IMO 1961 Problem 4
The problem is a pure proof problem.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 19m03s
Problem
In the interior of triangle $P_1 P_2 P_3$ a point P is given. Let $Q_1,Q_2,Q_3$ be the intersections of $PP_1, PP_2,PP_3$ with the opposing edges of triangle $ABC$. Prove that among the ratios $\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}$ there exists one not larger than 2 and one not smaller than 2.
Exploration
The problem is a pure proof problem. The statement asks for a comparison with the number $2$, and the configuration strongly suggests the use of ratios along cevians. Since $Q_i$ lies on the side opposite $P_i$, the quantity $\frac{PP_i}{PQ_i}$ compares how far $P$ is from the vertex $P_i$ relative to the whole cevian through $P$.
A first guess is that these ratios satisfy a linear relation. To test this, choose barycentric coordinates of $P$ with respect to triangle $P_1P_2P_3$. If $P=(\lambda_1,\lambda_2,\lambda_3)$ with $\lambda_1+\lambda_2+\lambda_3=1$ and $\lambda_i>0$, then the line $P_iP$ meets the opposite side where the $i$-th coordinate vanishes. Parametrizing the line suggests
$$\frac{PP_i}{PQ_i}=1-\lambda_i.$$
This formula must be checked carefully.
Testing small examples helps. If $P$ is the centroid, then $\lambda_1=\lambda_2=\lambda_3=\tfrac13$, hence
$$\frac{PP_i}{PQ_i}=\frac23$$
for all $i$. This cannot be correct because the statement predicts one ratio at least $2$. The parameterization must have been inverted.
Try again. Let $Q_1$ be on line $P_1P$. Since $P$ divides the segment $P_1Q_1$, write
$$P=(1-t)P_1+tQ_1.$$
If $Q_1$ lies on $P_2P_3$, then the first barycentric coordinate of $Q_1$ is $0$. Since $P=(\lambda_1,\lambda_2,\lambda_3)$,
$$\lambda_1=1-t.$$
Hence
$$t=1-\lambda_1.$$
Because
$$PP_1=t\cdot P_1Q_1, \qquad PQ_1=(1-t)\cdot P_1Q_1,$$
the ratio becomes
$$\frac{PP_1}{PQ_1} = \frac{t}{1-t} = \frac{1-\lambda_1}{\lambda_1}.$$
For the centroid this gives $2$, which fits the statement sharply.
The target now becomes transparent. Let
$$r_i=\frac{PP_i}{PQ_i}.$$
Then
$$r_i=\frac{1-\lambda_i}{\lambda_i}.$$
The condition $r_i\le 2$ becomes $\lambda_i\ge \tfrac13$, while $r_i\ge 2$ becomes $\lambda_i\le \tfrac13$. Since
$$\lambda_1+\lambda_2+\lambda_3=1,$$
at least one $\lambda_i\ge \tfrac13$ and at least one $\lambda_i\le \tfrac13$. This is the central idea.
A second possible route uses mass points or Menelaus/Ceva, expressing
$$\frac{P_iQ_i}{PQ_i}$$
through area ratios. That method appears longer and hides the essential averaging argument. The barycentric approach avoids unnecessary computation.
The most error-prone step is deriving the ratio formula from barycentric coordinates. Reversing the segment proportions produces the false formula discovered above.
Problem Understanding
We are given a point $P$ inside triangle $P_1P_2P_3$. For each vertex $P_i$, the line through $P_i$ and $P$ meets the opposite side at a point $Q_i$. The problem asks us to prove that among the three numbers
$$\frac{PP_1}{PQ_1}, \qquad \frac{PP_2}{PQ_2}, \qquad \frac{PP_3}{PQ_3},$$
at least one is at most $2$, and at least one is at least $2$.
This is a Type B problem, because the statement to be proved is fixed and no classification or extremal value is requested.
The geometric objects involved are a triangle, an interior point, and the cevians through that point. The difficulty lies in relating metric ratios on different cevians. A direct Euclidean computation quickly becomes cumbersome because the lengths belong to different lines. A naive approach comparing the three ratios independently misses the hidden relation connecting them.
The key observation is that barycentric coordinates of $P$ naturally encode how $P$ divides each cevian. Once the ratios are rewritten in terms of barycentric coordinates, the statement becomes a consequence of the fact that three positive numbers with sum $1$ cannot all exceed $\tfrac13$ and cannot all be smaller than $\tfrac13$.
Proof Architecture
We introduce barycentric coordinates of $P$ with respect to triangle $P_1P_2P_3$, writing
$$P=(\lambda_1,\lambda_2,\lambda_3), \qquad \lambda_1+\lambda_2+\lambda_3=1, \qquad \lambda_i>0.$$
Lemma 1. For each $i$,
$$\frac{PP_i}{PQ_i} = \frac{1-\lambda_i}{\lambda_i}.$$
The proof comes from parametrizing the line segment $P_iQ_i$ and identifying the coefficient of $P_i$ with the barycentric coordinate $\lambda_i$.
Lemma 2. At least one index satisfies $\lambda_i\ge \tfrac13$, and at least one index satisfies $\lambda_i\le \tfrac13$.
The proof uses only the equality
$$\lambda_1+\lambda_2+\lambda_3=1.$$
The hardest direction is translating geometry into the formula of Lemma 1. A careless parametrization can invert the ratio and produce the wrong expression.
Solution
Let
$$P=(\lambda_1,\lambda_2,\lambda_3)$$
be the barycentric coordinates of $P$ with respect to triangle $P_1P_2P_3$. Since $P$ lies in the interior of the triangle,
$$\lambda_1,\lambda_2,\lambda_3>0, \qquad \lambda_1+\lambda_2+\lambda_3=1.$$
Lemma 1
For each index $i\in{1,2,3}$,
$$\frac{PP_i}{PQ_i} = \frac{1-\lambda_i}{\lambda_i}.$$
Proof
It suffices to prove the formula for $i=1$, since the remaining cases are identical.
The point $Q_1$ lies on the side $P_2P_3$, hence its first barycentric coordinate equals $0$. Since $P$ lies on the segment $P_1Q_1$, there exists a number $t$ with $0<t<1$ such that
$$P=(1-t)P_1+tQ_1.$$
Because the first barycentric coordinate of $Q_1$ is $0$, the first barycentric coordinate of $P$ equals
$$\lambda_1=1-t.$$
Hence
$$t=1-\lambda_1.$$
The point $P$ divides the segment $P_1Q_1$ in the ratio
$$PP_1=t\cdot P_1Q_1, \qquad PQ_1=(1-t)\cdot P_1Q_1.$$
Dividing these equalities gives
$$\frac{PP_1}{PQ_1} = \frac{t}{1-t} = \frac{1-\lambda_1}{\lambda_1}.$$
The same argument for $i=2$ and $i=3$ yields
$$\frac{PP_i}{PQ_i} = \frac{1-\lambda_i}{\lambda_i}.$$
∎
This lemma converts the geometric ratios into algebraic expressions in barycentric coordinates, and any shortcut that skips the segment parametrization risks reversing the quotient.
Lemma 2
At least one of the numbers $\lambda_1,\lambda_2,\lambda_3$ is at least $\tfrac13$, and at least one is at most $\tfrac13$.
Proof
Suppose all three numbers satisfied
$$\lambda_i<\frac13.$$
Then
$$\lambda_1+\lambda_2+\lambda_3<1,$$
contradicting
$$\lambda_1+\lambda_2+\lambda_3=1.$$
Hence at least one index satisfies
$$\lambda_i\ge \frac13.$$
Suppose instead that all three numbers satisfied
$$\lambda_i>\frac13.$$
Then
$$\lambda_1+\lambda_2+\lambda_3>1,$$
again contradicting
$$\lambda_1+\lambda_2+\lambda_3=1.$$
Hence at least one index satisfies
$$\lambda_i\le \frac13.$$
∎
This lemma establishes the unavoidable comparison with $\tfrac13$, and any argument ignoring the condition that the coordinates sum to $1$ loses the key constraint.
