USAMO 2009/5

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tanmoy
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USAMO 2009/5

Unread post by tanmoy » Wed Feb 25, 2015 1:19 pm

Trapezoid $ABCD,$ with $\overline{AB}||\overline{CD}$, is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $\overline{AB}$ intersect $\overline{BD}$ and $\overline{BC}$ at points $R$ and $S$, respectively. Prove that quadrilateral $PQRS$ is cyclic if and only if $\overline{BG}$ bisects $\angle CBD$.
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Tahmid
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Re: USAMO 2009/5

Unread post by Tahmid » Wed Feb 25, 2015 11:24 pm

i have solved this . but my solution is too large :?
main part of my solution is to prove $DX$=$CY$ where $X=QR\cap w ; Y=PS\cap w$

tanmoy
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Re: USAMO 2009/5

Unread post by tanmoy » Thu Feb 26, 2015 3:25 pm

$\text{My proof}$:
$\text{If part}$:
Extend $QR$ to $T$.Join $B,T$.$BTRG$ is cyclic.We can also see that $PCSG$ is cyclic.We can prove that $T,G,C$ are collinear.$\therefore$ $\angle RQP+\angle PSR=180^{\circ}-\angle TBP+\angle PCT=180^{\circ}-\angle PCT+\angle PCT=180^{\circ}$.
$\text{Only if part}$:
The steps can be reversed. :)
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*Mahi*
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Re: USAMO 2009/5

Unread post by *Mahi* » Thu Feb 26, 2015 7:34 pm

Previously posted in IMO Marathon, you can see a few more solutions there -
viewtopic.php?p=13226#p13226
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tanmoy
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Re: USAMO 2009/5

Unread post by tanmoy » Thu Feb 26, 2015 8:43 pm

*Mahi* wrote:Previously posted in IMO Marathon, you can see a few more solutions there -
viewtopic.php?p=13226#p13226
Sorry for double posting.Actually I have searched but nothing was found.
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