USA TST 2017

For discussing Olympiad level Geometry Problems
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ahmedittihad
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USA TST 2017

Unread post by ahmedittihad » Sun Jan 29, 2017 2:18 am

Let $ABC$ be an acute scalene triangle with circumcenter $O$, and let $T$ be on line $BC$ such that $\angle TAO = 90^{\circ}$. The circle with diameter $\overline{AT}$ intersects the circumcircle of $\triangle BOC$ at two points $A_1$ and $A_2$, where $OA_1 < OA_2$. Points $B_1$, $B_2$, $C_1$, $C_2$ are defined analogously.
Prove that $\overline{AA_1}$, $\overline{BB_1}$, $\overline{CC_1}$ are concurrent.
Prove that $\overline{AA_2}$, $\overline{BB_2}$, $\overline{CC_2}$ are concurrent on the Euler line of triangle $ABC$.
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Raiyan Jamil
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Re: USA TST 2017

Unread post by Raiyan Jamil » Sun Jan 29, 2017 1:41 pm

$\text{Solution of (1):}$
Obviously, $T$ is the point where $A$ tangent of $\bigodot ABC$ meets $BC$. Now, $\bigodot ABC$ and $\bigodot AT$ are orthogonal. Let $OT$ meet $\bigodot AT$ for the second time for $T_1$. So, $T$ and $T_1$ are inverses respect to $\bigodot ABC$.So, applying inverion respect to $\bigodot ABC$ gives that $BT_1OC$ is cyclic. Therefore $T_1\equiv A_1$. Now, $TA_1A=AA_1O=90^\circ$ and $TA$ tangent to $\bigodot ABC$ gives that $AA_1$ is the $A$-symmedian of $ABC$. Similarly $BB_1$ and $CC_1$ are also symmedians. So, they're concurrent at the symmedian point of $\bigtriangleup ABC$. :D
$\text{Solution of (2):}$
(2)Let, the inverse of $A_2$ be $A_3$. Since, $\bigodot AT$ and $\bigodot ABC$ are orthogonal, so $A_3$ lies on $\bigodot AT$. Again, Since $BOCA_2$ is cyclic, so, $A_3$ lies on $B_C$. Therefore we conclude that $A_3$ is just the feet of perpendicular of $A$ on $BC$. So, Showing $AA_2$, $BB_2$, $CC_2$ concurrent on the Euler line of $\bigtriangleup ABC$ is equivalent to showing that $\bigodot AA_3O$,$\bigodot BB_3O$,$\bigodot CC_3O$ are coaxal and their radical axis is the Euler line of $\bigtriangleup ABC$.

Now, let $H$ be the orthocentre of $\bigtriangleup ABC$. Again let the Euler line meets $\bigodot AA_3O$ for the second time at $X$. Then, applying Power of point we get, $HX\times HO=HA\times HA_3=HB\times HB_3=HC\times HC_3$ . Which gives $BB_3OX$,$CC_3OX$ are cyclic, as desired. :D
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rah4927
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Re: USA TST 2017

Unread post by rah4927 » Tue Jan 31, 2017 2:44 am

Hints for the solution.
First part:
Try to prove that $AA1$ is a symmedian.
Second part:
Find the inverse of $A2$ with respect to the inversion about the circumcircle of $\triangle ABC$. What property does it satisfy?
This was a rather bland problem for a TST.

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