1960 IMO Problems/Problem 3

Problem

In a given right triangle $ABC$ , the hypotenuse $BC$ , of length $a$ , is divided into $n$ equal parts ( $n$ 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 $B$ .

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

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.

Solution 2

Let $P, Q, R$ be points on side $BC$ such that segment $PR$ contains midpoint $Q$ , with $P$ closer to $C$ and (without loss of generality) $AC \le AB$ . Then if $AD$ is an altitude, then $D$ is between $P$ and $C$ . Combined with the obvious fact that $Q$ is the midpoint of $PR$ (for $n$ is odd), we have $\tan {\angle PAR} = \tan (\angle RAD - \angle PAD) = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4nh}{(n^2-1)a}.$

Solution 3

Let $∠ A C B = x$ , and $∠ A B C = 90^{\circ} - x$ . Let $M$ be the midpoint on the hypotenuse $B C$ , and $Q$ and $P$ be points such that $P Q$ contains $B C$ , with $Q$ closer to $C$ and $P$ closer to $B$ . The midpoint will always be in the middle of line $Q P$ , unless $n$ is even or infinite, which it is not. Given such a triangle, we can express the altitude to the hypotenuse as:

$h = a \cos(x) \sin(x)$

Next, we shall denote line $A M$ as $f$ , where $A M$ is the median to the hypotenuse. This means that line $A M = B M = C M$ , and as $B M = \frac{a}{2}$ , we have:

$f = \frac{a}{2}$

We know that $∠ M A B = 90^{\circ} - x$ , and $∠ M A C = x$ . This means that $∠ A M B = 2 x$ and $∠ A M C = 180^{\circ} - 2 x$ . The length of $Q P$ is $\frac{a}{n}$ . Let $∠ Q A M = k$ and $∠ P A M = z$ , such that $∠ Q A P$ (or $α$ ) equals $k + z$ . This means that $∠ A Q M = 2 x - k$ , and $∠ A P M = 180^{\circ} - 2 x - z$ .

As $M$ is in the middle of $Q P$ , we have $Q M = P M = \frac{a}{2 n}$ . Applying the sine law on triangle $A Q M$ , we get:

$\frac{\sin(k)}{\frac{a}{2n}} = \frac{\sin(2x - k)}{\frac{a}{2}}$

Simplifying:

$\frac{2n \sin(k)}{a} = \frac{2 \sin(2x - k)}{a}$

$n \sin(k) = \sin(2x - k)$

Using the identity $\sin (2 x - k) = \sin (2 x) \cos (k) - \cos (2 x) \sin (k)$ , and since $\sin (2 x) = 2 \sin (x) \cos (x)$ , we substitute:

$\sin(2x) = \frac{2h}{a}$

Thus:

$n \sin(k) = \frac{2h}{a} \cos(k) - \cos(2x) \sin(k)$

Now, we know that:

$\cos(2x) = \frac{\sqrt{a^2 - 4h^2}}{a}$

Substituting this into the equation:

$n \sin(k) + \sin(k) \frac{\sqrt{a^2 - 4h^2}}{a} = \cos(k) \frac{2h}{a}$

Factoring out $\sin (k)$ :

$\sin(k) \left( n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos(k) \frac{2h}{a}$

Thus:

$\tan(k) = \frac{2h}{an + \sqrt{a^2 - 4h^2}}$

By performing similar steps with $\tan (z)$ , we can use the addition formula for $\tan (z + k)$ to find $\tan (α)$ , where $α = z + k$ .

Courtesy of Gordon Freeman

Solution 4

Let $D$ and $E$ be points on $BC$ such that $\alpha=\angle DAE$ , $D$ is closer to $B$ than to $C$ , and $E$ is closer to $C$ than $B$ . Furthermore, let $\alpha_1=\angle BAD$ and $\alpha_2=\angle CAE$ , and let $\beta=\angle B$ and $\gamma=\angle C$ . Applying Law of Sines on $\triangle BAD$ yields $\begin{aligned} \frac{\sin α_{1}}{\frac{a (n - 1)}{2 n}} & = \frac{\sin (180 - (α_{1} + β))}{c} \\ \frac{2 c n}{a (n - 1)} \sin α_{1} & = \sin (α_{1} + β) \\ \frac{2 c n}{a (n - 1)} \sin α_{1} & = \sin α_{1} \cos β + \cos α_{1} \sin β \\ \frac{2 c n}{a (n - 1)} \sin α_{1} & = \sin α_{1} \cdot \frac{c}{a} + \cos α_{1} \cdot \frac{b}{a} \\ \frac{2 c n}{n - 1} \sin α_{1} & = c \sin α_{1} + b \cos α_{1} \\ \frac{c n + c}{n - 1} \sin α_{1} & = b \cos α_{1} \\ \tan α_{1} & = \frac{b (n - 1)}{c (n + 1)} \end{aligned}$ Similarly, $\tan\alpha_2=\frac{c(n-1)}{b(n+1)}$ . Next, we use the sum of tangents formula: $\begin{aligned} \tan (α_{1} + α_{2}) & = \frac{\frac{b (n - 1)}{c (n + 1)} + \frac{c (n - 1)}{b (n + 1)}}{1 - \frac{b (n - 1)}{c (n + 1)} \cdot \frac{c (n - 1)}{b (n + 1)}} \\ = \frac{(\frac{b}{c} + \frac{c}{b}) \cdot \frac{n - 1}{n + 1}}{1 - \frac{(n - 1)^{2}}{(n + 1)^{2}}} \\ = \frac{(\frac{b}{c} + \frac{c}{b}) (n^{2} - 1)}{(n + 1)^{2} - (n - 1)^{2}} \\ = \frac{(\frac{b}{c} + \frac{c}{b}) (n^{2} - 1)}{4 n} \\ = \frac{(b^{2} + c^{2}) (n^{2} - 1)}{4 n b c} \end{aligned}$ Since $\alpha+\alpha_1+\alpha_2=90^\circ$ , we need to find $\tan\alpha=\tan\left(\frac{\pi}{2}-(\alpha_1+\alpha_2)\right)$ . This is equal to $\cot(\alpha_1+\alpha_2)$ , which is the reciprocal of $\tan(\alpha_1+\alpha_2)$ , or $\frac{4nbc}{(b^2+c^2)(n^2-1)}$ . Since $bc=ah$ due to triangle areas and $a^2=b^2+c^2$ due to the Pythagorean Theorem, we conclude that the expression is equal to $\frac{4nah}{a^2(n^2-1)}$ , which simplifies to $\frac{4nh}{(n^2-1)a}$ .