Let $I_A$ be the $A$-excenter; we claim that $I_A, P,O$ are collinear. To prove this, let $R\equiv I_CE\cap AB, S\equiv I_BF\cap AC, T\equiv PI_A\cap BC$; we will show that lines $AT, BS, CR$, concur. To do this, we use barycentric coordinates. Notice that cevian $AT$ has equation $$\begin{bmatrix} 0 & y & z\\ a^2S_A & b^2S_B & c^2S_C \\ -a & b & c \end{bmatrix}=0$$$$\Longrightarrow -y\cdot ac(aS_A+cS_C)+z\cdot ab(aS_A+bS_B)=0=\Longrightarrow \frac{y}{z}=\frac{b(aS_A+bS_B)}{c(aS_A+cS_C)}$$Cevian $BS$ has equation $$\begin{bmatrix} x & 0 & z\\ S_B & S_A & 0 \\ a & -b & c \end{bmatrix}=0$$$$\Longrightarrow x\cdot cS_A-z(bS_B+aS_A)=0 \Longrightarrow \frac{x}{z}=\frac{bS_B+aS_A}{cS_A}$$By symmetry, we can derive that cevian $CR$ has equation $$\frac{x}{y}=\frac{cS_C+aS_A}{bS_A}$$so that $$\frac{y}{z}\times\frac{z}{x}\times\frac{x}{y}=\frac{b(aS_A+bS_B)}{c(aS_A+cS_C)}\times\frac{cS_A}{aS_A+bS_B}\times\frac{aS_A+cS_C}{bS_A}=1$$hence we get the result by Ceva's Theorem. Now by the Cevian Nest Theorem, we have that $I_AT, I_CE, I_BF$ concur, which implies that $P, I_A, O$ are collinear, as desired. Lemma: Let $DEF$ be the orthic triangle in triangle $ABC$. Put $X\equiv DF \cap AC, Y\equiv ED \cap BA, Z \equiv FE \cap CB$. Then $X,Y,Z$ are collinear on a line perpendicular to the Euler line. Proof: Let $\mathcal{N}, \Omega, \omega_{a}$ denote the nine-point circle, the circumcircle of $\triangle ABC$ and the circle with diameter $\overline{BC}$ (and its cyclic variations), respectively. Then $$XF\cdot XD=XA\cdot XC$$hence $X$ lies on the radical axis of $\mathcal{N}$ and $\Omega$. By symmetry, so do $Y$ and $Z$, which gives the collinearity. Then since the centers of $\mathcal{N}$ and $\Omega$ are the nine-point center and the circumcenter, we get the desired perpendicularity. Now, let $I$ be the incenter of $\triangle ABC$. Since $I$ is the orthocenter of $\triangle I_AI_BI_C$, it follows that $I_A$ is the orthocenter of $\triangle II_BI_C$. Additionally, these two triangles share a nine-point circle, hence $O$ is the nine-point center of $\triangle II_BI_C$, which implies that $OI_A$ is the Euler line. Then since $XY$ is the orthic axis of $\triangle I_AI_BI_C$, we get that $OI_A\perp XY$ by our lemma, finishing the problem.$\:\blacksquare\:$

This problem was jointly written by me and Telv Cohl. Let $I_A$ denote the $A$-excenter and $I$ the incenter. Then let $D$ denote the foot of the altitude from $A$. Suppose the $A$-excircle is tangent to $\overline{BC}$, $\overline{AB}$, $\overline{AC}$ at $A_1$, $B_1$, $C_1$ and let $A_2$, $B_2$, $C_2$ denote the reflections of $I_A$ across these points. Let $S$ denote the circumcenter of $\triangle II_BI_C$. We begin with the following observation: points $D$, $I$, $A_2$ are collinear, as are points $E$, $I_C$, $C_2$ are collinear and points $F$, $I_B$, $B_2$ are collinear. This follows from the ``midpoints of altitudes'' lemma. Observe that $\overline{B_2C_2} \parallel \overline{B_1C_1} \parallel \overline{I_BI_C}$. Proceeding similarly on the other sides, we discover $\triangle II_BI_C$ and $\triangle A_2B_2C_2$ are homothetic. Hence $P$ is the center of this homothety (in particular, $D$, $I$, $P$, $A_2$ are collinear). Moreover, $P$ lies on the line joining $I_A$ to $S$, which is the Euler line of $\triangle II_BI_C$, so it passes through the nine-point center of $\triangle II_BI_C$, which is $O$. Consequently, $P$, $O$, $I_A$ are collinear as well. To finish, we need only prove that $\overline{OS} \perp \overline{YZ}$. In fact, we claim that $\overline{YZ}$ is the radical axis of the circumcircles of $\triangle ABC$ and $\triangle II_BI_C$. Actually, $Y$ is the radical center of these two circumcircles and the circle with diameter $\overline{II_B}$ (which passes through $A$ and $C$). Analogously $Z$ is the radical center of the circumcircles and the circle with diameter $\overline{II_C}$, and the proof is complete.

Attachments:

Darjn if only I knew this way written by vEnhance. I wouldn't have bashed.

