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:

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

Simplifying:

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

\[ n \sin(k) = \sin(2x - k) \]

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:

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

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:

\[ \tan(k) = \frac{2h}{a \left( n + \sqrt{a^2 - 4h^2} \right)} \]

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