Talk:1960 IMO Problems/Problem 3

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

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

We know that $\mathrm{\angle }MAB={90}^{\circ }-x$, and $\mathrm{\angle }MAC=x$. This means that $\mathrm{\angle }AMB=2x$ and $\mathrm{\angle }AMC={180}^{\circ }-2x$. The length of $QP$ is $\frac{a}{n}$. Let $\mathrm{\angle }QAM=k$ and $\mathrm{\angle }PAM=z$, such that $\mathrm{\angle }QAP$ (or $\alpha$) equals $k+z$. This means that $\mathrm{\angle }AQM=2x-k$, and $\mathrm{\angle }APM={180}^{\circ }-2x-z$.

As $M$ is in the middle of $QP$, we have $QM=PM=\frac{a}{2n}$. Applying the sine law on triangle $AQM$, we get:

Simplifying:

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

Thus:

Now, we know that:

Substituting this into the equation:

Factoring out $\sin (k)$:

Thus:

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