Difference between revisions of "1995 IMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
Draw <math>AE</math> and <math>BD</math> to make equilateral <math>\triangle EFA</math> and <math>\triangle BCD</math>, and draw points <math>I</math> and <math>J</math> such that <math>IA=IB</math>, <math>JD=JE</math>, directed angle <math>\measuredangle IAB=-\measuredangle CDB</math>, and directed angle <math>\measuredangle JDE=-\measuredangle FAE</math> to make equilateral <math>\triangle AIB</math> and <math>\triangle DJE</math>. Notice that <math>G</math> is on the circumcircle of <math>\triangle AIB</math> and <math>H</math> is on the circumcircle of <math>\triangle DJE</math>. By Ptolemy, <math>GA+GB=GI</math> and <math>HD+HE=HJ</math>. So, <cmath>AG+GB+GH+DH+HE=IG+GH+HJ.</cmath> Notice that octagon <math>AIBCDJEF</math> is symmetric about <math>\overline{BE}</math>. So, <math>IG+GH+HJ\ge IJ=CF</math>. --tigerzhang | Draw <math>AE</math> and <math>BD</math> to make equilateral <math>\triangle EFA</math> and <math>\triangle BCD</math>, and draw points <math>I</math> and <math>J</math> such that <math>IA=IB</math>, <math>JD=JE</math>, directed angle <math>\measuredangle IAB=-\measuredangle CDB</math>, and directed angle <math>\measuredangle JDE=-\measuredangle FAE</math> to make equilateral <math>\triangle AIB</math> and <math>\triangle DJE</math>. Notice that <math>G</math> is on the circumcircle of <math>\triangle AIB</math> and <math>H</math> is on the circumcircle of <math>\triangle DJE</math>. By Ptolemy, <math>GA+GB=GI</math> and <math>HD+HE=HJ</math>. So, <cmath>AG+GB+GH+DH+HE=IG+GH+HJ.</cmath> Notice that octagon <math>AIBCDJEF</math> is symmetric about <math>\overline{BE}</math>. So, <math>IG+GH+HJ\ge IJ=CF</math>. --tigerzhang | ||
+ | |||
+ | ==See Also== | ||
+ | {{IMO box|year=1995|num-b=4|num-a=6}} |
Latest revision as of 20:24, 4 July 2024
Problem
Let be a convex hexagon with and , such that . Suppose and are points in the interior of the hexagon such that . Prove that .
Solution
Draw and to make equilateral and , and draw points and such that , , directed angle , and directed angle to make equilateral and . Notice that is on the circumcircle of and is on the circumcircle of . By Ptolemy, and . So, Notice that octagon is symmetric about . So, . --tigerzhang
See Also
1995 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |