This was my problem Hope you guys liked it! EDIT: just for fun, here's the original I sent in, complete with picture: Quote: 令不等邊三角形 $\triangle ABC$ 的內切圓圓心為 $I$，且該內切圓分別為 $CA$, $AB$ 邊切於點 $E$, $F$。 $\triangle AEF$ 的外接圓在 $E$ 和 $F$ 的兩條切線的交點 $S$。 直線 $EF$ 與 $BC$ 交於點 $T$。 試證：以 $ST$ 為直徑的圓垂直於 $\triangle BIC$ 的九點圓。 [asy][asy]defaultpen(fontsize(9pt)); size(9cm); pair A = dir(150); pair B = dir(220); pair C = dir(320); draw(A--B--C--cycle, red+1); pair I = incenter(A, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); draw(incircle(A, B, C), dotted+red); pair T = extension(E, F, B, C); pair O = circumcenter(A, E, F); pair S = extension(F, F+(O-F)*dir(-90), E, E+(O-E)*dir(90)); draw(circumcircle(A, F, E), dashed+heavygreen); draw(E--S--F, dashed+heavygreen); draw(B--I--C, orange); draw(E--T--B); draw(circumcircle(midpoint(B--C), midpoint(B--I), midpoint(C--I)), blue); draw(CP(midpoint(S--T), S), blue); draw(S--T, blue+dashed); dot(midpoint(B--C)); dot(midpoint(B--I)); dot(midpoint(C--I)); dot("$A$", A, dir(A)); dot("$B$", B, dir(-90)); dot("$C$", C, dir(-90)); dot("$I$", I, dir(150)); dot("$E$", E, dir(50)); dot("$F$", F, dir(150)); dot("$T$", T, dir(T)); dot("$S$", S, dir(S)); /* Source generated by TSQ */ [/asy][/asy]

Let the incircle $(I)$ touch $BC$ at $D.$ $DI$ cuts $EF$ at $J$ and the perpendicular from $A$ to $DI$ cuts $DI$ and $EF$ at $X,P.$ Since $AX$ is the polar of $J$ WRT $(I),$ then $(E,F,J,P)=-1$ $\Longrightarrow$ $A(E,F,J,P)=A(B,C,J,\infty)=-1$ $\Longrightarrow$ $AJ$ cuts $BC$ at its midpoint $M.$ As $SJ$ is the polar of $P$ WRT $\odot(AEF),$ then $J(A,X,S,P)=-1$ $\Longrightarrow$ $(M,D,L,T)=-1,$ where $L \equiv JS \cap BC.$ On the other hand, if $EF$ cuts $BI,CI$ at $Y,Z,$ then it's known (easy to show by angle chase) that $CY \perp BI$ and $BZ \perp CI$ $\Longrightarrow$ $\omega \equiv \odot(DMYZ)$ is 9-point circle of $\odot(IBC).$ Now since $(Y,Z,J,T)=I(B,C,D,T)=-1$ and together with $(M,D,L,T)=-1$ previously found, we deduce that $\overline{SLJ}$ is the polar of $T$ WRT $\omega$ $\Longrightarrow$ $S,T$ are conjugate points WRT $\omega$ $\Longrightarrow$ circle with diameter $\overline{ST}$ is orthogonal to $\omega.$

My solution is very similar to that of Luis. [asy][asy] size(12cm); defaultpen(fontsize(9pt)); pair A = dir(150); pair B = dir(220); pair C = dir(320); pair I = incenter(A, B, C); pair M = midpoint(B--C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); draw(incircle(A, B, C), dotted+red); pair T = extension(E, F, B, C); pair X = extension(C, I, E, F); pair Y = extension(B, I, E, F); draw(A--B--C--cycle, red); draw(T--B, red); pair O = circumcenter(A, E, F); pair K = extension(E, F, I, D); pair S = extension(F, F+(O-F)*dir(-90), E, E+(O-E)*dir(90)); draw(circumcircle(A, F, E), dashed+mediumgreen); draw(circumcircle(D, X, Y), blue+1); pair N = midpoint(E--F); draw(E--S--F, mediumgreen); pair L = extension(K, S, B, C); draw(K--L, heavycyan+1); draw(K--D, heavycyan); draw(T--Y, heavycyan); draw(A--M, heavycyan); draw(B--Y, orange); draw(X--C, orange); draw(A--S, dashed+orange); dot("$A$", A, dir(A)); dot("$B$", B, dir(-90)); dot("$C$", C, dir(-90)); dot("$I$", I, dir(I)); dot("$M$", M, dir(M)); dot("$D$", D, dir(D)); dot("$E$", E, dir(60)); dot("$F$", F, dir(F)); dot("$T$", T, dir(T)); dot("$X$", X, dir(120)); dot("$Y$", Y, dir(60)); dot("$K$", K, dir(70)); dot("$S$", S, dir(S)); dot("$N$", N, dir(80)); dot("$L$", L, dir(-90)); /* Source generated by TSQ */ [/asy][/asy] Let $D$ be the foot from $I$ to $BC$. Let $X$, $Y$ denote the feet from $B$, $C$ to $CI$ and $BI$. We can show that $BIFX$, $CIEY$ are cyclic, so that $X$ and $Y$ lie on $EF$. Now let $M$ be the midpoint of $BC$, and $\omega$ the circumcircle of $DMXY$. The problem reduces to showing that $T$ lies on the polar of $S$ to $\omega$. Let $K$ be the intersection of $AM$ and $EF$. It's well known (say by SL 2005 G6) that points $K$, $I$, $D$ are collinear. Let $N$ be the midpoint of $EF$, and $L$ the intersection of $KS$ and $BC$. From \[ -1 = (A, I; N, S) \stackrel{K}{=} (T, L; M, D) \] and \[ -1 = (T, D; B, C) \stackrel{I}{=} (T, K; Y, X) \] we find that $T = MD \cap YX$ is the pole of line $KL$ with respect to $\omega$, completing the proof.

Consider inversion with center $I$ we can turn the original problem into this： A triangle $DEF$,and its circumcenter is $I$ $A,B,C$ are midpoints of $EF,FD,DE$ $N$ is the intersection of the tangents to $(DEF)$ in $E,F$ $IT$ is perpendicular to $DN$ at $T$ $S$ is the reflection of $N$ with respect to $EF$ $X,Y$ are the reflections of $I$ with respect to $DF,DE$ and we need to prove that $(DXY)$ is orthogonal to (STN) Suppose that $O_1,O_2$ are the circumcenters of $(DXY),(STN)$ It is obvious that $O_2$ lies on $EF$ we need to prove that $O_1D^2+O_2S^2=O_1O_2^2$ $\angle O_1DF=\angle O_1DX-\angle FDX=(90^o-\angle XYD)-\angle IDF=(180^o-\angle XIY)-\angle IDF=\angle IDE$ $\therefore DO_1 \perp EF$,Suppose $DH$ is perpendicular to $EF$ at $H$ $O_1O_2^2-O_2S^2-O_1D^2=(O_1O_2^2-O_2D^2)+(O_2D^2-O_2S^2)-O_1D^2=(O_1H^2-DH^2)+DT\cdot DN-O_1D^2$ $=DT\cdot DN-2\cdot DO_1\cdot DH$ Suppose that $(DXY)$ meets $DE$ at $D,P$ $\angle YXI=\angle YDP=\angle YXP\Rightarrow X,I,P$ are collinear $\Rightarrow \angle DPI=90^o-\angle EDF=\angle INE\Rightarrow I,P,E,N,F,T$ are concyclic $\Rightarrow DT\cdot DN=DP\cdot DE=2\cdot DO_1 \cdot DH$ which means that $O_1D^2+O_2S^2=O_1O_2^2$,as desired.

[asy][asy] unitsize(150); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair A, B, C, I, D, E, F, Ap, O, S, T, M, P, X, Y; A = dir(140); B = dir(230); C = dir(310); I = incenter(A, B, C); D = foot(I, B, C); E = foot(I, C, A); F = foot(I, A, B); Ap = (E + F) / 2; O = circumcenter(A, E, F); S = extension(F, F + (O - F) * dir(90), E, E + (O - E) * dir(90)); T = extension(B, C, E, F); M = (B + C) / 2; P = extension(A, M, E, F); X = extension(B, I, E, F); Y = extension(C, I, E, F); draw(C--A--B); draw(incircle(A, B, C), n_blue); draw(A--M^^D--P); draw(E--T^^B--X^^C--Y^^C--T); draw(circumcircle(A, E, F), n_red); draw(circumcircle(D, X, Y), n_orange); draw(E--S^^F--S); dot(A^^B^^C^^I^^D^^E^^F^^Ap^^S^^T^^M^^P^^X^^Y); label("$A$", A, dir(A)); label("$B$", B, dir(B)); label("$C$", C, dir(C)); label("$I$", I, dir(300)); label("$D$", D, dir(250)); label("$E$", E, dir(10)); label("$F$", F, dir(250)); label("$S$", S, dir(I - A)); label("$T$", T, dir(180)); label("$M$", M, dir(310)); label("$P$", P, dir(70)); label("$X$", X, dir(100)); label("$Y$", Y, dir(140)); label("$A'$", Ap, dir(A - I)); [/asy][/asy] Let the incircle touch $\overline{BC}$ at $D$, let $M$ be the midpoint of $\overline{BC}$, let $A'$ be the midpoint of $\overline{EF}$, and let $P = \overline{ID} \cap \overline{EF}$, $X = \overline{IB} \cap \overline{EF}$, $Y = \overline{IC} \cap \overline{EF}$. We will use the well-known results that (i) $P \in \overline{AM}$ and (ii) $\angle BXC = \angle BYC = 90^{\circ}$ (which implies that the nine-point circle $\omega$ of triangle $BIC$ contains $D$, $M$, $X$, and $Y$). Since $(B, C; D, T) = -1$, projecting through $I$ onto line $EF$ we get $(X, Y; P, T) = -1$, so $P$ lies on the polar of $T$ wrt $\omega$. Since $(A, I; S, A') = -1$ (as $A'$ lies on $EF$, the polar of $S$ wrt $(AEF)$), projecting through $P$, we get $(M, D; \overline{PS} \cap \overline{BC}, T) = -1$, so $\overline{PS} \cap \overline{BC}$ lies on the polar of $T$ wrt $\omega$. Thus, the polar of $T$ wrt $\omega$ is line $PS$; in particular, $S$ lies on the polar of $T$ wrt $\omega$, so $(ST)$ is orthogonal to $\omega$ as desired.

Excellent projective problem! wanwan4343 wrote: In a scalene triangle $ABC$ with incenter $I$, the incircle is tangent to sides $CA$ and $AB$ at points $E$ and $F$. The tangents to the circumcircle of triangle $AEF$ at $E$ and $F$ meet at $S$. Lines $EF$ and $BC$ intersect at $T$. Prove that the circle with diameter $ST$ is orthogonal to the nine-point circle of triangle $BIC$. Proposed by Evan Chen Let $A_1=\overline{AI} \cap \overline{EF}$, $L=\overline{ID} \cap \overline{EF}$ and $K=\overline{SL} \cap \overline{BC}$. Let $\overline{BI}$ and $\overline{CI}$ meet $\overline{EF}$ at $Y$ and $X$, respectively. Observe that $A, L, M$ are collinear, so$$-1=(A, I, A_1, S)\overset{L}{=} (M, D, T, K)$$and $$-1=(B,C, D,T) \overset{I}{=} (Y,X, L,T).$$It is known that $(MDXY)$ is the nine point circle $\mathcal{C}$ of $\triangle BIC$. Therefore, $K$ and $L$ lie on the polar of $T$ with respect to $\mathcal{C}$, hence line $\overline{KL}$ is the polar of $T$ in $\mathcal{C}$. Thus, $\odot(ST)$ and $\odot(MXY)$ are orthogonal because $S$ lies on $\overline{KL}$. $\blacksquare$

Here's my new cleaned up solution from August 14th, 2019. [asy][asy] unitsize(2.2inches); pair A=dir(160); pair B=dir(215); pair C=dir(-35); pair I=incenter(A,B,C); pair D=foot(I,B,C); pair E=foot(I,A,C); pair F=foot(I,A,B); pair T=extension(E,F,B,C); pair Y=extension(B,I,E,F); pair Z=extension(C,I,E,F); pair M=(E+F)/2; pair L=extension(E,F,D,I); pair S=extension(A,I,F,F+((A+I)/2-F)*dir(90)); pair X=extension(L,S,B,C); pair P=(B+C)/2; draw(A--B--C--cycle); draw(circumcircle(D,E,F)); draw(circumcircle(D,Y,Z),dotted); draw(Y--T); draw(T--B); draw(B--Y); draw(C--Z); draw(A--S); draw(L--X); draw(D--L); draw(A--P); dot("$A$",A,dir(A)); dot("$B$",B,dir(-110)); dot("$C$",C,dir(C)); dot("$D$",D,dir(-90)); dot("$E$",E,dir(90)); dot("$F$",F,dir(170)); dot("$T$",T,dir(180)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(130)); dot("$I$",I,2*dir(200)); dot("$L$",L,dir(90)); dot("$S$",S,dir(30)); dot("$X$",X,dir(-90)); dot("$P$",P,dir(-90)); dot("$M$",M,1.4*dir(80)); [/asy][/asy] We see that $S$ is the harmonic conjugate of $M$ in $AI$ where $M$ is the midpoint of $EF$. Let $Y=BI\cap EF$ and $Z=CI\cap EF$. Note that \[\angle EYI=\angle AFE-\angle ABI=\pi/2-A/2-B/2=C/2=\angle ECI,\]so $(EYCI)$ cyclic. The diameter is $CI$, so $Y$ is the foot from $C$ to $BI$, and similarly $Z$ is the foot from $B$ to $CI$. This is the so called Iran lemma. We see then that the nine point circle of $BIC$ is $\omega:=(XYZP)$, where $P$ is the midpoint of $BC$. It is well known that $DI$, $AP$, and $EF$ concur, and let the concurrency point be $L$. Let $X=LS\cap BC$. Note that \[-1=(BC;DT)\stackrel{I}{=}(YZ;LT)\]and \[-1=(AI;MS)\stackrel{L}{=}(PD;TX),\]so the polar of $S$ with respect to $\omega$ is $LX$. In particular, $T$ is on the polar of $S$ with respect to $\omega$, so $(ST)\perp\omega$, as desired.

Really cool problem. Let $G,K,L$ be the feet of the altitudes in $\triangle BCI$ to sides $\overline{BC},\overline{BI},\overline{CI}$, respectively. Clearly, if $M$ is the midpoint of $\overline{BC}$, then $K,L,M,G$ are concyclic following from the fact that they lie on the Nine-Point Circle of $\triangle BCI$. Moreover, it is well known that $K$ and $L$ lie on $\overline{EF}$. Let $J=\overline{AM} \cap \overline{EF}$. It is well known that $J,I,G$ are collinear. If $X=\overline{AS} \cap \overline{EF}$ and $R=\overline{JS} \cap \overline{BC}$, then $$ -1 = (A, I; X, S) \stackrel{K}{=} (T, L; M, G) $$ Also, $$ -1 = (T,G; B,C) \stackrel{I}{=} (T, J; K, L)$$ Thus, $T$ is the pole of $\overline{JS}$ with respect to inversion around the Nine-Point Circle of $\triangle{BIC}$. But this is equivalent to the given problem condition; thus, we are done. $\square$

nice but hard Let $D$, $P$, and $Q$ denote the feet of the altitudes from $I, B$, $C$ respectively. It is well known that $P, Q \in EF$. Now let $AI \cap EF = K$, $DI \cap EF = X$, and $AX \cap BC = M$ which is well known to be the midpoint of $BC$, so $(PQMD)$ is the Nine-Point circle of $\triangle BIC$. Let $XS \cap DM = Y$. We want to show that $T$ lies on the polar of $S$ with respect to $(PQMD)$. We have $$(A, I; K, S) \stackrel{X}{=} (M, D; T, Y) = -1$$and $$(T, D; B, C) \stackrel{I}{=} (T, Q; P, X) = -1$$so $T$ is the pole of $XY$ with respect to $(PQMD)$ and we are done by La Hire. $\blacksquare$

Really cool problem. Suppose circle $I$ touches $BC$ at $D$. Let $CI$, $BI$ intersect $EF$ at $X$ and $Y$ respectively. Let $M$ and $N$ be the midpoints of $EF$ and $BC$, and suppose $AN$ intersect $EF$ at $L$. It's well known that 1) $L,I,D$ are collinear; 2) $B,F,X,I$ and $C,Y,E,I$ are concyclic. Also, $M$ is on the circle with diameter $ST$ since $\angle SMT=90^{\circ}$. Use DDIT on quadrilateral $CEFB$ (and project to line $EF$), then there is a unique inversion swapping $(M,T)$, $(X,F)$, and $(E,Y)$. In fact, this inversion is centered at $L$, since $L$ lies on the radical axis of both $(BFXID)$ and $(CYEID)$. Now invert about $L$ fixing the circle with diameter $EF$, then $(XYD)$ goes to $(AFIE)$. Let the center of $(AFIE)$ be $O$, then we only need $OM\cdot OS=OF^2$, which is true. Edit on Feb.18, 2020: Here's another solution using cross ratios: Use the same notation as above. Let $J = LS \cap BC$. We have $(T,D;B,C) = -1$, which, projecting through $I$ to $EF$, gives $(T,L;X,Y) = -1$. We also have $(A,I;M,S) = -1$, which, projecting through $L$ to $BC$, gives $(T,J;D,N)=-1$. This implies that $LJ$ is the polar of $T$ wrt the nine-point circle of $BIC$. To finish, let $H$ be the other intersection point between $LS$ and the circle $(ST)$, and let $TH$ intersect the nine-point circle at $U,V$. Suppose $O'$ is the center of the nine-point circle, then $(T,H;U,V)=-1 \Longrightarrow OH\cdot OT = OK\cdot OL$, which implies that the two circles are orthogonal.

Really Beautiful Problem!!! 2015 Taiwan TST Round 3 Quiz 3 P2 wrote: In a scalene triangle $ABC$ with incenter $I$, the incircle is tangent to sides $CA$ and $AB$ at points $E$ and $F$. The tangents to the circumcircle of triangle $AEF$ at $E$ and $F$ meet at $S$. Lines $EF$ and $BC$ intersect at $T$. Prove that the circle with diameter $ST$ is orthogonal to the nine-point circle of triangle $BIC$. Proposed by Evan Chen Let the Nine-Point Circle of $\odot(BIC)=\omega$ and let $CI\cap EF=Q, BI\cap EF=P$ and $DI\cap EF=R$, then $\{P.Q\}\in\omega$ (Well known). Now, $-1=(TD;BC)\overset{I}{=}(TR;QP)\implies R\in\text{Polar}$ of $T$ WRT $\omega$. Let $X=AS\cap EF, RS\cap BC=N$ and let $M$ be the midpoint of $BC$. It's well known that $\overline{A-R-M}$. So, $-1=(AI;XS)\overset{R}{=}(MD;TN)\implies N\in\text{ Polar}$ of $T$ WRT $\omega$. Hence, $S\in\text{Polar }$ of $T$ WRT $\omega$. Hence, by Self Orthogonality Lemma we get that $\odot(ST)\perp\omega.$ $\blacksquare$

It suffices to show that $S$ is on the polar of $T$. Let $D$ be the tangency of the incircle wiht $BC$. Let $M$ be the midpoint of $BC$, and let $P = AI \cap EF$ and $N = AM \cap EF$. Also, let $R = NS \cap BC$. It is well known that $DI$ also passes through $N$. Now, let $BI$ meet $EF$ at $L$ and let $CI$ meet $EF $ at $K$. By the Iran lemma, $\angle BKC = \angle BLC = 90^{\circ}$, and therefore $(MDKL)$ is the nine-point circle of $BIC$. Therefore,\[-1 = (AI;PS) \stackrel{N}{=} (MD;TR).\]Also, we have \[-1 = (TD;BC) \stackrel{I}{=} (TN;LK).\]Thus, $T$ is on the polar of both $N$ and $R$, so $NS$ is the polar of $T$. This means that $S$ is on the polar of $T$, so the result follows.

Beautiful problem, but quite easy for the advertised "hardest problem in EGMO" Let $\omega$ denote the nine-point circle of $\triangle BIC$. Note that it suffices to show that $S$ is on the polar of $T$. Let $M$ be the midpoint of $BC$. Let $DI$, $AM$, and $EF$ concur at $N$. Let $BI$ and $EF$ intersect at $Q$, and let $CI$ and $EF$ intersect at $P$. By Iran Lemma, $P$ and $Q$ lie on $\omega$. Let $NS$ intersect $BC$ at $R$, and let $AI$ intersect $EF$ at $L$. Claim: $N$ lies on the polar of $T$. Proof: This is immediate from harmonic bundles. Note that $$-1 = (I,A;E,F) \stackrel{N}{=} (D,M;P,Q) \stackrel{M}{=} (T,N;P,Q).$$ Claim: $R$ lies on the polar of $T$. Proof: Recall that $SE$ and $SF$ are tangents, so $$-1 = (A,I;L,S) \stackrel{N}{=} (M,D;T,R) $$ Now the polar of $T$ is $NR$, which $S$ lies on, and we are done.

Watch out, here's some hot stuff done mostly by hand (I used computer only to sort coefficients in numerator of $s+t-2o_9$). I hope you appreciate my sacrifice. Consider the situation on complex plane such that $d,e,f$ are complex numbers lying on unit circle centered at $0$ and they represent points $D,E,F$. Let $K$ be the midpoint of segment $AI$, $O_9$ is the center of the nine-point circle of triangle $BIC$, $G$ is the centroid of triangle $BIC$, and $O$ is the center of circle circumscribed on triangle $BIC$. $$a=\frac{2ef}{e+f},b=\frac{2df}{d+f},c=\frac{2de}{d+e}, k=\frac{a}{2}=\frac{ef}{e+f}, t=\frac{d(2ef-de-df)}{ef-d^2}$$Using perpendicularity condition $$SI\perp EF\iff \overline{s}=\frac{s}{ef}$$hence $$SF\perp KF\implies s=\frac{ef(e+f)}{e^2+f^2}.$$Nine point circle passes through midpoints of sides of triangle $BIC$ $$o_9=\frac{\begin{vmatrix} \frac{b+c}{2} & \frac{b+c}{2} \cdot\overline{\left( \frac{b+c}{2} \right)} & 1\\ \frac{df}{d+f} & \frac{df}{d+f} \cdot\overline{\left(\frac{df}{d+f} \right)} & 1\\ \frac{de}{d+e} & \frac{de}{d+e} \cdot\overline{\left(\frac{de}{d+e} \right)} & 1 \end{vmatrix}}{\begin{vmatrix} \frac{b+c}{2} & \overline{\left( \frac{b+c}{2}\right)} & 1\\ \frac{df}{d+f} & \overline{\left( \frac{df}{d+f}\right)} & 1\\ \frac{de}{d+e} & \overline{\left(\frac{de}{d+e} \right)} & 1 \end{vmatrix}}=\frac{ef+df+de}{(d+f)(d+e)}\cdot d$$Because these matrices are terrible you can instead count the coordinates for circumcenter $\frac{2def}{(d+f)(d+e)}$ and centroid $\frac{b+c}{3}$ of triangle $BIC$ and then use Euler line theorem $o_9=\frac{b+c}{2}-\frac{def}{(d+f)(d+e)}$. That's certainly faster and cleaner. Circle with diameter $ST$ is orthogonal to the nine-point circle of triangle $BIC$ iff squared distance between their centers is equal to sum of square of their radii $$\left|\frac{s+t}{2}-o_9\right|^2=\left|\frac{s-t}{2}\right|^2+\left|\frac{b+c}{2}-o_9\right|^2$$The temperature is rising: $$s-t=\frac{d^2(e+f)(e^2+f^2-ef)-2def(e^2+f^2)+e^2f^2(e+f)}{(ef-d^2)(e^2+f^2)}$$$$\frac{s+t}{2}-o_9=\frac{-def}{(d+f)(d+e)}$$and now a horror $$s+t-2o_9=\frac{ef(e+f)(ef-d^2)(d+f)(d+e)+d(2ef-de-df)(d+f)(d+e)(e^2+f^2)-2d(ef+df+de)(e^2+f^2)(ef-d^2)}{(e^2+f^2)(ef-d^2)(d+f)(d+e)}=$$$$=\frac{d^4(e^3+f^3)+d^3(-e^4+e^3f-4e^2f^2+ef^3-f^4)-d^2ef(e+f)(e^2+f^2)+de^2f^2(e+f)^2+e^3f^3(e+f)}{(e^2+f^2)(ef-d^2)(d+f)(d+e)}$$We are left with proving that $$LHS=RHS$$where (I'm using here $|x|^2=x\cdot\overline{x}$) $$\frac{4|e^2+f^2|^2|ef-d^2|^2|d+f|^2|d+e|^2}{d^4e^4f^4}\cdot LHS=$$$$=\left(d^2(e+f)(e^2+f^2-ef)-2def(e^2+f^2)+e^2f^2(e+f)\right)\cdot \left((e+f)(e^2+f^2-ef)-2d(e^2+f^2)+d^2(e+f)\right)(d+f)^2(d+e)^2-4(e^2+f^2)^2(d^6-2d^4ef+d^2e^2f^2)ef$$How to handle such expression? Write down coefficients at $d$ $$[d^8]=(e+f)^2(e^2+f^2-ef)$$$$[d^7]=2(e+f)^3(e^2+f^2-ef)-2(e+f)(e^2+f^2)^2$$$$[d^6]=(e+f)^2(e^2+f^2-ef)(e^2+f^2+4ef)-4(e+f)^2(e^2+f^2)^2+(e+f)^2(e^2+f^2-ef)^2+e^2f^2(e+f)^2$$$$[d^5]=2ef(e+f)^3(e^2+f^2-ef)-2(e+f)(e^2+f^2)^2(e^2+f^2+4ef)-2ef(e+f)(e^2+f^2)^2+2(e+f)^3(e^2+f^2-ef)^2+8ef(e+f)(e^2+f^2)^2+2e^2f^2(e+f)^3$$$$[d^4]=2e^2f^2(e+f)^2(e^2+f^2-ef)-4ef(e+f)^2(e^2+f^2)^2+(e+f)^2(e^2+f^2-ef)^2(e^2+f^2+4ef)+4ef(e^2+f^2)^2(e^2+f^2+4ef)+e^2f^2(e+f)^2(e^2+f^2+4ef)-4ef(e+f)^2(e^2+f^2)^2$$$$[d^3]=-2e^2f^2(e+f)(e^2+f^2)^2+2ef(e+f)^3(e^2+f^2-ef)^2+8e^2f^2(e+f)(e^2+f^2)^2+2e^3f^3(e+f)^3$$$$[d^2]=e^2f^2(e+f)^2(e^2+f^2-ef)^2+e^4f^4(e+f)^2-4e^2f^2(e+f)^2(e^2+f^2)^2+e^2f^2(e+f)^2(e^2+f^2-ef)(e^2+f^2+4ef)$$$$[d^1]=2e^3f^3(e+f)^3(e^2+f^2-ef)-2e^3f^3(e+f)(e^2+f^2)^2$$$$[d^0]=e^4f^4(e+f)^2(e^2+f^2-ef)$$Let's go to the $$\frac{4|e^2+f^2|^2|ef-d^2|^2|d+f|^2|d+e|^2}{d^4e^4f^4}\cdot RHS=$$$$=\left(d^4(e^3+f^3)+d^3(-e^4+e^3f-4e^2f^2+ef^3-f^4)-d^2ef(e+f)(e^2+f^2)+de^2f^2(e+f)^2+e^3f^3(e+f)\right)\cdot$$$$\cdot\left(ef(e^3+f^3)+d(-e^4+e^3f-4e^2f^2+ef^3-f^4)-d^2(e+f)(e^2+f^2)+d^3(e+f)^2+d^4(e+f)\right)$$And the coefficients $$[d^8]=(e^3+f^3)(e+f)$$$$[d^7]=(e^3+f^3)(e+f)^2+(-e^4+e^3f-4e^2f^2+ef^3-f^4)(e+f)$$$$[d^6]=(e+f)^2(-e^4+e^3f-4e^2f^2+ef^3-f^4)-(e^3+f^3) (e+f)(e^2+f^2)-ef(e+f)^2(e^2+f^2)$$$$[d^5]=(e^3+f^3)(-e^4+e^3f-4e^2f^2+ef^3-f^4)-(-e^4+e^3f-4e^2f^2+ef^3-f^4)(e+f)(e^2+f^2)-(e+f)^2(e^2+f^2)ef$$$$[d^4]=ef(e^3+f^3)^2+(-e^4+e^3f-4e^2f^2+ef^3-f^4)^2+ef(e+f)^2(e^2+f^2)+e^2f^2(e+f)^4+e^3f^3(e+f)^2$$$$[d^3]=ef(e^3+f^3)(-e^4+e^3f-4e^2f^2+ef^3-f^4)-ef(e+f)(e^2+f^2)(-e^4+e^3f-4e^2f^2+ef^3-f^4)-e^2f^2(e+f)^3(e^2+f^2)+e^3f^3(e+f)^3$$$$[d^2]=-e^2f^2(e+f)(e^2+f^2)(e^3+f^3)+e^2f^2(e+f)^2(-e^4+e^3f-4e^2f^2+ef^3-f^4)-e^3f^3(e+f)^2(e^2+f^2)$$$$[d^1]=e^3f^3(e+f)^2(e^3+f^3)+e^3f^3(e+f)(-e^4+e^3f-4e^2f^2+ef^3-f^4)$$$$[d^0]=e^4f^4(e+f)(e^3+f^3)$$QED

Interpret the orthogonality condition as $S$ lying on the polar of $T$, or equivalently $T$ on polar of $S$. To handle the nine-point center of $\triangle BIC$, we add in $D$ the third intouch point, $M_A$ midpoint of $\overline{BC}$, $N = \overline{CI} \cap \overline{EF}$, and $M = \overline{BI} \cap \overline{EF}$. Recall by the Iran configuration that $\overline{BN} \perp \overline{CN}$, same with $M$. Next, let $\overline{AM_A} \cap \overline{EF} = Q$. Recall that $Q$ also lies on $\overline{DI}$. A few more points need to be added: $P$ is $\overline{EF} \cap \overline{AI}$, also the midpoint of $\overline{EF}$. Finally, let $G = \overline{QS} \cap \overline{BC}$. Now we can finally begin solving. By Ceva-Menelaus, we know $-1 = (T, D; B, C)$. Projection through $I$ yields $-1 = (T, Q; M, N)$, thus $Q$ lies on the polar of $T$. Next, we know $(A, I; E, F) = -1$, and projection at $M$ yields $(A, I; P, S) = -1$. Projection through $Q$ yields $-1 = (M_A, D; T, G) = -1$, so the polar of $T$ is $\overline{QG}$ and we are done.

Let $\omega$ be the nine-point circle of $\triangle{BIC}$, and let $B_1, C_1$ be the feet from $B, C$ to $CI, BI$; we desire to prove $S, T$ are polar wrt $\omega$. Also, let $D$ be the foot from $I$ to $BC$, $M$ be the midpoint of $BC$, and $H$ be the orthocenter of $\triangle{BIC}$. Note that by Iran Lemma, $B_1, C_1\in EF$, so thus $-1=(T, D; B, C)\overset{H}=(T, K; B_1, C_1)$, where $K=EF\cap DI$. Thus, $T, K$ are polar wrt $\omega$. It is well known that $K$ lies on $AM$ too. Next, let $N=AI\cap EF$, and note $(A, I; N, S)=-1$; thus, letting $L=SK\cap BC$, we have $-1=(A, I; N, S)\overset{K}=(M, D; T, L)$, so $T, L$ are polar wrt $\omega$. Hence, the polar of $T$ is $KL$, so since $S\in KL$ we are done.

Beautiful problem!! (Though similar to some of the above solutions, anyways, I will post. Storage ) 2015 Taiwan TST Round 3 Quiz 3 Problem 2 wrote: In a scalene triangle $ABC$ with incenter $I$, the incircle is tangent to sides $CA$ and $AB$ at points $E$ and $F$. The tangents to the circumcircle of triangle $AEF$ at $E$ and $F$ meet at $S$. Lines $EF$ and $BC$ intersect at $T$. Prove that the circle with diameter $ST$ is orthogonal to the nine-point circle of triangle $BIC$. Proposed by Evan Chen Solved with PUjnk Let $D,E,F$ denote the usual intouch points and $M$ the midpoint of $BC$. Now, by Incircle Concurrency lemma (SL 2005 G6) we know that $AM, EF, DI$ are concurrent, let this point be $R$. Now, let $X, Y$ denote the intersections of $\overline{BI}, \overline{CI}$ with $\overline{EF}.$ [asy][asy]import graph; size(12cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.387019027484143,xmax=14.792684989429183,ymin=-3.152769556025373,ymax=6.420169133192388; pen qqffff=rgb(0.,1.,1.), qqwuqq=rgb(0.,0.39215686274509803,0.), qqzzqq=rgb(0.,0.6,0.), ffqqff=rgb(1.,0.,1.), xfqqff=rgb(0.4980392156862745,0.,1.), ffxfqq=rgb(1.,0.4980392156862745,0.); pair A=(5.,5.), B=(4.,-1.), C=(11.,-1.), D=(6.298740696016423,-1.), F=(4.377910639215199,1.2674638352911898), T=(-2.697631631235331,-1.), I=(6.2987405780298245,0.947325412814344), M=(7.5,-1.), X=(8.07539645556875,2.4523787334125036), Y=(5.025126265847083,1.4748737341529161), R=(6.2987405605833295,1.8830226546000106); draw(A--B--C--cycle,linewidth(0.8)+qqffff); draw(A--B,linewidth(0.8)+qqffff); draw(B--C,linewidth(0.8)+qqffff); draw(C--A,linewidth(0.8)+qqffff); draw(circle(I,1.9473254128143442),linewidth(0.8)+dotted+red); draw(circle((5.64937033158966,2.9736627489819187),2.1278450325182),linewidth(0.8)+linetype("0 3 4 3")+green); draw(F--(6.766589118382218,-0.5125797340330547),linewidth(0.8)+qqwuqq); draw((6.766589118382218,-0.5125797340330547)--(7.675707582607741,2.3242924173922592),linewidth(0.8)+qqwuqq); draw(A--M,linewidth(0.8)+qqzzqq); draw(T--B,linewidth(0.8)+qqffff); draw(X--T,linewidth(0.8)+blue); draw(Y--C,linewidth(0.8)+ffqqff); draw(X--B,linewidth(0.8)+ffqqff); draw(circle((6.899370348008213,0.8742440133356122),1.9681327700071864),linewidth(0.8)+xfqqff); draw(R--D,linewidth(0.8)+qqzzqq); draw(A--(6.766589118382218,-0.5125797340330547),linewidth(0.8)+ffxfqq); draw(R--(6.861779734970601,-1.),linewidth(0.8)+ffxfqq); dot(A,linewidth(4.pt)+ds); label("$A$",(4.93221987315011,5.202410147991542),NE*lsf); dot(B,linewidth(4.pt)+ds); label("$B$",(3.9174207188160715,-1.427610993657508),NE*lsf); dot(C,linewidth(4.pt)+ds); label("$C$",(10.970274841437638,-1.376871035940806),NE*lsf); dot(D,linewidth(4.pt)+ds); label("$D$",(6.217632135306559,-1.4445243128964087),NE*lsf); dot((7.675707582607741,2.3242924173922592),linewidth(4.pt)+ds); label("$E$",(7.503044397463007,2.49627906976744),NE*lsf); dot(F,linewidth(4.pt)+ds); label("$F$",(4.238773784355184,1.3969133192388985),NE*lsf); dot(T,linewidth(4.pt)+ds); label("$T$",(-2.763340380549682,-1.3937843551797067),NE*lsf); dot(I,linewidth(4.pt)+ds); label("$I$",(6.3698520084566645,1.0755602536997864),NE*lsf); dot(M,linewidth(4.pt)+ds); label("$M$",(7.519957716701908,-1.4614376321353093),NE*lsf); dot((6.766589118382218,-0.5125797340330547),linewidth(4.pt)+ds); label("$S$",(6.725031712473578,-0.7849048625792837),NE*lsf); dot(X,linewidth(4.pt)+ds); label("$X$",(8.044270613107829,2.648498942917546),NE*lsf); dot(Y,linewidth(4.pt)+ds); label("$Y$",(4.847653276955607,1.6167864693446068),NE*lsf); dot(R,linewidth(4.pt)+ds); label("$R$",(6.2683720930232605,2.107272727272725),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Claim: $\odot(DMXY)$ is the Nine-point circle of $\triangle BIC.$ Proof. By Incircle Perpendicularity lemma (Iran lemma), we get $\overline{BY} \perp \overline{CI}$ and $\overline{CX} \perp \overline{BI}$. Hence, $\odot(DMXY)$ is the Nine-point circle of $\triangle BIC.$ $\quad \square$ Let us denote this by $\Omega.$ Claim: $S$ lies on the polar of $T$ w.r.t $\Omega$. Now, by Harmonic Quadrilateral lemma we get: $$-1=(A,I;\overline{AS}\cap\overline{EF},S)\stackrel{R}=(M,D;T,\underbrace{\overline{RS}\cap\overline{BC}}_G).$$ And by Cevians Induce Harmonic Bundles lemma we get: $$-1=(T, D; B, C)\stackrel{I}=(T, R; X, Y).$$ Hence, $T = \overline{MD} \cap \overline{XY}$ is the pole of $\overline{RG}$ w.r.t $\Omega.\quad\square$ Now, by Self-Polar Orthogonality, we get $\odot(ST)$ and $\Omega$ orthogonal. $\quad \blacksquare$

eduD_looC wrote: I have lots to learn. Wow thanks for this super instructive post on how to solve this problem! Thanks for blessing us all with your intelligence! Congratulations on gaining an extra post on aops! Please stop postfarming. It doesn't benefit anyone. Here is my solution. Lemma: Let $\omega$ be a circle and suppose $P$ and $Q$ are points such that $P$ lies on the pole of $Q$. Then, the circle $\gamma$ with diameter $PQ$ is orthogonal to $\omega$. Proof: Let $\omega$ and $\gamma$ intersect at $X$ and $Y$ and let the image of $Z$ under inversion with respect to $\omega$ be $Z^*,$ so \[\measuredangle PP^*Q=\measuredangle PQ^*Q=90^{\circ}\]\[\Rightarrow \gamma=(PXP^*QYQ^*)\]then $\gamma$ maps to itself, so $\gamma$ and $\omega$ are orthogonal, as desired. $\boxed{}$ Now, Let $Y=CI \cap EF$, $X=BI \cap EF$, $K=AM \cap EF$, $N=AI\cap EF$, $L=KS\cap BC$, $M$ be the midpoint of $BC$, and $(DMXY)=\omega$. Notice that $N$ is also the midpoint of $EF$, and that $DK$, $EF$, and $AM$ are concurrent at $K$ (EGMO Lemma 4.17). Also, $\omega$ is the nine point circle of $BIC$. Since $SE$ and $SF$ are tangent to $(AEF)$, then we have \[-1=(AI;NS)\overset{K}{=}(MD;TL)\]Notice that $AD$, $BE$, and $CF$ are concurrent, so \[-1=(TD;BC)\overset{I}{=}(TK;YX)\]This means that $T=MD \cap YX$ is the pole of $KL$ with respect to $\omega$. However, by the lemma, since $P$ is on $KL$ this means that the circle with diameter $TP$ is orthogonal to $\omega$, as desired.

Solution with Psyduck909 Let the incircle touch $BC$ at $D$. Drop the perpendiculars from $B$ to $IC$ and $C$ to $IB$, call them $X$ and $Y$ respectively. Now the 9-point circle of $\triangle BIC$ is simply the circle $\circ(DXY)$. By the Iran lemma We know that $X,Y$ lie on line $EF$. Now, by the polar-orthogonality lemma, it suffices to prove that $S$ lies on the polar of $T$ w.r.t. $\circ (DXY)$. Let $AM\cap EF=Z$, from a well known lemma we know that $D-I-Z$ -are collinear.Let line $SZ\cap BC=L$. From the definition of $S$ we know that $-1=(A,I;Q,S)$ 1)Now we have \[-1=(A,I;Q,S)\overset{Z}{=}(M,D;TL)\].This means that the polar of $T$ W.R.T $(DXY)$, pass through $L$. 2)Its well known that $-1=(T,D;B,C)$ Now projecting from $I$ (We need to use the fact that $B-I-Y$ and $C-I-X$, collinear), we get that \[-1=(T,D;B,C)\overset{I}{=}(T,Z;Y,X)\], which means that The polar of $T$ will pass through $Z$, So from (1) and (2) we get that the polar of $T$ W.R.T the nine point circle of $(BIC)$, will pass through $LZ$, which means tha will pass through $S$, and we finish from proposition 9.24 of EGMO. Proposition.9.24[\b].Let $AB$ be a chord of a circle w and select points $P$ and $Q$ on line $AB$ .Then $-1=(A,B;P,Q)$ if and only if P lies on the polar of $Q$

