Easy Mixtilinear Incircle Problem

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Easy Mixtilinear Incircle Problem

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Source: 2021 Taiwan TST Round 3 Independent Study 2-G

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USJL

#1 h May 1, 2021, 1:48 AM • 3 Y

Y by Rg230403, samrocksnature, itslumi

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 $\angle BAV=\angle DAC$ .

Proposed by usjl.

This post has been edited 1 time. Last edited by USJL, May 1, 2021, 8:02 AM

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#2 h May 1, 2021, 2:53 AM • 2 Y

Y by USJL, samrocksnature

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.

This post has been edited 1 time. Last edited by Rg230403, May 1, 2021, 2:53 AM

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USJL

#3 h May 1, 2021, 8:02 AM • 3 Y

Y by Rg230403, samrocksnature, Mango247

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!

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MP8148

#4 h May 1, 2021, 8:24 AM • 2 Y

Y by samrocksnature, MS_asdfgzxcvb

$[asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(145), B = dir(195), C = dir(345), I = incenter(A,B,C), M = dir(270), X = extension(A,I,B,C), H = foot(I,B,C), V = intersectionpoint(I--I+dir(dir(90)--I)*100,unitcircle), Tb = extension(I,I+dir(B--foot(B,A,I)),A,C), Tc = extension(I,I+dir(C--foot(C,A,I)),A,B), Q = circumcenter(Tb,Tc,V), O = origin, T = extension(A,A+dir(B--foot(B,A,O)),B,C), W = T+dir(T--V)*abs(T-B)*abs(T-C)/abs(T-V), W1 = 2*foot(W,O,M)-W, U = extension(A,M,W,V), M1 = extension(A,W1,V,M), Y = extension(X,X+dir(T--foot(T,A,I)),A,C), Z = extension(X,X+dir(T--foot(T,A,I)),A,B); draw(A--B--C--A^^unitcircle, heavyblue); draw(incircle(A,B,C)^^circumcircle(Tb,Tc,V), heavycyan); draw(W--A--W1--W^^A--M^^V--M, red); draw(A--T--B^^T--W, orange); draw(M1--U, red+dashed); draw(I--H^^Q--M1, heavygreen); draw(Y--Z--B^^circumcircle(A,Y,Z), blue+dotted); draw(Tb--Tc, blue+dotted); dot("$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$", Z, dir(225)); [/asy]$

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.

This post has been edited 3 times. Last edited by MP8148, May 20, 2021, 5:25 PM
Reason: Typo

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#5 h May 1, 2021, 1:22 PM • 3 Y

Y by Muaaz.SY, samrocksnature, Math-48

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 $NZ^*A=NY^*A=90$

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USJL

#6 h May 1, 2021, 2:00 PM • 1 Y

Y by samrocksnature

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?

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#7 h May 2, 2021, 12:14 AM • 4 Y

Y by khina, samrocksnature, KST2003, Mango247

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 $AI_a$

This post has been edited 1 time. Last edited by Muaaz.SY, May 2, 2021, 12:14 AM

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i3435

#8 h May 2, 2021, 12:55 AM • 2 Y

Y by samrocksnature, ehuseyinyigit

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 $TU$ .

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#9 h May 2, 2021, 3:38 AM • 1 Y

Y by itslumi

Solved with itslumi, khina, islander7, and me (although most of my contributions were "wait this works? what?").

Regardless, here's the solution:
$[asy] size(15cm); import geometry; point foot(pair A, pair B, pair C) { return projection(line(B,C))*A; } pair A=dir(155),B=dir(245),C=dir(325); triangle t=triangle(A,B,C); circle c=circumcircle(t); pair I=incenter(t),O=circumcenter(t),M=midpoint(B--C),M_A=extension(I,A,O,M),I_A=excenter(t.BC),D=foot(I_A,B,C),E=foot(I,B,C),T=intersectionpoint(tangents(c,A)[0],line(B,C)),X=extension(A,I_A,B,C),Y=intersectionpoint(perpendicular(X,line(A,I)),line(A,B)),Z=intersectionpoint(perpendicular(X,line(A,I)),line(A,C)),U=rotate(180,circumcenter(A,Y,Z))*A,AA=reflect(O,M_A)*A,V=intersectionpoints(T--U,c)[0],UU=foot(M,A,I); draw(A--B--C--cycle); clipdraw(c,red); draw(A--I_A,blue); draw(D--I_A^^I--E,purple); draw(A--T--B,brown); draw(Y--Z,deepgreen); draw(B--Y,dotted); clipdraw(circumcircle(A,Y,Z),orange); draw(A--AA,deepgreen); clipdraw(circumcircle(A,AA,D),dashed); draw(T--U,magenta); draw(circle(I,E,M),dashed+orange); dot(A); dot(AA); dot(B); dot(C); dot(D); dot(E); dot(I); dot(I_A); dot(M); dot(M_A); dot(T); dot(U); dot(UU); dot(V); dot(X); dot(Y); dot(Z); label("$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$",Z,N+NW); [/asy]$
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.

This post has been edited 2 times. Last edited by naman12, May 2, 2021, 3:39 AM

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L567

#10 h May 3, 2021, 6:32 AM • 1 Y

Y by Mango247

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!

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173 posts

#12 h May 12, 2021, 12:20 PM

Y by

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.

$[asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(140); pair B = dir(200); pair C = dir(340); pair I = incenter(A,B,C); pair O = circumcenter(A,B,C); pair M = 2*foot(O,A,I)-A; pair IA = 2*M-I; pair D = foot(IA,B,C); pair N = midpoint(B--C); pair D1 = 2*N-D; pair X = extension(A,I,B,C); pair T = extension(midpoint(A--X),bisectorpoint(A,X),B,C); pair Y = extension(rotate(90,X)*A,X,A,C); pair Z = extension(rotate(90,X)*A,X,A,B); pair U = 2*circumcenter(A,Y,Z)-A; pair T2 = 2*foot(A,M,N)-A; pair Y2 = foot(M,A,B); pair Z2 = foot(M,A,C); pair U2 = extension(A,M,Y2,Z2); draw(A--B--C--cycle, black+1); draw(A--IA); draw(A--T); draw(T--B); draw(A--IA); draw(Y--Z); draw(B--Y2); draw(IA--D, dotted); draw(I--D1, dotted); draw(Y2--Z2, mydash); draw(M--Y2, dotted); draw(M--Z2, dotted); draw(circumcircle(A,D1,D), mydash); draw(circumcircle(A,B,C)); draw(circumcircle(A,Y,Z), dotted); dot("$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$", U, dir(70)); [/asy]$

This post has been edited 2 times. Last edited by KST2003, May 12, 2021, 12:26 PM

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#13 h May 12, 2021, 6:05 PM • 5 Y

Y by Jerry37284, R.e.d.a.c.t.e.d, alexiaslexia, Mango247, R8kt

A solution that doesn’t use $\sqrt{bc}$ and thus making it 3 pages long :oops_sign:

:oops_sign:

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:

This post has been edited 4 times. Last edited by k12byda5h, May 13, 2021, 6:54 AM
Reason: pm me if there is typo

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#14 h Dec 23, 2024, 5:27 PM

Y by

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.

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#15 h Feb 7, 2025, 10:39 AM

Y by

Slightly different finishing, but basically same approach.
Let the line parallel to $BC$ passing through $A$ cuts $(ABC)$ at $T'$ , and $G$ be the tangency point of $BC$ and incircle. It's clear that $AT'DG$ is cyclic. Now $U'$ be the intersection of $(AT'DG)$ with line $XA$ . By some inversion stuff mentioned above it's sufficient to prove $AB*AC=AU*AU'$ .
Let $a,b,c$ be the side lengths of $\triangle ABC$ . And let $2*\alpha= \angle BAC$ .
$AU'*AU=(AX+XU')*AU=(AX+\frac{GX*XD}{AX})*AU=(AX^2+GX*XD)*\frac{AU}{AX}$ .
By some bash we get that: $GX=\frac{(b-c)(b+c+a)}{2(b+c)}$ , $DX=\frac{(b-c)(b+c-a)}{2(b+c)}$ , $AX^2=ab-\frac{bca^2}{(b+c)^2}$ $\implies$ $GX*XD+AX^2=\frac{(b+c+a)(b+c-a)}{4}$ . Note that $\frac{AX}{AU}=cos^2\alpha$ . So it's left to prove that: $(b+c+a)(b+c-a)=4bc*cos^2\alpha$ $\implies$ $(b+c)^2-a^2=4bc*cos^2\alpha$ $\implies$ $b^2+c^2-2bc(2cos^2\alpha-1)=a^2$ $\implies$ $b^2+c^2-2bc*cos(2\alpha)=a^2$ , which is clear by Cosine-Rule and we are done.

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everythingpi3141592

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everythingpi3141592

#16 h Mar 6, 2025, 4:49 AM

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Note that $\frac{BZ}{CY} = \frac{AB}{AC}$ (say by Menelaus or alternatively reflect $C$ and $T$ about $AI_a$ to get $ZXY$ bisects $BXC'$ and use angle bisector theorem twice). Now, note that $(ABC) \cap (AZY)$ is the spiral similarity centre sending $BZ$ to $CY$ , and thus $\frac{BN}{CN} = \frac{BZ}{CY} = \frac{AB}{AC}$ , and thus lies on the $A$ Apollonian circle. So this is $N$ , intersection of the $A$ symmedian with the circumcircle.

Now perform $\sqrt{bc}$ inversion, to see that $N$ goes to midpoint of $BC$ , $T$ becomes the intersection of a line through $A$ parallel to $BC$ with the circumcircle, and $U$ becomes the intersection of the internal angle bisector of $A$ with a line through $N$ perpendicular to the internal angle bisector of $A$ . It suffices to show that $ATDU$ is cyclic. Let $E$ be the touchpoint of incircle and since $ATDE$ is an isosceles trapezium, it suffices to show that $AEDU$ is cyclic. But for this let $F$ be the foot of the perpendicular from $A$ onto $BC$ , and noting that $AFUN$ is cyclic, we just need to show that $XF \cdot XN = XE \cdot XD$ ( $X$ is as defined in the problem) which we can do by length chasing. (Although it would be appreciated if someone has a nicer way to prove the last bit).

Also the configuration issues with the segment $TU$ can be fixed by noting that $B, E, D, C$ lie on a segment in this order.

This post has been edited 1 time. Last edited by everythingpi3141592, Mar 6, 2025, 4:53 AM
Reason: typo

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