IMO 1960 · Problem 3

In a given right triangle \(ABC\), the hypotenuse \(BC\), of length \(a\), is divided into \(n\) equal parts (\(n\) an odd integer). Let \(\alpha\) be the acute angle subtending, from \(A\), that segment which contains the midpoint of the hypotenuse. Let \(h\) be the length of the altitude to the hypotenuse of the triangle. Prove: \[ \tan\alpha=\frac{4nh}{(n^2-1)a}. \]

Official text of the problem. Official problem paper (PDF)

Solution

Let \(M\) be the midpoint of \(BC\), and let \(P,Q\), in that order along \(BC\), be the endpoints of the segment containing \(M\). Since \(n\) is odd, \[ MP=MQ=\frac{a}{2n}. \] The number \(n\) is positive. If \(n=1\), then \(P=B\), \(Q=C\), and the subtended angle is \(90^\circ\), contrary to the hypothesis that it is acute. Thus \(n\ge3\).

The midpoint of the hypotenuse of a right triangle is its circumcenter, so \[ AM=\frac a2. \] Write \[ r=AM=\frac a2,\qquad d=MP=MQ=\frac{a}{2n}. \]

Because \(MP\) and \(MQ\) are opposite rays, the angles \(\angle AMP\) and \(\angle AMQ\) are supplementary. Applying the cosine rule in triangles \(AMP\) and \(AMQ\) and adding therefore gives \[ AP^2+AQ^2=2r^2+2d^2. \] Now \(PQ=2d\) and \(\angle PAQ=\alpha\), so the cosine rule in triangle \(APQ\) yields \[ 2AP\cdot AQ\cos\alpha =AP^2+AQ^2-PQ^2 =2(r^2-d^2). \] Consequently, \[ AP\cdot AQ\cos\alpha=r^2-d^2>0. \]

The distance from \(A\) to the line \(PQ=BC\) is \(h\). Computing the area of triangle \(APQ\) in two ways gives \[ \frac12 AP\cdot AQ\sin\alpha =\frac12 PQ\cdot h=dh, \] hence \[ AP\cdot AQ\sin\alpha=2dh. \] Dividing the last two identities, we obtain \[ \tan\alpha =\frac{2dh}{r^2-d^2} =\frac{ah/n}{a^2/4-a^2/(4n^2)} =\boxed{\frac{4nh}{(n^2-1)a}}. \]

The area calculation remains valid whether the foot of the altitude lies inside or outside \(PQ\), so all possible positions are covered.

The solution was checked against John Scholes' IMO solution archive (Kalva).