Let $I$ and $I_A$ respectively be the incenter and $A$-excenter of $ABC$. The main claim is that $I_A$, $O$, and $P$ are collinear. To prove this, we will show by Trig Ceva on $I_AI_BI_C$ that $I_AO$, $I_BF$, and $I_CE$ concur. The distance from $O$ to $I_AI_B$ is $CO\sin\angle OCI_A=R\sin\left(90^\circ-A+90^\circ-\frac{C}{2}\right)=R\sin\left(A+\frac{C}{2}\right)$ and similarly the distance from $O$ to $I_AI_C$ is $R\sin\left(A+\frac{B}{2}\right)$. Hence, $\frac{\sin\angle OI_AI_B}{\sin\angle OI_AI_C}=\frac{\sin\left(A+\frac{C}{2}\right)}{\sin\left(A+\frac{B}{2}\right)}$. The distance from $F$ to $I_BI_C$ is $AF\sin\angle FAI_C=AF\sin\left(90^\circ-\frac{A}{2}\right)=AC\cos A\cos\frac{A}{2}$, and the distance from $F$ to $I_AI_B$ is $CF\sin\angle FCI_B=CF\sin\left(90^\circ-A+90^\circ-\frac{C}{2}\right)=CF\sin\left(A+\frac{C}{2}\right)=AC\sin A\sin\left(A+\frac{C}{2}\right)$, so $\frac{\sin\angle FI_BI_C}{\sin\angle FI_BI_A}=\frac{\cos A\cos\frac{A}{2}}{\sin A\sin\left(A+\frac{C}{2}\right)}$. Similarly we have that $\frac{\sin\angle EI_CI_A}{\sin\angle EI_CI_B}=\frac{\sin A\sin\left(A+\frac{B}{2}\right)}{\cos A\cos\frac{A}{2}}$. Multiplying the three ratios together clearly gives 1. Now, note that it suffices to prove that $I_AO\perp YZ$, which is a pretty famous problem. The idea is that by Power of a Point on cyclic quadriateral $IAI_BC$, $YA\cdot YC=YI\cdot YI_B$, so $Y$ lies on the radical axis of the circumcircles of $ABC$ and $II_BI_C$. $Z$ also has this property, so $YZ$ is the radical axis of those two circles. Since the circumcenter of $II_BI_C$, $O'$, is the reflection of the circumcenter of $I_AI_BI_C$, $Be$, over $I_BI_C$, we have that $I_AIO'Be$ is a parallelogram, so $I_AO'$ passes through the midpoint of $IBe$, which is $O$ by Euler line properties.

I see there are very similar solutions to mine, but anyway(I am little bit late ): Lemma 1(well known): Statement:If $\triangle XYZ$ is orthic triangle of $\triangle ABC$($X\in BC$, $Y\in AC$ and $Z\in AB$), their perspectrix(orthic axis) is perpendicular to $\textit{Euler}$ line of $\triangle ABC$. Proof: Let $\Gamma$ be circumcircle of $\triangle ABC$ and $\Gamma_1$ circumcircle of $\triangle XYZ$. Since circumcenter of $\Gamma_1$ is on $OH$, we get that $\textit{Euler}$ line of $\triangle ABC$ is perpendicular to radical axis of $\Gamma$ and $\Gamma_1$. Let $AB\cap YZ=\{ X'\}$, since $ABYZ$ is cyclic, by power of a point we get $X'A\cdot X'B=X'Y\cdot X'Z$, but since $Y,Z\in \Gamma_1$ and $A,B\in \Gamma$, we get that $X'$ is on radical axis of $\Gamma$ and $\Gamma_1$, similarly we get same for others. Lemma 2: Statement: If we denote everything same as in the main problem and let $I_A$ be $A$-excenter, then $I_AO\perp ZY$. Proof: It's obvious that $AI_A\perp I_CI_B$, similarly we get $I_BB\perp I_CI_A$ and $I_CC\perp I_AI_B\Longrightarrow$ $I$ is orthocenter of $\triangle I_AI_BI_C$ and since circumcircle $\omega$ of $\triangle ABC$ passes through foot of altitudes it follows that $\omega$ is $\textit{Euler}$ circle of $\triangle I_AI_BI_C$. Now we have that $I_A$ is orthocenter of $\triangle I_BI_CI$ and so $\omega$ is $\textit{Euler}$ circle of $\triangle I_BI_CI$ as well, thus $OI_A$ is $\textit{Euler}$ line of $\triangle I_BI_CI$ as well, and by Lemma 1 we get conclusion. ____________________________________________________________________________ Back to the main problem: By Lemma 2 we are left proving $I_A$, $O$ and $P$ are collinear. Then by Cevian Nest Theorem on we get desired. $QED$

PS

BTW

How difficult was this year's #3?

After experimenting with it during the test, I disagree with the idea that it was nice for bary. The coordinates of $P$ seem to be surprisingly gross with not a lot simplifying. It's probably doable, but under olympiad conditions it would be tough.

infiniteturtle wrote: After experimenting with it during the test, I disagree with the idea that it was nice for bary. The coordinates of $P$ seem to be surprisingly gross with not a lot simplifying. It's probably doable, but under olympiad conditions it would be tough. Couldn't agree more.. Trust me

I worked on it for two hours with complex and bary. P is incredibly nasty.

You're right about doing whole problem using Barycentrics(it's probably ugly...)), but for proving lemma that I used, or second part of problem, it is nice...

Considering that v_Enhance basically showed everyone how to bary bash almost any geometry problem, he would make a problem that is not bary-bashable

Congratulations to $Evan$ and $Telv$ for this probably very nice problem!

Barycentric: Let $f(a,b,c)=a^3+b^3+c^3-a^2b-ab^2-a^2c-ca^2-b^2c-bc^2$. The barycentric coordinates of $P$ are \[ (-af(a,b,c), bf(a,b,-c),cf(a,-b,c)). \]Let $S_a=b^2+c^2-a^2$. Then we can verify that \[ \begin{vmatrix} -a & b & c \\ a^2S_a & b^2S_b & c^2S_c\\ -af(a,b,c)& bf(a,b,-c)&cf(a,-b,c)\\ \end{vmatrix}=0 \] Complex: Let $A=x^2,B=y^2,C=z^2$. Then the complex coordinates of $P$ are $r(-yz+xy+yz)$, where \[ r=\frac{-x^2 y + x y^2 - x^2 z - y^2 z + x z^2 - y z^2}{ x^2 y - x y^2 + x^2 z - 4 x y z + y^2 z - x z^2 + y z^2} \]Since $r$ is invariant under conjugation, it is real, and we can verify it lies on $I_BF$, so by symmetry it also lies on $I_CE$.

How did you manage to do that? I tried but it was so ugly.

How many points would i get if i hve a full solution except that i quoted the OIa perpendicular to XY as well known

@above I don't think that $P$ is a sufficiently well known point for that to be accepted, in addition that was the majority of the problem because you can cite $OI_A$ perpendicular to $YZ$ as well known.

EDIT: Irrelevant now, was fixed

adamov1 wrote: @above I don't think that $P$ is a sufficiently well known point for that to be accepted, in addition that was the majority of the problem because you can cite $OI_A$ perpendicular to $YZ$ as well known. I realized I asked the question wrong. Wait so I can cite $OI_A\perp YZ$? sweet

Here's how I got the barycentric coordinates for $P$. We easily have $I_c = (a:b:-c), I_b = (a:-b:c), Z = (a:b:0), Y = (a:0:c),$ $E =(a\cos(C):0:c \cos(A)),$ $ F = (a\cos(B): b\cos(A):0)$. Then if we let $P=(x:y:z)$ we have \[ \begin{vmatrix} a & -b & c \\ a \cos(B) & b \cos(A) & 0\\ x& y &z\\ \end{vmatrix}=0 \]\[ \begin{vmatrix} a & b & -c \\ a\cos(C) &0 & c \cos(A)\\ x& y&z\\ \end{vmatrix}=0 \]This gives the two equations $$z(ab)\cos(A) +z(ab)\cos(B) +y(ac)\cos(B) -x(bc)\cos(A) =0$$$$y (ac) \cos(A) +z(ab) \cos(C) -x(bc)\cos(A) + y(ac) \cos(C) = 0$$We subtract them to get $$y(ac)[\cos(A)+\cos(C)-\cos(B)] = z(ab)[\cos(A)+\cos(B)-\cos(C)]$$Thus $$y = z \dfrac{b(\cos(A)+\cos(B)-\cos(C))}{c(\cos(A)+\cos(C)-\cos(B))}.$$We substitute this into the first equation and get: $$z(ab)(\cos(A)+\cos(B))+z \dfrac{b(\cos(A)+\cos(B)-\cos(C))}{c(\cos(A)+\cos(C)-\cos(B)} \cdot ac \cos(B) = x (bc)\cos(A)$$$$z\left[ ab(\cos(A)+\cos(B) + ab\cos(B) \dfrac{\cos(A)+\cos(B)-\cos(C)}{\cos(A)+\cos(C)-\cos(B)} \right] = x(bc)\cos(A)$$$$z\left[ a(\cos(A)+\cos(B)) + a\cos(B) \dfrac{\cos(A)+\cos(B)-\cos(C)}{\cos(A)+\cos(C)-\cos(B)} \right] = x(c)\cos(A)$$Therefore$$\dfrac{x}{z} = \dfrac{a(\cos(A)+\cos(B))(\cos(A)+\cos(C)-\cos(B))+\cos(B)(\cos(A)+\cos(B)-\cos(C)))}{c \cos(A)(\cos(A)+\cos(C)-\cos(B))}$$$$\dfrac{x}{z} = \dfrac{a(\cos^2(A)+\cos(A)\cos(C)+\cos(B)\cos(A))}{c(\cos(A)+\cos(C)-\cos(B))\cos(A)} = \dfrac{a(\cos(A)+\cos(B)+\cos(C))}{c(\cos(A)+\cos(C)-\cos(B)}.$$From before $$\dfrac{y}{z} = \dfrac{b(\cos(A)+\cos(B)-\cos(C))}{c(\cos(A)+\cos(C)-\cos(B))}.$$Therefore $$P = \left(a(\cos(A)+\cos(B)+\cos(C)):b(\cos(A)+\cos(B)-\cos(C)): c(\cos(A)+\cos(C)-\cos(B))\right).$$After this we substitute $P,O,Y,Z$ in the perpendicularity formula and it works after some lengthy computations.

Nothing new (the same idea is in above posts by RC., DapperPeppermint, and k12byda5h), but I guess I'll post for completion [asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -16.19, xmax = 16.19, ymin = -9.64, ymax = 9.64; /* image dimensions */ pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen zzttff = rgb(0.6,0.2,1); draw((-1.8925246636790645,7.336940534456325)--(5.202828571212935,8.111314284851739), linewidth(0.7) + blue); draw((-2.696690907293137,-8.550201340072649)--(-6.572828571212932,1.948207142803233), linewidth(0.7) + fuqqzz); draw((7.064965447187023,-6.301625135187665)--(5.202828571212935,8.111314284851739), linewidth(0.7) + zzttff); draw((-4.43,-0.46)--(-2.7713148788927335,2.6500346020761243), linewidth(0.7) + zzttff); 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label("$I_{A}$", (1.49,-7.88), NE * labelscalefactor); dot((-2.7871714287870657,1.0017928571967663),linewidth(4pt) + dotstyle); label("$I$", (-2.91,1.2), NE * labelscalefactor); dot((-1.8925246636790645,7.336940534456325),linewidth(4pt) + dotstyle); label("$O'$", (-1.71,7.6), NE * labelscalefactor); dot((-2.696690907293137,-8.550201340072649),linewidth(4pt) + dotstyle); label("$F'$", (-3.01,-9.4), NE * labelscalefactor); dot((7.064965447187023,-6.301625135187665),linewidth(4pt) + dotstyle); label("$E'$", (7.15,-6.14), NE * labelscalefactor); dot((0.307481140238021,-2.2138590288606363),linewidth(4pt) + dotstyle); label("$P'$", (0.49,-2.16), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Observe that since $YZ$ is the orthic axis of $\triangle II_BI_C,$ it is well-known to be perpendicular to $I_AO$ (the proof of this is just showing that $YZ$ is the radical axis of the nine-point circle $(ABC)$ and circumcircle $(I_AI_BI_C)$); thus, it suffices to show that $I_AO,I_BF,I_CE$ concur. However, we can observe that $BE,BO$ and $CF,CO$ are isogonal with respect to angles $\angle ABC$ and $\angle ACB,$ respectively, so this motivates the construction of points $E',F',$ and $O'$ outside of $\triangle I_AII_BI_C$ such that $\triangle I_AI_BE'\stackrel{-}{\sim}\triangle ABE, \triangle I_BF'I_A\stackrel{-}{\sim}\triangle BFA,$ and $\triangle O'I_CI_B\stackrel{-}{\sim}\triangle OCB,$ as Jacobi's Theorem then implies that $I_AO',I_BE',I_CF'$ concur at a point $P'$. However, since $AB$ and $I_AI_B$ are antiparallel with respect to angle $\angle I_AI_CI_B,$ lines $I_CE$ and $I_CE'$ must be isogonal with respect to the same angle. Symmetrically, it then follows that $I_AO,I_BF,I_CE$ concur at the isogonal conjugate of $P'$ in $\triangle I_AI_BI_C,$ which is just $P$ as desired. $\blacksquare$

