# 1960 IMO, Problem 3

In a given right triangle ABC, the hypotenuse BC of length a is divided into n equal parts (n an 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:

tan\alpha = \frac {4nh}{(n² - 1) a}.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Nguyen Huu Huan)

Let the two left and right segments meeting at A to make up angle \alpha meet BC at P and Q, respectively. We have PQ = \frac{a}{n}. Note also that, since ΔABC is right, AM = BC/2 = a/2.

Let H be the foot of A to BC, so that AH = h. From P drop a perpendicular PJ to AQ.

Draw a circle with diameter PQ and radius r = a/2n. From A draw a tangent to the circle to meet it at F.

We have tan\alpha = \frac{PJ}{AJ} = \frac{PJ\times AQ}{AJ\times AQ} = {PJ \times AQ}/AF^{2} = {PJ \times AQ} / (AM^{2} - r^{2}).

But PJ \times AQ is twice the area of ΔAPQ which is also equal to AH \times PQ. We then have

tan\alpha = {AH \times PQ} / (AM^{2} - r^{2}) = \frac {ha/n} {[(a/2)^{2} - (a/2n)^{2}]} = \frac {4nh} {(n² - 1) a}.