2010 IMO Problems/Problem 2

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 $IF$ .

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.

2010 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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2010 IMO Problems/Problem 2

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