Since $\overline{OI_A}\perp\overline{YZ}$ (assume you know this by reading page 50 of a helpful handout with a nicely designed cover by Aritra12), we can generalize the problem to the following. In a triangle $ABC$, let $I_A,I_B,I_C,I$ be the $A,B,C$ excenters and the incenter. Let $R,Q$ be isogonal conjugates. Let $\overline{BQ}\cap\overline{AC}=E$ and $\overline{CQ}\cap\overline{AB}=F$. Then if $P=\overline{I_CE},\overline{I_BF}$, $P-R-I_A$ concur. (This can actually be generalized further, $\triangle I_AI_BI_C$ is the anticevian triangle of $I$ wrt $\triangle ABC$ and $R,Q$ are isoconjugates wrt $I$ and $\triangle ABC$). Note that by isogonal conjugation, $\overline{IR}$ is tangent to $\mathcal{C}(A,B,C,I,Q)$. In the involution created by $ABCQ$ and $\overline{IR}$ through DIT, $I$ is a fixed point. Since $-1=(B,\overline{BI}\cap\overline{AC};I,I_B)$, projecting this through $E$ we get that $\overline{I_BE}\cap\overline{IR}$ is the other fixed point of the involution. By symmetry $\overline{I_CF}\cap\overline{I_BE}$ is on $\overline{IR}$. $-1=B(R,Q;I,I_C)=(R,Q;\overline{BI}\cap\overline{PQ},\overline{I_AI_C}\cap\overline{RQ})$. By DIT, there is an involution created by $I_AI_BI_CI$ and $\overline{RQ}$. Using the harmonic bundle and symmetry, $R,Q$ must be the fixed points of this involution. Thus $\overline{RQ}$ is tangent to $\mathcal{C}(I_A,I_B,I_C,I,R)$. Let $\overline{AR}$ intersect this conic again at $R'$. Then $(I_B,I_C) (I,I_A) (R,R')$ are pairs under an involution, and unprojecting this through $R$ we have that $(\overline{RI_B},\overline{RI_C}) (\overline{RI},\overline{RI_A}) (\overline{RQ},\overline{RA})$ are pairs under an involution. DDIT on $I_BII_AI_C$ and $R$ gives that $(\overline{RB},\overline{RC})$ is also a pair under this involution. DDIT on $ABQC$ gives that $(\overline{RE},\overline{RF})$ is a pair under this involution as well. By DDIT on $I_BFI_CE$, we know that $(\overline{RI},\overline{RP})$ is a pair under this involution as well. Thus $P-R-I_A$ as desired.

proposed by evan chen O.O

Viewing from the point of view of $\triangle II_BI_C$, $\triangle ABC$ is the orthic triangle, $OI_A$ is the Euler line, and $YZ$ is the orthic axis. So it suffices to show $P$ lies on $OI_A$, or in other words Lemma: In a triangle $\triangle ABC$, if $\triangle XYZ$ is the orthic triangle of the orthic triangle, then $AX, BY, CZ$ meet on the Euler line. Proof: Let $\triangle DEF$ be the orthic triangle, $H$ be the orthocenter, $O$ be the circumcenter, $T$ be the midpoint of $DX$, and $\triangle PQR$ be the intouch triangle of $\triangle DEF$. Note shapes $\triangle PQRH$ and $\triangle ABCO$ are homothetic from a point $K \in OH$. Since $APT$ are well-known to be collinear, projecting through $A$ $$-1=(DX;T\infty_{DX}) = (HW;KO)$$where $W=AX \cap OH$. So the desired concurrence is at $W$.

Could someone check this??? We begin with a lemma Lemma: Let $ABC$ be a triangle and $I$ a point in its interior. Let $\ell_A$ be the line through $A$ with $(AB,AC;AI,\ell_A)=-1$ and define $\ell_B$, $\ell_C$ similarly. Let the triangle formed by $\ell_A$, $\ell_B$, and $\ell_C$ be $\triangle I_AI_BI_C$. Now, let $P,Q$ be points such that $(\ell_A, AI;AP,AQ)=-1$ and similar relations hold for $B,C$. Now, let $BP\cap AC=E$, $CP\cap AB=F$, and $I_CE\cap I_BF=X$. Then, $X, Q, I_A$ are collinear Proof. The statement is purely projective so take a homography sending $ABC$ to an equilateral triangle and $I$ its center. We proceed with barycentric coordinates. Let $P=(x:y:z)$ so $Q=\left(\frac 1x, \frac 1y, \frac 1z\right)$. Also, $I_A=(-1,1,1)$, $I_B=(1,-1,1)$, and $I_C=(1,1,-1)$. Then, $E=(x:0:z)$ $F=(x:y:0)$ $X=\left(\frac 1x + \frac 1y + \frac 1z:\frac 1x + \frac 1y - \frac 1z : \frac 1x + \frac 1y - \frac 1z\right)$. Then, we check \[ \begin{vmatrix} \frac 1x + \frac 1y + \frac 1z & \frac 1x + \frac 1y - \frac 1z & \frac 1x + \frac 1y - \frac 1z\\ \frac 1x & \frac 1y & \frac 1z\\ -1 & 1 & 1\\ \end{vmatrix}=0 \]Proving the claim. $\blacksquare$ Applying the lemma to $ABC$ with $I$ being the incenter and $P$ being $O$ and $Q$ being $H$ we get $P, O, I_A$ are collinear. Now, observe that $AY\cdot YB=YI\cdot YI_B$ so $YZ$ is the radical axis of $(II_BI_C)$ and $(ABC)$. Thus it suffices to show $I_AO$ passes through the center of $(II_BI_C)$. But this is immediate by homothety at $I_A$ with scale factor $2$ since $(ABC)$ is the nine point circle of $\triangle I_AI_BI_C$. We are thus done. EDIT: @venhance pointed out that the homography is not necessary. Instead, the computation I did could be done in trillinear coordinates and it would still be valid.

what the goat Claim. $\overline{OSI_A} \perp \overline{YZ}$. Proof. $(AI_CBI)$ is cyclic along $(I_CI)$, so $Z$ lies on the radical axis of $(II_BI_C)$ and $(ABC)$; it follows that $\overline{YZ}$ is the radical axis of the two circles. Then $\overline{SOI_A}$, the Euler line of triangle $II_BI_C$, is perpendicular to the radical axis $\overline{YZ}$. $\blacksquare$ Denote by $A_1, B_1, C_1$ the touchpoints of $(I_A)$ and $A_2, B_2, C_2$ the reflections of $I_A$ over these respective points. Claim. $\overline{DIA_2}$, $\overline{FI_CC_2}$, $\overline{EI_BB_2}$ are all collinear. Proof. Let $M_A$ be the midpoint of the $A$-altitude and so on. By midpoint of altitude lemma, $I, M_A, A_1$ are collinear, hence $D, I, A_2$ are collinear by homothety. Similarly, the homothety taking $\overline{CF}$ to the $B$-excircle diameter $\ell$ takes $M_C$ to $I_B$, hence $\overline{CM_CI_B}$ is collinear. By the same homothety argument using $\overline{I_ACI_B}$ we get $\overline{C_2FI_B}$ collinear, as needed. $\blacksquare$ Thus as $\overline{B_2C_2} \parallel \overline{B_1C_1} \parallel \overline{I_BI_C}$ and $\angle B_1C_1A_1 = \angle I_BII_C$, it follows that triangles $A_2B_2C_2$ and $II_BI_C$ are homothetic at $P$. As $P$ is the circumcenter of $A_2B_2C_2$, it follows that $\overline{SI_AP}$ is collinear, implying the problem by the previous result.

took just over an hour, no calculator/wolfram/paper used. i think even if you straight up expand the EFFT expression this is very doable in contest and probably takes <2 hours

Let $I_A,I,$ and $H$ be the $A$-excenter, incenter, and orthocenter of $\triangle{}ABC$, respectively. We claim that $I,I_A,B,C,O,$ and $H$ are coconic. Let the conic through $I,I_A,B,C,$ and $O$ be $\gamma$. It suffices that the involution taking a point $K$ on $\gamma$ and returning the intersection of the line through $B$ and the isogonal conjugate of $K$ with $\gamma$ other than $B$ and the involution taking a point $K$ on $\gamma$ and returning the intersection of the line through $C$ and the isogonal conjugate of $K$ with $\gamma$ other than $C$ are the same, but they both send $I$ and $I_A$ to themselves, proving the claim. Also, notice that these involutions send $B$ to $B'$ the second intersection of $\overline{CA}$ and $\gamma$ and send $C$ to $C'$ the second intersection of $\overline{AB}$ and $\gamma$. We claim that $P,O,$ and $I_A$ are collinear. Define $D$ and $X$ as the foot from $A$ to $\overline{BC}$ and the intersection of $\overline{AI}$ and $\overline{BC}$. By Dual of Desargues Involution Theorem on $I_BEI_CF$, it suffices that there exists an involution swapping lines $\overline{I_AI_B}$ and $\overline{I_AI_C}$, lines $\overline{I_AE}$ and $\overline{I_AF}$, and lines $\overline{I_AD}$ and $\overline{I_AO}$ since $\overline{I_AD},\overline{I_BE},$ and $\overline{I_CF}$ concur by the Cevian Nest Theorem. By Dual of Desargues Involution Theorem on $BECF$, it suffices that there exists an involution swapping lines $\overline{I_AI_B}$ and $\overline{I_AI_C}$, lines $\overline{I_AA}$ and $\overline{I_AH}$, and lines $\overline{I_AD}$ and $\overline{I_AO}$. Let $J$ be the intersection of $\overline{I_AD}$ and $\gamma$ other than $I_A$. Projecting onto $\gamma$ gives that we want that there exists an involution swapping points $B$ and $C$, points $I$ and $H$, and points $J$ and $O$. Notice that $\overline{HD}$ and $\overline{OX}$ intersect on $\gamma$ because the involution taking a point $K$ on $\gamma$ and returning the intersection of the line through $A$ and the isogonal conjugate of $K$ with $\gamma$ other than the isogonal conjugate of $K$ swaps points $B$ and $C$, points $B'$ and $C'$, and the intersection of $\overline{HD}$ with $\gamma$ other than $H$ and $O$. However, we see that $\overline{BC}\cap\overline{B'C'}=X$ by Brokard's Theorem since $\overline{BB'}\cap\overline{CC'}=\overline{II}\cap\overline{I_AI_A}$ which is the pole of $\overline{II_A}$, so $\overline{HD}$ and $\overline{OX}$ intersect at a point $L$ on $\gamma$. By Pascal's Theorem on $II_AJOLH$ we see that $D,X,$ and $\overline{IH}\cap\overline{JO}$ are collinear, so $\overline{BC},\overline{IH},\overline{JO}$ concur, meaning that there exists an involution swapping points $B$ and $C$, points $I$ and $H$, and points $J$ and $O$, proving the claim. Then, we see that $\overline{I_AO}\perp\overline{YZ}$ by barycentric coordinates.

