AoPS - Problem

Problem

Source: TSTST 2021/6

Tags: USA TSTST, geometry

AwesomeYRY

13.12.2021 20:00

Triangles $ABC$ and $DEF$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A,F,B,D,C,$ and $E$ occur in this order along $\Omega$. Let $\Delta_A$ be the triangle formed by lines $AB,AC,$ and $EF,$ and define triangles $\Delta_B, \Delta_C, \ldots, \Delta_F$ similarly. Furthermore, let $\Omega_A$ and $\omega_A$ be the circumcircle and incircle of triangle $\Delta_A$, respectively, and define circles $\Omega_B, \omega_B, \ldots, \Omega_F, \omega_F$ similarly. (a) Prove that the two common external tangents to circles $\Omega_A$ and $\Omega_D$ and the two common external tangents to $\omega_A$ and $\omega_D$ are either concurrent or pairwise parallel. (b) Suppose that these four lines meet at point $T_A$, and define points $T_B$ and $T_C$ similarly. Prove that points $T_A,T_B$, and $T_C$ are collinear. Nikolai Beluhov

JNEW

13.12.2021 20:02

Really good geometry problem, appropriate for its placement. Or maybe I just like the techniques used. 6a. Note that the intersection of the two common external tangent to two circles is the center of the unique positive homothety mapping one circle to the other. Let $f_A$ be the positive homothety mapping $\Omega_A$ to $\Omega$, and define $f_D$ similarly. It suffices to show that $f_A$ and $f_D$ both map $\omega_A$ and $\omega_D$, respectively, to the same circle. Let $\triangle{A'B'C'}$ be the image of $\Delta_A$ under $f_A$ and define $\triangle{D'E'F'}$ similarly. Let their incenters be $J_A$ and $J_D$. Let $BC$ intersect $EF$ at $X$. Let $l = XI$ be the interior angle bisector of these two lines. Also let $X'$ be the circumecenter of $\triangle{AID}$. Our first claim is that $X'$ lies on $l$. This follows since if we let $l'$ be the interior angle bisector of $BF$ and $CE$, then angle chasing immediately shows that $l \perp l'$ but also that $l'$ is parallel to $A_1D_1$ where $A_1$ and $D_1$ are the midpoints of "minor" arcs $\overset{\huge\frown}{BC}$ and $\overset{\huge\frown}{EF}$ respectively. This shows $A_1D_1 \perp l$. But we also know $AA_1$ intersects $DD_1$ at $I$ and that from isogonal conjugates (read: angle chasing) in similar $\triangle{A_1ID_1}$ and $\triangle{AID}$ that $A_1D_1 \perp l$ implies that $X'$ lies on $l$ as claimed. Define $B_1$ and its variants in a similar fashion. We have from $CA || C'A'$ that $B_1$ is still the "minor" arc midpoint of $\overset{\huge\frown}{C'A'}$. As a combined corollary, $B_1C_1$ is the perpendicular bisector of $A'J_A$. We also already know $B_1C_1$ is the perpendicular bisector of $AI$ and hence passes through $X'$. From $AB || A'B'$ and $CA || C'A'$ we have that $BC' || B'C$ and moreover that both latter lines are parallel to the "interior" angle bisector of $BC$ and $B'C'$. But $B'C' || EF$, so $B'C || XI$. Now create a dummy point $N$ at the intersection of $AI$ and $BC$. We get: \begin{align*}\angle{IAA'} &= 180^{\circ} - \angle{ACA_1} - \angle{ACA'} = 180^{\circ} - \angle{ACA_1} - \angle{B'CB} \\ &= 180^{\circ} - \angle{ACA_1} - \angle{B'CB} = 180^{\circ} - \angle{INX} - \angle{IXN} \\ &= \angle{AIX'} \\ &= \angle{IAX'}\end{align*}So $A'$ also lies on $AX'$ which also gives that $J_A$ lies on $l$. Similarly $J_D$ lies on $l$. Recall that within our angle chase we showed $\overset{\huge\frown}{AA'} = \angle{BXE}$. Similarly we also have $\overset{\huge\frown}{DD'} = \angle{BXE}$. Thus, $IJ_A = AA' = DD' = IJ_D$, which proves $J_A = J_D$. From Poncelet's Porism, since the incircles of $\triangle{A'B'C'}$ and $\triangle{D'E'F'}$ have the same center, they are necessarily identical, which is exactly what we wanted to show. $\blacksquare$ 6b. Suppose that the common tangents to $\omega_A$ and $\omega_D$ meet at $P$. Let $I_A$ be the incenter of $\Delta_A$ and define $I_D$ similarly. In the spirit of 6a, let $f'_A$ be the positive homothety mapping $\omega_A$ to $\omega$, and define $f'_D$ similarly. They have centers at $A$ and $D$ respectively. Let the image of $\Delta_A$ under $f'_A$ be $\triangle{AB^*C^*}$ and define $\triangle{DE^*F^*}$ similarly. From 6a, we are already given that these share a circumcircle, $\Gamma$. Let $A_1^*$ be the intersection of $AI$ with $\Gamma$ and define $D_1^*$ similarly. Also let $I_A^*$ be the image of the $A$-excenter of $\Delta$ under $f'_A$ and define $I_D^*$ similarly. We know that $A_1^*$ is the midpoint of $II_A^*$. Define $f(X) = \text{Pow}(X, \omega) - \text{Pow}(X, \Omega)$. Linearity of Power of a Point states this function is linear. We have $\frac{I_AP}{I_DP} = \frac{r_A}{r_D} = \frac{r_A}{r} : \frac{r_D}{r} = \frac{AI_A}{AI} : \frac{DI_D}{DI}$ so after applying Menelaus twice we have that $P$ lies on $AD$ and $\frac{AP}{DP} = \frac{AI_A}{I_AI} : \frac{DI_D}{I_DI}$. So now we compute: \begin{align*} F(P) &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{I_AI}{AI_A} \cdot F(A) - \frac{I_DI}{DI_D} \cdot F(D)\right) \\ &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{I_AI}{AI_A} \cdot \text{Pow}(A, \omega) - \frac{I_DI}{DI_D} \cdot \text{Pow}(D, \omega)\right) \\ &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{I_AI}{AI_A} \cdot (AI^2 - r^2) - \frac{I_DI}{DI_D} \cdot (DI^2 - r^2)\right) \\ &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{I_AI}{AI_A} \cdot AI^2 - \frac{I_DI}{DI_D} \cdot DI^2\right) - r^2 \\ &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{II_A^*}{AI} \cdot AI^2 - \frac{II_D^*}{DI} \cdot DI^2\right) - r^2 \\ &= \frac{2}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(IA_1^* \cdot AI - ID_1^* \cdot DI\right) - r^2 \\ &= \frac{2}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\text{Pow}(I, \Gamma) - \text{Pow}(I, \Gamma)\right) - r^2 \\ &= -r^2 \end{align*}The locus of points where $F$ is constant is necessarily a line, so we are done. $\blacksquare$

VulcanForge

13.12.2021 20:25

Part (a) is implied by the following, which I haven't been able to prove: Quote: Let $B', C'$ be the points on $AB, AC$ such that $\overline{B'C'}$ is parallel to $\overline{EF}$ and tangent to $\omega$ (i.e. $\triangle AB'C'$ is the image of $\Delta_A$ under the homothety at $A$ sending $\omega_A$ to $\omega$). Similarly define $E',F'$. Then $AB'C'DE'F'$ are concyclic. can someone help? (amusingly, this statement is its own converse, if we consider $AB'C'DE'F'$ as the original hexagon)

Idio-logy

13.12.2021 20:46

@above maybe something like the following: Construct B’, C’ instead such that B’ lies on AB, C’ lies on AC, $\angle EDC = \angle FDB’$, and $\angle FDB = \angle EDC’$. By Pascal B’C’ is parallel to EF. The goal is to show this is the same definition as yours, i.e. B’C’ is tangent to $\omega$. Let X = B’C’ cap BC. We use DDIT on the point D and the quadrilateral ABXC’ and an inconic tangent to its four sides plus the side DE. (This inconic exists except when three of the five sides concur; if that happens, replace DE with DF and DF with DE in the argument below.) Then the involution is just “reflection across internal angle bisector of EDF”, so DF is tangent to the said inconic as well, so this inconic is just the incircle $\omega$, which shows that B’C’ is tangent to it. The rest is just angle chasing to prove AB’C’D concyclic (D is the Miquel point of ABXC'). A similar argument shows AE’F’D concyclic; finally, because $\angle F'AB' = \angle EAC = \angle EDC = \angle F'DB'$, the six points lie on the same circle.

a1267ab

14.12.2021 04:52

There was a lot of work put into the official solutions for this problem. I hope you enjoy some of them. Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_A$ and $I_A$ be the circumcenter and incenter of triangle $\Delta_A$, and define $O_B$, $I_B, \dotsc, I_F$ similarly. Let $\omega$ touch $EF$ at $A_1$, and define $B_1$, $C_1, \dotsc, F_1$ similarly.

