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 angel subtending, from
, that segment which contains the mdipoint of the hypotenuse. Let
be the length of the altitude to the hypotenuse fo the triangle. Prove that:
\[
\tan{\alpha}=\dfrac{4nh}{(n^2-1)a}.
\]