2000 IMO Problems/Problem 1

Problem

Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$ . Let $AB$ be the line tangent to these circles at $A$ and $B$ , respectively, so that $M$ lies closer to $AB$ than $N$ . Let $CD$ be the line parallel to $AB$ and passing through the point $M$ , with $C$ on $G_1$ and $D$ on $G_2$ . Lines $AC$ and $BD$ meet at $E$ ; lines $AN$ and $CD$ meet at $P$ ; lines $BN$ and $CD$ meet at $Q$ . Show that $EP=EQ$ .

Solution

 Given a triangle,  and a point  in it's interior, assume that the circumcircles of  and     are tangent to . Prove that ray  bisects .
 Let the intersection of  and  be . By power of a point,  and , so .

$\textbf{Proof of problem:}$ Let ray $NM$ intersect $AB$ at $X$ . By our lemma, $\textit{(the two circles are tangent to AB)}$ , $X$ bisects $AB$ . Since $\triangle{NAX}$ and $\triangle{NPM}$ are similar, and $\triangle{NBX}$ and $\triangle{NQM}$ are similar implies $M$ bisects $PQ$ .

By simple parallel line rules, $\angle{ABM}=\angle{EBA}$ . Similarly, $\angle{BAM}=\angle{EAB}$ , so by $\textit{ASA}$ criterion, $\triangle{ABM}$ and $\triangle{EAB}$ are congruent.

Now, $\angle{EBA} = \angle{ABM} = \angle{BDM}$ and $\angle{ABM} = \angle{BMD}$ since $CD$ is parallel to $AB$ . But $AB$ is tangent to the circumcircle of $\triangle{BMD}$ hence $\angle{ABM} = \angle{BDM}$ and that implies $\angle{BMD} = \angle{BDM} .$ So $\triangle{BMD}$ is isosceles and $\angle{BMD}=\angle{BDM}$ .

Join points $E$ and $M$ , $EM$ is perpendicular on $PQ$ (why?), previously we proved $MP = MQ$ , hence $\triangle{EPQ}$ is isoscles and $EP = EQ$ .

2000 IMO (Problems) • Resources
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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2000 IMO Problems/Problem 1

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