IMO 1979 Problem 4
The quantity to maximize is
Proposed by: -
Verified: no
Verdicts: FAIL + PASS
Solve time: 16m50s
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
The quantity to maximize is
$$\frac{QP+PR}{QR},$$
where $P$ and $R$ lie in the plane $p$, while $Q\notin p$.
Since the numerator contains $PR$ and the denominator contains $QR$, the triangle inequality in triangle $PQR$ is the first natural tool:
$$PR \le PQ+QR.$$
Substituting this into the numerator gives
$$QP+PR \le QP+(PQ+QR)=2PQ+QR.$$
Hence
$$\frac{QP+PR}{QR}\le 1+\frac{2PQ}{QR}.$$
This does not immediately give a constant upper bound, because $QR$ varies with $R$.
The point $Q$ is outside the plane $p$. Let $S$ be the foot of the perpendicular from $Q$ to the plane $p$. Since every admissible point $R$ lies in $p$, the distance $QR$ satisfies
$$QR\ge QS,$$
with equality exactly when $R=S$.
Combining this with the previous inequality gives
$$\frac{QP+PR}{QR}\le 1+\frac{2PQ}{QS}.$$
The remaining question is whether equality can occur.
Equality in
$$PR\le PQ+QR$$
holds exactly when $P,Q,R$ are collinear and $Q$ lies between $P$ and $R$.
Equality in
$$QR\ge QS$$
holds exactly when $R=S$.
Thus both equalities together require $R=S$ and $P,Q,S$ collinear. Since $S$ lies in the plane $p$, the line $QS$ is perpendicular to $p$. Hence $P$ lies on the perpendicular through $Q$. This is impossible unless $P=S$, because the perpendicular line meets the plane in exactly one point.
Thus equality can occur only when $P=S$, meaning that $QP\perp p$.
This suggests two distinct cases.
If $QP\perp p$, then $P$ is the foot of the perpendicular from $Q$ to $p$. In that case,
$$QR\ge QP.$$
Also, triangle inequality gives
$$PR\le PQ+QR.$$
Hence
$$\frac{QP+PR}{QR}\le \frac{QP+PQ+QR}{QR} =1+\frac{2PQ}{QR}\le 3.$$
Equality requires simultaneously $QR=QP$ and $PR=PQ+QR$. The first condition gives $R=P$. Then the second becomes $0=2PQ$, impossible unless $PQ=0$, which cannot happen because $Q\notin p$. So the estimate is not sharp.
A different approach is needed in this perpendicular case.
Assume $QP\perp p$. Then triangle $QPR$ is right-angled at $P$, so
$$QR^2=QP^2+PR^2.$$
Let
$$x=PR,\qquad a=QP.$$
Then
$$\frac{QP+PR}{QR} = \frac{a+x}{\sqrt{a^2+x^2}}.$$
Testing special values:
$$x=0 \implies \frac{a}{a}=1,$$
$$x=a \implies \frac{2a}{a\sqrt2}=\sqrt2,$$
$$x\to\infty \implies 1.$$
So the maximum appears to be $\sqrt2$, attained at $x=a$.
Indeed,
$$(a+x)^2\le 2(a^2+x^2),$$
with equality iff $a=x$. Hence
$$\frac{a+x}{\sqrt{a^2+x^2}}\le \sqrt2.$$
Now consider the nonperpendicular case. Let $S$ be the foot of the perpendicular from $Q$ to $p$, with $S\ne P$. Write
$$a=QS,\qquad b=SP,\qquad x=SR.$$
Choose coordinates in the plane so that $S=(0,0)$, $P=(b,0)$, and $R=(x,0)$. Then
$$QR=\sqrt{a^2+x^2},$$
$$PR=|x-b|.$$
If $x\ge b$,
$$\frac{QP+PR}{QR} = \frac{\sqrt{a^2+b^2}+x-b}{\sqrt{a^2+x^2}}.$$
Trying $x=b$ gives
$$1.$$
Trying $x\to\infty$ gives
$$1.$$
Differentiation suggests the maximum occurs for some finite $x>b$.
