USA TST 2007

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FahimFerdous
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USA TST 2007

Unread post by FahimFerdous » Sun Apr 15, 2012 1:08 pm

Triangle $ABC$ is inscribed in circle $W$. The tangent lines to $W$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS$ is perpendicular to $AT$. points $B_1$ and $C_1$ lies on ray $ST$ (with $C_1$ in between $B_1$ and $S$) such that $B_1T$=$BT$=$C_1T$. Prove that triangles $ABC$ and $AB_1C_1$ is similar.
Last edited by FahimFerdous on Sun Apr 15, 2012 1:25 pm, edited 1 time in total.
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*Mahi*
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Re: USA TST 2007

Unread post by *Mahi* » Sun Apr 15, 2012 1:16 pm

FahimFerdous wrote: such that $B_1T$=$BT$=$CT_1$.
Shouldn't that be $C_1T$? Please clarify.
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FahimFerdous
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Re: USA TST 2007

Unread post by FahimFerdous » Sun Apr 15, 2012 1:26 pm

Sorry, edited. :)
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Re: USA TST 2007

Unread post by *Mahi* » Sun Apr 15, 2012 2:19 pm

Hint:
Lemma 3 in "Lemmas in Euclidean Geometry By Yufei Zhao"
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Re: USA TST 2007

Unread post by Phlembac Adib Hasan » Fri May 18, 2012 4:52 pm

Here is my proof.Please somebody check it if there is any mistake.(As it is in my nature)
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Re: USA TST 2007

Unread post by *Mahi* » Sat May 19, 2012 12:39 am

It seems quite okay to me, but you should try reading http://yufeizhao.com/olympiad/cyclic_quad.pdf
You can use the lemmas found there to prove it much easily.
$A$ becomes the miquel point of cyclic $BB_1C_1C$ :)
(Hint to the shortest proof {provided you gave the link to cyclic quads})
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tanmoy
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Re: USA TST 2007

Unread post by tanmoy » Thu Oct 01, 2015 12:20 pm

$\text{Another solution}$:
Let $M$ be the midpoint of the line $BC$.By Zhao's lemma,se know that $AT$ is a symmedian of $\Delta ABC$.So,$\angle BAT=\angle CAM$,$\angle TBA=\pi-\angle ACB$.$\frac{BT} {TA}=\frac{C_{1}T} {TA}=\frac{CM}{MA}$.Now,$\angle TMS=\angle TAS=\frac{\pi }{2}$.
So,$T,M,A,S$ are concyclic.$\therefore \angle AMS=\angle AMC=\angle ATS=\angle ATC_{1}$.
$\therefore \Delta AMC \sim ATC_{1}$...........(1)
Again,$\frac{B_{1}T} {TA}=\frac{BM} {MA}$ and $ \angle AMB=\pi-\angle AMC=\pi-\angle ATS=\angle ATB_{1}$.
$\therefore \Delta AMB \sim ATB_{1}$......(2)
Combining (1) and (2),we get that $\angle BAC=\angle BA_{1}C_{1},\angle ABC=\angle AB_{1}C_{1},\angle ACB=\angle AC_{1}B_{1}$.
$\therefore \Delta ABC \sim AB_{1}C_{1}$ :D .
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