^2=\frac{3c^2}{2}.$
Since $s>0$,
$s=c\sqrt{\frac32}.$
The numbers $a$ and $b$ are roots of
$x^2-sx+p=0,$
that is,
$x^2-c\sqrt{\frac32},x+\frac{c^2}{4}=0.$
Its discriminant equals
$\Delta=s^2-4p=\frac{3c^2}{2}-c^2=\frac{c^2}{2},$
so
$\sqrt{\Delta}=\frac c{\sqrt2}.$
Applying the quadratic formula gives
$x=\frac{s\pm\sqrt{\Delta}}2.$
Substituting the values of $s$ and $\sqrt{\Delta}$,
$$\begin{aligned} x &=\frac{1}{2}\left(c\sqrt{\frac32}\pm\frac c{\sqrt2}\right) \ &=\frac c2\left(\frac{\sqrt3}{\sqrt2}\pm\frac1{\sqrt2}\right) \ &=\frac c{2\sqrt2}(\sqrt3\pm1). \end{aligned}$$
Thus
$a=\frac c{2\sqrt2}(\sqrt3+1), \qquad b=\frac c{2\sqrt2}(\sqrt3-1).$
Both are positive, so such a triangle may exist.
Solution
Construct a segment $AB$ of length $c$.
Define the lengths
$a=\frac c{2\sqrt2}(\sqrt3+1), \qquad b=\frac c{2\sqrt2}(\sqrt3-1).$
A direct computation gives
$$\begin{aligned} ab &=\frac{c^2}{8}(\sqrt3+1)(\sqrt3-1) \ &=\frac{c^2}{8}(3-1) \ &=\frac{c^2}{4}. \end{aligned}$$
Also,
$$\begin{aligned} a^2+b^2 &=\frac{c^2}{8}\left((\sqrt3+1)^2+(\sqrt3-1)^2\right) \ &=\frac{c^2}{8}\left((4+2\sqrt3)+(4-2\sqrt3)\right) \ &=\frac{c^2}{8}\cdot 8 \ &=c^2. \end{aligned}$$
Now draw the circle centered at $A$ with radius $a$, and the circle centered at $B$ with radius $b$.
We verify that these two circles intersect. Indeed,
$$\begin{aligned} a+b &=\frac c{2\sqrt2}\big((\sqrt3+1)+(\sqrt3-1)\big) \ &=\frac{2\sqrt3,c}{2\sqrt2} \ &=c\sqrt{\frac32} \ &>c. \end{aligned}$$
Also,
$$\begin{aligned} |a-b| &=\frac c{2\sqrt2}\big((\sqrt3+1)-(\sqrt3-1)\big) \ &=\frac{2c}{2\sqrt2} \ &=\frac c{\sqrt2} \ &<c. \end{aligned}$$
Therefore
$|a-b|<AB<c<a+b.$
Hence the two circles intersect in two symmetric points. Let $C$ be one of their intersection points. Then, by construction,
$AC=a, \qquad BC=b.$
Since
$AC^2+BC^2=a^2+b^2=c^2=AB^2,$
the converse of the Pythagorean theorem implies that
$\angle ACB=90^\circ.$
Thus $ABC$ is a right triangle with hypotenuse $AB=c$.
Let $M$ be the midpoint of $AB$. In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices, so
$CM=\frac{AB}{2}=\frac c2.$
From the computation above,
$\sqrt{ab}=\sqrt{\frac{c^2}{4}}=\frac c2.$
Therefore
$CM=\sqrt{ab}.$
The constructed triangle satisfies all required conditions.
Verification of Key Steps
We verify carefully that every condition holds.
The triangle is right angled because
$AC^2+BC^2=AB^2.$
Indeed,
$$\begin{aligned} AC^2+BC^2 &=a^2+b^2 \ &=c^2 \ &=AB^2. \end{aligned}$$
Hence, by the converse of the Pythagorean theorem,
$\angle ACB=90^\circ.$
The hypotenuse has the prescribed length because
$AB=c$
by construction.
The legs have the required lengths because the point $C$ was chosen as an intersection point of the circles centered at $A$ and $B$ with radii $a$ and $b$. Therefore
$AC=a, \qquad BC=b.$
The circles do intersect because
$|a-b|<AB<a+b,$
which is the necessary and sufficient condition for two circles with centers $A,B$ and radii $a,b$ to intersect.
The median to the hypotenuse equals half the hypotenuse. Since $M$ is the midpoint of $AB$,
$CM=AM=BM=\frac{AB}{2}=\frac c2.$
The geometric mean of the legs equals the same quantity:
$$\begin{aligned} \sqrt{AC\cdot BC} &=\sqrt{ab} \ &=\sqrt{\frac{c^2}{4}} \ &=\frac c2. \end{aligned}$$
Therefore
$CM=\sqrt{AC\cdot BC}.$
All required properties have been verified.
Alternative Approaches
A shorter synthetic approach is possible once one observes that the median to the hypotenuse in a right triangle always equals half the hypotenuse.
Thus the condition
$CM=\sqrt{ab}$
immediately gives
$\sqrt{ab}=\frac c2,$
so
$ab=\frac{c^2}{4}.$
Together with
$a^2+b^2=c^2,$
we obtain
$$\begin{aligned} (a+b)^2 &=a^2+b^2+2ab \ &=c^2+\frac{c^2}{2} \ &=\frac{3c^2}{2}. \end{aligned}$$
Hence
$a+b=c\sqrt{\frac32}.$
The legs are therefore the two roots of
$x^2-(a+b)x+ab=0,$
that is,
$x^2-c\sqrt{\frac32},x+\frac{c^2}{4}=0.$
Solving this quadratic again yields
$a,b=\frac c{2\sqrt2}(\sqrt3\pm1).$
One may then construct the triangle directly from the three side lengths $a,b,c$ by the standard ruler and compass construction for a triangle with prescribed side lengths.