Posted before at: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=386935 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=489896 For a generalization see: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=325959

Quote: Hi, can you solve Romania TST 2010 Day 6 Q2 with barycentrics? Yup. Without inversion, too! Let $A = (1,0,0)$, $B=(0,1,0)$ and $C=(0,0,1)$ and define $a,b,c$ in the usual fashion. Then, we get \[ A_1 = (0:s-b:s-c) \] and its cyclic variants, as well as $I = (a:b:c)$. Let us calculate $\omega_A = (AIA_1)$ and its cyclic variants. Upon using the generic circle form $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)$ we find $u=0$ and the system \begin{align*} abc &= vb + wc \\ a(s-b)(s-c) &= v(s-b) + w(s-c) \end{align*} Solving, we find that $v = \frac{ac(s-c)(2b-s)}{s(b-c)}$ and $w = \frac{ab(s-b)(2c-s)}{s(c-b)}$. Symmetrically, so we obtain the equations: \begin{align*} \omega_A : 0 &= -a^2yz-b^2zx-c^2xy + (x+y+z)\left( \frac{ac(s-c)(2b-s)}{s(b-c)}y + \frac{ab(s-b)(2c-s)}{s(c-b)}z \right) \\ \omega_B : 0 &= -a^2yz-b^2zx-c^2xy + (x+y+z)\left( \frac{ba(s-a)(2c-s)}{s(c-a)}z + \frac{bc(s-c)(2a-s)}{s(a-c)}x \right) \\ \omega_C : 0 &= -a^2yz-b^2zx-c^2xy + (x+y+z)\left( \frac{cb(s-b)(2a-s)}{s(a-b)}x + \frac{ca(s-a)(2b-s)}{s(b-a)}y \right) \\ \end{align*} Now the radical axis $\ell_{AB}$ of $\omega_A$ and $\omega_B$ is given by subtracting the RHS's: \[ \frac{bc(s-c)(2a-s)}{s(c-a)}x + \frac{ac(s-c)(2b-s)}{s(b-c)}y + \left( \frac{ab(s-b)(2c-s)}{s(c-b)} - \frac{ba(s-a)(2c-s)}{s(c-a)} \right)z \] We can compute \begin{align*} \frac{ab(s-b)(2c-s)}{s(c-b)} - \frac{ba(s-a)(2c-s)}{s(c-a)} &= \frac{ab(2c-s)}{s} \left( \frac{s-b}{c-b} - \frac{s-a}{c-a} \right) \\ &= \frac{ab(2c-s)}{s} \frac{(s-b)(c-a) - (s-a)(c-b)}{(c-b)(c-a)} \\ &= \frac{ab(2c-s)}{s} \frac{-(a-b)(s-c)}{(c-b)(c-a)} \end{align*} So, the equation of the radical axis is just \[ \ell_{AB}: 0 = \frac{ca(s-c)(2a-s)}{s(c-a)} x + \frac{ca(s-c)(2b-s)}{s(b-c)}y + \frac{ab(s-c)(2c-s)(a-b)}{(b-c)(c-a)}z \] or, after multiplying through by some common factors: \[ \ell_{AB}: 0 = ca(2a-s)(b-c) x + ab(2b-s)(c-a) y + bc(2c-s)(a-b) z \] which is symmetric, so the equations for $\ell_{BC}$ and $\ell_{CA}$ are the same, and hence the conclusion holds.