We now complete the proof. By Lemma 2, choose an index $i$ such that
$$\lambda_i\ge \frac13.$$
Applying Lemma 1,
$$\frac{PP_i}{PQ_i} = \frac{1-\lambda_i}{\lambda_i} \le \frac{1-\tfrac13}{\tfrac13} = 2.$$
Similarly, choose an index $j$ such that
$$\lambda_j\le \frac13.$$
Lemma 1 gives
$$\frac{PP_j}{PQ_j} = \frac{1-\lambda_j}{\lambda_j} \ge \frac{1-\tfrac13}{\tfrac13} = 2.$$
Thus among the three ratios there exists one not larger than $2$ and one not smaller than $2$.
This completes the proof.
∎
Verification of Key Steps
The most delicate step is the derivation of
$$\frac{PP_i}{PQ_i} = \frac{1-\lambda_i}{\lambda_i}.$$
Re-derive it independently for $i=1$. Since $P$ lies on segment $P_1Q_1$, write
$$P=P_1+s(Q_1-P_1).$$
Then
$$PP_1=s\cdot P_1Q_1, \qquad PQ_1=(1-s)\cdot P_1Q_1.$$
Because $Q_1$ has first barycentric coordinate $0$, the first coordinate of $P$ equals $1-s$. Hence
$$\lambda_1=1-s, \qquad s=1-\lambda_1.$$
Substituting gives
$$\frac{PP_1}{PQ_1} = \frac{s}{1-s} = \frac{1-\lambda_1}{\lambda_1}.$$
The common mistake is to interchange $s$ and $1-s$, which produces the incorrect reciprocal formula.
Another delicate step is converting the threshold $2$ into a condition on $\lambda_i$. Starting from
$$\frac{1-\lambda_i}{\lambda_i}\le 2,$$
multiply by the positive number $\lambda_i$:
$$1-\lambda_i\le 2\lambda_i.$$
Hence
$$1\le 3\lambda_i, \qquad \lambda_i\ge \frac13.$$
The positivity of $\lambda_i$ matters. Ignoring it would make the inequality transformation unjustified.
A third point requiring care is the averaging argument. The conclusion does not follow from symmetry alone. The equality
$$\lambda_1+\lambda_2+\lambda_3=1$$
forces the existence of a coordinate at least $\tfrac13$ and one at most $\tfrac13$. If all coordinates exceeded $\tfrac13$, the sum would exceed $1$; if all were smaller, the sum would be below $1$.
Alternative Approaches
A substantially different proof uses areas. Let
$$[P_iP_jP]$$
denote the area of triangle $P_iP_jP$. Since $P$ lies inside the triangle, define
$$\lambda_1=\frac{[PP_2P_3]}{[P_1P_2P_3]}, \quad \lambda_2=\frac{[P_1PP_3]}{[P_1P_2P_3]}, \quad \lambda_3=\frac{[P_1P_2P]}{[P_1P_2P_3]}.$$
These numbers are positive and sum to $1$. Along cevian $P_1Q_1$, triangles $PP_2P_3$ and $P_1P_2P_3$ share the same altitude to line $P_2P_3$, so
$$\lambda_1=\frac{PQ_1}{P_1Q_1}.$$
Since
$$P_1Q_1=PP_1+PQ_1,$$
one obtains
$$\frac{PP_1}{PQ_1} = \frac{1-\lambda_1}{\lambda_1},$$
and the remainder proceeds exactly as above.
The barycentric proof is preferable because it isolates the essential structure immediately. The area method introduces additional geometric bookkeeping before reaching the same algebraic statement.