All solutions to part (a) will prove the stronger claim that \[(\Omega_A\cup \omega_A)\sim (\Omega_D\cup \omega_D).\]The four lines will concur at the homothetic center of these figures (possibly at infinity). Solution 1 (author)Let the second tangent to $\omega$ parallel to $EF$ meet lines $AB$ and $AC$ at $P$ and $Q$, respectively, and let the second tangent to $\omega$ parallel to $BC$ meet lines $DE$ and $DF$ at $R$ and $S$, respectively. Furthermore, let $\omega$ touch $PQ$ and $RS$ at $U$ and $V$, respectively. Let $h$ be inversion with respect to $\omega$. Then $h$ maps $A$, $B$, and $C$ onto the midpoints of the sides of triangle $D_1E_1F_1$. So $h$ maps $k$ onto the Euler circle of triangle $D_1E_1F_1$. Similarly, $h$ maps $k$ onto the Euler circle of triangle $A_1B_1C_1$. Therefore, triangles $A_1B_1C_1$ and $D_1E_1F_1$ share a common nine-point circle $\gamma$. Let $K$ be its center; its radius equals $\frac{1}{2}r$. Let $H$ be the reflection of $I$ in $K$. Then $H$ is the common orthocenter of triangles $A_1B_1C_1$ and $D_1E_1F_1$. Let $\gamma_U$ of center $K_U$ and radius $\frac{1}{2}r$ be the Euler circle of triangle $UE_1F_1$, and let $\gamma_V$ of center $K_V$ and radius $\frac{1}{2}r$ be the Euler circle of triangle $VB_1C_1$. Let $H_U$ be the orthocenter of triangle $UE_1F_1$. Since quadrilateral $D_1E_1F_1U$ is cyclic, vectors $\overrightarrow{HH_U}$ and $\overrightarrow{D_1U}$ are equal. Consequently, $\overrightarrow{KK_U}=\frac{1}{2}\overrightarrow{D_1U}$. Similarly, $\overrightarrow{KK_V}=\frac{1}{2}\overrightarrow{A_1V}$. Since both of $A_1U$ and $D_1V$ are diameters in $\omega$, vectors $\overrightarrow{D_1U}$ and $\overrightarrow{A_1V}$ are equal. Therefore, $K_U$ and $K_V$ coincide, and so do $\gamma_U$ and $\gamma_V$. As above, $h$ maps $\gamma_U$ onto the circumcircle of triangle $APQ$ and $\gamma_V$ onto the circumcircle of triangle $DRS$. Therefore, triangles $APQ$ and $DRS$ are inscribed inside the same circle $\Omega_{AD}$. Since $EF$ and $PQ$ are parallel, triangles $\Delta_A$ and $APQ$ are homothetic, and so are figures $\Omega_A \cup \omega_A$ and $\Omega_{AD}\cup \omega$. Consequently, we have \[(\Omega_A\cup \omega_A)\sim (\Omega_{AD}\cup \omega)\sim (\Omega_D\cup \omega_D),\]which solves part (a). Solution 2 (Michael Ren)As in the previous solution, let the second tangent to $\omega$ parallel to $EF$ meet $AB$ and $AC$ at $P$ and $Q$, respectively. Let $(APQ)$ meet $\Omega$ again at $D'$, so that $D'$ is the Miquel point of $\{AB, AC, BC, PQ\}$. Since the quadrilateral formed by these lines has incircle $\omega$, it is classical that $D'I$ bisects $\angle PD'C$ and $BD'Q$ (e.g. by DDIT). Let $\ell$ be the tangent to $\Omega$ at $D'$ and $D'I$ meet $\Omega$ again at $M$. We have \[ \measuredangle(\ell, D'B) = \measuredangle D'CB = \measuredangle D'QP = \measuredangle(D'Q, EF). \]Therefore $D'I$ also bisects the angle between $\ell$ and the line parallel to $EF$ through $D'$. This means that $M$ is one of the arc midpoints of $EF$. Additionally, $D'$ lies on arc $BC$ not containing $A$, so $D'=D$. Similarly, letting the second tangent to $\omega$ parallel to $BC$ meet $DE$ and $DF$ again at $R$ and $S$, we have $ADRS$ cyclic. Lemma. There exists a circle $\Omega_{AD}$ tangent to $\Omega_A$ and $\Omega_D$ at $A$ and $D$, respectively. Proof. (This step is due to Ankan Bhattacharya.) It is equivalent to have $\measuredangle OAO_A = \measuredangle O_DDO$. Taking isogonals with respect to the shared angle of $\triangle ABC$ and $\Delta_A$, we see that \[\measuredangle OAO_A = \measuredangle(\perp EF, \perp BC) = \measuredangle (EF, BC).\](Here, $\perp EF$ means the direction perpendicular to $EF$.) By symmetry, this is equal to $\measuredangle O_DDO$. The circle $(ADPQ)$ passes through $A$ and $D$, and is tangent to $\Omega_A$ by homothety. Therefore it coincides with $\Omega_{AD}$, as does $(ADRS)$. Like the previous solution, we finish with \[(\Omega_A\cup \omega_A)\sim (\Omega_{AD}\cup \omega)\sim (\Omega_D\cup \omega_D).\] Solution 3 (Andrew Gu)Construct triangles homothetic to $\Delta_A$ and $\Delta_D$ (with positive ratio) which have the same circumcircle; it suffices to show that these copies have the same incircle as well. Let $O$ be the center of this common circumcircle, which we take to be the origin, and $M_{XY}$ denote the point on the circle such that the tangent at that point is parallel to line $XY$ (from the two possible choices, we make the choice that corresponds to the arc midpoint on $\Omega$, e.g. $M_{AB}$ should correspond to the arc midpoint on the internal angle bisector of $ACB$). The first diagram below shows the original triangles $ABC$ and $DEF$, while the second diagram shows the triangles homothetic to $\Delta_A$ and $\Delta_D$. [asy][asy] size(6cm); pair O = origin; pair Mbc = dir(-90); pair Mca = dir(40); pair Mab = dir(160); pair A = -Mca*Mab/Mbc; pair B = -Mab*Mbc/Mca; pair C = -Mbc*Mca/Mab; pair I = Mbc+Mca+Mab; pair Mef = dir(75); pair D = 2*foot(O, I, Mef)-Mef; pair X = midpoint(D--I); pair Y = X+5*dir(90)*(X-D); pair Mfd = intersectionpoints(X--Y, unitcircle)[0]; pair Mde = 2*foot(O, X, Mfd)-Mfd; pair E = -Mde*Mef/Mfd; pair F = -Mef*Mfd/Mde; draw(unitcircle); draw(A--B--C--cycle, red); draw(D--E--F--cycle, blue); draw(incircle(A, B, C)); string[] names = {"$A$", "$B$", "$C$", "$M_{BC}$", "$M_{CA}$", "$M_{AB}$", "$I$", "$D$", "$E$", "$F$", "$M_{FD}$", "$M_{DE}$", "$M_{EF}$"}; pair[] pts = {A, B, C, Mbc, Mca, Mab, I, D, E, F, Mfd, Mde, Mef}; pair[] labels = {A, B, C, Mbc, Mca, Mab, I, D, E, F, Mfd, Mde, Mef}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] [asy][asy] size(6cm); pair O = origin; pair Mbc = dir(-90); pair Mca = dir(40); pair Mab = dir(160); pair Mef = dir(75); pair oldI = Mbc+Mca+Mab; pair I = -Mef+Mca+Mab; pair oldD = 2*foot(O, oldI, Mef)-Mef; pair X = midpoint(oldD--oldI); pair Y = X+5*dir(90)*(X-oldD); pair Mfd = intersectionpoints(X--Y, unitcircle)[0]; pair Mde = 2*foot(O, X, Mfd)-Mfd; pair A = Mca*Mab/Mef; pair B = Mab*Mef/Mca; pair C = Mef*Mca/Mab; pair D = Mfd*Mde/Mbc; pair E = Mde*Mbc/Mfd; pair F = Mbc*Mfd/Mde; draw(unitcircle); draw(C--A--B, red); draw(B--C, blue); draw(F--D--E, blue); draw(E--F, red); draw(incircle(A, B, C)); string[] names = {" ", " ", " ", "$M_{BC}$", "$M_{CA}$", "$M_{AB}$", "$I$", " ", " ", " ", "$M_{FD}$", "$M_{DE}$", "$M_{EF}$"}; pair[] pts = {A, B, C, Mbc, Mca, Mab, I, D, E, F, Mfd, Mde, Mef}; pair[] labels = {A, B, C, Mbc, Mca, Mab, I, D, E, F, Mfd, Mde, Mef}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] Using the fact that the incenter is the orthocenter of the arc midpoints, the incenter of $\Delta_A$ in this reference frame is $M_{AB}+M_{AC}-M_{EF}$ and the incenter of $\Delta_D$ in this reference frame is $M_{DE}+M_{DF}-M_{BC}$. Since $ABC$ and $DEF$ share a common incenter, we have \[M_{AB}+M_{BC}+M_{CA}=M_{DE}+M_{EF}+M_{FD}.\]Thus the copies of $\Delta_A$ and $\Delta_D$ have the same incenter, and therefore the same incircle as well (Euler's theorem determines the inradius).

We present several solutions for this part of the problem. Solutions 3 and 4 require solving part (a) first, while the others do not. Solutions 1, 4, and 5 define $T_A$ solely as the exsimilicenter of $\omega_A$ and $\omega_D$, whereas solutions 2 and 3 define $T_A$ solely as the exsimilicenter of $\Omega_A$ and $\Omega_D$. Solution 1 (author)By Monge's theorem applied to $\omega$, $\omega_A$, and $\omega_D$, points $A$, $D$, and $T_A$ are collinear. Therefore, $T_A= AD\cap I_AI_D$. Let $p$ be pole-and-polar correspondence with respect to $\omega$. Then $p$ maps $A$ onto line $E_1F_1$ and $D$ onto line $B_1C_1$. Consequently, $p$ maps line $AD$ onto $X_A=B_1C_1\cap E_1F_1$. Furthermore, $p$ maps the line that bisects the angle formed by lines $AB$ and $EF$ and does not contain $I$ onto the midpoint of segment $A_1F_1$. Similarly, $p$ maps the line that bisects the angle formed by lines $AC$ and $EF$ and does not contain $I$ onto the midpoint of segment $A_1E_1$. So $p$ maps $I_A$ onto the midline of triangle $A_1E_1F_1$ opposite $A_1$. Similarly, $p$ maps $I_D$ onto the midline of triangle $D_1B_1C_1$ opposite $D_1$. Consequently, $p$ maps line $I_AI_D$ onto the intersection point $Y_A$ of this pair of midlines, and $p$ maps $T_A$ onto line $X_AY_A$. As in the solution to part (a), let $H$ be the common orthocenter of triangles $A_1B_1C_1$ and $D_1E_1F_1$. Let $H_A$ be the foot of the altitude from $A_1$ in triangle $A_1B_1C_1$ and let $H_D$ be the foot of the altitude from $D_1$ in triangle $D_1E_1F_1$. Furthermore, let $L_A=HA_1\cap E_1F_1$ and $L_D=HD_1\cap B_1C_1$. Since the reflection of $H$ in line $B_1C_1$ lies on $\omega$, $A_1H \cdot HH_A$ equals half the power of $H$ with respect to $\omega$. Similarly, $D_1H \cdot HH_D$ equals half the power of $H$ with respect to $\omega$. Then $A_1H \cdot HH_A = D_1H \cdot HH_D$ and $A_1HH_D\sim D_1HH_A$. Since $\angle HH_DL_A = 90^\circ = \angle HH_AL_D$, figures $A_1HH_DL_A$ and $D_1HH_AL_D$ are similar as well. Consequently, \[ \frac{HL_A}{L_AA_1}=\frac{HL_D}{L_DD_1}=s \]as a signed ratio. Let the line through $A_1$ parallel to $E_1F_1$ and the line through $D_1$ parallel to $B_1C_1$ meet at $Z_A$. Then $HX_A/X_AZ_A=s$ and $Y_A$ is the midpoint of segment $X_AZ_A$. Therefore, $H$ lies on line $X_AY_A$. This means that $T_A$ lies on the polar of $H$ with respect to $\omega$, and by symmetry so do $T_B$ and $T_C$. Solution 2 (author)As in the first solution to part (a), let $h$ be inversion with respect to $\omega$, let $\gamma$ of center $K$ be the common Euler circle of triangles $A_1B_1C_1$ and $D_1E_1F_1$, and let $H$ be their common orthocenter. Again as in the solution to part (a), $h$ maps $\Omega_A$ onto the nine-point circle $\gamma_A$ of triangle $A_1E_1F_1$ and $\Omega_D$ onto the nine-point circle $\gamma_D$ of triangle $D_1B_1C_1$. Let $K_A$ and $K_D$ be the centers of $\gamma_A$ and $\gamma_D$, respectively, and let $\ell_A$ be the perpendicular bisector of segment $K_AK_D$. Since $\gamma_A$ and $\gamma_D$ are congruent (both of them are of radius $\frac{1}{2}r$), they are reflections of each other across $\ell_A$. Inversion $h$ maps the two common external tangents of $\Omega_A$ and $\Omega_D$ onto the two circles $\alpha$ and $\beta$ through $I$ that are tangent to both of $\gamma_A$ and $\gamma_D$ in the same way. (That is, either internally to both or externally to both.) Consequently, $\alpha$ and $\beta$ are reflections of each other in $\ell_A$ and so their second point of intersection $S_A$, which $h$ maps $T_A$ onto, is the reflection of $I$ in $\ell_A$. Define $\ell_B$, $\ell_C$, $S_B$, and $S_C$ similarly. Then $S_B$ is the reflection of $I$ in $\ell_B$ and $S_C$ is the reflection of $I$ in $\ell_C$. As in the solution to part (a), $\overrightarrow{KK_A}=\frac{1}{2}\overrightarrow{D_1A_1}$ and $\overrightarrow{KK_D}=\frac{1}{2}\overrightarrow{A_1D_1}$. Consequently, $K$ is the midpoint of segment $K_AK_D$ and so $K$ lies on $\ell_A$. Similarly, $K$ lies on $\ell_B$ and $\ell_C$. Therefore, all four points $I$, $S_A$, $S_B$, and $S_C$ lie on the circle of center $K$ that contains $I$. (This is also the circle on diameter $IH$.) Since points $S_A$, $S_B$, and $S_C$ are concyclic with $I$, their images $T_A$, $T_B$, and $T_C$ under $h$ are collinear, and the solution is complete. Solution 3 (Ankan Bhattacharya)From either of the first two solutions to part (a), we know that there is a circle $\Omega_{AD}$ passing through $A$ and $D$ which is (internally) tangent to $\Omega_A$ and $\Omega_D$. By Monge's theorem applied to $\Omega_A, \Omega_D$, and $\Omega_{AD}$, it follows that $A, D$, and $T_A$ are collinear. The inversion at $T_A$ swapping $\Omega_A$ with $\Omega_D$ also swaps $A$ with $D$ because $T_A$ lies on $AD$ and $A$ is not homologous to $D$. Let $\Omega_A$ and $\Omega_D$ meet $\Omega$ again at $L_A$ and $L_D$. Since $ADL_AL_D$ is cyclic, the same inversion at $T_A$ also swaps $L_A$ and $L_D$, so $T_A=AD\cap L_AL_D$. By Taiwan TST 2014, $L_A$ and $L_D$ are the tangency points of the $A$-mixtilinear and $D$-mixtilinear incircles, respectively, with $\Omega$. Therefore $AL_A\cap DL_D$ is the exsimilicenter $X$ of $\Omega$ and $\omega$. Then $T_A, T_B,$ and $T_C$ all lie on the polar of $X$ with respect to $\Omega$. Solution 4 (Andrew Gu)We show that $T_A$ lies on the radical axis of the point circle at $I$ and $\Omega$, which will solve the problem. Let $I_A$ and $I_D$ be the centers of $\omega_A$ and $\omega_D$ respectively. By the Monge's theorem applied to $\omega$, $\omega_A$, and $\omega_D$, points $A$, $D$, and $T_A$ are collinear. Additionally, these other triples are collinear: $(A, I_A, I), (D, I_D, I), (I_A, I_D, T)$. By Menelaus's theorem, we have \[\frac{T_AD}{T_AA}=\frac{I_AI}{I_AA}\cdot\frac{I_DD}{I_DI}.\]If $s$ is the length of the side opposite $A$ in $\Delta_A$, then we compute \begin{align*} \frac{I_AI}{I_AA} &=\frac{s/\cos(A/2)}{r_A/\sin(A/2)} \\ &=\frac{2R_A\sin(A)\sin(A/2)}{\cos(A/2)} \\ &=\frac{4R_A\sin^2(A/2)}{r_A} \\ &=\frac{4R_Ar^2}{r_AAI^2}. \end{align*}From part (a), we know that $\frac{R_A}{r_A}=\frac{R_D}{r_D}$. Therefore, doing a similar calculation for $\frac{I_DD}{I_DI}$, we get \begin{align*} \frac{T_AD}{T_AA} &=\frac{I_AI}{I_AA}\cdot\frac{I_DD}{I_DI} \\ &=\frac{4R_Ar^2}{r_AAI^2}\cdot \frac{r_DDI^2}{4R_Dr^2} \\ &=\frac{DI^2}{AI^2}. \end{align*}Thus $T_A$ is the point where the tangent to $(AID)$ at $I$ meets $AD$ and $T_AI^2=T_AA\cdot T_AD$. This shows what we claimed at the start. Solution 5 (Ankit Bisain)As in the previous solution, it suffices to show that $\frac{I_AI}{AI_A}\cdot \frac{DI_D}{I_DI} = \frac{DI^2}{AI^2}$. Let $AI$ and $DI$ meet $\Omega$ again at $M$ and $N$, respectively. Let $\ell$ be the line parallel to $BC$ and tangent to $\omega$ but different from $BC$. Then \[ \frac{DI_D}{I_DI}=\frac{d(D, BC)}{d(BC, \ell)} = \frac{DB\cdot DC/2R}{2r} = \frac{MI^2-MD^2}{4Rr}. \]Since $IDM\sim IAN$, we have \[\frac{DI_D}{I_DI}\cdot \frac{I_AI}{AI_A} = \frac{MI^2-MD^2}{NI^2-NA^2}=\frac{DI^2}{AI^2},\]as desired.

