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BdMO Online Forum • View topic - IMO 2000/1

IMO 2000/1

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IMO 2000/1

Post Number:#1  Unread postby tanmoy » Thu Sep 03, 2015 3:52 pm

Two circles $\Gamma _{1}$ and $\Gamma _{2}$ intersect at $M$ and $N$.Let $l$ be the common tangent to $\Gamma _{1}$ and $\Gamma _{2}$ so that $M$ is closer to $l$ than $N$ is.Let $l$ touch $\Gamma _{1}$ at $A$ and $\Gamma _{2}$ at $B$.Let the line through $M$ parallel to $l$ meet the circle $\Gamma _{1}$ again at $C$ and the circle $\Gamma _{2}$ again at $D$.Lines $CA$ and $DB$ meet at $E$;lines $AN$ and $CD$ meet at $P$;lines $BN$ and $CD$ meet at $Q$.Show that $EP=EQ$.
[Though this is an IMO problem,this is not so hard!]
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Re: IMO 2000/1

Post Number:#2  Unread postby Prosenjit Basak » Fri Sep 04, 2015 12:00 pm

As $AB || CD$, we get $\angle EBA = \angle EDC$.
Again by Alternate Segment Theorem, $\angle ABM = \angle BDM$.
So, $\angle EBA = \angle ABM$.
Similarly we get $\angle EAB = \angle BAM$.

So,we get $\triangle EBA \cong \triangle ABM$ by the ASA theorem.
Then, we get $EM \perp AB.$

Now, let the segment $NM$ intersect $AB$ at $F$.
Now, as $FA$ is tangent to $\Gamma_{1}$ we get, $FA^2 = FM.MN$.
Similarly we get $FB^2 = FM.MN$.
So, $FA = FB$.
Now as $PQ || AB$, we get $PM = MQ$.

So, finally as $\triangle EPM \cong \triangle EQM$ by the SAS theorem, we get $EP = EQ$.
Last edited by Prosenjit Basak on Fri Sep 04, 2015 11:11 pm, edited 2 times in total.
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Re: IMO 2000/1

Post Number:#3  Unread postby tanmoy » Fri Sep 04, 2015 2:59 pm

Prosenjit Basak wrote:So,we get $\triangle EBA \cong \triangle ABN$.

I think this is $\triangle EBA \cong \triangle ABM$.
BTW,I have got the same solution :)
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Re: IMO 2000/1

Post Number:#4  Unread postby Tariq Hasan Rizu » Fri Sep 04, 2015 9:13 pm

Prosenjit Basak wrote:As $AB || CD$, we get $\angle EBA = \angle EDC$.
Again by Alternate Segment Theorem, $\angle ABM = \angle BDM$.
So, $\angle EBA = \angle ABM$.
Similarly we get $\angle EAB = \angle BAM$.

So,we get $\triangle EBA \cong \triangle ABN$ by the ASA theorem.
Then, we get $EN \perp AB.$

Now, let the segment $NM$ intersect $AB$ at $F$.
Now, as $FA$ is tangent to $\Gamma_{1}$ we get, $FA^2 = FM.MN$.
Similarly we get $FB^2 = FM.MN$.
So, $FA = FB$.
Now as $PQ || AB$, we get $PM = MQ$.

So, finally as $\triangle EPM \cong \triangle EQM$ by the SAS theorem, we get $EP = EQ$.


But, when the area of $\Gamma_{1}$ and $\Gamma_{2}$ aren't equal, EN isn't perpendicular to AB
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Re: IMO 2000/1

Post Number:#5  Unread postby tanmoy » Fri Sep 04, 2015 10:56 pm

Tariq Hasan Rizu wrote:
Prosenjit Basak wrote:As $AB || CD$, we get $\angle EBA = \angle EDC$.
Again by Alternate Segment Theorem, $\angle ABM = \angle BDM$.
So, $\angle EBA = \angle ABM$.
Similarly we get $\angle EAB = \angle BAM$.

So,we get $\triangle EBA \cong \triangle ABN$ by the ASA theorem.
Then, we get $EN \perp AB.$

Now, let the segment $NM$ intersect $AB$ at $F$.
Now, as $FA$ is tangent to $\Gamma_{1}$ we get, $FA^2 = FM.MN$.
Similarly we get $FB^2 = FM.MN$.
So, $FA = FB$.
Now as $PQ || AB$, we get $PM = MQ$.

So, finally as $\triangle EPM \cong \triangle EQM$ by the SAS theorem, we get $EP = EQ$.


But, when the area of $\Gamma_{1}$ and $\Gamma_{2}$ aren't equal, EN isn't perpendicular to AB

I think he had made a mistake.It should be $EM \perp AB$.
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Re: IMO 2000/1

Post Number:#6  Unread postby Prosenjit Basak » Fri Sep 04, 2015 11:06 pm

Yeah, I meant $EM$. And I meant $\triangle ABM$ too. It was just a typo.I don't know what happened to my typing. :( I edited it though. Thanks for pointing out the mistakes. :)
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Re: IMO 2000/1

Post Number:#7  Unread postby Jon's Ghost » Thu Sep 10, 2015 12:15 pm

You guys missed the fact that the Power of Point Theorem actually shows that $FA^2 = FM.FN $.
Not $FA^2 = FM.MN$ :roll:
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