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>. | + | 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>. |
Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>. | Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>. | ||
+ | |||
Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well. | Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well. | ||
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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. | 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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+ | ==Solution 2== | ||
+ | [[File:2010 IMO 2.png|330px|right]] | ||
+ | Let <math>A'</math> be A-excenter <math>\triangle ABC \implies</math> | ||
+ | <cmath>DI = DB = DC = DA', AB \cdot AC = AI \cdot AA'.</cmath> | ||
+ | |||
+ | <cmath>\angle BAF = \angle EAC, \angle ABF = \angle ABC = \angle AEC \implies \triangle ABF \sim \triangle AEC \implies</cmath> | ||
+ | |||
+ | <cmath>\frac {AB}{AF} = \frac {AE}{AC} \implies AF \cdot AE = AB \cdot AC = AI \cdot AA' \implies \frac {AF}{AA'} = \frac {AI}{AE}.</cmath> | ||
+ | |||
+ | <cmath>\angle FAA' = \angle IAE \implies \triangle FAA' \sim \triangle IAE \implies \angle FA'A = \angle IEA.</cmath> | ||
+ | |||
+ | <cmath>IG = GF, ID = DA' \implies GD || FA' \implies \angle GDA = \angle FA'A = \angle IEA \implies </cmath> | ||
+ | the intersection of lines <math>EI</math> and <math>DG</math> lies on <math>\Gamma</math>. | ||
+ | |||
+ | After <i><b>pavel kozlov</b></i> '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
== See Also == | == See Also == |
Latest revision as of 13:52, 16 July 2023
Contents
[hide]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.
Solution 2
Let be A-excenter
the intersection of lines and lies on .
After pavel kozlov vladimir.shelomovskii@gmail.com, vvsss
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 |