Remark 1 (Author comments on generalization of part (b) with a circumscribed hexagram)Let triangles $ABC$ and $DEF$ be circumscribed about the same circle $\omega$ so that they form a hexagram. However, we do not require anymore that they are inscribed in the same circle. Define circles $\Omega_A$, $\omega_A, \dotsc, \omega_F$ as in the problem. Let $T^\text{Circ}_A$ be the intersection point of the two common external tangents to circles $\Omega_A$ and $\Omega_D$, and define points $T^\text{Circ}_B$ and $T^\text{Circ}_C$ similarly. Also let $T^\text{In}_A$ be the intersection point of the two common external tangents to circles $\omega_A$ and $\omega_D$, and define points $T^\text{In}_B$ and $T^\text{In}_C$ similarly. Then points $T^\text{Circ}_A$, $T^\text{Circ}_B$, and $T^\text{Circ}_C$ are collinear and points $T^\text{In}_A$, $T^\text{In}_B$, and $T^\text{In}_C$ are also collinear. The second solution to part (b) of the problem works also for the circumcircles part of the generalisation. To see that segments $K_AK_D$, $K_BK_E$, and $K_CK_F$ still have a common midpoint, let $M$ be the centroid of points $A$, $B$, $C$, $D$, $E$, and $F$. Then the midpoint of segment $K_AK_D$ divides segment $OM$ externally in ratio $3 : 1$, and so do the other two midpoints as well. For the incircles part of the generalisation, we start out as in the first solution to part (b) of the problem, and eventually we reduce everything to the following: Let points $A_1$, $B_1$, $C_1$, $D_1$, $E_1$, and $F_1$ lie on circle $\omega$. Let lines $B_1C_1$ and $E_1F_1$ meet at point $X_A$, let the line through $A_1$ parallel to $B_1C_1$ and the line through $D_1$ parallel to $E_1F_1$ meet at point $Z_A$, and define points $X_B$, $Z_B$, $X_C$, and $Z_C$ similarly. Then lines $X_AZ_A$, $X_BZ_B$, and $X_CZ_C$ are concurrent. Take $\omega$ as the unit circle and assign complex numbers $u$, $v$, $w$, $x$, $y$, and $z$ to points $A_1$, $F_1$, $B_1$, $D_1$, $C_1$, and $E_1$, respectively, so that when we permute $u$, $v$, $w$, $x$, $y$, and $z$ cyclically the configuration remains unchanged. Then by standard complex bash formulas we obtain that each two out of our three lines meet at $\varphi/\psi$, where \[\varphi = \sum_\text{Cyc} u^2vw(wx - wy + xy)(y - z)\]and \[\psi = {} - u^2w^2y^2 - v^2x^2z^2 - 4uvwxyz + \sum_\text{Cyc} u^2(vwxy - vwxz + vwyz - vxyz + wxyz).\] (But the calculations were too difficult for me to do by hand, so I used SymPy.) Remark 2 (Author comments on generalization of part (b) with an inscribed hexagram)Let triangles $ABC$ and $DEF$ be inscribed inside the same circle $\Omega$ so that they form a hexagram. However, we do not require anymore that they are circumscribed about the same circle. Define points $T^\text{Circ}_A$, $T^\text{Circ}_B$, $\ldots$, $T^\text{In}_C$ as in the previous remark. It looks like once again points $T^\text{Circ}_A$, $T^\text{Circ}_B$, and $T^\text{Circ}_C$ are collinear and points $T^\text{In}_A$, $T^\text{In}_B$, and $T^\text{In}_C$ are also collinear. However, I do not have proofs of these claims. Remark 3 (Further generalization from Andrew Gu)Let $ABC$ and $DEF$ be triangles which share an inconic, or equivalently share a circumconic. Define points $T^\text{Circ}_A$, $T^\text{Circ}_B, \dotsc, T^\text{In}_C$ as in the previous remarks. Then it is conjectured that points $T^\text{Circ}_A$, $T^\text{Circ}_B$, and $T^\text{Circ}_C$ are collinear and points $T^\text{In}_A$, $T^\text{In}_B$, and $T^\text{In}_C$ are also collinear. (Note that extraversion may be required depending on the configuration of points, e.g. excircles instead of incircles.) Additionally, it appears that the insimilicenters of the circumcircles lie on a line perpendicular to the line through $T^\text{Circ}_A$, $T^\text{Circ}_B$, and $T^\text{Circ}_C$.

iman007

14.12.2021 09:27

the vertices of $\Delta_A$ are $A,X_A,Y_A$ in clockwise order, $O_A$ and $I_A$ are the circumcenter and incenter of $\Delta_A$ resp. denote the center of mixtilinear incircle associated with $A$ and $\triangle ABC$ with $M_A$. Part a:

Let $P=AD' \cap A'D$. It is obvious that $P$ is fixed (since in (1) we already proved that $(A',D')$ and $(A,D)$ are anti-homologous pairs) so $T_A$ is on the polar of $P$ wrt $\Omega$. $\blacksquare$

[1] https://artofproblemsolving.com/community/q3h186117p1022749

chirita.andrei

24.04.2022 09:26

a) Let $T$ be the exsimilicenter of $\omega_A$ and $\omega_D$ and let $\delta$ be the homothety centered at $T$ which sends $\omega_D$ to $\omega_A$. Claim 1. $T$ lies on $AD$ ProofApply Monge Theorem on $ \omega_A,\omega_D,\omega$. Claim 2. $\delta$ sends the $A$ mixtlinear incircle of triangle $ABC$ (call it $ \Gamma_A$) to the $D$ mixtlinear incircle of triangle $DEF$ (call it $\Gamma_D$). ProofBy applying Monge Theorem on $\Gamma_A$, $\Gamma_D$ and $\omega$ it follows that the exsimilicenter of $\Gamma_A$ and $ \Gamma_D$ lies on $AD$. If $U_A,I_A$ are the centers of $ \Gamma_A,\omega_A$ (define $U_D,I_D$ similarly), then it suffices to prove that $ U_AU_D,I_AI_D,AD$ concur. From Menelaus Theorem, this is equivalent to $$ \frac{U_AI}{U_AA}\cdot\frac{DU_D}{U_DI}=\frac{I_AI}{I_AA}\cdot \frac{DI_D}{I_DI}$$ Note that the LHS is equal to $ \frac{AI^2}{DI^2} $ (using the fact that the incenter lies on the chord formed by the tangency points of the mixtlinear incircle and the sides). Also, note that $ \frac{DI_D}{I_DI}=\frac{d(D,BC)}{d} $, where $d$ is the diameter of $\omega$, hence the relation boils down to $$ \frac{d(D,BC)}{d(A,EF)}=\frac{AI^2}{DI^2}$$ Consider an inversion w.r.t. $\omega$. Let $A',D'$ be the images of $A,D$ under this inversion. The Euler circles of the intouch triangles of $ABC$ and $DEF$ are mapped to $\Omega$, hence they coincide. That means their orthocenters coincide. If $K_A,K_D$ are the tangency points of $\omega$ with $EF,BC$ respectively, then in complex numbers we have $2d'+k_a=2a'+k_d$ or $ 2a'-k_a=2d'-k_d $ i.e. the reflections of $K_A,K_D$ w.r.t. $A',B'$ respectively coincide. Call the point $S$. If $X,Y$ are the feet of perpendiculars from $I$ to $ SK_A,SK_D $ respectively, our relation becomes $$ \frac{XA'}{YB'}= \frac{SK_D}{SK_A} $$But note that if $Z$ is the midpoint of segment $K_AK_D$, then it's easy to prove that triangles $ ZA'X $ and $ ZB'Y$ are similar and this gives the conclusion. We will use Taiwan TST 2014 as a lemma several times. Let $AD$ meet for the second time $\Omega_D$ at $M$ and $\Omega_A$ at $N$. Claim 3. $\delta$ maps $M$ to $A$ and $D$ to $N$. ProofEnough to prove for $M,A$. Let $\lambda_D$ be the homothety sending $ \omega_D $ to $\omega$. If $K,H$ are the images of the intersection points $DE,DF$ with $BC$, we have to prove that $\lambda_D$ sends $M$ to $A$ or $ A,D,H,K$ are cocyclic. The line $KH$ is tangent to $\omega$ and parallel to $BC$. If $\Delta$ is the triangle whose incircle and circumcircle are $ \omega,\Omega $ respectively , and has one side parallel to $BC$ (exists by Poncelet Porism) then from Taiwan TST 2014 applied to $A,\Delta$ it follows that $A$ is the $\Delta$ mixtlinear incircle touchpoint with $\Omega$. Now if lines $ KH,BC $ intersect at $O$, then by Taiwan TST 2014 we have $AOFH$ and $AOKE$ are cyclic. Thus $A$ is the Miquel point of $KHFE$ and hence it lies on $(DKH)$ Let $\Omega$ be tangent to $\Gamma_A,\Gamma_D$ at $P,L$ respectively. From Taiwan TST 2014 it follows that $P,L$ lie on $\Omega_D,\Omega_A$ respectively. Claim 4. $ T,L,P $ collinear. ProofApply Monge Theorem on $ \Gamma_A,\Gamma_D,\Omega $ and use claim 2. Let $LP$ meet $\Omega_D$ for the second time at $V$. From Reim's Theorem we obtain that $MV$ and $AL$ are parallel, hence $\delta$ sends $V$ to $L$. Therefore $\delta$ maps triangle $VMD$ to triangle $LAN$, so it maps $\Omega_D$ to $ \Omega_A$ and we are done. b) Let $G$ be the exsimilicenter of $ \Omega,\omega$. From Monge Theorem on $ \Gamma_A,\Omega,\omega $ it follows that $A,G,P$ are collinear. Similarly, $D,G,L$ are collinear. Thus, from Brokard Theorem on $ADPL$ it follows that $T$ lies on the polar of $G$ w.r.t. $ \Omega $ and that gives the desired conclusion.

