1960 IMO Problem 3

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  • Vinuthan S
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    1960 IMO Problem 3

     

    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 angel 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: \[ \tan{\alpha}=\dfrac{4nh}{(n^2-1)a}. \]

     

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