AoPSWiki
Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!

1960 IMO Problems/Problem 3

From AoPSWiki

Problem

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 that:

\tan{\alpha}=\frac{4nh}{(n^2-1)a}.

Solution

Using coordinates, let A=(0,0), B=(b,0), and C=(0,c). Also, let PQ be the segment that contains the midpoint of the hypotenuse with P closer to B.

size(8cm);pair A,B,C,P,Q;A=(0,0);B=(4,0);C=(0,3);P=(2.08,1.44);Q=(1.92,1.56);dot(A);dot(B);dot(C);dot(P);dot(Q);label("A...

Then, P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right), and Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right).

So, \text{slope}(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}, and \text{slope}(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}.

Thus, \tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1... = \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}.

Since [ABC]=\frac{1}{2}bc=\frac{1}{2}ah, bc=ah and \tan{\alpha}=\frac{4nh}{(n^2-1)a} as desired.

See Also

1960 IMO (Problems)
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us