PHSH

27.08.2022 21:21

Great problem! It took me a lot of time but it worthwhile! First notice that by Poncelet's Porism, we can take $D$ as any point. Now make some construction to caracterize the extimilicenter of $(\Omega_A,\Omega_D).$ Let $E'\in DE, F'\in DF$ such that $E'F'\parallel BC$ and $E'F'$ is tangent to $\omega$ (i.e. the intersections of the reflection of $BC$ over $I.$ Similarly, define points $B'$ and $C'.$ Let $I$ be the center of $\omega.$ Claim 1: Points $A,E',F'$ and $D'$ are concyclic. Proof: Notice that since $EE'$ and $F'F$ concurr at $D$ and $A \in (DEF),$ our claim is equivalent to show that $A$ is the Miquel Point of lines $\overline{DE'E}, \overline{DFF'},\overline{EF}$ and $\overline{E'F'}.$ Therefore, if we let $H:=EF \cap E'F',$ it suffies to show that points $AHEF'$ are concyclic. In other words, we have to show that $$\measuredangle (AH,AE) =\measuredangle (F'H,F'F) = \measuredangle (E'F', FD) = \measuredangle (BC,FD) = $$$$=\measuredangle CAF + \measuredangle BAD $$On the other hand $$ \measuredangle (AH,AF) = \measuredangle HAC + \measuredangle CAF $$So we need to show that $\measuredangle BAD = \measuredangle HAC,$ i.e., that $AH$ and $AD$ are isogonal conjugate with respect to $\angle BAC.$ Now we use unheterred moving points. Fix$\triangle ABC$ and let $S_A$ be the tangency point of the $A$-mixtilinear incircle with $\Omega$ and let $X_A$ be the tangency point of $\omega$ with $\overline{BC}.$ Animate point $D$ over $\Omega,$ and let $Q$ be the second intersection of $DX_A$ with $\Omega$ and let $P:=\overline{EF} \cap \overline{BC}.$ First notice that by DDIT $P,Q$ and $A$ are colinear. Then $D$ has degree 2, then $P$ also has degree $2$ (since the map $D\rightarrow P$ is projective); then $Q$ has degree $2$ and therefore line $EF$ (the reflection of $\overline{BC}$ over $\overline{DI}$ has degree $2.$ Therefore $H$ has degree at most $2.$ On the other hand the isogonal of $AD$ has degree $1.$ Then we only have to check that $H$ lies on the isogonal conjugate of $AD$ for $4$ choices of $D.$ The degenerate cases $D=A,B,C$ are trivial; to finish, consider the case $D=S_A,$ then $\overline{EF}\parallel \overline{BC}$ (since $\overline{S_A}$ passes through the arc midpoint of $BAC$) and $H$ is the reflection of $X_A$ over $I;$ thus indeed $AS_A$ and $AH$ are isogonal due to the well known facts that $AH$ passes through $Y_A,$ the tangency point of the $A$-excircle with $\overline{BC},$ and the fact that $AT_A$ and $AY_A$ are isogonal. $\square.$ Similarly we show that $A,B',C',D$ are concyclic. Further, let $K:=BC\cap B'C'$ and let $D'$ and $A'$ be the second intersections of lines $AH$ and $DK,$ respectively. Then, we have that $D'$ is the reflection of $D$ over the perpendicualr bisector of $BC$ and similarly $A'$ is the reflection of $A$ over the perpendicular bisector of $EF.$ Now let $O_A$ and $O_D$ be the centers of $\Omega_A$ and $\Omega_D,$ respectively. We then claim the following Claim 2: Lines $AO_A,DO_D$ and the perpendicuar bisector of $AD$ concurr. Proof: Our claim is equivalent to show that $$\measuredangle (AD,AO_A) = \measuredangle (DO_D,AD)$$Fist note that since $AD$ and $AH$ are isogonal conjugate with respect to $\angle BAC$ and $AO_A$ is isogonal to the perpendicular from $A$ to $EF$ with respect to $\angle BAC,$ we have that $$\measuredangle (AD,AO_A) = 90 + \measuredangle (EF, AH) = 90 + \measuredangle = 90 + \measuredangle BAF + \measuredangle D'AB + \measuredangle ABE $$Similarly, $$\measuredangle (DO_D, AD) = 90 + \measuredangle CBD + \measuredangle CDA' + \measuredangle A'DA$$Therefore it suffies to show $$\measuredangle BAF + \measuredangle D'AB + \measuredangle ABE = \measuredangle CBD + \measuredangle BDF + \measuredangle FDA'$$which is obvious by the isogonal definition of $A'$ and $D'.~ \square$ Now let $R$ be the intersection of $AO_A, DO_D$ and the perpendicular bisector of $AD.$ Considering the homothety with center $A$ sending ${\triangle}_A$ to $\triangle AB'C',$ we must have that the center of $(AB'C')$ must lie on $AO_A$ and at the perpendicular bisector of $AD,$ (by the homothety and by Claim 1) therefore, $R$ is the center of $(AB'C'D).$ Analogously, we have that $R$ is the center of $(AE'F'D)$ and then the six points $A,D,B',C',E',F'$ must lie on a circle centered at $R.$ Considering the homothetys again we can state the following: Lemma wrote: Let $I_A$ and $I_D$ are the centers of ${\omega}_A$ and ${\omega}_D.$ We have that $(i)$ The points $A,D,B',C',E',F'$ lie on a circle ${\Gamma}_A$ which is tangent to ${\omega}_A$ at $A$ and to ${\omega}_D$ at $D.$ $(ii)$ The lines $O_AI_A, RI$ and $O_DI_D$ are pairwise parallel (equivalently, they concurr at the line at infinity). Now we are able to establish part (a); apply Monge's Theorem to $({\Omega}_A, {\Omega}_D, {\Gamma}_A)$ and $({\omega}_A, {\omega}_D, \omega)$ to conclude that the extimilicenters of $({\Omega}_A,{\Omega}_D)$ and $({\omega}_A,{\omega}_D)$ both lies on $AD.$ Then, it suffies to show that lines $O_AO_D,I_AI_D$ and $AD$ concurr; indeed, this follows from part $(ii)$ of the Lemma applying Desargue's Theorem to $\triangle AO_AI_A$ and $\triangle DO_DI_D$. Now we have that $T_A$ lies on $AD;$ we need to identify a second line. Let $S_D$ be the tangency point of the $D$-mixtilinear incircle of $\triangle DEF.$ We then have the following: Claim 3: $T_A$ lies on $S_AS_D.$ Proof: The key idea is to notice that the line $E'F'$ is the line passing through the intersections of $\Omega$ with the tangents from $S_A$ to $\omega;$ then we can find nice relations by swaping $A\leftrightarrow S_A$ and $D\leftrightarrow S_D$ and apply the Lemma. In which follows, dentoe by $\triangle (P,Q)$ the corresponding of $\Omega_A$ by swapping $A\leftrightarrow P$ and $D \leftrightarrow Q$ for any points $P,Q \in \Omega$ and further let $\Omega (P,Q)$ be the circumcircle of $\triangle P,Q$ (note that this is posible by Poncelet's Porism; for intance we have $\triangle (D,A)={\triangle}_A$.) Now consider the configuration generated by $D \leftrightarrow S_D; $ then it follows by applying the part (i) of the Lemma that $\Omega (S_D,S_A) = \Omega (A,D);$ similarly, we have that $\Omega (S_A,S_D) = \Omega (D,A).$ Then from applying the part (a) of the problem to the configuration $(A,D) \leftrightarrow (S_D,S_A)$ we have that the extimilincenter of $(\Omega (S_A,S_D), \Omega (S_D,S_A))$ lies on $S_AS_D;$ but by we have just shown that $(\Omega (S_A,S_D), \Omega (S_D,S_A))= ({\Omega}_D, {\Omega}_A). \square.$ Then we finally have that $T_A = AD \cap S_AS_D$ (the case $D=S_A$ is done by continuity). To finish, let $P:=AT_A \cap DT_D;$ therefore $T_A$ lies on the polar of $P$ with respect to $\Omega.$ Now notice that $P$ is a point not depending at $A$ or $D$ (i.e. it can be determined uniquely by $\omega$ and $\Omega$). Indeed, by applying Monge's Theorem to the incircle, circumcircle and mixtilinear incircle, $P$ is the extimilincenter of $\omega$ and $\Omega.$ Another way to see that $P$ is fixed, is to notice that the map $A\Rightarrow S_A$ is projective and furthermore an involution (since $S_A$ can be defined as the point such that the second intersection of $AI$ and $S_AI$ cut $\Omega$ in diametrically oppositive points), and therefore $AS_A$ must pass through a fixed point.

