1960 IMO Problems

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Problems of the 2nd IMO 1960 Romania.

Day I

Problem 1

Solution

Problem 2

Solution

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 mdipoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse fo the triangle. Prove that: \[ \tan{\alpha}=\dfrac{4nh}{(n^2-1)a}. \]


Solution

Day II

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

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