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IMO 1959 – Question 4

Construct a right triangle with given hypotenuse c such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.

Solution

Consider the right/right-angled triangle inscribed in a semicircle represented in the diagram below:

A right(-angled) triangle labeled/labelled A B C is inscribed in a semicircle. Its sides are labeled/labelled a, b and c. Its height is the line segment labeled/labelled C D and h. The right(-angled) triangle labeled/labelled B C D and the right(-angled) triangle labeled/labelled A C D share the same side labeled/labelled C D.

The triangles ABCABC and CBDCBD are similar.

Then,

bc=ha\frac{b}{c}=\frac{h}{a}
h=abc(1)h=\frac{ab}{c}\; \left(1\right)

We know that the radius of the semicircle is the median of the triangle.

Then,

ab=(c2)2(2)ab=\left(\frac{c}{2}\right)^2\; \left(2\right)

Substituting (2) in (1), we get:

h=(c2)2ch=\frac{\left(\frac{c}{2}\right)^2}{c}
h=c4h=\frac{c}{4}

So, our right/right-angled triangle ABCABC has to have height c4\frac{c}{4} from the hypotenuse.

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