1979 IMO Problems/Problem 4

Problem

We consider a point $P$ in a plane $p$ and a point $Q \not\in p$ . Determine all the points $R$ from $p$ for which $\frac{QP+PR}{QR}$ is maximum.

Solution

Let $T$ be the orthogonal projection of the point $Q$ on the plane $p$ . Then, the line $PT$ is the orthogonal projection of the line $PQ$ on the plane $p$ , and thus forms the least angle with the line $PQ$ among all lines through the point $P$ which lie in the plane $p$ ; hence, $\measuredangle RPQ\geq\measuredangle TPQ$ , and equality holds if and only if the point $R$ lies on the ray $PT$ (the only exception is when $PQ\perp p$ ; in this case, $P = T$ , so the ray $PT$ is undefined, and equality holds for all points $R$ in the plane $p$ , since we always have $\angle RPQ = \angle TPQ = 90^{\circ}$ ).

From $\measuredangle RPQ\geq\measuredangle TPQ$ , it follows that $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$ ; also, since the angles $\angle RPQ$ and $\angle TPQ$ are $180^{\circ}$ , their half-angles $\frac{\measuredangle RPQ}{2}$ and $\frac{\measuredangle TPQ}{2}$ are < 90°, so that from $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$ we can conclude that $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ . Equality holds, as in the above, if and only if the point $R$ lies on the ray $PT$ (and, respectively, for all points $R$ in the plane $p$ if $PQ\perp p$ ).

On the other hand, we obviously have $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ with equality if and only if $\frac{\measuredangle QRP-\measuredangle PQR}{2}=0^{\circ}$ , i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base $QR$ , i. e. if and only if $PR = PQ$ , i. e. if and only if the point $R$ lies on the sphere with center $P$ and radius $PQ$ .

Now, applying the Mollweide theorem in triangle $QPR$ , we get

$\frac{QP+PR}{QR}=\frac{\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}}{\sin\frac{\measuredangle RPQ}{2}}$ $\leq\frac{1}{\sin\frac{\measuredangle RPQ}{2}}$ (since $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ ) $\leq\frac{1}{\sin\frac{\measuredangle TPQ}{2}}$ (since $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ ),

and equality holds here if and only if equality holds in both of the inequalities $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ and $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ that we have used, i. e. if and only if the point $R$ lies both on the ray $PT$ (this condition should be ignored if $PQ\perp p$ ) and on the sphere with center $P$ and radius $PQ$ .

Hence, the point $R$ for which the ratio $\frac{QP+PR}{QR}$ is maximum is the point of intersection of the ray $PT$ with the sphere with center $P$ and radius $PQ$ (or, respectively, it can be any arbitrary point on the intersection of the plane $p$ with the sphere with center $P$ and radius $PQ$ if $PQ\perp p$ ).

This solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [1]

1979 IMO (Problems) • Resources
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1979 IMO Problems/Problem 4

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