1960 IMO Problems/Problem 3

Problem

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:

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

Solution

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

$[asy] size(8cm); pair A,B,C,P,Q; A=(0,0); B=(4,0); C=(0,3); P=(2.08,1.44); Q=(1.92,1.56); dot(A); dot(B); dot(C); dot(P); dot(Q); label("A",A,SW); label("B",B,SE); label("C",C,NW); label("P",P,ENE); label("Q",Q,NNE); draw(A--B--C--cycle); draw(A--P); draw(A--Q); [/asy]$

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

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

1960 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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1960 IMO Problems/Problem 3

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