Talk:1960 IMO Problems/Problem 3
A proof of the 3rd question from the 1960 IMO
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Let \( \angle ACB = x \) and \( \angle ABC = 90^\circ - x \). Let \( M \) be the midpoint of the hypotenuse \( BC \), and let \( Q \) and \( P \) be points such that \( PQ \) contains \( BC \), with \( Q \) closer to \( C \) and \( P \) closer to \( B \). The midpoint \( M \) will always be in the middle of segment \( QP \), unless \( n \) is even or infinite, which it is not.
Given such a triangle, we can express the altitude to the hypotenuse as: \[ h = a \cos x \sin x \]
Now, let us denote the median \( AM \) by \( f \). Since \( AM \) is the median to the hypotenuse, we have: \[ AM = BM = CM \] Since \( BM = \frac{a}{2} \), it follows that \( f = \frac{a}{2} \).
Angles in the triangle give us: \[ \angle MAB = 90^\circ - x, \quad \angle MAC = x \] \[ \angle AMB = 2x, \quad \angle AMC = 180^\circ - 2x \]
The length of \( QP \) is given by: \[ QP = \frac{a}{n} \]
Define \( \angle QAM = k \) and \( \angle PAM = z \), so that: \[ \angle QAP = \alpha = k + z \] From angle properties, we get: \[ \angle AQM = 2x - k, \quad \angle APM = 180^\circ - 2x - z \]
Since \( M \) is the midpoint of \( QP \), we know: \[ QM = PM = \frac{a}{2n} \]
- Applying the Sine Rule in \( \triangle AQM \):
\[ \frac{\sin k}{a/2n} = \frac{\sin(2x - k)}{a/2} \]
Rearranging: \[ \frac{2n \sin k}{a} = \frac{2 \sin(2x - k)}{a} \]
\[ n \sin k = \sin(2x - k) \]
Using the sine subtraction identity: \[ \sin(2x - k) = \sin 2x \cos k - \sin k \cos 2x \]
Since \( \sin 2x = 2 \sin x \cos x \) and given that \( h = a \cos x \sin x \), we can substitute: \[ 2 \sin x \cos x = \frac{2h}{a} \] \[ \cos 2x = \frac{\sqrt{a^2 - 4h^2}}{a} \]
Substituting in our equation: \[ n \sin k + \sin k \frac{\sqrt{a^2 - 4h^2}}{a} = \cos k \frac{2h}{a} \]
Factorizing: \[ \sin k \left(n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos k \frac{2h}{a} \]
Dividing both sides by \( \cos k \): \[ \tan k = \frac{2h}{an + \sqrt{a^2 - 4h^2}} \]
Similarly, solving for \( \tan z \), we can apply the tangent sum identity: \[ \tan (z + k) = \frac{\tan z + \tan k}{1 - \tan z \tan k} \] to find the value of \( \tan \alpha \), where \( \alpha = z + k \).
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