We use barycentric coordinates freely, with reference $\triangle A BC$. Claim: $\overline{POI_A}$ collinear. Proof. Note, $O = (a^2S_A : b^2S_B : c^2S_C)$ $I_A = ( -a : b : c)$, $I_B = (a : - b : c)$, $I_C = (a : b : - c)$ $E = (S_C : 0 : S_A)$, $F = (S_B : S_A : 0)$ Now we first compute the equation of $\overline{I_BF}$. We must solve the system, \begin{align*} u_1a - v_1b + w_1c &= 0\\ u_1S_B + v_1S_A &= 0 \end{align*}Thus we find, \begin{align*} \frac{u_1}{v_1} &= -\frac{S_A}{S_B}\\ w_1 &= \frac{v_1b - u_1a}{c} \end{align*}The equation of the lines are then, \begin{align*} \overline{I_BF} \colon -S_Acx + S_Bcy + (aS_A + bS_B)z &= 0\\ \overline{I_CE} \colon -S_Abx + (aS_A + cS_C)y + S_Cbz &= 0 \end{align*}Next we compute the equation of $\overline{I_AO}$. We wish to solve the system, \begin{align*} -u_2a + v_2b + w_2c &= 0\\ u_2a^2S_A + v_2b^2S_B + w_2c^2S_C &= 0 \end{align*}Multiply the first equation by $aS_A$ and add to the second to find, \begin{align*} v_2(b^2S_B + abS_A) + w_2(c^2S_C + acS_A) &= 0\\ \iff \frac{v_2}{w_2} &= -\frac{c^2S_C + acS_A}{b^2S_B + abS_A} \end{align*}Then we find that, \begin{align*} \overline{I_AO} \colon (b^2cS_B - bc^2S_C)x - (a^2cS_A + ac^2S_A) + (a^2bS_A + ab^2S_B) &= 0 \end{align*}Now they three lines are concurrent at $P$ if and only if we find that the determinant, \begin{align*} 0 &= \begin{vmatrix} -S_Ac & S_Bc & aS_A + bS_B \\ -S_Ab & aS_A + cS_C & S_Cb \\ b^2cS_B - bc^2S_C & -a^2cS_A - ac^2S_C & a^2bS_A + ab^2S_B \end{vmatrix}\\ &= ab^2c^2 \begin{vmatrix} -S_A & S_B & \frac{a}{c}S_A + \frac{b}{c}S_B\\ -S_A & \frac{a}{b}S_A + \frac{c}{b}S_C & S_C\\ \frac{b}{a}S_B - \frac{c}{a}S_C & -\frac{a}{b}S_A - \frac{c}{b}S_C & \frac{a}{c}S_A + \frac{b}{c}S_B \end{vmatrix}\\ &= bc \begin{vmatrix} -aS_A & bS_B & aS_A + bS_B\\ -aS_A & aS_A + cS_C & cS_C\\ bS_B - cS_C & -aS_A - cS_C & aS_A + bS_B \end{vmatrix}\\ \end{align*}Substitute $x = aS_A$, $y = bS_B$ and $z = cS_C$ to find we require, \begin{align*} 0 &= bc\begin{vmatrix} -x & y & x + y\\ -x & x + z & z\\ y - z & -x - z & x + y \end{vmatrix}\\ &= bc\begin{vmatrix} 0 & -x + y - z & x + y - z\\ -x & x + z & z\\ y - z & -x -z & x + y \end{vmatrix}\\ &= bc\begin{vmatrix} 0 & -2x & x + y - z\\ -x & x & z\\ y - z & -2x - y - z & x + y \end{vmatrix}\\ &= -x(-2x(x + y) - (x + y - z)(-2x - y - z)) - (y - z)(-2xz - x(x + y - z))\\ &= -x(-2x^2 - 2xy - (-2x^2 - 3xy - y^2 + 2yz + xz + z^2)) - (y - z)(-2xz - x^2 - xy + xz)\\ &= -x(xy + y^2 - 2yz - xz - z^2) - x(y - z)(- x - y - z)\\ &= x(y - z)(- x - y - z) - x(y - z)( - x - y - z)\\ &= 0 \end{align*}This proves the claim. $\square$ However it is well known, or straightforward with barycentric coordinates to prove, $\overline{I_AO} \perp \overline{YZ}$ so we're done. $\square$ Remark: If you recognize $\overline{I_AO} \perp \overline{YZ}$ this problem is pretty straightforward to just compute; it doesn't take too long either, maybe twenty to thirty minutes if you don't screw up.

This problem asserts that the orthocenter $H$ of $\triangle ABC$ has the following property: If $E \equiv BH \cap AC, \; F \equiv CH \cap AB$ and $P \equiv I_BF \cap I_CE$, then $PO \perp YZ$. One might ask: what is the locus of such points $H$? The answer is a conic through $A, B, C, I_A$. Moreover, we have the following significant generalization of the original problem. Generalization: Given an arbitrary triangle $\triangle I_AI_BI_C$, let $A, B, C$ be distinct points on the respective sidelines of $\triangle I_AI_BI_C$ such that the lines $AI_A, BI_B, CI_C$ concur. Let $\mathcal{C}$ be a smooth conic through $A, B, C, I_A$. For a variable point $H \in \mathcal{C}$, denote $E \equiv BH \cap AC, \; F \equiv CH \cap AB$ and $P \equiv I_BF \cap I_CE$. Then as $H$ varies on $\mathcal{C}$, $P$ moves along a fixed line, namely the tangent line $\ell$ to $\mathcal{C}$ at $I_A$. How the generalization solves the original problem: Take $\mathcal{C}$ to be the (unique) conic through $A, B, C, I_A, H$. By the generalization, $P$ lies on the tangent line $\ell$ to $\mathcal{C}$ at $I_A$. We claim that $\ell$ coincides with the isogonal conjugate $\ell'$ of $\mathcal{C}$ with respect to $\triangle ABC$ (recall that the isogonal conjugate of a line is a circumconic and vice versa). Indeed, the set $\ell' \cap \mathcal{C}$ is fixed under isogonal conjugation and contains two points counting multiplicity (this is a very special case of Bézout's theorem), at least one of which is $I_A$. It follows easily that $\ell' \cap I_A$ is a double point at $I_A$, hence $\ell'$ is tangent to $\mathcal{C}$ at $I_A$, i.e. $\ell' \equiv \ell$. Finally, as $\ell$ is the isogonal conjugate of $\mathcal{C}$, it contains the isogonal conjugate $O$ of $H \in \mathcal{C}$. Therefore, $\ell \equiv I_AO$. As explained in many above solutions, $I_AO \perp YZ$, e.g. using radical axes. It follows that $PO \perp YZ$, as desired. $\blacksquare$ Proof of Generalization: We must show that lines $I_BF, I_CE, \ell$ concur. We can rephrase this concurrence in a way amenable to projective geometry as follows. Define $P_1 \equiv I_BF \cap \ell$ and $P_2 \equiv I_CE \cap \ell$. The desired concurrence is equivalent to $P_1 \equiv P_2$. Since the maps $\mathcal{C} \to \ell$ given by $H \mapsto F \mapsto P_1$ and $H \mapsto E \mapsto P_2$ are homographies, to show that $P_1 \equiv P_2$, it suffices to show that $P_1 \equiv P_2$ for three choices of $H \in \mathcal{C}$. When $H \equiv I_A$, we quickly find that $I_BF \equiv I_BI_A$ and $I_CE \equiv I_CI_A$, hence $P_1 \equiv P_2 \equiv I_A$. When $H \equiv C$, we make use of the assumption that lines $AI_A, BI_B, CI_C$ concur, say at $X$. Note that $E \equiv C$, and hence line $CF$ is tangent to $\mathcal{C}$ at $C$. A cross-ratio computation gives \[ (C, I_C, X, P_2) \stackrel{I_A}{=} (C, B, A, I_A) \stackrel{C}{=} (F, B, A, I_AI_B \cap AB) \stackrel{I_B}{=} (P, X, I_C, C) = (C, I_C, X, P). \]Therefore $P \equiv P_2$, hence $P_1 \equiv P_2$ as well. $H \equiv B$ is analogous. We conclude that $H \mapsto P_1$ and $H \mapsto P_2$ are in fact the same homography $\mathcal{C} \to \ell$. 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I'd like to point out there is a Very interesting generalization of this problem. Namely: Let $A,B,C,I_A,I_B,I_C$ as usual. Let $(U,V)$ be a pair of isogonal conjugates. Let $BU$ cut $AC$ in $E$ and let $CU$ cut $AB$ in $F$. Then, $I_AV, I_BF$ and $I_CE$ are concurrent. It's pretty funny how I camed up with that. Basically, when I saw this problem, I recalled the lemma that someone told me that $YZ$ (in the original problem) is perpendicular to $OI_A$; then problem was equivalent to showing $I_AO, I_BF,I_CE$ are concurrent. I thought all these lines where pretty nice with barycentric coordinates. So I settled Conway's Notation, namely $O=(a^2S_a:b^2S_B:c^2S_C)$, $E=(S_C:0:S_A)$ and $F=(S_B:S_A:0)$; and then I made the computations. The funny part is that in no moment of the computation I used something about $S_A,S_B,S_C$; in other words, we could take any choice of $S_A,S_B,S_C$, i.e, $O$ can be "any point" and we can define $E,F$ as the isogonals and problem still holds. I've got no idea how to prove this synthetically up to know. PS: I've checked this holds on geogebra, so that I don't make errors in the computation.

PHSH wrote: I'd like to point out there is a Very interesting generalization of this problem. Namely: Let $A,B,C,I_A,I_B,I_C$ as usual. Let $(U,V)$ be a pair of isogonal conjugates. Let $BU$ cut $AC$ in $E$ and let $CU$ cut $AB$ in $F$. Then, $I_AV, I_BF$ and $I_CE$ are concurrent. It's pretty funny how I camed up with that. Basically, when I saw this problem, I recalled the lemma that someone told me that $YZ$ (in the original problem) is perpendicular to $OI_A$; then problem was equivalent to showing $I_AO, I_BF,I_CE$ are concurrent. I thought all these lines where pretty nice with barycentric coordinates. So I settled Conway's Notation, namely $O=(a^2S_a:b^2S_B:c^2S_C)$, $E=(S_C:0:S_A)$ and $F=(S_B:S_A:0)$; and then I made the computations. The funny part is that in no moment of the computation I used something about $S_A,S_B,S_C$; in other words, we could take any choice of $S_A,S_B,S_C$, i.e, $O$ can be "any point" and we can define $E,F$ as the isogonals and problem still holds. I've got no idea how to prove this synthetically up to know. PS: I've checked this holds on geogebra, so that I don't make errors in the computation. Yes, this follows by group law (cayley bacharach also probably works) on the isogonal pivotal cubic with pivot $U$.

The main claim is that $P,O,I_A$ are collinear. To show this, construct the pivotal isogonal cubic $\mathcal C$ with pivot at the orthocenter $H,$ and thus passing through vertices $D,E,F$ of the orthic triangle, incenter $I$ and excenters $I_A,I_B,I_C,$ triangle vertices $A,B,C,$ and $H$ and its isogonal conjugate $O.$ We first show that $P$ lies on this cubic as well. To do this, notice that the two cubics $\overline{PEI_C}\cup\overline{BII_B}\cup\overline{CHF}$ and $\overline{PFI_B}\cup\overline{CII_C}\cup\overline{BHE}$ meet at the nine points $B,C,E,F,I_B,I_C,I,H,P.$ Since $\mathcal C$ passes through the first eight of these, it must pass through the ninth as well, by Cayley-Bacharach. Now to show $P,O,I_A$ collinear, we use the group law. First notice that the tangent to $\mathcal C$ at $H$ passes through $O,$ since as a point on the cubic approaches $H,$ its isogonal conjugate approaches $O,$ and the line through these two passes through $H$ so the line from $H$ to the point has to approach line $OH.$ Thus $2H+O=0.$ Then we have \begin{align*}0&=(P+I_C+E)+(O+2H)+(I_A+B+I_C)+(A+B+F)+(C+I_A+I_B)\\&-(B+H+E)-(C+H+F)-(A+I_B+I_C)-(B+I_A+I_C)\\&=P+O+I_A.\end{align*} Now it suffices to show that $OI_A\perp YZ.$ To do this, we invert about the $A$-excircle. Suppose the $A$-excircle is tangent to sides $BC,AC,AB$ at $R,S,T$ respectively. Then the images $B',C'$ of $B,C$ under this inversion are the midpoints of $RT,RS.$ The circumcircle is sent to the nine-point circle of $RST.$ Now notice line $BI$ is sent to the circle with diameter $B'I_A$ since $BI\perp BI_A.$ Then line $AC$ is sent to the circle with diameter $SI_A.$ Thus the image $Y'$ of $Y$ must be the intersection of these two circles, which is the foot from $I_A$ to $B'S.$ Similarly, $Z'$ is the foot from $I_A$ to $C'T.$ Now the conclusion $YZ\perp OI_A$ is equivalent to showing that $I_A$ is collinear with the centers of the nine-point circle of $RST$ and the circumcircle of $I_AY'Z'.$ But the antipode of $I_A$ on $(I_AY'Z')$ must be the intersection of $B'S$ and $C'T,$ which is the centroid of $RST.$ Thus these centers are collinear along the Euler line of $RST,$ so we are done.

