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2000 IMO Problems/Problem 1

Revision as of 14:12, 21 June 2024 by Alexanderruan (talk | contribs) (→‎Solution)


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

$\textbf{Lemma:}$ Given a triangle, $ABC$ and a point $P$ in its interior, assume that the circumcircles of $\triangle{ACP}$ and    $\triangle{ABP}$ are tangent to $BC$. Prove that ray $AP$ bisects $BC$.
$\textbf{Proof:}$ Let the intersection of $AP$ and $BC$ be $D$. By power of a point, $BD^2=AD(PD)$ and $CD^2=AD(PD)$, so $BD=CD$.

$\textbf{Proof of problem:}$ 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 $PQ$.

Now, $\angle{ABM} = \angle{BMD}$ 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 $BM=BD$.

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

We know that$ (Error compiling LaTeX. Unknown error_msg)BE=BM=BD$so a circle with diameter$ED$can be circumscribed around$\triangle{EMD}$. Join points$E$and$M$,$EM$is perpendicular on$PQ$, previously we proved$MP = MQ$, hence$\triangle{EPQ}$is isoscles and$EP = EQ$ .

See Also

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