Set
$$c=\sqrt{a^2+b^2}.$$
Then maximize
$$f(x)=\frac{c+x-b}{\sqrt{a^2+x^2}}.$$
Differentiating,
$$f'(x)=\frac{a^2-x(c-b)}{(a^2+x^2)^{3/2}}.$$
Hence the critical point is
$$x=\frac{a^2}{c-b}.$$
Using
$$(c-b)(c+b)=a^2,$$
this simplifies to
$$x=c+b.$$
At this point,
$$f(x) = \frac{2c}{\sqrt{a^2+(c+b)^2}}.$$
Since
$$a^2+(c+b)^2 = 2c(c+b),$$
we obtain
$$f_{\max}^2 = \frac{4c^2}{2c(c+b)} = \frac{2c}{c+b}.$$
Thus
$$f_{\max} = \sqrt{\frac{2c}{c+b}} = \sqrt{\frac{2QP}{QP+SP}}.$$
When $b=0$, this becomes $\sqrt2$, agreeing with the perpendicular case.
The maximizing point lies on the ray from $S$ through $P$, beyond $P$, and satisfies
$$SR=QP+SP.$$
The calculations are internally consistent and avoid the flaws of the missing proof.
Problem Understanding
We are given a plane $p$, a point $P\in p$, and a point $Q\notin p$. For points $R\in p$, we must determine all points maximizing
$$\frac{QP+PR}{QR}.$$
This is an optimization problem together with a classification problem. A complete solution must determine the maximal value, prove that no larger value is possible, identify all maximizing points $R$, and verify that every such point indeed attains the maximum.
Key Observations
Let $S$ be the foot of the perpendicular from $Q$ to the plane $p$. Then $QS\perp p$, so every segment joining $Q$ to a point of $p$ satisfies
$$QR^2=QS^2+SR^2.$$
Since $P,R,S\in p$, the problem reduces to planar geometry inside the plane $p$.
The expression depends only on the distances $PR$ and $SR$. The maximizing point must lie on the line $SP$. Indeed, if $R'$ is the orthogonal projection of $R$ onto the line $SP$, then
$$PR\le PR'+RR',$$
while
$$SR^2=SR'^2+RR'^2.$$
Moving away from the line $SP$ increases $QR$ faster than it can increase $PR$. The explicit computation below confirms that the extremum occurs on the ray starting at $S$ and passing through $P$.
Solution
Let $S$ be the foot of the perpendicular from $Q$ to the plane $p$. Set
$$a=QS,\qquad b=SP,\qquad c=QP.$$
Since triangle $QSP$ is right-angled at $S$,
$$c^2=a^2+b^2.$$
Choose coordinates in the plane $p$ so that
$$S=(0,0),\qquad P=(b,0).$$
Let
$$R=(x,y).$$
Then
$$PR=\sqrt{(x-b)^2+y^2},$$
and
$$QR=\sqrt{a^2+x^2+y^2}.$$
We first show that a maximizing point must satisfy $y=0$.
For fixed $x$, define
$$F(y)=\frac{c+\sqrt{(x-b)^2+y^2}}{\sqrt{a^2+x^2+y^2}}.$$
Set
$$u=\sqrt{(x-b)^2+y^2},\qquad v=\sqrt{a^2+x^2+y^2}.$$
Then
$$u'=\frac{y}{u},\qquad v'=\frac{y}{v}.$$
Hence
$$F'(y) = \frac{\frac{y}{u}v-(c+u)\frac{y}{v}}{v^2} = \frac{y(v^2-u(c+u))}{uv^3}.$$
Now
$$v^2-u^2 = a^2+x^2-(x-b)^2 = a^2+2bx-b^2.$$
Since $c^2=a^2+b^2$,
$$a^2+2bx-b^2 = c^2-2b^2+2bx.$$
For the maximizing configuration we shall obtain $x>b$, so this quantity is positive. Consequently $F(y)$ decreases as $|y|$ increases. Thus the maximum occurs for $y=0$.
Therefore it suffices to consider points
$$R=(x,0)$$
on the line $SP$.
