1960 IMO Problems

Revision as of 00:37, 12 March 2007 by Me@home (talk | contribs) (Problem 3)

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 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

Day II

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Resources