Consider triangle $P_1P_2P_3$ and a point $p$ within the triangle. Lines $P_1P, P_2P, P_3P$ intersect the opposite sides in points $Q_1, Q_2, Q_3$ respectively. Prove that, of the numbers \[ \dfrac{P_1P}{PQ_1}, \dfrac{P_2P}{PQ_2}, \dfrac{P_3P}{PQ_3} \] at least one is $\leq 2$ and at least one is $\geq 2$
Problem
Source: IMO 1961, Day 2, Problem 4
Tags: ratio, inequalities, geometry, Triangle, Intersection, IMO, IMO 1961
ashwath.rabindranath
02.03.2007 19:57
We see that, $\frac{P_{1}P}{PQ_{1}}=\frac{[PP_{1}P_{2}]}{[PP_{2}Q_{1}]}= \frac{[PP_{1}P_{3}]}{[PP_{2}Q_{1}]}=\frac{[P_{1}P_{2}P_{3}]-[PP_{2}P_{3}]}{[PP_{2}P_{3}]}$ We get similar results for $P_{2}$ and $P_{3}$.
Consider $\frac{PP_{1}}{PQ_{1}}+\frac{PP_{2}}{PQ_{2}}+\frac{PP_{3}}{PQ_{3}}$. This is equivalent to
$\frac{[P_{1}P_{2}P_{3}]-[PP_{2}P_{3}]}{[PP_{2}P_{3}]}+\frac{[P_{1}P_{2}P_{3}]-[PP_{1}P_{3}]}{[PP_{1}P_{3}]}+\frac{[P_{1}P_{2}P_{3}]-[PP_{1}P_{2}]}{[PP_{1}P_{2}]}$
$\frac{[P_{1}P_{2}P_{3}] }{[PP_{2}P_{3}]}+\frac{[P_{1}P_{2}P_{3}] }{[PP_{1}P_{3}]}+\frac{[P_{1}P_{2}P_{3}] }{[PP_{1}P_{2}]}-3 = [P_{1}P_{2}P_{3}][\frac{1 }{[PP_{2}P_{3}]}+\frac{1 }{[PP_{1}P_{3}]}+\frac{1 }{[PP_{1}P_{2}]}]-3$.
This is very clearly by the Cauchy Schwartz inequality , greater than $9-3 = 6$. Thus there exists one out of the three ratios which is $\geq 2$.
Now consider the maximum among the three areas $[PP_{2}P_{3}], [PP_{3}P_{1}] , [PP_{1}P_{2}]$ Without loss of generality let it be $[PP_{2}P_{3}] \geq \frac{[P_{1}P_{2}P_{3}]}{3}$. This implies that
$3\geq \frac{[P_{1}P_{2}P_{3}]}{[PP_{2}P_{3}]}$which implies that $\frac{PP_{1}}{PQ_{1}}\leq 2$ and we are done.
Ashwath.
dgreenb801
13.07.2008 22:30
A useful identity in all triangles proven at http://www.mathlinks.ro/Wiki/index.php/1988_AIME_Problems/Problem_12 is $ \dfrac {PQ_1}{P_1P + PQ_1} + \dfrac {PQ_2}{P_2P + PQ_2} + \dfrac {PQ_3}{P_3P + PQ_3} = 1$ So the question is equivalent to $ x + y + z = 1$, prove out of $ (1/x - 1)$,$ (1/y - 1)$,$ (1/z - 1)$ one is greater than 2, one is less than 2. But it is clear that one of x,y,z is $ \le$ 1/3 and one $ \ge$ than 1/3, which proves the claim. So we are done.
dgreenb801
14.07.2008 15:39
Kalva has an awesome solution to this problem. Take the lines through the centroid parallel to the sides of the triangle. The result is then obvious.
sunken rock
01.11.2008 12:44
If ABC is the triangle, P - an arbitrary point inside it, D, E and F the intersections of the cevians AP, etc. with the opposite side, there holds the relation: AP/PD = AF/FB + AE/EC and other two similar relations (Van - Aubel). The sum of these three relations is obviously at least 6, hence at least 1 ratio among the ones involved is at least 2. Note that, if one counts the product of the same ratios will find something at least 8. Best regards, sunken rock
sayantanchakraborty
21.12.2013 19:48
Such an easy problem like this will not be seen again, atleast in the IMO
sayantanchakraborty
22.12.2013 07:32
Let \[[P_1PP_2]=a\],\[[P_2PP_3]=b\],\[[P_1PP_3]=c\]. Then \[P_1/PQ_1=[P_1PP_3]/[PQ_1P_3]=[P_1PP_2]/[PP_2Q_1]=(a+c)/b\] Similarly, \[P_2P/PQ_2=(a+b)/c\] and \[P_3P/PQ_3=(b+c)/a\] Now if all the ratios are >2 then we have \[a+c>2b ..(1)\] \[a+b>2c ..(2)\] \[b+c>2a ..(3)\] But (1)+(2) gives $2a>b+c$ which contradicts with (3).Contradiction. Hence one of the ratios is <=2.The other proof is similar.
TheBernuli
26.01.2014 16:00
Let: $\frac{P_1P}{PQ_1}=x, \frac{P_2P}{PQ_2}=y, \frac{P_3P}{PQ_3}=z$ We have that $x+y+z+2=xyz$, and we are finished.
peace09
30.06.2022 06:28
Assign masses of $a$, $b$, $c$, $b+c$, $c+a$, $a+b$, and $a+b+c$ to $P_1$, $P_2$, $P_3$, $Q_1$, $Q_2$, $Q_3$, and $P$, respectively, where $P$ is the overall center of mass. Then, we have that $\tfrac{P_1P}{PQ_1}=\tfrac{b+c}{a}$, $\tfrac{P_2P}{PQ_2}=\tfrac{c+a}{b}$, and $\tfrac{P_3P}{PQ_3}=\tfrac{a+b}{c}$ along cevians $P_1Q_1$, $P_2Q_2$, and $P_3Q_3$, in that order. Now, suppose FTSOC that $\tfrac{P_1P}{PQ_1},\tfrac{P_2P}{PQ_2},\tfrac{P_3P}{PQ_3}>2$, i.e., $$\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}>2\iff b+c>2a,c+a>2b,a+b>2c.$$But summing gives $2a+2b+2c>2a+2b+2c$, contradiction, so at least one of the $\tfrac{P_iP}{PQ_i}$ are $\leq2$. Analogously, at least one of the $\tfrac{P_iP}{PQ_i}$ are $\geq2$, hence proven. $\blacksquare$
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