Talk:1960 IMO Problems/Problem 3

\section*{Proof for 1960 IMO Problem 3}

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 \( B \). Since \( BC \) is divided into \( n \) equal parts, where \( n \) is odd, the length of each segment is:

\[ \frac{a}{n} \]

Because \( M \) is the midpoint of \( BC \), it must lie between two consecutive division points, meaning \( M \) is at the center of segment \( QP \).

1. 1. 1. Step 1: Altitude to the Hypotenuse

From right triangle trigonometry, the altitude to the hypotenuse is given by:

\[ h = a \cos x \sin x \]

1. 1. 1. Step 2: Median to the Hypotenuse

Since \( AM \) is the median to the hypotenuse, we have:

\[ AM = BM = CM = \frac{a}{2} \]

We also know that:

\[ \angle MAB = 90^\circ - x, \quad \angle MAC = x \] \[ \angle AMB = 2x, \quad \angle AMC = 180^\circ - 2x \]

1. 1. 1. Step 3: Length of \( QP \) and Triangle Relations

The length of \( QP \) is:

\[ QP = \frac{a}{n} \]

Define:

\[ \angle QAM = k, \quad \angle PAM = z, \quad \angle QAP = \alpha = k + z \]

From angle properties:

\[ \angle AQM = 2x - k, \quad \angle APM = 180^\circ - 2x - z \]

Since \( M \) is the midpoint of \( QP \), we have:

\[ QM = PM = \frac{a}{2n} \]

1. 1. 1. Step 4: Applying the Law of Sines

Applying the sine rule in \( \triangle AQM \):

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

Multiplying both sides by \( \frac{2}{a} \):

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

which simplifies to:

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

Using the identity:

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

and the double-angle formula:

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

we substitute:

\[ \cos 2x = \frac{\sqrt{a^2 - 4h^2}}{a} \]

into our equation:

\[ n \sin k + \sin k \frac{\sqrt{a^2 - 4h^2}}{a} = \cos k \frac{2h}{a} \]

Factoring:

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

Dividing both sides by \( \cos k \):

\[ \tan k = \frac{2h}{an + \sqrt{a^2 - 4h^2}} \]

1. 1. 1. Step 5: Finding \( \tan \alpha \)

Using a similar derivation for \( \tan z \), we apply the tangent sum identity:

\[ \tan (z + k) = \frac{\tan z + \tan k}{1 - \tan z \tan k} \]

to find \( \tan \alpha \), where \( \alpha = z + k \).