It's also bashable by trilinear coordinates! Claim: $I_AO\perp YZ$. Proof: Let $YZ$ intersect $(ABC)$ at $K,L$. Since $ZI.ZI_C=ZA.ZB=ZK.ZL$ and $YI.YI_B=YA.YC=YK.YL$, we see that $I,I_B,I_C,K,L$ are concyclic. Under the inversion centered at $I_A$ with radius $\sqrt{I_AI.I_AA}$, $(II_BI_CKL)$ swaps with $(ABC)$ hence $I_AK=I_AL$. Combining this with $OK=OL$ implies that $I_AO$ is the perpendicular bisector of $KL$ thus, $I_AO\perp YZ$.$\square$ Now we will show that $I_A,O,P$ are collinear. We work on trilinear coordinates. Let $P=(p:q:r)$. Note that $E=(\cos C: 0 : \cos A)$ and $F=(\cos B: \cos A: 0)$. \[ \begin{vmatrix} p&q&r \\ 1&-1&1\\ \cos B&\cos A &0\\ \end{vmatrix}=0 \implies q\cos B+r\cos A=-r\cos B+p\cos A\]\[ \begin{vmatrix} p&q&r \\ 1&1&-1\\ \cos C& 0& \cos A\\ \end{vmatrix}=0 \implies p\cos A-q\cos C=r\cos C+q\cos A\]We have $(q+r)\cos B=(p-r)\cos A$ and $(q+r)\cos C=(p-q)\cos A$. \[ \begin{vmatrix} p&q&r \\ -1&1&1\\ \cos A&\cos B&\cos C\\ \end{vmatrix}\overset{?}{=}0 \iff -r\cos B+p\cos C+q\cos A\overset{?}{=} p\cos B-q\cos C+r\cos A \]\[(p+r)\cos B\overset{?}{=}(q-r)\cos A+(p+q)\cos C=(q+r)(\cos B-\cos C)+(p+q)\cos C\]\[\iff (p+r)\cos B\overset{?}{=}q\cos B+r\cos B-r\cos C+p\cos C\iff (p-q)\cos B\overset{?}{=}(p-r)\cos C\]Which is true since $\frac{p-q}{p-r}=\frac{(p-q)\cos A}{(p-r)\cos A}=\frac{\cos C}{\cos B}$ as desired.$\blacksquare$

Let $I_A$ be the $A$-excenter and $B_1$ and $C_1$ be the $A$-extouchpoints on $AC$ and $AB$, respectively. Claim: $OI_A\perp YZ$ Let $BC=a$, $CA=b$, and $AB=c$. Denote $(ABC)$ by $\Omega$ and the $A$-excircle by $\omega_A$. We have \begin{align*} OZ^2-OY^2 &=\text{Pow}_{\Omega}(Z)-\text{Pow}_{\Omega}(Y) \\ &= ZA\cdot ZB-YA\cdot YC \\ &= \left(\frac{bc}{a+b}\right)\left(\frac{ac}{a+b}\right)-\left(\frac{bc}{a+c}\right)\left(\frac{ab}{a+c}\right) \end{align*}whereas \begin{align*} ZI_A^2-YI_A^2 &= \text{Pow}_{\omega_A}(Z)-\text{Pow}_{\omega_A}(Y) \\ &= ZC_1^2 - YB_1^2 \\ &= \left(\frac{ac}{a+b}+\frac{a+b-c}{2}\right)^2-\left(\frac{ab}{a+c}+\frac{a+c-b}{2}\right)^2 \end{align*}Verifying the identity then becomes a matter of expansion, which will be omitted. It suffices to prove that $P$, $O$, $I_A$ are collinear. Claim: $FC$ bisects $I_BFI_A$, $EB$ bisects $I_CFI_A$. Let $AB$ intersect $I_AI_B$ at $X$. We have $(X,C;I_A,I_B)=-1$ by the cevian configuration. Since $\angle CFX=90^\circ$, $FC$ must bisect $\angle I_BFI_A$. The other claim follows similarly. Let $B_2$ and $C_2$ be reflections of $I_A$ across $AC$ and $AB$, respectively. Note that $I_B$, $F$, $C_2$ are collinear, and $I_C$, $E$, $B_2$ are collinear. We also have $B_1C_1\parallel I_BI_C$ because they're both perpendicular to $AI_A$. Let $A_1$ be the $A$-extouchpoint on $BC$. Note that $A_1B_1$ and $I_CI$ are both perpendicular to $I_AI_B$, so they are parallel. Similarly, $A_1C_1$ and $I_BI$ are parallel. Therefore, there is a homothety taking $\triangle II_BI_C$ to $\triangle A_1C_1B_1$. Let $A_2$ be the reflection of $I_A$ across $BC$. There is a homothety taking $\triangle A_1C_1B_1$ to $\triangle A_2C_2B_2$. Therefore, there is a homothety taking $\triangle II_BI_C$ to $\triangle A_2C_2B_2$, whose center must be $P$. Now, consider the fact that $I_AC_2=I_AA_2=I_AB_2=2r_A$ where $r_A$ is the $A$-exradius. Indeed, $I_A$ is the circumcenter of $\triangle A_2C_2B_2$. $I_AP$ passes through the circumcenter of $II_BI_C$. Noting that $\triangle ABC$ is the orthic triangle of of $\triangle I_AI_BI_C$, we take a force overlaid inversion centered at $I_A$ that swaps $B$ with $I_C$, $A$ with $I$, and $C$ with $I_B$. Although centers of circles aren't necessarily preserved, $O$, the center of $\triangle ABC$ and the center of $\triangle II_CI_B$ are collinear with $I_A$, and we're done.