Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 IMO Problems/Problem 1"

(→‎Solution)
Line 16: Line 16:
  
 
Join points <math>E</math> and <math>M</math>, <math>EM</math> is perpendicular on <math>PQ</math> (why?), previously we proved <math>MP = MQ</math>, hence <math>\triangle{EPQ}</math> is isoscles and <math>EP = EQ</math> .
 
Join points <math>E</math> and <math>M</math>, <math>EM</math> is perpendicular on <math>PQ</math> (why?), previously we proved <math>MP = MQ</math>, hence <math>\triangle{EPQ}</math> is isoscles and <math>EP = EQ</math> .
 +
 +
==See Also==
 +
 +
{{IMO box|year=2000|after=First Question|num-a=2}}

Revision as of 23:03, 18 November 2023


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 it's 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 $PQ$.

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 $EP = EQ$ .

See Also

2000 IMO (Problems) • Resources
Preceded by
[[2000 IMO Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]] 		1 • 2 • 3 • 4 • 5 • 6 Followed by
First Question
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_IMO_Problems/Problem_1&oldid=204474"