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

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<math>\textbf{Proof of problem:}</math> Let ray <math>NM</math> intersect <math>AB</math> at <math>X</math>. By our lemma, <math>\textit{(the two circles are tangent to AB)}</math>, <math>X</math> bisects <math>AB</math>. Since <math>\triangle{NAX}</math> and <math>\triangle{NPM}</math> are similar, and <math>\triangle{NBX}</math> and <math>\triangle{NQM}</math> are similar implies <math>M</math> bisects <math>PQ</math>.
 
<math>\textbf{Proof of problem:}</math> Let ray <math>NM</math> intersect <math>AB</math> at <math>X</math>. By our lemma, <math>\textit{(the two circles are tangent to AB)}</math>, <math>X</math> bisects <math>AB</math>. Since <math>\triangle{NAX}</math> and <math>\triangle{NPM}</math> are similar, and <math>\triangle{NBX}</math> and <math>\triangle{NQM}</math> are similar implies <math>M</math> bisects <math>PQ</math>.
  
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By simple parallel line rules, <math>\angle{ABM}=\angle{EBA}</math>. Similarly, <math>\angle{BAM}=\angle{EAB}</math>, so by <math>\textit{ASA}</math> criterion, <math>\triangle{ABM}</math> and <math>\triangle{EAB}</math> are congruent.  
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By simple parallel line rules, <math>\angle{ABM}=\angle{EBA}</math>. Similarly, <math>\angle{BAM}=\angle{EAB}</math>, so by <math>\textit{ASA}</math> criterion, <math>\triangle{ABM}</math> and <math>\triangle{ABE}</math> are congruent.  
  
 
Now, <math>\angle{EBA} = \angle{ABM} = \angle{BDM}</math> and <math>\angle{ABM} = \angle{BMD}</math> since <math>CD</math> is parallel to <math>AB</math>. But <math>AB</math> is tangent to the circumcircle of <math>\triangle{BMD}</math> hence <math>\angle{ABM} = \angle{BDM}</math> and that implies <math>\angle{BMD} = \angle{BDM} . </math>So<math>\triangle{BMD}</math> is isosceles and <math>\angle{BMD}=\angle{BDM}</math> .
 
Now, <math>\angle{EBA} = \angle{ABM} = \angle{BDM}</math> and <math>\angle{ABM} = \angle{BMD}</math> since <math>CD</math> is parallel to <math>AB</math>. But <math>AB</math> is tangent to the circumcircle of <math>\triangle{BMD}</math> hence <math>\angle{ABM} = \angle{BDM}</math> and that implies <math>\angle{BMD} = \angle{BDM} . </math>So<math>\triangle{BMD}</math> is isosceles and <math>\angle{BMD}=\angle{BDM}</math> .
  
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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> .
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<math>BE=BM=BD</math> so a circle with diameter <math>ED</math> can be circumscribed around <math>\triangle{EMD}</math>. Join points <math>E</math> and <math>M</math>, <math>EM</math> is perpendicular on <math>PQ</math>, previously we proved <math>MP = MQ</math>, hence <math>\triangle{EPQ}</math> is isoscles and <math>EP = EQ</math> .
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=2000|before=First Question|num-a=2}}
 
{{IMO box|year=2000|before=First Question|num-a=2}}

Revision as of 14:04, 21 June 2024


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 $\triangle{ABE}$ 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 $\angle{BMD}=\angle{BDM}$ .

$BE=BM=BD$ 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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