Talk:1960 IMO Problems/Problem 3

Let $\angle ACB = x$, and $\angle ABC = 90^\circ - x$. 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:

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

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:

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

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

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

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