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1979 IMO Problems/Problem 4

Revision as of 22:01, 29 January 2021 by Hamstpan38825 (talk | contribs) (Created page with "==Problem== We consider a point <math>P</math> in a plane <math>p</math> and a point <math>Q \not\in p</math>. Determine all the points <math>R</math> from <math>p</math> for...")
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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 $PQ\perp p$).

See Also

1979 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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