In fact, with a little trick it is not even necessary to compute pairwise radical axes! mavropnevma wrote: Let $ABC$ be a scalene triangle, let $I$ be its incentre, and let $A_1$, $B_1$ and $C_1$ be the points of contact of the excircles with the sides $BC$, $CA$ and $AB$, respectively. Prove that the circumcircles of the triangles $AIA_1$, $BIB_1$ and $CIC_1$ have a common point different from $I$. Cezar Lupu & Vlad Matei Call the circles $\omega_A, \omega_B, \omega_C$ in the usual fashion. Now let $\omega_A$ meet $\odot(ABC)$ again at $A_2$; likewise define $B_2, C_2$. Lemma. Lines $\overline{AA_2}, \overline{BB_2}, \overline{CC_2}$ concur. (Proof) Suppose $u_A,v_A,w_A$ are parameters for which $$\omega_A \overset{\text{def}}{=} a^2yz+b^2zx+c^2xy=(u_Ax+v_Ay+w_Az)(x+y+z).$$Plugging $A=(1,0,0), I=(a:b:c), A_1=(0:s-c:s-b)$, we obtain $$\begin{cases} u_A=0 \\ abc=v_Ab+w_Ac \\ a(s-b)(s-c)=v_A(s-b)+w_A(s-c). \end{cases}$$Solving, we obtain $v_A=\frac{c(s-c)(2b-s)}{s(b-c)}$ while $w_A=\frac{b(s-b)(2c-s)}{s(c-b)}$ so $$\overline{AA_2} \overset{\text{def}}{=} \frac{y}{z}=\frac{-v_A}{w_A}=\frac{c(s-c)(s-2b)}{b(s-b)(s-2c)} $$because $\overline{AA_2}$ is the radical axis of $\odot(ABC), \omega_A$. Consequently, the claim follows by Ceva's Theorem. $\blacksquare$ Now let $P$ be the concurrency point. Then $$\text{Pow}(P, \omega_A)=\overline{PA} \cdot \overline{PA_2}=OP^2-R^2$$which is symmetric in all three variables; hence proving $\overline{PI}$ is the common radical axis. $\blacksquare$ Remark. The only reason I could successfully solve this problem was because the lemma is way nicer. It'd probably take a miracle for me to pull off a direct computation

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/cercles%20coaxiaux2.pdf p. 21-25. Sincerely Jean-Louis

I have almost literally the exact same solution as v_Enhance, but I'll post it here for storage anyways.

v_Enhance wrote: Quote: Hi, can you solve Romania TST 2010 Day 6 Q2 with barycentrics? Yup. Without inversion, too! . How does one use inversion? Inverting about the incircle seems to send $A_1,B_1,C_1$ to ugly points...

Here is a synthetic solution. Let $I_A, I_B, I_C$ be the excenters of $\triangle ABC$. Consider the inversion about $I$ with power $-II_A\cdot IA$. It swaps $(A,I_A)$, $(B,I_B)$, $(C,I_C)$. Let $X$ be the image of $A_1$. Notice that $II_BI_CX$ and $II_AA_1X$ are cyclic quadrilateral. Now, invert at $I_A$ with power $I_AI\cdot I_AA$. This inversion swaps $(I,A)$, $(C,I_B)$, $(B,I_C)$ thus it swaps $\odot(II_BI_C)\mapsto\odot(ABC)$. However, $A_1$ maps to $I_A$-antipode, $I_A'$ of $\odot(I_AI_BI_C)$. Thus if $M_A,M_B,M_C$ are midpoints of $I_BI_C$, $I_CI_A$, $I_AI_B$, then $\odot(II_AA_1)\mapsto IM_A$. Therefore $X$ maps to $X'=IM_A\cap \odot(ABC)$. Since $I_A,X,X'$ are colinear, it suffices to show that $I_AX', I_BY', I_CZ'$ are concurrent. This actually the special case of USA TST 2015 P6 (applied on $\triangle I_AI_BI_C$) so we are done.

Bary wrt ABC. Let $A=-a+b+c$, $B=a-b+c$, $C=a+b-c$, $S=a+b+c$. Obviously $A=(0:B:C)$ and $I=(a:b:c)$. In $(AIA_1)$, we have $u=0$ and $$bv+cw=abc, Bv+Cw=\frac{1}{2}aBC.$$Adding two times the first equation to the second and dividing by $S$, we get $$v+w=\frac{a}{2S}(4bc+BC)$$and upon subtracting $c$ times this equation from the first equation we get $$v=\frac{ac}{2(b-c)S}(2Sb-4bc-BC)=\frac{ac}{2(b-c)S}(2bC-BC)=\frac{Cac(2b-B)}{2(b-c)S}$$Getting similar values of $u,v,w$ for $(BIB_1)$ by symmetry, we get the radical axis of $(AIA_1)$ and $(BIB_1)$ as $$\left(\frac{bc(2a-A)C}{2(a-c)S}\right)x+\left(\frac{ac(2b-B)C}{2(c-b)S}\right)y+\left(\frac{ab(2c-C)}{2S}(\frac{A}{c-a}-\frac{B}{c-b})\right)z=0$$or $$\frac{C}{2S(a-c)(c-b)}\left(bc(2a-A)(c-b)x+ac(2b-B)(a-c)y+ab(2c-C)(b-a)z\right)=0.$$Because $C>0$ by triangle inequality the first factor is clearly nonzero, and so after dividing out, the second factor is symmetric in $a,b,$ and $c$, done.