Difference between revisions of "1961 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | {{ | + | Since triangles <math>P_1P_2P_3</math> and <math>PP_2P_3</math> share the base <math>P_2Q_2</math>, we have <math>\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{P_1Q_1}{PQ_1}</math>, where <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Similarly, <math>\frac{[PP_1P_3]}{[P_1P_2P_3]}=\frac{P_2Q_2}{PQ_2}, \frac{[PP_1P_2]}{[P_1P_2P_3]}=\frac{P_3Q_3}{PQ_3}</math>. Adding all of these gives <math>\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{P_2Q_2}{PQ_2}+\frac{P_3Q_3}{PQ_3}+\frac{P_1Q_1}{PQ_1}</math>, or <math>\frac{P_2Q_2}{PQ_2}+\frac{P_3Q_3}{PQ_3}+\frac{P_1Q_1}{PQ_1}=\frac{[PP_1P_2]+[PP_2P_3]+[PP_3P_1]}{[P_1P_2P_3]}=1</math> |
+ | We see that we must have at least one of the three fractions less than or equal to <math>\frac{1}{3}</math>, and at least one greater than <math>\frac{1}{3}</math>. These correspond to ratios <math>\frac{PP_i}{PQ_i}</math> being less than or equal to <math>2</math>, and greater than or equal to <math>2</math>, respectively, so we are done. | ||
{{IMO box|year=1961|num-b=3|num-a=5}} | {{IMO box|year=1961|num-b=3|num-a=5}} |
Revision as of 01:23, 8 February 2009
Problem
In the interior of triangle a point is given. Let be the intersections of with the opposing edges of triangle . Prove that among the ratios there exists one not larger than and one not smaller than .
Solution
Since triangles and share the base , we have , where denotes the area of triangle . Similarly, . Adding all of these gives , or We see that we must have at least one of the three fractions less than or equal to , and at least one greater than . These correspond to ratios being less than or equal to , and greater than or equal to , respectively, so we are done.
1961 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |