Hi, there must be some typo in the required condition. Could you please fix it? USJL wrote: Let $ABC$ be a triangle with $AB<AC$, and let $I_a$ be its $A$-excenter. Let $D$ be the projection of $I_a$ to $BC$. Let $X$ be the intersection of $AI_a$ and $BC$, and let $Y,Z$ be the points on $AC,AB$, respectively, such that $X,Y,Z$ are on a line perpendicular to $AI_a$. Let the circumcircle of $AYZ$ intersect $AI_a$ again at $U$. Suppose that the tangent of the circumcircle of $ABC$ at $A$ intersects $BC$ at $T$, and the segment $TU$ intersects the circumcircle of $ABC$ at $V$. Show that $\angle BAC=\angle DAC$. Proposed by usjl.

Rg230403 wrote: Hi, there must be some typo in the required condition. Could you please fix it? USJL wrote: Let $ABC$ be a triangle with $AB<AC$, and let $I_a$ be its $A$-excenter. Let $D$ be the projection of $I_a$ to $BC$. Let $X$ be the intersection of $AI_a$ and $BC$, and let $Y,Z$ be the points on $AC,AB$, respectively, such that $X,Y,Z$ are on a line perpendicular to $AI_a$. Let the circumcircle of $AYZ$ intersect $AI_a$ again at $U$. Suppose that the tangent of the circumcircle of $ABC$ at $A$ intersects $BC$ at $T$, and the segment $TU$ intersects the circumcircle of $ABC$ at $V$. Show that $\angle BAC=\angle DAC$. Proposed by usjl. Thank you. The typo is fixed now!

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(145), B = dir(195), C = dir(345), I = incenter(A,B,C), M = dir(270), X = extension(A,I,B,C), H = foot(I,B,C), V = intersectionpoint(I--I+dir(dir(90)--I)*100,unitcircle), Tb = extension(I,I+dir(B--foot(B,A,I)),A,C), Tc = extension(I,I+dir(C--foot(C,A,I)),A,B), Q = circumcenter(Tb,Tc,V), O = origin, T = extension(A,A+dir(B--foot(B,A,O)),B,C), W = T+dir(T--V)*abs(T-B)*abs(T-C)/abs(T-V), W1 = 2*foot(W,O,M)-W, U = extension(A,M,W,V), M1 = extension(A,W1,V,M), Y = extension(X,X+dir(T--foot(T,A,I)),A,C), Z = extension(X,X+dir(T--foot(T,A,I)),A,B); draw(A--B--C--A^^unitcircle, heavyblue); draw(incircle(A,B,C)^^circumcircle(Tb,Tc,V), heavycyan); draw(W--A--W1--W^^A--M^^V--M, red); draw(A--T--B^^T--W, orange); draw(M1--U, red+dashed); draw(I--H^^Q--M1, heavygreen); draw(Y--Z--B^^circumcircle(A,Y,Z), blue+dotted); draw(Tb--Tc, blue+dotted); dot("$A$", A, dir(135)); dot("$B$", B, dir(230)); dot("$C$", C, dir(350)); dot("$I$", I, dir(30)); dot("$M$", M, dir(270)); dot("$X$", X, dir(330)); dot("$H$", H, dir(225)); dot("$V$", V, dir(225)); dot("$Q$", Q, dir(90)); dot("$W$", W, dir(300)); dot("$W'$", W1, dir(240)); dot("$M'$", M1, dir(225)); dot("$U$", U, dir(45)); dot("$T$", T, dir(180)); dot("$Y$", Y, dir(45)); dot("$Z$", Z, dir(225)); [/asy][/asy] Let $M = \overline{AI} \cap (ABC)$ and $H$ be the foot from $I$ to $\overline{BC}$. Redefine $V$ to be the $A$-mixtilinear intouch point so that $\angle BAV = \angle CAD$; let $U = \overline{TV} \cap \overline{AM}$ and $W = \overline{TV} \cap (ABC)$, so that we want to show that $AYZU$ is cyclic. Let $Q$ be the center of $\omega_A$, the $A$-mixtilinear incircle. It is well-known that $I$ lies on the line through the tangency points of $\omega_A$ with $\overline{AB}$ and $\overline{AC}$, so it suffices to show that the homothety at $A$ taking $I$ to $X$ also sends $Q$ to $U$. To characterize $W$, note that by ratio lemma (with the usual $f(P) = PB/PC$) $$f(W) = \frac{f(T)}{f(V)} = \frac{f(A)^2}{f(A)f(H)} = \frac{f(A)}{f(H)}.$$Thus if $W'$ is the point on $(ABC)$ with $\overline{WW'} \parallel \overline{BC}$, we have $f(W') = f(W)^{-1} = f(H)/f(A)$, meaning $W' = \overline{AH} \cap (ABC)$. Observe that by the homothety at $A$ taking the incircle to $\omega_A$ and the homothety at $V$ taking the circumcircle to $\omega_A$, the point $M' = \overline{AW'} \cap \overline{VM}$ is the bottom point on $\omega_A$ (with $\overline{QM'} \parallel \overline{IH}$). By Pascal's on $AW'WVMM$, we have $\overline{UM'} \parallel \overline{BC} \equiv \overline{HX}$. Therefore $\triangle IHX$ and $\triangle QM'U$ are homothetic with center $A$, implying the conclusion.

Redefine $V$ as the $A-$mixtilinear point and $U=TV \cap AI$ we'll prove that $AZUY$ is cyclic and $N$ is the midpoint of arc $BC$ do $\sqrt{AB.AC}$ inversion then $V^*=D$ and $T^*$ on $(ABC)$ such that $AT^* \parallel BC$ so $U^*=AI \cap (AD'D)$ where $D'$ is the in-touch point of the incircle. let $M $ be the midpoint of $BC$ claim: $\angle MU^*A=90$ $\angle U^*D'D=\angle U^*AD=U^*IM$ so $ID'U^*D$ is cyclic $\blacksquare$ so it suffices to show that $Y^*Z^*$ are on the line throw $M$ perpendicular to $AI$ but that's trivial since $NZ^*A=NY^*A=90$

The solutions above show that one of the intersections of line $TU$ and the circumcircle of $ABC$ satisfies the condition. Is there any quick but rigorous way to show that the one lying between $T$ and $U$ is the point?

USJL wrote: The solutions above show that one of the intersections of line $TU$ and the circumcircle of $ABC$ satisfies the condition. Is there any quick but rigorous way to show that the one lying between $T$ and $U$ is the point? It's impossible to be the second point since $T$ and $D$ lie on different sides wrt $AI_a$

Similar to sqrt(bc) inversion of problem $\sqrt{bc}$ invert. Then we want to show the following: Let $M_A$ be the arc midpoint of $\widehat{BC}$. Let $Y,Z$ be the feet from $M_A$ to $\overline{AC},\overline{AB}$, and let $\overline{YZ}$ intersect $\overline{AI_a}$ at $U$. Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$. Prove that $A,A',D,U$ are concyclic. Let $M$ be the midpoint of $BC$. $\overline{YZ}$ is the Simson Line of $M_A$, so $M$ is on it. It is well known that $\overline{AD}||\overline{IM}$. Let $D_1$ be the foot from $I$ to $\overline{BC}$. $A,A',D_1,D$ are obviously cyclic, and so are $I,D_1,U,M$. $\measuredangle UD_1D=\measuredangle UD_1M=\measuredangle UIM=\measuredangle UAD$, as desired, unless you are pedantic. $\angle XAA'$ is acute and $D$ lies on segment $XC$ by the angle condition. Thus segment $AD$ is in between segments $AX$ and $AA''$. $\sqrt{bc}$ inverting gives the desired intersection point on $TU$.

Solved with itslumi, khina, islander7, and me (although most of my contributions were "wait this works? what?"). Regardless, here's the solution: [asy][asy] size(15cm); import geometry; point foot(pair A, pair B, pair C) { return projection(line(B,C))*A; } pair A=dir(155),B=dir(245),C=dir(325); triangle t=triangle(A,B,C); circle c=circumcircle(t); pair I=incenter(t),O=circumcenter(t),M=midpoint(B--C),M_A=extension(I,A,O,M),I_A=excenter(t.BC),D=foot(I_A,B,C),E=foot(I,B,C),T=intersectionpoint(tangents(c,A)[0],line(B,C)),X=extension(A,I_A,B,C),Y=intersectionpoint(perpendicular(X,line(A,I)),line(A,B)),Z=intersectionpoint(perpendicular(X,line(A,I)),line(A,C)),U=rotate(180,circumcenter(A,Y,Z))*A,AA=reflect(O,M_A)*A,V=intersectionpoints(T--U,c)[0],UU=foot(M,A,I); draw(A--B--C--cycle); clipdraw(c,red); draw(A--I_A,blue); draw(D--I_A^^I--E,purple); draw(A--T--B,brown); draw(Y--Z,deepgreen); draw(B--Y,dotted); clipdraw(circumcircle(A,Y,Z),orange); draw(A--AA,deepgreen); clipdraw(circumcircle(A,AA,D),dashed); draw(T--U,magenta); draw(circle(I,E,M),dashed+orange); dot(A); dot(AA); dot(B); dot(C); dot(D); dot(E); dot(I); dot(I_A); dot(M); dot(M_A); dot(T); dot(U); dot(UU); dot(V); dot(X); dot(Y); dot(Z); label("$A$",A,dir(A)); label("$A'$",AA,dir(AA)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,2*SE); label("$E$",E,2*SW); label("$I_A$",I_A,dir(I_A)); label("$I$",I,NE); label("$M$",M,2*NE); label("$M_A$",M_A,S); label("$T$",T,S); label("$U$",U,2*N); label("$U^*$",UU,S); label("$V$",V,S); label("$X$",X,2*N); label("$Y$",Y,S); label("$Z$",Z,N+NW); [/asy][/asy] We take a $\sqrt{bc}$ inversion, and denote the inverse of a point with $X^*$. We can note that the line $AT$ is tangent to $(ABC)$ and $T\in BC$, so thus $AT^* \parallel BC$ and $T^*\in (ABC)$, so thus if we represent $A'$ with the intersection of the line through $A$ parallel to $BC$ and $(ABC)$, we get $T^*=A'$. Now, as the line $AI$ is fixed, $X$ goes to the midpoint of arc $BC$, $M_A$. Note $(AM_AY^*)$ is orthogonal to $AM_A$, so $A$ and $M_A$ are antipodes in $(AM_AY^*)$, so thus $Y^*$ is the foot from $M_A$ to $AC$. Thus, by symmetry, $Z^*$ is the foot from $M_A$ to $AB$, and $Y^*Z^*$ is the Simson line of $M_A$ with respect to $\triangle ABC$, and thus passes through the midpoint of $BC$, $M$. Thus, as $AY=AZ$ and $\angle YAU=\angle ZAU$, we get $U$ is the antipode of $A$ in $(AYZ)$. Thus, we get by orthogonality, $U^*M\perp AI$. Thus, we have that $\angle MU^*I=\angle MEI=\frac\pi2$, implying that $IEU^*M$ is cyclic. Since $IM$ and $AD$ are parallel, by Reim's Theorem, we have $AEU^*D$ is cyclic by Reim's Theorem. However, we note that $AEDA'$ is an isosceles trapezoid, and thus $AA'DU^*$ is cyclic. Back to the problem, this means \[V^*=BC\cap(AA'U^*)=D\]so thus, upon inversion, the desired result is found.

Very similar to i3435's solution, but in more detail probably. Do a $\sqrt{bc}$ inversion. Because I'm bad at inversion, I assume everyone reading this is, so I'm going to explicitly write out which points go where. $BC$ and $(ABC)$ are swapped and $AI_a$ stays $AI_a$ and the incenter $I$ gets swapped with $I_a$ and vice versa. Since $X = AI_a \cap BC$, $X' = AI \cap (ABC)$, which is midpoint of minor arc $BC$. Since lines $AB, AC$ are fixed and $\angle AXY = 90^\circ$, $Y'$ is the foot of perpendicular from $X'$ onto $AC$ and similarly with $Z'$. Since $T$ lies on $BC$, $T'$ lies on $(ABC)$ and since its reflected across angle bisector, $T'$ is the point on $(ABC)$ with $AT' || BC$. Finally, since $U = (AYZ) \cap AI_a$, $U = Y'Z' \cap AI$. Since $Y'Z'$ is just the simson line of $X'$ on $\triangle ABC$ and so it passes through the foot of perpendicular from $X'$ onto $BC$, which is just the midpoint of $BC$, call it $M$. So, $U'$ is just the foot of perpendicular from $M$ onto $AI$. Since we want to prove $V$ is the mixtilinear touchpoint of $ABC$, we need to prove that $V'$ is the ex-touch point, and so, we can rewrite the problem as: In $\triangle ABC$ with incenter $I$, $M$ midpoint of $BC$, $E$ is the extouch point. $T$ is a point on $(ABC)$ such that $AT || BC$ and $U$ is the foot of perpendicular from $M$ onto $AI$. If $X$ is midpoint of minor arc $BC$, prove that $TEXU$ is cyclic. If we let $D$ on $BC$ such that $ID \perp BC$. Obviously, since $D,E$ are symmetric across $M$, $(ATE)$ passes through $D$ and so it suffices to show $ADUE$ is cyclic. Note that due to the right angles, $IDUM$ is cyclic and $IM || AE$. So, $\angle UAE = \angle UIM = \angle UDM = \angle UDE$ and we're done!

This is probably similar to the above solutions, but I'm gonna post it anyway because why not? Let $M,N$ be the midpoints of arc $BC$ and segment $BC$ respectively. Perform a $\sqrt{bc}$ inversion at $A$. Then $X\rightarrow M$, so $Y^\ast$ and $Z^\ast$ are foots of perpendiculars from $M$ to $\overline{AB}$ and $\overline{AC}$ respectively. $U^\ast$ is the intersection of $\overline{Y^\ast Z^\ast}$ with $\overline{AM}$. $T^\ast$ is the point on $(ABC)$, such that $AT^\ast\parallel BC$. We wish to show that $V\rightarrow D$. This is equivalent to showing that $A, T^\ast, U^\ast, D$ are concyclic. Since $\overline{Y^\ast Z^\ast}$ is the Simson line of $M$, it must pass through $N$, and by symmetry it is perpendicular to $\overline{AM}$. Therefore, $U^\ast$ must be the foot of perpendicular from $N$ to $\overline{AM}$. Let $I$ and $D'$ be the incenter and $A$-intouch point in $\triangle ABC$. Then obviously $I,D',U^\ast,N$ are concyclic due to right angles. It is known that $IN\parallel AD$, so by Reim's theorem, it follows that $A,D',D,U^\ast$ are concyclic. Meanwhile, $AD'DT^\ast$ is cyclic since it is an isosceles trapezoid, so we are done. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(140); pair B = dir(200); pair C = dir(340); pair I = incenter(A,B,C); pair O = circumcenter(A,B,C); pair M = 2*foot(O,A,I)-A; pair IA = 2*M-I; pair D = foot(IA,B,C); pair N = midpoint(B--C); pair D1 = 2*N-D; pair X = extension(A,I,B,C); pair T = extension(midpoint(A--X),bisectorpoint(A,X),B,C); pair Y = extension(rotate(90,X)*A,X,A,C); pair Z = extension(rotate(90,X)*A,X,A,B); pair U = 2*circumcenter(A,Y,Z)-A; pair T2 = 2*foot(A,M,N)-A; pair Y2 = foot(M,A,B); pair Z2 = foot(M,A,C); pair U2 = extension(A,M,Y2,Z2); draw(A--B--C--cycle, black+1); draw(A--IA); draw(A--T); draw(T--B); draw(A--IA); draw(Y--Z); draw(B--Y2); draw(IA--D, dotted); draw(I--D1, dotted); draw(Y2--Z2, mydash); draw(M--Y2, dotted); draw(M--Z2, dotted); draw(circumcircle(A,D1,D), mydash); draw(circumcircle(A,B,C)); draw(circumcircle(A,Y,Z), dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(225)); dot("$C$", C, dir(C)); dot("$I_A$", IA, dir(270)); dot("$I$", I, dir(30)); dot("$M$", M, dir(250)); dot("$D$", D, dir(315)); dot("$N$", N, dir(270)); dot("$D'$", D1, dir(45)); dot("$X$", X, dir(70)); dot("$T$", T, dir(180)); dot("$Y$", Y, dir(30)); dot("$Z$", Z, dir(180)); dot("$Y^\ast$", Y2, dir(180)); dot("$Z^\ast$", Z2, dir(30)); dot("$T^\ast$", T2, dir(T2)); dot("$U^\ast$", U2, 2*dir(250)); dot("$U$", U, dir(70)); [/asy][/asy]

