Difference between revisions of "2014 IMO Problems/Problem 4"
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Points <math>P</math> and <math>Q</math> lie on side <math>BC</math> of acute-angled <math>\triangle{ABC}</math> so that <math>\angle{PAB}=\angle{BCA}</math> and <math>\angle{CAQ}=\angle{ABC}</math>. Points <math>M</math> and <math>N</math> lie on lines <math>AP</math> and <math>AQ</math>, respectively, such that <math>P</math> is the midpoint of <math>AM</math>, and <math>Q</math> is the midpoint of <math>AN</math>. Prove that lines <math>BM</math> and <math>CN</math> intersect on the circumcircle of <math>\triangle{ABC}</math>. | Points <math>P</math> and <math>Q</math> lie on side <math>BC</math> of acute-angled <math>\triangle{ABC}</math> so that <math>\angle{PAB}=\angle{BCA}</math> and <math>\angle{CAQ}=\angle{ABC}</math>. Points <math>M</math> and <math>N</math> lie on lines <math>AP</math> and <math>AQ</math>, respectively, such that <math>P</math> is the midpoint of <math>AM</math>, and <math>Q</math> is the midpoint of <math>AN</math>. Prove that lines <math>BM</math> and <math>CN</math> intersect on the circumcircle of <math>\triangle{ABC}</math>. | ||
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Revision as of 21:41, 28 February 2016
Problem
Points and lie on side of acute-angled so that and . Points and lie on lines and , respectively, such that is the midpoint of , and is the midpoint of . Prove that lines and intersect on the circumcircle of .
Solution 1
We are trying to prove that the intersection of and , call it point , is on the circumcircle of triangle . In other words, we are trying to prove . Let the intersection of and be point , and the intersection of and be point . Let us assume . Note: This is circular reasoning. If , then should be equal to and . We can quickly prove that the triangles , , and are similar, so . We also see that . Also because angles and and are equal, the triangles and , and must be two pairs of similar triangles. Therefore we must prove angles and and are equal. We have angles . We also have , . Because the triangles and are similar, we have , so triangles and are similar. So the angles and and are equal and we are done.
Solution 2
Let be the midpoint of . Easy angle chasing gives . Because is the midpoint of , the cotangent rule applied on triangle gives us Hence, by the cotangent rule on , we have Because the period of cotangent is , but angles are less than , we have
Similarly, we have Hence, if and intersect at , then by the Angle Sum in a Triangle Theorem. Hence, is cyclic, which is equivalent to the desired result.
--Suli 23:27, 7 February 2015 (EST)
Solution 3
Let be the midpoint of . By AA Similarity, triangles and are similar, so and . Similarly, , and so triangle is isosceles. Thus, , and so . Dividing both sides by 2, we have , or But we also have , so triangles and are similar by similarity. In particular, . Similarly, , so . In addition, angle sum in triangle gives . Therefore, if we let lines and intersect at , by Angle Sum in quadrilateral concave , and so convex , which is enough to prove that is cyclic. This completes the proof.
--Suli 10:38, 8 February 2015 (EST)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2014 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |