1960 IMO Problems/Problem 3

Problem

In a given right triangle $ABC$ , the hypotenuse $BC$ , of length $a$ , is divided into $n$ equal parts ( $n$ and odd integer). Let $\alpha$ be the acute angle subtending, from $A$ , that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:

$\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$

Solution

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1960 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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1960 IMO Problems/Problem 3

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