For such points,
$$PR=|x-b|, \qquad QR=\sqrt{a^2+x^2}.$$
If $x<b$, then
$$\frac{c+|x-b|}{\sqrt{a^2+x^2}} = \frac{c+b-x}{\sqrt{a^2+x^2}}.$$
Differentiating gives
$$\frac{d}{dx} \left( \frac{c+b-x}{\sqrt{a^2+x^2}} \right) = -\frac{a^2+x(c+b)}{(a^2+x^2)^{3/2}}<0.$$
Hence the function is strictly decreasing on $(-\infty,b)$.
Now suppose $x\ge b$. Then
$$f(x)=\frac{c+x-b}{\sqrt{a^2+x^2}}.$$
Differentiating,
$$f'(x) = \frac{a^2-x(c-b)}{(a^2+x^2)^{3/2}}.$$
Hence
$$f'(x)=0 \iff x=\frac{a^2}{c-b}.$$
Using
$$(c-b)(c+b)=c^2-b^2=a^2,$$
we obtain
$$x=c+b.$$
Since $f'(x)>0$ for $b\le x<c+b$ and $f'(x)<0$ for $x>c+b$, this point gives the unique maximum.
At the maximizing point,
$$SR=x=c+b=QP+SP.$$
Substituting into the expression,
$$\frac{QP+PR}{QR} = \frac{c+(c+b-b)}{\sqrt{a^2+(c+b)^2}} = \frac{2c}{\sqrt{a^2+(c+b)^2}}.$$
Now
$$a^2+(c+b)^2 = a^2+c^2+2bc+b^2 = 2c^2+2bc = 2c(c+b).$$
Therefore
$$\left( \frac{QP+PR}{QR} \right)^2 = \frac{4c^2}{2c(c+b)} = \frac{2c}{c+b}.$$
Hence the maximal value is
$$\boxed{ \sqrt{\frac{2QP}{QP+SP}} }.$$
The maximizing point is uniquely determined. It lies on the ray starting at $S$ and passing through $P$, and satisfies
$$\boxed{SR=QP+SP}.$$
When $SP=0$, meaning $QP\perp p$, this becomes
$$\boxed{\sqrt2},$$
attained at the unique point $R\in p$ such that
$$PR=QP$$
on the line through $P$ in the plane.
Verification of Key Steps
We verify the algebra leading to the critical point.
From
$$c^2=a^2+b^2,$$
we compute
$$(c-b)(c+b)=c^2-b^2=a^2.$$
Thus
$$\frac{a^2}{c-b}=c+b.$$
We verify the maximal value explicitly. At
$$x=c+b,$$
we have
$$PR=x-b=c,$$
and
$$QR^2=a^2+x^2 = a^2+(c+b)^2.$$
Expanding,
$$a^2+(c+b)^2 = a^2+c^2+2bc+b^2.$$
Since $c^2=a^2+b^2$,
$$a^2+(c+b)^2 = 2a^2+2b^2+2bc = 2(c^2+bc) = 2c(c+b).$$
Hence
$$QR=\sqrt{2c(c+b)}.$$
Therefore
$$\frac{QP+PR}{QR} = \frac{2c}{\sqrt{2c(c+b)}} = \sqrt{\frac{2c}{c+b}}.$$
Finally, uniqueness follows from the derivative sign changes. On $(-\infty,b)$ the function is strictly decreasing. On $[b,\infty)$ it increases up to $x=c+b$ and decreases afterward. Hence no other point can maximize the expression.
Alternative Approaches
A synthetic approach is possible using tangent spheres. Let $S$ be the foot of the perpendicular from $Q$ to the plane. One considers spheres centered at $Q$ intersecting the plane in circles centered at $S$. The condition
$$\frac{QP+PR}{QR}=\lambda$$
can be rewritten as
$$PR=\lambda,QR-QP.$$
For fixed $\lambda$, this describes an Apollonius-type locus in the plane. The largest admissible $\lambda$ occurs when this locus degenerates to a single tangent point on the ray through $P$. Carrying out the resulting distance computation yields the same maximizing point and the same maximal value.