MathLuis

17.05.2023 23:04

Cant believe this geo was once something i called impossible and now here i'm :O, i'm pretty sure this is one of the best geos i've ever seen in my whole life so far (May 17th of 2023), congrats to Nikolai Beluhov for making such a good problem . Let $B',C'$ points in $AB,AC$ respectivily such that $B'C' \parallel EF$ and $B'C'$ tangent to $\omega$, let $E'F'$ points in $DE,DF$ respectivily such that $E'F' \parallel BC$ and $E'F'$ tangent to $\omega$, now let $DE$ hit $BC,AC$ at $D_2,E_2$ respectivily and let $DF$ hit $AB,BC$ at $F_2,D_1$ respectivily and finally let $EF$ hit $AB,AC$ at $F_1,E_1$ respectivily. Now let $\Omega_A \cap \Omega=A'$ and $\Omega_D \cap \Omega=D'$. Claim 1: $(\Omega_A \cap \omega_A) \sim (\Omega_D \cap \omega_D)$ (this implies part a) Proof: Let $D''$ the miquel point of complete quad $\{B'C',BC, AB, AC \}$, now clearly $\angle BD''B'=\angle (BC, B'C')=\angle CD''C'$ so $D''B, D''C'$ is isogonal w.r.t. $\angle B'D''C$, now by DDIT over that complete quad and inconic $\omega$ we get that $D''I$ bisects $\angle B'D''C, \angle BD''C'$, let $\ell_{D''}$ the tangent from $D''$ to $AB'D''C'$, we have $\angle (\ell_{D''}, D''B')=\angle B'C'D''=\angle BCD''= \angle CD''\infty_{E'F'}$ so $D''I$ bisects $\angle E'DF'$ as well this means that $D''=D$ so $AB'DC'$ is cyclic, now clearly $(AB'DC')$ is tangent to $\Omega_A$ but we can also show it is tangent to $\Omega_D$, notice that this is equivalent to $\angle D_1D_2D-\angle B'DC'=\angle D_1DB'$ but this holds as $\angle D_1D_2D-\angle B'DC'=\angle BD_2D-\angle BDC=\angle EDC=\angle D_1DB'$ thus $(AB'DC')$ is tangent to $\Omega_D$ as well, now notice that by the same process u get that $AE'DF'$ is cyclic and $(AE'DF')$ tangent to both $\Omega_A$ and $\Omega_D$ so they are basically the same circle, hence $AF'B'DC'E'$ is cyclic and $(AF'B'DC'E')$ tangent to $\Omega_A$ and $\Omega_D$ so now to finish we have $(\Omega_A \cap \omega_A) \sim ((AF'B'DC'E') \cap \omega) \sim (\Omega_D \cap \omega_D)$ thus we are done. Claim 2: $T_A, T_B, T_C$ all lie in the polar of $X_{57}$ w.r.t. $\Omega$ (implies part b) Proof: Notice that here $\triangle ABC$ and $\triangle DEF$ share same $X_{57}$, now by Taiwan TST 2014 R3 P3 we have that $D'$ is the A-mixtilinear intouch point of $\triangle ABC$ and $A'$ is the D-mixtilinear intouch point of $\triangle DEF$, by monge over $\Omega_A, \Omega_D, (AF'B'DC'E')$ we have that $AD$ passes through $T_A$ so make a $\sqrt{T_AA \cdot T_AD}$ inversion with center $T_A$, this sends $\Omega_A \to \Omega_D$ but since $ADD'A'$ is cyclic we have that $A' \to D'$ so $A'D' \cap AD=T_A$, now by monge its known that $AD' \cap DA'=X_{57}$ so by Brokard over $\Omega$ we have that $T_A$ lies in the polar of $X_{57}$, but now u can do the same for all $T_B, T_C$ so $T_A,T_B,T_C$ make a line which is the polar of $X_{57}$ w.r.t. $\Omega$ thus we are done . Wow, what a journey, waiting for more problems like this coming soon .

aleksijt

18.05.2023 15:11

Denote with $O_\bullet$ and $I_\bullet$ the centers of $\Omega_\bullet$ and $\omega_\bullet$, respectively. We first prove the following problem: Lemma wrote: Let $\triangle ABC$ have incircle $\omega$ and let $P$ be some point on the circumcircle $\Omega$. If $\triangle PQR$ is the triangle which has incircle $\omega$ and $QR\parallel BC$ prove that $APQR$ is cyclic.

Now we begin with some Claims. Claim 1: $A$ and $D$ are antihomologous points w.r.t. $\Omega_A$ and $\Omega_D$. Proof. Let $AD\cap\Omega_A=\left\{A,D'\right\}$, it then suffices $\angle AD'O_A=\angle ADO_D$ \[\Longleftrightarrow \angle EDA-\angle EDO_D=\angle CAD-\angle CAO_A\Longleftrightarrow \angle EDA-\angle CAD=\angle ABC-\angle DFE\]which is seen to be correct. Claim 2: $\left\{\Omega_A\cup\omega_A\right\}\sim\left\{\Omega_D\cup\omega_D\right\}$ which implies a). Proof. Taking the homothety centered at $A$, $\chi:\omega_A\rightarrow\omega$ we get from the lemma that $D\in\chi(\Omega_A)$ and hence $\chi(D')=D\Longrightarrow D'I_A\parallel DI_D$ and similarly $A'I_D\parallel AI_A$ where $A'$ is analogously defined. Let $T_A$ be the homothetic center of $\triangle AD'I_A\sim\triangle A'DI_D$ and we see that $T_A$ is the exsimilicenter of $\Omega_A$ and $\Omega_D$ and by Monge, the exsimilicenter of $\omega_A$ and $\omega_D$ is on $AD$ hence is $T_A$. From the Taiwan TST 2014 problem we know that $S_A$, the $A$-mixtilinear intouch point of $\triangle ABC$, is on $\Omega_D$ and similarly $S_D\in\Omega_A$. Claim 3: $S_A$ and $S_D$ are antihomologous w.r.t. $T_A$. Proof. This is true because of Claim 1 and the fact that pairs of antihomologous points form cyclic quadrilaterals (here $AS_DS_AD$ is cyclic). And for the final blow: $T_A$ is on the polar of the exsimilicenter $X$ of $\Omega$ and $\omega$ w.r.t. $\Omega$. This is due Monge $X=AS_A\cap DS_D$ and by Claims 1 and 3, $T_A=S_AS_D\cap AD$ hence Brokard's theorem on $AS_DS_AD$ finishes.