IMO-2006-G9

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Tahmid Hasan
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IMO-2006-G9

Unread post by Tahmid Hasan » Sun May 13, 2012 9:47 am

Points $A_1,B_1,C_1$ are chosen on the sides $BC,CA,AB$ of a triangle $ABC$ respectively.The circumcircles of triangles $AB_1C_1,BC_1A_1,CA_1B_1$ intersect the circumcircle of triangle $ABC$ again at points $A_2,B_2,C_2$ respectively.($A_2 \neq A,B_2 \neq B,C_2 \neq C$).Points $A_3,B_3,C_3$ are symmetric to $A_1,B_1,C_1$ with respect to the midpoints of the sides $BC,CA,AB$ respectively.Prove that the triangles
$A_2B_2C_2$ and $A_3B_3C_3$ are similar.
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Tahmid Hasan
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Re: IMO-2006-G9

Unread post by Tahmid Hasan » Sun May 13, 2012 9:49 am

Here's a sketch of my solution
Be careful of the orientation as all triangles in my solution are directly similar.
first we use lemma-3 Yufei Zhao to prove $A_2$ is the center of spiral simiarity that maps $C_1B_1$ to $BC$.
So $\triangle A_2C_1B_1 \sim \triangle A_2BC,\triangle A_2C_1B \sim \triangle A_2B_1C$
so $\frac {A_2B}{A_2C}=\frac {A_2C_1}{A_2B_1}=\frac {BC_1}{B_1C}=\frac {AC_3}{AB_3}$[The last argument follows from the symmetry]
since $\angle C_3AB_3=\angle C_1A_2B_1$[sine $A,B,C,A_2$ are cyclic]
we conclude $\triangle A_2BC \sim \triangle A_2C_1B_1 \sim \triangle A_2C_3B_3$
we can do the same reasoning with the other two circles.
Now $\angle A_3C_3B_3=180^{\circ}-\angle AC_3B_3-\angle BC_3A_3$
$=180^{\circ}-\angle A_2BC-\angle B_2AC=180^{\circ}-\angle A_2AC-\angle B_2AC=\angle A_2C_2B_2$
We can do the same with the other angles and prove $A_2B_2C_2 \sim A_3B_3C_3$
[N:B:I shouldn't take much credit as the crux move was shown to me by Sourav vaiya :oops:.
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