^^thanks @yayps for the diagram wanwan4343 wrote: In a scalene triangle $ABC$ with incenter $I$, the incircle is tangent to sides $CA$ and $AB$ at points $E$ and $F$. The tangents to the circumcircle of triangle $AEF$ at $E$ and $F$ meet at $S$. Lines $EF$ and $BC$ intersect at $T$. Prove that the circle with diameter $ST$ is orthogonal to the nine-point circle of triangle $BIC$. Proposed by Evan Chen Let $L,M,Y,Z=ID,AS,BI,CI\cap EF$ respectively and $X=LS\cap BC$ and $P$ midpoint of $BC$.Its well known that $A,L,P$ are collinear. [asy][asy] currentpicture=new picture; unitsize(2.2inches); pair A=dir(160); pair B=dir(215); pair C=dir(-35); pair I=incenter(A,B,C); pair D=foot(I,B,C); pair E=foot(I,A,C); pair F=foot(I,A,B); pair T=extension(E,F,B,C); pair Y=extension(B,I,E,F); pair Z=extension(C,I,E,F); pair M=(E+F)/2; pair L=extension(E,F,D,I); pair S=extension(A,I,F,F+((A+I)/2-F)*dir(90)); pair X=extension(L,S,B,C); pair P=(B+C)/2; filldraw(A--B--C--cycle,0.1*purple+0.9*white); filldraw(circumcircle(D,E,F),0.1*magenta+0.9*white); draw(circumcircle(D,Y,Z),dotted); draw(Y--T,purple); draw(T--C,purple); draw(B--Y,purple); draw(C--Z,purple); draw(A--S,magenta); draw(L--X,magenta); draw(D--L,magenta); dot("$A$",A,dir(A)); dot("$B$",B,dir(-110)); dot("$C$",C,dir(C)); dot("$D$",D,dir(-90)); dot("$E$",E,dir(90)); dot("$F$",F,dir(170)); dot("$T$",T,dir(180)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(130)); dot("$I$",I,2*dir(200)); dot("$L$",L,dir(90)); dot("$S$",S,dir(30)); dot("$X$",X,dir(-90)); dot("$P$",P,dir(-90)); dot("$M$",M,1.4*dir(80));[/asy][/asy] Next note that $-1=(T,D,C,B)\overset{I}{=}(T,L,Z,Y)$ so $L$ lies on polar of $T$ wrt to $(DPYZ)$. Similarly $-1=(A,I,M,S)\overset{L}{=}(P,D,X,T)$ which gives $X$ lies on polar of $T$ wrt $(DPYZ)$. In conclusion $LX$ is polar of $T$ wrt $(DPYZ)$ which conlcudes the problem.$\blacksquare$

Lemma: [Self-Polar Orthogonality] Let $\omega$ be a circle and suppose $P$ and $Q$ are points such that $P$ lies on the pole of $Q$. This is equivalent to $(PQ)$ being orthogonal to $\omega$. Proof: Let $Q^*$ be the inverse of $Q$ in $\omega$. We have $Q^* \in (PQ)$ by right angles, so if $T=(PQ)\cap (\omega)$, then $\overline{OT}$ is tangent to $(PQ)$, where $O$ is the center of $(PQ)$. Hence $\omega \perp (PQ)$. $\blacksquare$ Let $M$ be the midpoint of $\overline{BC}$ and $D$ the foot from $I$ to $\overline{BC}$. Let $X$ and $Y$ the feet from $B$ to $\overline{CI}$ and $C$ to $\overline{BI}$ respectively. Let $D' = \overline{DI}\cap \overline{EF}$. By the Iran Lemma, $X,Y\in \overline{EF}$. It is well-known that $D'\in \overline{AM}$. By the lemma, it suffices to show $S$ is on the polar of $T$ with respect to $(MDXY)$. We claim $D'$ lies on this polar. Indeed, \[ (TD';XY) \stackrel{I}{=} (TD;CB)=-1.\] We claim $\overline{D'S}\cap \overline{BC}$ lies this polar as well. Indeed, \[ (T, \overline{D'S}\cap \overline{BC}) \stackrel{D'}{=}(\overline{AI}\cap \overline{EF},S;I,A)=-1\]since $S=\overline{EE}\cap \overline{FF}$. Combining the above two implies $\overline{D'S}$ is the polar of $T$, which $S$ lies on.

Beautiful and difficult. The problem is equivalent to $S$ being on the polar of $T$ with respect to the nine pt circle. Label the intouch points as $D,E,F$ accordingly. Let the feet of perpendicularity from $B,C$ to $CI,BI$ be $Y,Z$, respectively. By Iran lemma, we know that $Y,Z$ lie on $EF$. Set $X=DI\cap EF$. We can get $(T,X;F,E)\stackrel{I}{=}(T,D;B,C)=-1$ by the existence of the Gergonne point and the ceva/menelaus config. By EGMO lemma 4.17, we know that $AM$ also passes through $X$, where $M$ is the midpoint of $BC$. We have $(T,XS\cap BC; D,M)=(AS\cap EF,S;I,A)=-1$ by $IEAF$ being a harmonic quadrilateral. From here, we can see that the line through $XS\cap BC$ and $X$ is the polar of $T$, which obviously passes through $S$ and we are done.

Let the incircle be tangent to $BC$ at $D$, and let $M$ be the midpoint of $BC$. Let $X = CI \cap EF$, $Y = BI \cap EF$, and let $\omega$ be the nine-point circle of $\triangle BIC$. Let $T$ be tangent to $\omega$ at $P$ and $Q$. Finally, let $G$ be $DI \cap EF$, and let $K$ be $AS \cap EF$. First, note that the desired result is equivalent to proving that $S$ lies on the polar of $T$ wrt $\omega$ by the Self-Polar Orthogonality Lemma. Also note that it is well known that $DI$, $AM$, and $EF$ concur, so $AM$ passes through $G$, and note that $BX \perp CI$ and $CX \perp BI$ by the Iran Lemma. Now, we see that $(T, D; B, C) = -1$ by the Ceva/Menelaus picture, and that $(T, PQ\cap XY; B, C) = -1$ by harmonic quadrilateral properties on $XPYQ$. Then \[-1 = (T, D; B, C) \stackrel{I}{=} (T, G; Y, X)\] so $PQ\cap XY = G$, which implies that $G$ lies on $PQ$. Finally, note that $(A, I; K, S) = -1$, so \[-1 = (A, I; K, S) \stackrel{G}{=} (M, D; T, GS \cap BC).\] However, we also see that $(M, D; PQ \cap BC) = -1$ by harmonic quadrilateral properties on $MPDQ$. Then since $G$ lies on $PQ$, we know that $S$ lies on $PQ$ as well, so $S$ lies on the polar of $T$ wrt $\omega$ as desired. $\square$

Let the incircle touch $BC$ at $D$, $M$ be the midpoint of $BC$, $H$ be the orthocenter of $BIC$, $X = BI \cap EF$, $Y = CI \cap EF$, and $S^*$ denote the inverse of $S$ wrt $(AEIF)$, which coincides with the midpoint of $EF$. Trivially, $A, I, S, S^*$ are collinear. By the Iran Lemma, $\angle BXC = \angle BYC = 90^{\circ}$, so the Nine-Point Circle of $BIC$ is just $(DMXY)$. In addition, we also know $CX, BY, CI$ concur at $H$. Claim: $S$ lies on the polar of $T$ wrt $(DMXY)$. Proof. By Ceva-Menelaus, $$-1 = (T, D; B, C) \overset{H}{=} (T, K; Y, X)$$where $K = DI \cap EF$. Now, it suffices to show $(T, KS \cap BC; D, M)$ is harmonic. The Incircle Concurrency Lemma implies $K \in AM$. Hence, $$-1 = (S^*, S; I, A) \overset{K}{=} (T, KS \cap BC; D, M)$$as desired. $\square$ Now, the Self-Polar Orthogonality Lemma yields the desired result, and we're done. $\blacksquare$ Remark: The circle with diameter $ST$ is also orthogonal to $(AEIF)$, as $\angle SS^*T = 90^{\circ}$.

Diagram taken from Evan's solution. [asy][asy] size(12cm); defaultpen(fontsize(9pt)); pair A = dir(150); pair B = dir(220); pair C = dir(320); pair I = incenter(A, B, C); pair M = midpoint(B--C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); draw(incircle(A, B, C), dotted+red); pair T = extension(E, F, B, C); pair X = extension(C, I, E, F); pair Y = extension(B, I, E, F); draw(A--B--C--cycle, red); draw(T--B, red); pair O = circumcenter(A, E, F); pair K = extension(E, F, I, D); pair S = extension(F, F+(O-F)*dir(-90), E, E+(O-E)*dir(90)); draw(circumcircle(A, F, E), dashed+mediumgreen); draw(circumcircle(D, X, Y), blue+1); pair N = midpoint(E--F); draw(E--S--F, mediumgreen); pair L = extension(K, S, B, C); draw(K--L, heavycyan+1); draw(K--D, heavycyan); draw(A--M, heavycyan); draw(B--Y, orange); draw(X--C, orange); draw(A--S, dashed+orange); dot("$A$", A, dir(A)); dot("$B$", B, dir(-90)); dot("$C$", C, dir(-90)); dot("$I$", I, dir(I)); dot("$M$", M, dir(M)); dot("$D$", D, dir(D)); dot("$E$", E, dir(60)); dot("$F$", F, dir(F)); dot("$T$", T, dir(T)); dot("$X$", X, dir(120)); dot("$Y$", Y, dir(60)); dot("$L$", K, dir(70)); dot("$S$", S, dir(S)); dot(N); dot("$J$", L, dir(-90)); pair X1 = extension(D, K, (0, 50), (50, 50)); pair X2 [] = intersectionpoints(K--X1, circumcircle(A, E, F)); pair X3 = X2[0]; pair K1 = extension(A, X3, F, E); draw(A--K1, orange+dashed); dot("$K$", K1, dir(90)); draw(T--K1, heavycyan); /* Taken from Evan Chen */ [/asy][/asy] First the foot of altitudes of $C, B$ to $\overline{BI}, \overline{CI}$ which we will call $Y, X$ respectively lie on $\overline{EF}$ by Right Angles on Incircle Chord. Also define $L = \overline{AM} \cap \overline{EF}$ which also lies on $\overline{ID}$ by EGMO 4.17. Verify that $$(TK;XY) \stackrel I= (TD;BC) = -1,$$so $K$ lies on the polar of $T$. Now let $J$ be a point on $\overline{BC}$ such that $(TJ;DM)=-1$. Again denote by $K$ our ubiquitous $\overline{EF} \cap \overline{AP_\infty}$. Then $$-1 = (TJ;DM) \stackrel L= (K, \overline{LJ} \cap \overline{AP_\infty}; \overline{LD} \cap \overline{AP_\infty}, A).$$Observe that $\overline{LD} \cap \overline{AP_\infty}$ lies on $(AEF)$. Henceforth $\overline{LJ} \cap \overline{AP_\infty}$ lies on the polar of $K$ with respect to $(AEF)$. However, by La Hire $S$ also lies on the polar of $K$, and $L$ lies on the polar of $K$ with respect to $(AEF)$ becuase $L$ lies on the polar of $K$ with respect to the incircle by Sharygin 2013, implying $$(KL;EF)=-1,$$or $L$ also lies on the polar of $K$ with respect to $(AEF)$. Thus, $\overline{LJ} \cap \overline{AP_\infty}, L, S$ are collinear, so $L, S, J$ are collinear. Therefore, $\overline{LJ}$ is the polar of $T$ with respect to the nine-point circle of $(BIC)$, so $S$ lies on the polar of $T$ and we are done by self-polar orthogonality.

Let the incircle of $\triangle ABC$ be $\omega,$ the nine point circle of $\triangle BIC$ be $\Omega,$ the center of $\overline{BC}$ be $M,$ and $D=\overline{BC}\cap\omega.$ Let $\overline{EF}$ intersect $\overline{AI},\overline{BI},\overline{CI},$ and $\overline{DI}$ at $N,K,L,$ and $P,$ respectively. Finally, let $Q=\overline{PS}\cap\overline{BC}.$ By the Incircle Concurrency lemma, we know let $P$ lies on $\overline{AM}.$ Notice that $M$ lies on $\Omega,$ and by the Right Angles on Incircle Chord lemma, we know that $\angle BLC=\angle BKC=90,$ meaning that $K$ and $K$ lie on $\Omega.$ Notice that $$-1=(T,D;B,C)\stackrel{I}{=}(T,P;K,L)$$and $$-1=(A,I;N,S)\stackrel{P}{=}(M,D;T,Q)$$so $P$ and $Q$ lie on the polar of $T.$ Hence, $S$ lies on the polar of $T$ and we are done by Self Polar Orthogonality. $\square$

Let $\omega$ be the nine point circle of $\triangle BIC$, $M$ be the midpoint of $\overline{BC}$, $EF\cap AM=X$ (Also, $DI$, $AM$ and $EF$ concur at $X$ by Incircle Concurrency lemma), $BI\cap EF=P$, $CI\cap EF=Q$, $AI\cap EF=R$ (Also, $A-R-I-S$ since $S$ is the intersection of tangents to $\omega$ at $E$ and $F$). By Iran TST 2009/9, $(DMQP)\equiv \omega$. Claim - $X$ and $Y$ lies on the polar of $T$ wrt $\omega$. Proof - This follows as $$-1=(TD, CB) \stackrel{I}{=} (TX, PQ)$$and $$-1=(AI, RS)\stackrel{X}{=} (TY, DM)$$ Hence, $S$ lies on polar of $T$. We finish using Self-Polar Orthogonality lemma.

By symmetry it follows $S\in AI.$ Let $R=AI\cap EF,K=DI\cap EF,L=KS\cap BC$. Also let $M$ denotes midpoint of $BC$ and $D$ denotes touch point of $BC$ with incircle. It's well-known that $K\in AM$ and by the Iran lemma points $X=BI\cap EF,Y=CI\cap EF$ are projections of $B$ and $C$ respectively onto $BI$ and $CI,$ so $\omega =\odot (XYDM)$ is the nine-point circle of $BIC.$ From $(XYKT)\stackrel{I}{=} (BCDT)=-1=(IASR)\stackrel{K}{=} (DMLT)$ it follows that $KL$ is a polar of $T$ wrt $\omega,$ so with $S\in KL$ we finish by Self-Polar Orthogonality lemma.

projective spam just like all the other sols lol [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=lightblue; pen sec=heavycyan; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=palecyan; pair A,B,C,D,E,F,P,S,I,O,R,U,W,M,T,A1,B1,X,Y; O=(0,0); A=dir(130); B=dir(210); C=dir(330); A1=dir(270); B1=dir(50); I=intersectionpoint(A--A1,B--B1); D=foot(I,B,C); E=foot(I,A,C); F=foot(I,A,B); T=extension(E,F,B,C); M=(B+C)/2; W=foot(M,E,F); U=extension(D,I,M,W); O=(M+U)/2; X=extension(O,I,E,F); S=extension(A,I,M,X); P=intersectionpoint(F--E,D--U); Y=intersectionpoint(A--I,E--F); filldraw(A--B--C--cycle,tfil,pri); draw(I--A1--M--U--cycle,sec); draw(E--T,pri); draw(T--B,pri); draw(A--M,pri); draw(D--I,pri); draw(T--U,pri); draw(F--S,tri); draw(E--S,tri); draw(A--I,pri); draw(M--X,dashed+sec); draw(O--X,dashed+sec); filldraw(circumcircle(A,B,C),fil,pri); filldraw(circle(I,distance(I,E)),fil,pri); filldraw(circumcircle(A,E,F),fil,tri); label("$A$",A,dir(120)); label("$B$",B,dir(210)); label("$C$",C,dir(330)); label("$D$",D,dir(270)); label("$E$",E,dir(60)); label("$F$",F,dir(60)); label("$P$",P,dir(270)); label("$S$",S,dir(270)); label("$I$",I,dir(270)); label("$N_9$",O,dir(270)); label("$U$",U,dir(90)); label("$W$",W,dir(60)); label("$M$",M,dir(270)); label("$T$",T,dir(270)); label("$A_1$",A1,dir(270)); label("$X$",Y,dir(120)); [/asy][/asy] $N_9$ is 9-point center $M$ is midpoint of $BC$ $A_1$ is midpoint of $\widehat{BC}$ $U$ is antipode of $M$ wrt the 9-point circle Claim 1: $IA_1MU$ is a parallelogram. Proof: $A_1$ is the circumcenter of $\triangle{BIC}$, the reflection of $I$ over $U$ is the orthocenter of $\triangle{BIC}$, and homothety of factor $-2$ at the centroid of $\triangle{BIC}$ finishes. By the Iran Lemma $DI, EF, AM$ concur at point $P$ which is the orthocenter of $\triangle{TUM}$. It suffices to show that $S, T$ lie on each others' polars wrt $(MU)$; by Brocard on the quadrilateral formed by $U, D, M, MP \cap TU$ the pole of $T$ wrt $(MU)$ is the line through $P$ perp to $TN_9$, so it suffices to show that $PS \perp TN_9$. Claim 2: $PS \perp TN_9$. Because $SI \perp TW$, it suffices to show that if $K \in EF$ such that $SK \perp BC$, then $XP : PK = WN_9 : N_9M$, which is the same as $XI : IS = WN_9 : N_9M$, so it suffices to show that $EF, IN_9, SM$ concur. $S, I, X, A$ is harmonic; projecting from $M$ onto $EF \implies MS \cap EF, MI \cap EF, X, P$ harmonic $U, N_9, M, P_{\infty}$ is harmonic; projecting from $I$ onto $EF \implies MI \cap EF, IN_9 \cap EF, P, X$ harmonic So $IN_9 \cap EF = SM \cap EF \implies EF, IN_9, SM$ concur $\implies PS \perp TN_9$ and we are done. $\square$

Let $\Omega$ be the nine-point circle of $\triangle BIC$. By EGMO lemma 9.27, it suffices to show that $S$ lies on the polar of $T$ wrt $\Omega$. Now, construct all of the following points: $D$, the point where the incircle of $\triangle ABC$ meets $\overline{BC}$; evidently this lies on $\Omega$ $X = \overline{BI} \cap \overline{EF}$ and $Y = \overline{CI} \cap \overline{EF}$; by Iran lemma, these are the feet of the altitudes from $C$ and $B$ in $\triangle BIC$, respectively, so they also lie on $\Omega$ $M$, the midpoint of $\overline{BC}$, which lies on $\Omega$ $P$, the point where lines $AM$, $DI$, and $EF$ are known to concur by EGMO Chapter 4 $Q = \overline{PS} \cap \overline{BC}$ $Z = \overline{AI} \cap \overline{EF}$ Then since $AD$, $BE$, and $CF$ concur at the Gergonne point of $\triangle ABC$, by Ceva-Menelaus we get $$-1 = (T, D; B, C) \overset{I}{=} (T, P; X, Y),$$so $P$ lies on the polar of $T$ wrt $\Omega$. Also, by symmetry, we find that $S$ lies on $AI$ and quadrilateral $AEIF$ is harmonic. So, we get $$-1 = (A, I; E, F) \overset{F}{=} (A, I; Z, S) \overset{P}{=} (M, D; T, Q).$$Therefore $Q$ also lies on the polar of $T$ wrt $\Omega$, so this polar is just line $PQ$. Since $S$ lies on this line, we are done.

solved with too many hints Let $P$ and $Q$ be the feet from $B$ to $CI$ and the feet from $C$ to $BI$. Claim: $P$ and $Q$ line on $EF$. Proof: Consider just $P$, the other one follows by symmetry. Note that $FPIDB$ is cyclic because $\angle BFI=\angle BPI=\angle BDI=90$. Thus, $$\angle DFP=\angle DBP=90-\frac{\gamma}{2}=\angle DFE,$$hence proven. The nine point circle is just $(PQMD)$, where $M$ is the midpoint of $BC$. Let $A^{\ast}$ denote the midpoint of $EF$ (I am not inverting, but this is what I usually name this point because I do that a lot). Then, we have $(TD;BC)=-1$ since $AD,BE,CF$ concur at the Gergonne point. It is well known that $EF,AM,ID$ concur at a point, say $K$, so going through $I$ onto $EF$ we have $(TK;QP)=-1$. Thus, $K$ is on the polar of $T$. Furthermore, due to the tangencies at $E$ and $F$ to $S$, we have that $(AI;A^{\ast}S)=-1$. Let $X$ be the intersection of $KS$ and $BC$. If we project this from $K$ onto $BC$, it becomes $(MD;TX)=-1,$ which means that $Y$ is also on the polar of $T$. Since $K,S,X$ are collinear, $S$ is on the polar of $T$. Hence, by the polar orthogonality lemma, we are done.

Note that it is equivalent to show that $S$ lies on the polar of $T$ wrt to the nine-pont circle. Let $D$, $G$, and $H$ be the feet of the altitudes from $I$, $B$, and $C$ respectively wrt to $\triangle BIC$. Let $M$ be the midpoint of $BC$. Note that $(DGHM)$ is thus the nine-point circle and that $G$ and $H$ lie on $\overline{EF}$ by Iran Lemma. and let $N = \overline{EF} \cap \overline{AM} \cap \overline{ID}$. Then, by Ceva-Menelaus it follows that \[ (TD;BC) \overset{I}= (TN;HG) = -1 \]which implies that $N$ lies on the polar of $T$ wrt to $(DGHM)$. Let $P$ be the midpoint of $EF$. Then, \[ (PS;AI) \overset{N}= (T,\overline{NS}\cap\overline{BC};MD) = -1 \]which implies that $\overline{NS} \cap \overline{BC}$ lies on the polar. However, this simply means that $NS$ is the polar of $T$.

