1961 IMO Problems/Problem 4


In the interior of triangle $P_1P_2P_3$ 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 $P_1P_2P_3$. 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 $2$.

Video Solution

https://youtu.be/3SQKgeFlMiA?si=5vhw28fTN2L4qRqr [Video Solution by little-fermat]

Solution 1

Let $[ABC]$ denote the area of triangle $ABC$.

Since triangles $P_1P_2P_3$ and $PP_2P_3$ share the base $P_2P_3$, we have $\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_1}{P_1Q_1}$.

Similarly, $\frac{[PP_1P_3]}{[P_1P_2P_3]}=\frac{PQ_2}{P_2Q_2}, \frac{[PP_1P_2]}{[P_1P_2P_3]}=\frac{PQ_3}{P_3Q_3}$.

Adding all of these gives $\frac{[PP_1P_3]}{[P_1P_2P_3]}+\frac{[PP_1P_2]}{[P_1P_2P_3]}+\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_2}{P_2Q_2}+\frac{PQ_3}{P_3Q_3}+\frac{PQ_1}{P_1Q_1}=1$.

We see that we must have at least one of the three fractions not greater than $\frac{1}{3}$, and at least one not less than $\frac{1}{3}$. These correspond to ratios $\frac{PP_i}{PQ_i}$ being less than or equal to $2$, and greater than or equal to $2$, respectively, so we are done.

Solution 2

Let $K_1=[P_2PP_3], K_2=[P_3PP_1],$ and $K_3=[P_1PP_2].$ Note that by same base in triangles $P_2PP_3$ and $P_2P_1P_3,$ \[\frac{P_1P}{PQ_1}+1=\frac{P_1Q_1}{PQ_1}=\frac{[P_2P_1P_3]}{[P_2PP_3]}=\frac{K_1+K_2+K_3}{K_1}.\] Thus, \[\frac{P_1P}{PQ_1}=\frac{K_2+K_3}{K_1}\]\[\frac{P_2P}{PQ_2}=\frac{K_1+K_3}{K_2}\] \[\frac{P_3P}{PQ_3}=\frac{K_1+K_2}{K_3}.\] Without loss of generality, assume $K_1\leq K_2\leq K_3.$ Hence, \[\frac{P_1P}{PQ_1}=\frac{K_2+K_3}{K_1}\geq \frac{K_1+K_1}{K_1}=2\] and \[\frac{P_3P}{PQ_3}=\frac{K_1+K_2}{K+3}\leq \frac{K_3+K_3}{K_3}=2,\] as desired. $\blacksquare$

1961 IMO (Problems) • Resources
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
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