Difference between revisions of "2010 IMO Problems/Problem 2"

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''Authors: Tai Wai Ming and Wang Chongli, Hong Kong''
 
''Authors: Tai Wai Ming and Wang Chongli, Hong Kong''
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== Solution ==
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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.
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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 ==
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{{IMO box|year=2010|num-b=1|num-a=3}}
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[[Category:Olympiad Geometry Problems]]

Latest revision as of 18:50, 11 November 2015

Problem

Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE< \frac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the intersection of lines $EI$ and $DG$ lies on $\Gamma$.

Authors: Tai Wai Ming and Wang Chongli, Hong Kong

Solution

Note that it suffices to prove alternatively that if $EI$ meets the circle again at $J$ and $JD$ meets $IF$ at $G$, then $G$ is the midpoint of $IF$. Let $JD$ meet $BC$ at $K$.

Observation 1. D is the midpoint of arc $BDC$ because it lies on angle bisector $AI$.

Observation 2. $AI$ bisects $\angle{FAE}$ as well.

Key Lemma. Triangles $DKI$ and $DIJ$ are similar. Proof. Because triangles $DKB$ and $DBJ$ are similar by AA Similarity (for $\angle{KBD}$ and $\angle{BJD}$ both intercept equally sized arcs), we have $BD^2 = BK \cdot BJ$. But we know that triangle $DBI$ is isosceles (hint: prove $\angle{BID} = \angle{IBD}$), and so $BI^2 = BK \cdot BJ$. Hence, by SAS Similarity, triangles $DKI$ and $DIJ$ are similar, as desired.

Observation 3. As a result, we have $\angle{KID} = \angle{IJD} = \angle{DAE} = \angle{FAD}$.

Observation 4. $IK // AF$.

Observation 5. If $AF$ and $JD$ intersect at $L$, then $AJLI$ is cyclic.

Observation 6. Because $\angle{ALI} = \angle{AJE} = \angle{AJC} + \angle{CJE} = \angle{B} + \angle{AEC} = \angle{B} + \angle{BAF} = \angle{AFC}$, we have $LI // FK$.

Observation 7. $LIKF$ is a parallelogram, so its diagonals bisect each other, so $G$ is the midpoint of $FI$, 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
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