IMO 1960 - Problem 3

Problem: A right - angled triangle $ABC$ is given for which the hypotenuse $BC$ has length $a$ and is divided into $n$ equal segments, where $n$ is odd. Let $\alpha$ be the angle with which the point $A$ sees segment containing the middle of the hypotenuse. Prove that: $tan(\alpha)=\frac{4nh}{(n^2-1)a}$.
Solution: Let $B'C'$ be the middle of the $n=2k+1$ segments and let $D$ be the foot of the perpendicular from $A$ to the hypotenuse. Let us assume $\mathscr{B}(C,D,C',B',B)$. Then from $CD<BD,CD+BD=a$, and $CD.BD=h^2$ we have $CD^2-a.CD+h^2=0\implies CD=\frac{a-\sqrt{a^2-4h^2}}{2}$. Let us define $\angle{DAC'}=\gamma$ and $\angle{DAB'}=\beta$; then $tan(\beta)=\frac{DB'}{h}$ and $tan(\gamma)=\frac{DC'}{h}$. Since $DB'=CB'-CD=\frac{(k+1)a}{2k+1}-\frac{c-\sqrt{c^2-4h^2}}{2}$ and $DC'=\frac{ka}{2k+1}-\frac{c-\sqrt{c^2-4h^2}}{2}$, we have : $tan(\alpha)=tan(\beta-\gamma)=\frac{tan(\beta)-tan(\gamma)}{1+tan(\beta).tan(\gamma)}=\frac{\frac{a}{(2k+1)h}}{1+\frac{a^2-4h^2}{4h^2}-\frac{a^2}{4h^2(2k+1)^2}}=\frac{4h(2k+1)}{4ak(k+1)}=\frac{4nh}{(n^2-1)a}$.