1960 IMO Problems
Problems of the 2nd IMO 1960 Romania.
Contents
Day I
Problem 1
Problem 2
Problem 3
In a given right triangle , the hypotenuse
, of length
, is divided into
equal parts (
and odd integer). Let
be the acute angle subtending, from
, that segment which contains the midpoint of the hypotenuse. Let
be the length of the altitude to the hypotenuse of the triangle. Prove that:
![$\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$](http://latex.artofproblemsolving.com/f/2/7/f27044f40ce377a2b7dda5eed1e1c050c9b86bd9.png)