Let $U$ and $V$ be the feet of $C$ and $B$ onto $BI$ and $CI$, respectively. Iran Lemma gives $T$, $E$, $F$, $U$, $V$ collinear. Note $BDIVF$ and $CDIUE$ are cyclic, and To show $ABCI \stackrel -\sim DVUI$, observe \[\measuredangle UDV = \measuredangle UDI + \measuredangle IDV = \measuredangle UEI + \measuredangle IFV = \measuredangle EIF = \measuredangle CAB,\]\[\measuredangle DUV = \measuredangle IUV + \measuredangle DUI = \measuredangle BCV + \measuredangle DCI = \measuredangle BAC,\]analogously $\measuredangle UVD = \measuredangle ABC$, and \[\measuredangle IUV = \measuredangle BCI, \measuredangle UVI = \measuredangle IBC.\] Further, observe that $DT$ is the external angle bisector of $\angle UDV$. ---- Now, we restate the problem in terms of triangle $DUV$ (relabeled as $ABC$): Let $ABC$ be a triangle with incenter $I$, and let $T\in BC$ be the point so that $AT$ is the $A$-external angle bisector. $(AIB)$ and $(AIC)$ intersect $BC$ again at $P$ and $Q$, respectively (noting $IP = IQ$), and let $S$ be the point on the same side of $BC$ as $I$ such that $\angle SPQ = \angle SQP = \angle A$. Show that $(ST)$ and $(ABC)$ are orthogonal. To this end, note that this is equivalent to $S$ lying on the polar of $T$ with respect to $(ABC)$. Let $O$ be the circumcenter, let $D$ be the $A$-intouch point, and let $AI\cap BC = D'$. Since $(BC;D'T) = -1$, $D'$ lies on the polar of $T$, and thus it suffices to show that $TO\perp SD'$. Let $N$ be the midpoint of arc $BAC$, and let $M$ be the midpoint of $BC$. Let $L$ be the point on $BC$ such that $SL\parallel AI$. It suffices to show that $TNOM\sim SLD'D$. This is not hard to check -- we have $\angle TNM = \angle C + \angle A/2$ and $\angle SMD = \angle AD'D = \angle C + \angle A/2$. Also, $DD'/ D'M = ID / SI = PD/PS = \cos A$, and $OM/ON = OM/OB = \cos A$ as well, so we're done.

Another terrible geometry problem. What else is new? Clearly we are being asked to prove $S$ lies on the polar of $T$ wrt the 9-point circle. Let $D$ be the $A$-intouch point, $M$ be the midpoint of $\overline{BC}$, $X,Y$ the intersections of $\overline{CI}$ and $\overline{BI}$ with $\overline{EF}$ respectively, $R=\overline{EF} \cap \overline{AS}$ the midpoint of $\overline{EF}$, and $K$ the well-known concurrency point of $\overline{AM},\overline{DI},\overline{EF}$. Finally let $\overline{KS} \cap \overline{BC}=L$. It is known that $X,Y$ are the projections of $B,C$ onto $\overline{CI},\overline{BI}$ respectively. We have $(T,K;X,Y)\stackrel{I}{=}(T,D;B.C)=-1$ by Ceva-Menelaus and the existence of the Gergonne point, hence $K$ lies on the polar of $T$. We also have $(T,U;D,M)\stackrel{K}{=}(R,S;I,A)=-1$, hence $U$ lies on the polar as well, hence $S$ does. $\blacksquare$

Here's a complex bash that is by far doable by hand. Compared to the complex bash earlier in the thread, I perform the final computation in a different way, such as by canceling $s\overline s$ and $t\overline t$ from both sides. Use $(DEF)$ as the unit circle. We have $T=DD\cap EF$, so $t=\frac{d^2(e+f)-ef(d+d)}{d^2-ef}=\frac{d(de+df-2ef)}{d^2-ef}$. Also, $\overline t=\frac{2d-e-f}{d^2-ef}$. By angle chasing, we have $\measuredangle FSE=2\measuredangle FIE$ (in directed angles), so $\frac{s-e}{s-f}=-\frac{e^2}{f^2}$, where the negative sign is to deal with config issues. Solving, $s=\frac{ef(e+f)}{e^2+f^2}$ so $\overline s=\frac{e+f}{e^2+f^2}$. Let $O$ be the center of $(BIC)$. Since $I$ is the origin, we have \begin{align*} o&=\frac{bc(\overline c-\overline b)}{b\overline c-\overline bc}\\ &=\frac{\frac{2fd}{f+d}\cdot\frac{2de}{d+e}(\frac2{d+e}-\frac2{f+d})}{\frac{2fd}{f+d}\cdot\frac2{d+e}-\frac2{f+d}\cdot\frac{2de}{d+e}}\\ &=\frac{8d^2ef(\frac1{d+e}-\frac1{f+d})}{4fd-4de}\\ &=\frac{2def}{(f+d)(d+e)}. \end{align*} Thus, the radius $R$ of the nine-point circle of $\triangle BIC$, which is half the radius of $(BIC)$, satisfies \[4R^2=m\overline m=\frac{4d^2ef}{(f+d)^2(d+e)^2}.\] Let $o_9$ be the nine point center of $BIC$. By ratios on the Euler line, \begin{align*} o_9&=\frac{b+j+c-m}2\\ &=\frac{fd}{f+d}+\frac{de}{d+e}-\frac{def}{(f+d)(d+e)}\\ &=\frac{d(ef+fd+de)}{(f+d)(d+e)} \end{align*}and $\overline{o_9}=\frac{d+e+f}{(f+d)(d+e)}$. To show that the two circles are orthogonal, it suffices to show that the square of the distance between their centers equals the sum of the squares of their radii: \begin{align*} \left(\frac{s+t}2-o_9\right)\left(\overline{\frac{s+t}2-{o_9}}\right)&\overset?=R^2+\left(\frac{s-t}2\right)\left(\overline{\frac{s-t}2}\right)\\ (s+t-2o_9)(\overline s+\overline t-2\overline{o_9})&\overset?=4R^2+(s-t)(\overline s-\overline t)\\ 2(s\overline t+\overline st)-(4R^2-4o_9\overline{o_9})&\overset?=2o_9(\overline s+\overline t)+2\overline{o_9}(s+t) \end{align*} The second term of the left side is \begin{align*} 4R^2-4o_9\overline{o_9}&=\frac{4d^2ef}{(f+d)^2(d+e)^2}-\frac{4d(d+e+f)(ef+fd+de)}{(f+d)^2(d+e)^2}\\ &=\frac{-4d(e+f)(f+d)(d+e)}{(f+d)^2(d+e)^2}\\ &=\frac{-4d(e+f)}{(f+d)(d+e)}. \end{align*} The first term of the left side is \begin{align*} 2(s\overline t+\overline st)&=2\left(\frac{ef(e+f)}{e^2+f^2}\cdot\frac{2d-e-f}{d^2-ef}+\frac{e+f}{e^2+f^2}\cdot\frac{d(de+df-2ef)}{d^2-ef}\right)\\ &=2\left(\frac{ef(e+f)(2d-e-f)+(e+f)(d)(de+df-2ef)}{(d^2-ef)(e^2+f^2)}\right)\\ &=\frac{2(e+f)^2(d^2-ef)}{(d^2-ef)(e^2+f^2)}\\ &=\frac{2(e+f)^2}{e^2+f^2}.\\ \end{align*} The first term of the right side is \begin{align*} 2o_9(\overline s+\overline t)&=\frac{2d(ef+fd+de)}{(f+d)(d+e)}\left(\frac{e+f}{e^2+f^2}+\frac{2d-e-f}{d^2-ef}\right)\\ &=\frac{2d(ef+fd+de)(-e^3-2e^2f-2ef^2-f^3+d^2e+d^2f+2de^2+2df^2)}{(f+d)(d+e)(e^2+f^2)(d^2-ef)}. \end{align*} Adding it to its conjugate (the second term), we get \begin{align*} &\frac{2d(ef+fd+de)(-e^3-2e^2f-2ef^2-f^3+d^2e+d^2f+2de^2+2df^2)}{(f+d)(d+e)(e^2+f^2)(d^2-ef)}-\frac{2(d+e+f)(-d^2f^3-2d^2ef^2-2d^2e^2f-d^2e^3+e^2f^3+e^3f^2+2def^3+2de^3f)}{(f+d)(d+e)(e^2+f^2)(d^2-ef)}\\ &=\frac{2d(ef+fd+de)(-e^3-2e^2f-2ef^2-f^3+d^2e+d^2f+2de^2+2df^2)-2(d+e+f)(-d^2f^3-2d^2ef^2-2d^2e^2f-d^2e^3+e^2f^3+e^3f^2+2def^3+2de^3f)}{(f+d)(d+e)(e^2+f^2)(d^2-ef)}\\ &=\frac{10def(d^2f+d^2e-ef^2-e^2f)+6d(d^2e^3+d^2f^3-e^4f-ef^4)+4(d^4ef-e^3f^3)+2(d^4e^2+d^4f^2-e^2f^4-e^4f^2)}{(f+d)(d+e)(e^2+f^2)(d^2-ef)}, \end{align*}where the last step is a painless $48$-term expansion. It suffices to show that this equals $\frac{2(e+f)^2}{e^2+f^2}+\frac{4d(e+f)}{(f+d)(d+e)}$. Equivalently, we need to show that the numerator equals $(d^2-ef)(e+f)(2(e+f)(f+d)(d+e)+4d(e^2+f^2))=(d^2-ef)(e+f)(2d^2e+2d^2f+6de^2+6df^2+4def+2e^2f+2ef^2)$. This is true by a $28$-term expansion (in fact, only $20$ terms if you begin by multiplying the first and third factors).