A solution that doesn’t use $\sqrt{bc}$ and thus making it 3 pages long Let $M,N$ be the midarc of $BC$ contain, not containing $A$, $MN \cap BC = M'$, $I$ is the incenter of $\odot{ABC}$. It suffices to show that if $V$ is the touch point of $A-$ mixtillinear incircle and $\odot{ABC}$ ($\overline{M-I-V}$) then $\overline{V-U-T}$. Since $\frac{MM'}{M'N} = \frac{AX}{XU}$. By spiral similarity, equivalent to $\odot(ENU)$ is tangent to $MN$. Let $MX \cap \odot{ABC} = \{M,E\}$ and $F$ be the orthocenter of $\triangle MXN$. We rephrase the problem by setting $\triangle FMN$ as the main triangle as follows. Let $\triangle ABC$ be a triangle with feet of altitudes $D,E,F$, and orthocenter $H$. $I$ is the point on segment $BE$ such that $BI^2 = BD \cdot BC = BF \cdot BA$. $CI$ intersects $\omega = \odot(BCFE)$ again at $J$. $K$ is the point on $BE$ such that $\odot(BFK)$ is tangent to $BC$. $M$ is the midpoint of $AH$. Prove that $\overline{M-J-K}$ By angle chasing, $\angle EFK = 90^{\circ}$. Let $E'$ be the antipode of $E$ wrt. $\omega$. Let $N,P$ be the midpoint of $BC$ and $AB \cap NE$ respectively. By Pascal theorem $\omega$ with $EEBFFE'$, $\overline{P-M-K}$. Let $\Omega:= \odot(B,BI), \omega'=\odot{ACDF}$. Hence $I \in \Omega$. The line pass through $B$ and tangent to $\omega'$ at $Q$ (closer to $A$ than $C$) and $R$ (the other one). By Brokard theorem, $\overline{Q-R-H}$. By angle chasing, $\angle CQD = \angle CQB - \angle BQD = \angle QAC - \angle BCQ = \angle QAC - \angle DAQ = \angle DAC = \angle CBE$. Let $CQ \cap BE = I' \implies \odot{BDQI'}$. $\angle BI'D = \angle BCQ \implies I'=I$ and $\overline{C-I-Q}$, similarly, $\overline{A-I-R}$. Therefore $J$ is the midpoint of $QI$. Let $S = BH \cap AQ \cap CR$ and $T$ be the midpoint of $SR$. By Gauss line, $\overline{M-J-T}$. It suffices to show that $TJ, NE, BF$ are concurrent at $P$ $B,E,C,T,J \in \omega$. $(B,E;J,E')=C(B,E;I,\infty_{BE})=A(B,E;I,\infty_{BE}) \overset{\perp}{=} C(F,E';T,E)$. Done. This is a great example that shows how important inversion is.

Attachments:

We apply $\sqrt{bc}$ inversion at $A$ in $\triangle ABC$. And now note that $\bullet$ $X$ gets swapped with $N=\overline{AI} \cap (ABC)$. $\bullet$ $I_A$ gets swapped with the incenter $I$. $\bullet$ $T$ gets swapped with $A'$ which lies on $(ABC)$ and $\overline{AA'} \parallel \overline{BC}$. $\bullet$ Let $M$ be the midpoint of $\overline{BC}$. $\bullet$ Let $E$ be foot of $I$ onto $\overline{BC}$. See that $\overline{Y^*Z^*}$ is the Simson line of $N$ w.r.t $(ABC)$ so $M \in \overline{Y^*Z^*}$. Also note that $AEDA'$ is an isosceles trapezoid (as $M$ is the midpoint of $\overline{DE}$) and hence cyclic. Also see that $\measuredangle IEM=\measuredangle IU^*M=90^{\circ}$ (as $U^*$ is the midpoint of $\overline{Y^*Z^*}$). Now since $\overline{IM} \parallel \overline{AD}$ we have \[\frac{XE}{XU^*}=\frac{XI}{XM}=\frac{XA}{XD}\]and so $(ADEU^*)$ is cyclic and we are done.