Difference between revisions of "2010 IMO Problems/Problem 2"
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== Solution == | == Solution == | ||
− | {{ | + | Note that it suffices to prove alternatively that if <math>EI</math> meets the circle again at <math>J</math> and <math>JD</math> meets <math>IF</math> at <math>G</math>, then <math>G</math> is the midpoint of <math>IF</math>. Let <math>JD</math> meet <math>BC</math> at <math>K</math>. |
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+ | Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>. | ||
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+ | Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well. | ||
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+ | Key Lemma. Triangles <math>DKI</math> and <math>DIJ</math> are similar. | ||
+ | Proof. Because triangles <math>DKB</math> and <math>DBJ</math> are similar by AA Similarity (for <math>\angle{KBD}</math> and <math>\angle{BJD}</math> both intercept equally sized arcs), we have <math>BD^2 = BK \cdot BJ</math>. But we know that triangle <math>DBI</math> is isosceles (hint: prove <math>\angle{BID} = \angle{IBD}</math>), and so <math>BI^2 = BK \cdot BJ</math>. Hence, by SAS Similarity, triangles <math>DKI</math> and <math>DIJ</math> are similar, as desired. | ||
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+ | Observation 3. As a result, we have <math>\angle{KID} = \angle{IJD} = \angle{DAE} = \angle{FAD}</math>. | ||
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+ | Observation 4. <math>IK // AF</math>. | ||
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+ | Observation 5. If <math>AF</math> and <math>JD</math> intersect at <math>L</math>, then <math>AJLI</math> is cyclic. | ||
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+ | Observation 6. Because <math>\angle{ALI} = \angle{AJE} = \angle{AJC} + \angle{CJE} = \angle{B} + \angle{AEC} = \angle{B} + \angle{BAF} = \angle{AFC}</math>, we have <math>LI // FK</math>. | ||
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+ | Observation 7. <math>LIKF</math> is a parallelogram, so its diagonals bisect each other, so <math>G</math> is the midpoint of <math>FI</math>, as desired. | ||
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== See Also == | == See Also == | ||
{{IMO box|year=2010|num-b=1|num-a=3}} | {{IMO box|year=2010|num-b=1|num-a=3}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 17:50, 11 November 2015
Problem
Given a triangle , with as its incenter and as its circumcircle, intersects again at . Let be a point on arc , and a point on the segment , such that . If is the midpoint of , prove that the intersection of lines and lies on .
Authors: Tai Wai Ming and Wang Chongli, Hong Kong
Solution
Note that it suffices to prove alternatively that if meets the circle again at and meets at , then is the midpoint of . Let meet at .
Observation 1. D is the midpoint of arc because it lies on angle bisector .
Observation 2. bisects as well.
Key Lemma. Triangles and are similar. Proof. Because triangles and are similar by AA Similarity (for and both intercept equally sized arcs), we have . But we know that triangle is isosceles (hint: prove ), and so . Hence, by SAS Similarity, triangles and are similar, as desired.
Observation 3. As a result, we have .
Observation 4. .
Observation 5. If and intersect at , then is cyclic.
Observation 6. Because , we have .
Observation 7. is a parallelogram, so its diagonals bisect each other, so is the midpoint of , as desired.
See Also
2010 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |