63. A nice problem with radical center (livetolove212).

by Virgil Nicula, Jul 17, 2010, 9:28 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=298310
livetolove212 wrote:
Given triangle $ ABC$ with circumcircle $ (O)$ and incircle $ (I)$ which touches $ BC, CA, AB$ at $ X, Y, Z$, respectively. Denote the midpoints $ M, N, P$ of $ BC, CA, AB$ respectively and $\{A_1, A_2\}=NP\cap (O)$ . Similar for $ B_1$ , $B_2$ , $ C_1$ , $C_2$ . Prove that $ I$ is the radical center of the circumcircles of the triangles $ (XA_1A_2)$ , $(YB_1B_2)$ , $(ZC_1C_2)$ .

Proof (metric). Denote the circumcircle $w_a=\mathrm C(P,\rho )$ of $\triangle A_1XA_2$ , the intersection $N\in A_1A_2\cap OM$ and $NP=x$ , the projection $R$ of $I$ on $OM$ and $S\in OM\cap (O)$ so that the sideline $BC$ separates $A$ , $S$ .

$\blacktriangleright$ $A_1A_2^2=4\left(OA_1^2-ON^2\right)=$ $4\left[R^2-\left(\frac {h_a}{2}-R\cos A\right)^2\right]=$ $4\left(R^2\sin^2A-\frac {h_a^2}{4}+h_aR\cos A\right)=$ $a^2-h_a^2+4h_aR\cos A=$ $a^2-h_a^2+b^2+c^2-a^2$ $\implies$ $\boxed {\ A_1A_2^2=b^2+c^2-h_a^2\ }$ .

$\blacktriangleright$ $\left\|\begin{array}{c} \rho^2=AP^2=AN^2+NP^2=\frac 14\cdot\left(b^2+c^2-h_a^2\right)+x^2\\\\ \rho^2=XP^2=XM^2+MP^2=\left(\frac {b-c}{2}\right)^2+\left(\frac {h_a}{2}+x\right)^2\end{array}\right\|\implies$ $4\rho^2-4x^2=b^2+c^2-h_a^2=(b-c)^2+h_a^2+4xh_a$ $\implies$

$\blacksquare\ x=\frac {bc-h_a^2}{2h_a}=\frac {2Rh_a-h_a^2}{2h_a}$ $\implies$ $\boxed {\ R=x+\frac {h_a}{2}\ }$ $\Longleftrightarrow$ $\underline{MP=R}$ , i.e. $\underline{OP=MS}\ \ \mathrm{!!}$
$\blacksquare$ Since $XM=\frac {|b-c|}{2}$ and $MP=R$ obtain $\rho^2=XM^2+MP^2$ $\implies$ $\boxed {\rho^2=\left(\frac {b-c}{2}\right)^2+R^2}$ .

$\blacktriangleright$ The power of $I$ w.r.t. the circle $w_a$ is $p_{w_a}(I)=IP^2-\rho^2=$ $XM^2+RP^2-\rho^2=$ $\left(\frac {b-c}{2}\right)^2+\left(\frac {h_a}{2}-r+x\right)^2-\left(\frac {b-c}{2}\right)^2-R^2=(R-r)^2-R^2$ , $\implies$ $\boxed {\ p_{w_a}(I)=-r(2R-r)\ }$ .

Since the expression of the power $p_{w_a}(I)$ is symmetrically w.r.t $\triangle ABC$ obtain the conclusion of this nice proposed problem.
This post has been edited 2 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:11 PM

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