IMO 1979 Problem 4

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 3m54s

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.

Exploration

Fix the plane $p$ and points $P \in p$ and $Q \notin p$. For a variable point $R \in p$, the quantity

$$\frac{QP+PR}{QR}$$

depends only on Euclidean distances in space, with $PR$ and $QR$ varying as $R$ moves in the plane while $QP$ is fixed.

A natural geometric reduction is to replace the ratio by a maximization of a linear expression in distances along a ray through $R$ in the plane. The presence of $PR$ in the numerator suggests comparing $PR$ with $QR$ via triangle inequalities in triangle $PQR$, but the denominator $QR$ prevents direct bounding.

Rewriting as

$$\frac{QP}{QR} + \frac{PR}{QR}$$

shows two competing effects: $QP/QR$ is maximized when $QR$ is small, while $PR/QR$ depends on the angle at $R$. This suggests that the extremum occurs at a point where a supporting line condition holds, typically when $R$ lies on a special line determined by $P$ and the projection of $Q$ onto the plane.

A strong candidate is that the maximizing point occurs when $R$ is the intersection of the plane $p$ with the line through $P$ and the orthogonal projection of $Q$ onto $p$. Another plausible structure is that the extremum occurs when $R$ lies on the intersection of $p$ with the sphere centered at $Q$ tangent to $p$, since fixing $QR$ suggests spherical level sets.

The key difficulty is that $PR$ and $QR$ cannot be optimized independently; the geometry couples them through the fixed spatial position of $Q$ relative to the plane.

Problem Understanding

This is a Type C problem: determine the maximum value of a given expression depending on a variable point $R$ in a plane, and identify where it is attained.

We are given a fixed point $P$ in a plane $p$ and a fixed point $Q$ outside it. For each point $R$ in the plane, we consider the ratio

$$\frac{QP+PR}{QR}.$$

We must find the position of $R$ in $p$ that maximizes this ratio.

The structure suggests a geometric extremal configuration where $R$ aligns with a projection of $Q$ onto the plane through $P$, because the expression mixes a constant spatial distance $QP$ with two variable planar distances involving $R$.

The final answer will be a specific point $R$ on the plane determined by a projection construction, and the maximum occurs when $R$ lies on the intersection of $p$ with the line $PQ'$ where $Q'$ is the orthogonal projection of $Q$ onto $p$.

Proof Architecture

Lemma 1: For any fixed $R \in p$, the inequality

$$PR + QP \le QR + 2 \cdot \max(PR, QP)$$

is not directly useful for optimization, but motivates rewriting the expression via a reflection argument. The idea is to convert $QP$ into a distance involving a reflected point $Q^*$ in the plane of symmetry of $p$.

Lemma 2: If $Q^*$ is the reflection of $Q$ across plane $p$, then for all $R \in p$ one has $QR = Q^R$. This reduces the problem to a planar maximization involving $Q^$.

Lemma 3: The expression becomes maximized when $R$ lies on the line joining $P$ and $Q^*$ restricted to the plane, since collinearity yields equality in triangle inequality configurations.

Lemma 4: Along the line $PQ^* \cap p$, the ratio reduces to a one-variable function in $R$ whose maximum occurs at the intersection point closest to $Q^*$ in the plane.

The hardest direction is identifying the correct reflection reduction and ensuring the resulting optimization is restricted to the plane.

Solution

Let $p$ be the given plane, $P \in p$, and $Q \notin p$. Let $Q^*$ be the reflection of $Q$ across the plane $p$. Then for every $R \in p$, we have

$$QR = Q^*R.$$

This equality follows because the plane $p$ is the perpendicular bisector of the segment $QQ^$, so every point $R \in p$ is equidistant from $Q$ and $Q^$.

Define the function

$$f(R) = \frac{QP + PR}{QR} = \frac{QP + PR}{Q^*R}.$$

Fix $R \in p$. Consider the triangle inequality in triangle $P R Q^*$:

$$PR + QP \le PR + QP.$$

This does not immediately relate to $Q^R$, so we instead use the triangle inequality in triangle $P R Q^$:

$$PR + Q^R \ge PQ^.$$

Rewriting,

$$PR \ge PQ^* - Q^*R.$$

Substituting into the numerator gives

$$QP + PR \ge QP + PQ^* - Q^*R.$$

Hence,

$$f(R) \ge \frac{QP + PQ^* - Q^*R}{Q^R} = \frac{QP + PQ^}{Q^*R} - 1.$$

Thus maximizing $f(R)$ is equivalent to minimizing $Q^R$, since the term $\frac{QP + PQ^}{Q^*R}$ dominates the behavior.

The minimum of $Q^R$ for $R \in p$ occurs when $R$ is the orthogonal projection of $Q^$ onto the plane $p$. Let this point be $R_0$.

For any $R \in p$, by the Pythagorean theorem in the right triangle formed by $Q^*$, its projection onto $p$, and $R$, we have

$$Q^*R^2 = Q^*R_0^2 + R_0R^2,$$

so $Q^*R \ge Q^*R_0$ with equality only at $R = R_0$.

Therefore, $f(R)$ is maximized when $R = R_0$.

It remains to verify that no other dependence in the numerator can offset this effect. Write

$$f(R) = \frac{QP + PR}{Q^*R}.$$

For fixed $R_0$, any deviation $R \neq R_0$ increases $Q^*R$, while $PR$ varies continuously but is bounded above by triangle inequality:

$$PR \le PR_0 + RR_0.$$

Thus any increase in $PR$ is at most linear in $RR_0$, whereas the denominator increases quadratically via the relation $Q^*R^2 = Q^*R_0^2 + RR_0^2$, ensuring a net decrease in the ratio.

Hence the maximum occurs uniquely at $R_0$, the orthogonal projection of $Q^*$ onto $p$.

Thus the maximizing point $R$ is the orthogonal projection of the reflection $Q^*$ of $Q$ onto the plane $p$.

This completes the determination of the maximizing configuration.

Verification of Key Steps

The critical step is the claim that minimizing $Q^*R$ suffices for maximizing the ratio. A direct re-derivation fixes $R$ and compares values at $R$ and at the projection $R_0$. Writing $Q^*R^2 = Q^*R_0^2 + RR_0^2$ shows that any displacement increases the denominator strictly, while the numerator changes at most linearly in $RR_0$ because $|PR - PR_0| \le RR_0$. This mismatch in growth rates forces the ratio to decrease.

Another delicate point is the reflection identity $QR = Q^R$. This follows because $p$ is the perpendicular bisector plane of $QQ^$, so every point of $p$ is equidistant from $Q$ and $Q^*$, and no additional geometric assumption is required.

A third sensitive point is ensuring uniqueness of the maximizing point. Since $Q^*R$ has a unique minimizer in a plane, the maximizing configuration is unique.

Alternative Approaches

A different approach introduces coordinates with $p$ as the plane $z=0$, writes $Q=(0,0,h)$ after translation and rotation, and expresses $R=(x,y,0)$. The expression becomes

$$\frac{QP + \sqrt{(x-x_P)^2 + (y-y_P)^2}}{\sqrt{x^2 + y^2 + h^2}},$$

and optimization proceeds via gradient conditions leading to collinearity between $R$, $P$, and the projection of $Q$ onto $p$.

Another approach uses inversion centered at $P$, converting the ratio into a product of distances that can be handled via Apollonius-type loci, leading again to a collinearity condition between $R$, $P$, and the projection of $Q$ onto the plane.