From self polar-orthogonality we find that it is equivalent to show that $S$ lies on the polar of $T$ with respect to the nine point circle of $\triangle BIC$, which from here on I will refer to as $\Gamma$. Let $X$ be the foot of the altitude from $C$ to $\overline{BI}$ and define $Y$ similarly. Claim: $X$ and $Y$ lie on $\overline{EF}$. Proof. This is simply the statement of the Iran Lemma. $\blacksquare$ Now let the polar of $T$ meet $\Gamma$ at $P$ and $Q$, with $P$ closer to $A$. Also let $G = \overline{DI} \cap \overline{EF}$. It is well known that $G \in \overline{AM}$. Claim: $G \in \overline{PQ}$. Proof. Projecting we find, \begin{align*} -1 = (TM, BC) \overset{I}{=} (TG,XY) \end{align*}as desired. $\blacksquare$ Now finally let $N$ be the midpoint of $\overline{EF}$. Claim: $S \in \overline{PQ}$. Proof. Denote by $K = \overline{GS} \cap \overline{BC}$. Projecting once more we have, \begin{align*} - 1 = (AI,NS) &\overset{G}{=} (MD, TK)\\ &\overset{P}{=} (M, D; P, \overline{PK} \cap \Gamma) \end{align*}However noting that $Q$ satisfies $-1 = (MD,PQ)$, we must have $\overline{PK} \cap \Gamma = Q$. Then $Q$ lies on $\overline{PK}$, or equivalently $\overline{PS}$. $\blacksquare$

Sweet problem. I love how everything can be quoted as well known now a days lmao. Let $M$ be the midpoint of $BC$, $J$ the midpoint of $EF$, $P,Q$ the feet of altitudes from $B,C$ to $\overline{CI}, \overline{BI}$, and $K = AM \cap EF.$ Finally $X = BC \cap \overline{KS}.$ Note the 9-pt circle is $(PQDM)$. By Iran Lemma, $P, Q \in \overline{EF}.$ By EGMO 4.17, $K \in \overline{DI}.$ By Gregonne point and Ceve-Menelaus config, $(T, D; B, C) = -1.$ Also $(A, I; J, S) = -1$. So we have two harmonic bundles. Projecting through $I$, $(T, K; P, Q) = (T, D; B, C) = -1.$ Projecting through $K,$ $(M, D; T, X) = (A, I; J, S) = -1$. Hence $K, X$ lie on the polar of $T$ w.r.t the 9-pt circle. Since $K, S, X$ linear, $S$ does as well. By EGMO 9.27, that means the desired circles are orthogonal.

Let $K$ and $L$ be the feet from $C$ and $B$ to $BI$ and $CI$, respectively. Also denote $M$ as the midpoint of $BC$, $N$ as the midpoint of $EF$, and $D$ as the incircle touch point to $BC$. $(KLMD)$ is the 9-point circle of $\triangle BIC$. $K$ and $L$ lie on $EF$ by Iran lemma. $AM$, $EF$, and $DI$ concur at a point, which we call $X$. $BK$, $CL$, and $DI$ concur at the orthocenter of $\triangle BIC$, which we call $Y$. We then use the harmonic bundles \begin{align*} -1 &= (TD;BC) \overset{Y}{=} (TX;KL) \\ -1 &= (NS;IA) \overset{X}{=} (T, XS \cap BC;DM) \end{align*} to get that $S$ lies on the polar of $T$, which finishes. $\blacksquare$

Nice and cute problem (while I say that, this took me almost two hours to solve.... WHY AM I SO BAD AT GEO??? ) We define some new points. $D$ is the $A$-intouch point. $K \coloneqq \overline{EF} \cap \overline{BI}$ and $L \coloneqq \overline{EF} \cap \overline{CI}$. $M$ is the midpoint of $\overline{BC}$ and $N$ is the midpoint of $\overline{EF}$. $X \coloneqq \overline{AM} \cap \overline{EF} \cap \overline{DI}$ (its existence is well known). Remember this? (lol) Iran Lemma: The nine-point circle of $\triangle BIC$ is actually $(KLDM)$. Now see that \[(T,X;K,L) \overset{I}{=} (T,D;B,C)=-1 \text{ and }(T,\overline{BC} \text{ }\cap \text{ } \overline{SX};M,D) \overset{X}{=}(N,S;A,I)=-1\]which proves that $\overline{SX}$ is polar of $T$ with respect to the nine-point circle of $\triangle BIC$ which implies what is asked in the problem statement.

What we want to prove is equivalent to showing S lies on the polar of T, wrt. to the 9th point circle of $\triangle BIC$. Let BK be the altitude from B to IC and let CL be the altitude from C to IB. Now by the Iran lemma we know that $K \in EF$ and $L \in EF$. Let D be the intouch point of the incircle to BC. Let M be the midpoint of BC. Now we have that (KLDM) is the 9th point circle of $\triangle BIC$. Let (KLDM) be $\omega$. Let $EF \cap AM = X$, by lemma 4.17 $X \in DI$. Let N be the midpoint of EF. We have that $(N,S;A,I) = -1$. Now let $EF \cap BC = T$, its obvious by Ceva that $(T,D;B,C) = -1$. Projecting trough I we get $(T,D;B,C)\stackrel{I}{=}(T,X;L,K) = -1$. Also projecting $(N,S;A,I)$ trough X, we get $(N,S;A,I)\stackrel{X}{=}(T,SX \cap BC;M,D) = -1$. So we got $(T,X;L,K) = -1$ and $(T,SX \cap BC;M,D) = -1$. From $(T,X;L,K) = -1$, since L and K lie on $\omega$, then X lies on the polar of T. Let $SX \cap BC = Z$. From $(T,SX \cap BC;M,D) = -1$, since M and D lie on $\omega$, then Z lies on the polar of T $\Rightarrow$ XZ is the polar of T and since $S \in XZ$ by definition, then S lies on the polar of T, which is exactly what we wanted to prove so we are ready.

The conclusion can be rephrased as $S \in polar(T)$ (circles are orthogonal iff inverse of circle maps to itself. Circles are orthogonal implies $T$ maps to $T'$, still on the circle, which implies $\angle TT'S=90$, so $T'S$ is polar of $T.$ Conversely, if $S \in polar(T)$ and $T \in polar(S)$ then $T'$ and $S'$ are in the original circle. Furthermore, the points of intersection between two circles are unchanged, which implies circle maps to itself). Letting $R$ and $T$ be foot of altitudes from $B$ and $C$ to $CI$ and $BI$ respectively, by Iran lemma $R, T \in EF.$ Denote $D$ by tangency between $BC$ and incircle. Since lines $AD, BE, CF$ are concurrent (by Ceva) and points $T,E,F$ are collinear, $(T,D;B,C)=-1.$ Taking perspectivity at $I$ gives $(T, X; T, R)$, where $X = DI \cap EF \cap AM$ (these are indeed concurrent; follows from simson lines and homothethy). Thus, $X \in polar(T).$ Let $N$ be midpoint of $EF.$ We know $(I, A; N, S)=-1$, and perspectivity at $X$ gives $(D, M; XS \cap BC, T)=-1,$ which means $XS \cap BC \in polar(T).$ Combining this with $X \in polar(T)$ gives $S \in polar(T),$ so we are done. \qed

Denote $D$ as the remaining intouch point on $\triangle ABC$; $X$ and $Y$ as the feet of the altitudes from $B$ and $C$ in $\triangle BIC$, respectively; $M$ and $N$ as the midpoints of $\overline{BC}$ and $\overline{EF}$, respectively; $K$ as the concurrency point of $\overline{AM}$, $\overline{EF}$, $\overline{DI}$, which is well-known to exist; and $L$ as intersection of $\overline{KS}$ and $\overline{BC}$. Self-polar orthogonality implies we simply need $S$ to lie on the polar of $T$ with respect to the nine-point circle of $\triangle BIC$ (we will henceforth refer to this circle as $\omega$). Claim 1: $X$ and $Y$ lie on $\overline{EF}$ Proof: Notice that $X$ lies on the $C$-bisector of $\triangle ABC$, and that $\overline{BX} \perp \overline{CX}$. Hence, the Iran Lemma states that $X$ also lies on $\overline{EF}$. A similar argument works on $Y$ so we are done. $\square$ Claim 2: $K$ and $L$ lie on the polar of $T$ with respect to $\omega$ Proof: Observe that $I$ and $N$ lie on $\overline{AS}$ since $AE=AF$. Therefore, $AEIF$ is harmonic and we find \[-1 = (A,I;N,S) \overset{K}{=} (M,D;T,L).\] Moreover, it is well-known that $(T,D;B,C) = -1$. So, we have \[-1 = (T,D;B,C) \overset{I}{=} (T,K;Y,X).\] The first bundle implies $L$ lies on the polar of $T$ and the second bundle implies the same for point $K$. $\square$ Claim 2 implies that $\overline{KL}$ is the desired polar of $T$, but $S$ lies on $\overline{KL}$ by definition, hence done. $\square$

Invert about the incircle. The problem becomes: Quote: Let $\triangle DEF$ have circumcenter $I$. Let $P$ be such that $PE$ and $PF$ are tangent to the circumcircle of $\triangle DEF$. Let $T$ be the foot from $I$ to $PD$. Let $M$ and $N$ be the reflections of $I$ over $DF$ and $DE$, respectively. Prove that the circumcircle of $\triangle MDN$ is orthogonal to the circle through $P$ and $T$ whose center lies on line $EF$. Invert about point $D$ with arbitrary radius. The problem becomes: Quote: In $\triangle DEF$ let $I$ be the reflection of $D$ over $EF$ and let $T$ be the reflection of $D$ over the midpoint of $EF$. Let $P$ be the $D$-Humpty point. Let $M$ and $N$ be the reflections of $I$ over $DF$ and $DE$, respectively. Prove that line $MN$ is a diameter of the circle through $P$ and $T$ orthogonal to the circumcircle of $\triangle DEF$. At this point, let $(EFITP)$ meet $DF$ and $DE$ again at $X$ and $Y$, respectively. Then we have \[\measuredangle NYE=\measuredangle EYI=\measuredangle EFI=\measuredangle DFE=\measuredangle XFE=\measuredangle XYE,\]so line $XY$ passes through $N$, and similarly it must also pass through $M$. Invert about point $P$ with arbitrary radius. The problem becomes: Quote: In $\triangle PEF$ let $D$ be such that $DE$ and $DF$ are tangent to its circumcircle. Let $DP$ meet $EF$ at $T$. The circumcircle of $\triangle DPF$ meets $EF$ at $X\ne F$, and the circumcircle of $\triangle DPE$ meets $EF$ at $Y\ne E$. Prove that $T$ lies on the line joining the circumcenters of $\triangle DEF$ and $\triangle PXY$. Invert about $T$ with radius $\sqrt{TE\cdot TY}=\sqrt{TD\cdot TP}=\sqrt{TF\cdot TX}$. Then $(DEF)$ is sent to $(PXY)$, so $T$ is a similicenter of the two circles, implying the result.

Let the incircle touch $BC$ at $D$, the midpoint of $BC$ be $M$ ,let $BI \cap EF=K$,$CI \cap EF=G$ , let $BG \cap CK=H$, $HI \cap GK=J$ and $AI \cap EF=N$. Then we need to prove that $S$ lies on the polar of $T$ with respect to $(GDMK)$. Since $(T,J;G,K)=-1$, We get that $J$ lies on the polar of $T$ with respect to the required circle. Let $JS \cap BC=L$ then if we can prove that $(T,L;M,D)=-1$, we'd be done. Now note that $(A,I;N,S) \stackrel{J}{=} (M,D;T,L)=-1$ since it is well known that $\overline{A-J-M}$ are collinear. Now since $S$ lies on the polar of $T$ with respect to the nine point circle, we are done by self polar orthogonality.