Difference between revisions of "1961 IMO Problems/Problem 4"
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In the interior of [[triangle]] <math>P_1P_2P_3</math> a [[point]] <math>P</math> is given. Let <math>Q_1,Q_2,Q_3</math> be the [[intersection]]s of <math>PP_1, PP_2,PP_3</math> with the opposing [[edge]]s of triangle <math>P_1P_2P_3</math>. Prove that among the [[ratio]]s <math>\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}</math> there exists one not larger than <math>2</math> and one not smaller than <math>2</math>. | In the interior of [[triangle]] <math>P_1P_2P_3</math> a [[point]] <math>P</math> is given. Let <math>Q_1,Q_2,Q_3</math> be the [[intersection]]s of <math>PP_1, PP_2,PP_3</math> with the opposing [[edge]]s of triangle <math>P_1P_2P_3</math>. Prove that among the [[ratio]]s <math>\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}</math> there exists one not larger than <math>2</math> and one not smaller than <math>2</math>. | ||
− | ==Solution== | + | ==Solution 1== |
− | + | Let <math>[ABC]</math> denote the area of triangle <math>ABC</math>. | |
+ | Since triangles <math>P_1P_2P_3</math> and <math>PP_2P_3</math> share the base <math>P_2P_3</math>, we have <math>\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_1}{P_1Q_1}</math>. | ||
+ | |||
+ | Similarly, <math>\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}</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{PQ_2}{P_2Q_2}+\frac{PQ_3}{P_3Q_3}+\frac{PQ_1}{P_1Q_1}=1</math>. | ||
+ | |||
+ | We see that we must have at least one of the three fractions not greater than <math>\frac{1}{3}</math>, and at least one not less 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. | ||
+ | ==Solution 2== | ||
+ | Let <math>K_1=[P_2PP_3], K_2=[P_3PP_1],</math> and <math>K_3=[P_1PP_2].</math> Note that by same base in triangles <math>P_2PP_3</math> and <math>P_2P_1P_3,</math> <cmath>\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}.</cmath> Thus, <cmath>\frac{P_1P}{PQ_1}=\frac{K_2+K_3}{K_1}</cmath><cmath>\frac{P_2P}{PQ_2}=\frac{K_1+K_3}{K_2}</cmath> <cmath>\frac{P_3P}{PQ_3}=\frac{K_1+K_2}{K_3}.</cmath> | ||
+ | Without loss of generality, assume <math>K_1\leq K_2\leq K_3.</math> Hence, <cmath>\frac{P_1P}{PQ_1}=\frac{K_2+K_3}{K_1}\geq \frac{K_1+K_1}{K_1}=2</cmath> and <cmath>\frac{P_3P}{PQ_3}=\frac{K_1+K_2}{K+3}\leq \frac{K_3+K_3}{K_3}=2,</cmath> as desired. <math>\blacksquare</math> | ||
{{IMO box|year=1961|num-b=3|num-a=5}} | {{IMO box|year=1961|num-b=3|num-a=5}} |
Latest revision as of 20:04, 5 December 2018
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 1
Let denote the area of triangle .
Since triangles and share the base , we have .
Similarly, .
Adding all of these gives .
We see that we must have at least one of the three fractions not greater than , and at least one not less than . These correspond to ratios being less than or equal to , and greater than or equal to , respectively, so we are done.
Solution 2
Let and Note that by same base in triangles and Thus, Without loss of generality, assume Hence, and as desired.
1961 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |