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Source: 2019 Taiwan TST Round 3

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#1 h Apr 2, 2020, 7:22 AM • 2 Y

Y by amar_04, takachonn

Given a triangle $\triangle{ABC}$ with circumcircle $\Omega$ . Denote its incenter and $A$ -excenter by $I, J$ , respectively. Let $T$ be the reflection of $J$ w.r.t $BC$ and $P$ is the intersection of $BC$ and $AT$ . If the circumcircle of $\triangle{AIP}$ intersects $BC$ at $X \neq P$ and there is a point $Y \neq A$ on $\Omega$ such that $IA = IY$ . Show that $\odot\left(IXY\right)$ tangents to the line $AI$ .

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#3 h Apr 2, 2020, 10:31 AM • 3 Y

Y by AmirKhusrau, amar_04, enzoP14

I saw this problem about a year ago, then it suddenly got deleted after I solved it, quite surprise to see it again

Let $L$ be the reflection of $I$ with $O$
Let $D$ be the midpoint of arc $BC$ of $(O)$ not contain $A$ and $(J)$ tangent to $BC$ at $E$
From Euler's formula $JO^2=R^2+2RR_a$ , with $R, R_a$ circumradius of $(ABC)$ and $(J)$
Or $JO^2-R^2=R.JE$ , or $JD.JA=R.JT$
note that $JL = 2R$ and $JI = 2JD$ $\Rightarrow$ $JI.JA=JL.JT$ so $A, I, L, T$ are concyclic
By Miquel theorem $I, X, E, L$ are concyclic so $IX \perp IO$
From the definition of $Y$ we have $AY \perp IO$ too
Let $DY$ meet $BC$ at $K$
$\angle DKB = \angle YBD = \angle YAD = \angle IYA = \angle YIX$ so $K, I, X, Y$ are concyclic
We have $DI^2 = DY. DK$ $\Rightarrow$ $DI$ is tangent to $(IXY)$

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#4 h Apr 2, 2020, 10:49 AM • 2 Y

Y by amar_04, MS_asdfgzxcvb

Nice! Here's my solution: Let $M$ be the midpoint of $IJ$ , and let $I'$ be the reflection of $I$ in line $BC$ . Also suppose lines $AI,MY$ meet line $BC$ at $L,K$ . Invert about $M$ with radius $MI$ . Using Shooting Lemma, $Y \mapsto K$ and $A \mapsto L$ . Then, using inversive distance formula ( $r=MI$ ) and the fact that $IA=IY$ , we have $\frac{IK}{IL}=\frac{r^2 \cdot IY}{MI \cdot MY} \div \frac{r^2 \cdot IA}{MI \cdot MA}=\frac{MA}{MY}=\frac{MK}{ML}$ Since $L \in MI$ , so this means that $KL$ bisects $\angle IKM$ . In particular, point $I'$ is the reflection of $I$ in the angle bisector of $\angle IKM$ , and so $I' \in MK$ . Suppose $\odot (IKY)$ meets line $BC$ again at $X'$ . We claim that $X' \equiv X$ . Note that this finishes the problem, since we have $MK \cdot MY=MI^2$ , i.e. $\odot (IKY)$ is tangent to $MI$ . Thus it suffices to prove that $X' \in \odot (AIP)$ .

Let $H$ be the orthocenter of $\triangle JBC$ , and let $H'$ be its reflection in line $BC$ . Note that $HH'II'$ is a rectangle, and $M$ lies on the common perpendicular bisector of $HI',H'I$ . In particular, this gives us $\measuredangle YIX'=\measuredangle YKX'=\angle (I'M,BC)=\angle (BC,MH)$ Also, $\sqrt{JB \cdot JC}$ inversion about $J$ followed by antiparallelism in $\angle BJC$ gives that $JM \cdot JT=JB \cdot JC=JH \cdot JA \Rightarrow \frac{JM}{JA}=\frac{JH}{JT}$ where we use the fact that $\{H,A\}$ and $\{M,T\}$ are swapped in this process. This gives that $AT \parallel MH$ , and so $\measuredangle APX'=\angle (BC,MH)=\measuredangle YIX'$ Finally, since $KX'$ bisects $\angle IKY$ , so by Fact 5, $X'I=X'Y$ . Thus, we have $\measuredangle AIX'=\measuredangle X'YI=\measuredangle YIX'=\measuredangle APX' \Rightarrow X' \in \odot (AIP) \quad \blacksquare$
NOTE: The first paragraph is basically this KoMaL problem.

This post has been edited 2 times. Last edited by math_pi_rate, Apr 2, 2020, 11:02 AM

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#5 h Apr 6, 2020, 11:11 AM • 4 Y

Y by amar_04, sameer_chahar12, mueller.25, MS_asdfgzxcvb

$[asy] import olympiad; size(12cm); pair A=dir(108); pair B=dir(198); pair C=dir(343); draw(A--B--C--A,blue); draw(circumcircle(A,B,C),orange); pair O=circumcenter(A,B,C); pair I=incenter(A,B,C); pair M=circumcenter(B,I,C); pair J=2*M-I; draw(incircle(A,B,C),blue); pair T=reflect(B,C)*J; draw(T--J,dotted+purple); pair P=extension(A,T,B,C); path alice=circumcircle(I,A,P); pair X=intersectionpoints(alice,Line(B,C))[1]; draw(circumcircle(I,A,P),green+dotted); pair Y=reflect(O,I)*A; pair H=extension(O,I,J,T); draw(T--P,red); draw(P--B--C,blue); draw(I--Y,pink); draw(I--A,pink+pink+pink); draw(I--O--H,green); draw(circumcircle(I,X,Y),red); draw(A--I--M--J,yellow+orange); draw(O--M,purple); draw(circumcircle(I,A,H),dotted+blue); draw(I--X,brown); draw(A--Y,brown); path roo=circumcircle(I,X,Y); pair G=intersectionpoints(roo,Line(A,Y))[0]; dot("$A$",A,N); dot("$B$",B,W); dot("$C$",C,E); dot("$I$",I,E); dot("$J$",J,E); dot("$T$",T,E); dot("$X$",X,S); dot("$O$",O,E); dot("$Y$",Y,W); dot("$P$",P,S); dot("$H$",H,E); dot("$M$",M,S); dot("$G$",G,NW); [/asy]$
Proof: Define $M$ as the midpoint of $\widehat{BC}$ . We break the problem into several claims.

Claim 1: If $H= \overline{IO} \cap \overline{JT}$ then $\overline{IO}=\overline{OH}$ .

Proof: Notice that since $\overline{JT} \perp \overline{BC}$ and also $\overline{OM} \perp \overline{BC}$ $\implies$ $\overline{OM} \parallel \overline{JH}$ . Now by fact 5 we have that $M$ is the midpoint of $\overline{IJ}$ and so by thales theorem we are done $\square$ .

Claim 2: $(AIHT)$ is cyclic.

Proof: Denote by $R$ the circumradius. For this firstly notice that $\begin{cases} JI=4R \sin \left(\frac{A}{2} \right) \\ JA=4R\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)+4R \sin \left(\frac{A}{2} \right) \\ JH=2R \\ JT=8R \sin \left(\frac{A}{2} \right) \cos \left (\frac{B}{2} \right) \cos \left(\frac{C}{2} \right) \end{cases}$ Now we are all set to bash. Notice that we just need to show
$JI \cdot JA=JH \cdot JT$ $\iff 4R \sin \left(\frac{A}{2} \right) \times \left(4R\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)+4R \sin \left(\frac{A}{2} \right)\right)=2R \times 8R \sin \left(\frac{A}{2} \right) \cos \left (\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)$ $\iff\sin \left(\frac{A}{2} \right)=\cos \left(\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)-\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)$ $\iff 4\sin \left(\frac{A}{2} \right) \cos \left(\frac{A}{2} \right)=4 \cos \left(\frac{A}{2} \right) \cos \left(\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)-4\cos \left(\frac{A}{2} \right) \sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)$ $\iff 4\sin \left(\frac{A}{2} \right) \cos \left(\frac{A}{2} \right)=\sin A+\sin B+\sin C- \sin B-\sin C+\sin A$ which is trivial. $\square$ .

Now back to the main problem. Observe that $\angle XIJ=\angle APX=90^\circ-\angle PTJ=90^\circ-\angle HIJ=\angle AYI=\angle GYI$ where $G=\overline{AY} \cap \odot(IYX)$ . Now this implies that $\overline{GX} \parallel \overline{AI}$ $\implies$ $\angle YGX=\angle YAI=\angle IYA=\angle IYG$ $\implies$ $\overline{YG} \parallel \overline{IX}$ . This implies the required tangency $\blacksquare$ .
Edit

This post has been edited 2 times. Last edited by GeoMetrix, Apr 8, 2020, 6:15 PM

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#6 h Apr 8, 2020, 5:51 PM

Y by

@above Why does this imply tangency?

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#7 h May 25, 2020, 9:08 AM

Y by

Nice! Here is a synthetic solution

Firstly, we are going to prove the following $\textbf{lemma.}$
Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$ . Let $M=AI \cap (ABC)$ and $L$ be the reflection of $I$ with respect to $O$ . If $K$ was the $A$ -excircle touch point with $BC$ , then $ALKM$ is cyclic.
$\textcolor{black}{\textit{\textbf{proof.}}}$
Let $R=AI \cap BC$ , $N=MO \cap (ABC)$ and $J$ the $A$ -excenter. The perpendicular to $AI$ at $I$ meet $BC$ at $Q$ , finally $S=QM \cap (ABC)$ .
If $D$ was the projection of $I$ on $BC$ then clearly $K$ is the reflection of $D$ with respect to $MN$ , thus $LK \perp BC$ which implies that $L,K,J$ are collinear.
Some angle chaising gives
$\angle{QIB} = 90 - \angle{QIM} = 90 - \frac{\angle{A}}{2} - \frac{\angle{B}}{2} = \frac{\angle{C}}{2} = \angle{ICB}$ Hence $QI^2 = QB \cdot QC = QS \cdot QM$ , therefore
$\angle{MSI} = 180 - \angle{QSI} = 180 - \angle{QIM} = 90 = \angle{MSN}$ giving that $N,I,S$ are collinear. But notice that $O$ is the midpoint of $IL,MN$ thus $MINL$ is a parallelogram, so we get $ML \perp MQ$ and
$\Longrightarrow \angle{QML} = 90 = \angle{QKL}$ yields that $QMKL$ is cyclic.
It is clear that $\angle{QIJ} = 90 = \angle{QKJ}$ , so $QIKJ$ is cyclic giving that
$QR \cdot RK = \underbrace{IR \cdot RJ = BR \cdot RC}_{IBJC \cdots cyclic} = AR \cdot RM$ Thus $QMKA$ is cyclic, as we proved earlier $QMKL$ is cyclic, so both $A,L \in (PMK)$ and therefore $ALKM$ is cyclic, as required. $\blacksquare$
Now let $T$ be the reflection of $J$ with respect to $BC$ , clearly $K$ is the midpoint of $JT$ , but $M$ is the midpoint of $JI$ , thus $IT \parallel MK$ . Some angle chaising gives
$\angle{JTI} = \underbrace{\angle{JKM} = \angle{JAL}}_{AMKL \cdots cyclic}$ Eventually, $AILT$ is cyclic.
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Going back to the main problem with the result that $AILT$ is cyclic.
Since $IA=IY$ , $OA=OY$ , then $OI$ is the perpendicular bisector of $AY$ . Some angle chaising gives
$\angle{TPC} = 90 - \angle{ATL} = 90 - (180 - \angle{AIL}) = \angle{AIL} - 90 = \frac{360-\angle{AIY}}{2} - 90 = \frac{180-\angle{AIY}}{2} = \angle{IYA} =\angle{IAY}$ Now let the circle that passes through $Y,I$ and tangent to $AI$ meet $BC$ at $X'$ , and define $Z = MX' \cap (ABC)$ .
It is easy to get that
$\angle{MZB} = \frac{\angle{A}}{2} = \angle{MBX'}$ Thus $MI^2 = MB^2 = MX' \cdot MZ$ .
Hence $MI$ is tangent to $(X'IZ)$ , but $MI$ by defenition is tangent to $(X'IY)$ , therefore $X'YZI$ is cyclic.
By angle chaising
$\angle{X'IY} = \angle{X'ZY} = \angle{MZY} = \angle{MAY} = \angle{IAY}$ Thus
$\angle{APX'} + \angle{AIX'} = \angle{TPC} + \angle{YIA} + \angle{X'IY} = \angle{IAY}+ (180 - 2\angle{IAY}) + \angle{IAY} = 180$ So $X' \equiv (AIP) \cap BC \equiv X$ , as desired. $\blacksquare$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -394.07964514897344, xmax = 3123.5977929092946, ymin = -1732.9666354926225, ymax = 663.187963228231; /* image dimensions */ pen wwqqcc = rgb(0.4,0,0.8); pen ffdxqq = rgb(1,0.8431372549019608,0); pen qqccqq = rgb(0,0.8,0); pen ffqqtt = rgb(1,0,0.2); /* draw figures */ draw((-190.97556956311763,-53.03741002300586)--(-90.56751623206671,175.74320132216192), linewidth(0.8)); draw((-190.97556956311763,-53.03741002300586)--(191.06492092018888,-53.331761279518844), linewidth(0.8)); draw((191.06492092018888,-53.331761279518844)--(-90.56751623206671,175.74320132216192), linewidth(0.8)); draw(circle((0.08529317897995087,-0.4668565823161425), 198.16133921836283), linewidth(0.8)); draw((-56.481232916628954,34.7386829188209)--(56.65181927458886,-35.67239608345318), linewidth(0.8)); draw((56.65181927458886,-35.67239608345318)--(56.346464091924226,-431.9949568864847), linewidth(0.8)); draw((-0.06738441235236081,-198.62813698383192)--(56.65181927458886,-35.67239608345318), linewidth(0.8)); draw((-190.97556956311763,-53.03741002300586)--(-56.481232916628954,34.7386829188209), linewidth(0.8)); draw((-56.481232916628954,34.7386829188209)--(191.06492092018888,-53.331761279518844), linewidth(0.8)); draw(circle((90.56225737523314,144.90707117963547), 183.73584791542046), linewidth(0.8) + wwqqcc); draw((-90.56751623206671,175.74320132216192)--(56.65181927458886,-35.67239608345318), linewidth(0.8)); draw((-198.04476394896832,3.0540556987585767)--(-56.481232916628954,34.7386829188209), linewidth(0.8) + ffdxqq); draw((56.65181927458886,-35.67239608345318)--(56.930121985307444,325.5385780612209), linewidth(0.8)); draw((-315.7441641161175,-52.9412793941985)--(56.930121985307444,325.5385780612209), linewidth(0.8)); draw((-315.7441641161175,-52.9412793941985)--(-190.97556956311763,-53.03741002300586), linewidth(0.8)); draw((-111.14907630760793,-53.09891404987762)--(-198.04476394896832,3.0540556987585767), linewidth(0.8)); draw((-111.14907630760793,-53.09891404987762)--(-56.481232916628954,34.7386829188209), linewidth(0.8)); draw(circle((-213.35072768227604,71.43946723890106), 161.10551187882564), linewidth(0.8) + linetype("2 2") + qqccqq); draw(circle((-126.99367896883765,17.693078702614773), 72.54348813891694), linewidth(0.8) + ffqqtt); draw((-0.06738441235236081,-198.62813698383192)--(-191.10075891570028,51.64657590169209), linewidth(0.8)); draw((56.346464091924226,-431.9949568864847)--(-56.481232916628954,34.7386829188209), linewidth(0.8)); draw((-56.481232916628954,34.7386829188209)--(-90.56751623206671,175.74320132216192), linewidth(0.8) + ffdxqq); draw((-198.04476394896832,3.0540556987585767)--(56.65181927458886,-35.67239608345318), linewidth(0.8)); /* dots and labels */ dot((-90.56751623206671,175.74320132216192),linewidth(3pt) + dotstyle); label("$A$", (-82.12022055512265,186.63644958486628), NE * labelscalefactor); dot((-190.97556956311763,-53.03741002300586),linewidth(3pt) + dotstyle); label("$B$", (-183.55488008563856,-41.11306898364936), NE * labelscalefactor); dot((191.06492092018888,-53.331761279518844),linewidth(3pt) + dotstyle); label("$C$", (199.21742002951575,-41.11306898364936), NE * labelscalefactor); dot((-0.06738441235236081,-198.62813698383192),linewidth(3pt) + dotstyle); label("$M$", (7.831269971938604,-186.56654302740725), NE * labelscalefactor); dot((56.346464091924226,-431.9949568864847),linewidth(3pt) + dotstyle); label("$J$", (63.333253488635975,-420.0576460976502), NE * labelscalefactor); dot((-56.481232916628954,34.7386829188209),linewidth(3pt) + dotstyle); label("$I$", (-49.58457504533454,46.92456004283568), NE * labelscalefactor); dot((0.08529317897995087,-0.4668565823161425),linewidth(3pt) + dotstyle); label("$O$", (7.831269971938604,10.561191531896206), NE * labelscalefactor); dot((56.65181927458886,-35.67239608345318),linewidth(3pt) + dotstyle); label("$L$", (65.24711498921175,-23.888315478467504), NE * labelscalefactor); dot((-198.04476394896832,3.0540556987585767),linewidth(3pt) + dotstyle); label("$Y$", (-191.21032608794164,14.38891453304773), NE * labelscalefactor); dot((56.930121985307444,325.5385780612209),linewidth(3pt) + dotstyle); label("$T$", (65.24711498921175,337.8315081303515), NE * labelscalefactor); dot((-315.7441641161175,-52.9412793941985),linewidth(3pt) + dotstyle); label("$P$", (-307.9558776230637,-41.11306898364936), NE * labelscalefactor); dot((-111.14907630760793,-53.09891404987762),linewidth(3pt) + dotstyle); label("$X'$", (-103.17269706145615,-41.11306898364936), NE * labelscalefactor); dot((-191.10075891570028,51.64657590169209),linewidth(3pt) + dotstyle); label("$Z$", (-183.55488008563856,62.235452047441775), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This post has been edited 1 time. Last edited by Fouad-Almouine, May 25, 2020, 10:52 AM

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#8 h Aug 8, 2020, 5:04 PM • 5 Y

Y by Jerry37284, Afternonz, Tudor1505, Muaaz.SY, R8kt

I use " $J'$ " instead of " $T$ "

Let $M=AI \cap \odot(ABC) \ne A$ , $I'$ be the reflection of $I$ across $BC$ , $A'$ be the antipode of $A$ and incircle touch $BC$ at $D$ . Since, $\frac{AI}{AA'}=\frac{ID}{IM}$ We get $\triangle AIA' \sim \triangle IDM$ then $\triangle AIO \sim \triangle II'M$ . By angle-chasing, $I',M,Y$ are collinear. Let $S=AI \cap BC$ . By power of points, $A,I',M'J'$ are concyclic. $X'=MI' \cap BC$ . $\angle SAP = \angle SI'X'=\angle SIX'$ . Therefore, $AP \parallel IX'$ . By power points, $\frac{SI}{SX}=\frac{SP}{SA}=\frac{SX'}{SI},SI^2=SX \cdot SX'$ . So, $\omega_1=\odot(IXX')$ tangents to $AI$ and $\omega_2=\odot(IYX)$ tangents to $AI$ ( $MI^2=MX' \cdot MY$ ). So, $\omega_1=\omega_2$ . Therefore, $\odot(IXY)$ tangents to $AI$ .

Attachments:

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435 posts

#10 h Mar 12, 2021, 3:32 PM • 3 Y

Y by Mango247, Mango247, Mango247

GeoMetrix wrote:

$[asy] import olympiad; size(12cm); pair A=dir(108); pair B=dir(198); pair C=dir(343); draw(A--B--C--A,blue); draw(circumcircle(A,B,C),orange); pair O=circumcenter(A,B,C); pair I=incenter(A,B,C); pair M=circumcenter(B,I,C); pair J=2*M-I; draw(incircle(A,B,C),blue); pair T=reflect(B,C)*J; draw(T--J,dotted+purple); pair P=extension(A,T,B,C); path alice=circumcircle(I,A,P); pair X=intersectionpoints(alice,Line(B,C))[1]; draw(circumcircle(I,A,P),green+dotted); pair Y=reflect(O,I)*A; pair H=extension(O,I,J,T); draw(T--P,red); draw(P--B--C,blue); draw(I--Y,pink); draw(I--A,pink+pink+pink); draw(I--O--H,green); draw(circumcircle(I,X,Y),red); draw(A--I--M--J,yellow+orange); draw(O--M,purple); draw(circumcircle(I,A,H),dotted+blue); draw(I--X,brown); draw(A--Y,brown); path roo=circumcircle(I,X,Y); pair G=intersectionpoints(roo,Line(A,Y))[0]; dot("$A$",A,N); dot("$B$",B,W); dot("$C$",C,E); dot("$I$",I,E); dot("$J$",J,E); dot("$T$",T,E); dot("$X$",X,S); dot("$O$",O,E); dot("$Y$",Y,W); dot("$P$",P,S); dot("$H$",H,E); dot("$M$",M,S); dot("$G$",G,NW); [/asy]$
Proof: Define $M$ as the midpoint of $\widehat{BC}$ . We break the problem into several claims.

Claim 1: If $H= \overline{IO} \cap \overline{JT}$ then $\overline{IO}=\overline{OH}$ .

Proof: Notice that since $\overline{JT} \perp \overline{BC}$ and also $\overline{OM} \perp \overline{BC}$ $\implies$ $\overline{OM} \parallel \overline{JH}$ . Now by fact 5 we have that $M$ is the midpoint of $\overline{IJ}$ and so by thales theorem we are done $\square$ .

Claim 2: $(AIHT)$ is cyclic.

Proof: Denote by $R$ the circumradius. For this firstly notice that $\begin{cases} JI=4R \sin \left(\frac{A}{2} \right) \\ JA=4R\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)+4R \sin \left(\frac{A}{2} \right) \\ JH=2R \\ JT=8R \sin \left(\frac{A}{2} \right) \cos \left (\frac{B}{2} \right) \cos \left(\frac{C}{2} \right) \end{cases}$ Now we are all set to bash. Notice that we just need to show
$JI \cdot JA=JH \cdot JT$ $\iff 4R \sin \left(\frac{A}{2} \right) \times \left(4R\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)+4R \sin \left(\frac{A}{2} \right)\right)=2R \times 8R \sin \left(\frac{A}{2} \right) \cos \left (\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)$ $\iff\sin \left(\frac{A}{2} \right)=\cos \left(\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)-\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)$ $\iff 4\sin \left(\frac{A}{2} \right) \cos \left(\frac{A}{2} \right)=4 \cos \left(\frac{A}{2} \right) \cos \left(\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)-4\cos \left(\frac{A}{2} \right) \sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)$ $\iff 4\sin \left(\frac{A}{2} \right) \cos \left(\frac{A}{2} \right)=\sin A+\sin B+\sin C- \sin B-\sin C+\sin A$ which is trivial. $\square$ .

Now back to the main problem. Observe that $\angle XIJ=\angle APX=90^\circ-\angle PTJ=90^\circ-\angle HIJ=\angle AYI=\angle GYI$ where $G=\overline{AY} \cap \odot(IYX)$ . Now this implies that $\overline{GX} \parallel \overline{AI}$ $\implies$ $\angle YGX=\angle YAI=\angle IYA=\angle IYG$ $\implies$ $\overline{YG} \parallel \overline{IX}$ . This implies the required tangency $\blacksquare$ .
Edit

Why is $90^\circ-\angle HIJ=\angle AYI$ ????

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#11 h May 25, 2021, 6:04 PM • 1 Y

Y by SK_pi3145

Beautiful problem!

Let $\triangle LMN$ be the circumcevian triangle of $I$ , and let $D$ and $D'$ be the incircle and $A$ -excircle touch points. Let $Q$ be the reflection of $D$ over $I$ and let $R$ be the reflection of $I$ over $Q$ . Finally, let $E$ be the reflection of $X$ over $I$ .

Claim 1: $E$ lies on $\overline{MN}$ .

Proof. Consider the homothety $\mathcal{H}$ centered at $A$ taking the incircle to $A$ -excircle. This takes $\{I,Q,R\}$ to $\{J,D',T\}$ , so $A,R,T$ are collinear. It is also easy to see that $E$ lies on the perpendicular bisector of $IR$ . Now angle chasing shows that
$\measuredangle IEQ = \measuredangle IXD = \measuredangle IAR$ which means that $E$ is in fact the circumcenter of $\triangle AIR$ . Hence it lies on the perpendicular bisector of $AI$ which is $\overline{MN}$ by the incenter lemma. $\square$

Let $O$ be the circumcenter of $\triangle ABC$ . Since $IE=IX$ , by Butterfly theorem it follows that $OI \perp XE$ . Since $Y$ is the reflection of $A$ over $\overline{OI}$ , angle chasing finally gives us
$\measuredangle IYX = -\measuredangle IAE = \measuredangle AIE = \measuredangle XIJ$ and we're done. $\blacksquare$
$[asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(123); pair B = dir(228); pair C = dir(312); pair I = incenter(A,B,C); pair O = circumcenter(A,B,C); pair L = 2*foot(O,A,I)-A; pair M = 2*foot(O,B,I)-B; pair N = 2*foot(O,C,I)-C; pair J = 2*L-I; pair T = 2*foot(J,B,C)-J; pair P = extension(A,T,B,C); pair D = foot(I,B,C); pair D1 = foot(J,B,C); pair Q = 2*I-D; pair R = 2*Q-I; pair E = circumcenter(A,R,I); pair X = 2*I-E; pair Y = 2*foot(A,O,I)-A; draw(A--B--C--cycle, black+1); draw(A--J); draw(D--R); draw(J--T); draw(A--P); draw(E--X); draw(B--M, mydash); draw(C--N, mydash); draw(M--N); draw(C--P); draw(E--Q); draw(O--I); draw(circumcircle(I,X,Y)); draw(circumcircle(A,B,C)); draw(circle(I, abs(I-D))); draw(circumcircle(B,I,C), dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(315)); dot("$I$", I, dir(180)); dot("$O$", O, dir(180)); dot("$L$", L, dir(250)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$J$", J, dir(270)); dot("$T$", T, dir(60)); dot("$P$", P, dir(0)); dot("$D$", D, dir(270)); dot("$Q$", Q, dir(135)); dot("$R$", R, dir(90)); dot("$E$", E, dir(90)); dot("$X$", X, dir(270)); dot("$Y$", Y, dir(Y)); dot("$D'$", D1, dir(315)); [/asy]$

This post has been edited 1 time. Last edited by KST2003, May 25, 2021, 6:05 PM

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#12 h Jun 6, 2024, 6:35 PM • 1 Y

Y by GeoKing

Beautiful problem! similar solution to above, posting for storage

Let $O$ denote the circumcenter and $\Delta DEF$ be the intouch triangle. Reflect $D$ across $I$ to $S$ and reflect $I$ across $S$ to $K$ . Note that the homothety sending the incircle to the $A$ -excircle sends $K$ to $T$ , so $K$ lies on $AT$ (and we can erase $T,J$ from the diagram). Let $Z$ be the midpoint of $AI$ , then by midpoint theorem we have $ZS \parallel AP$ . Furthermore let $L$ be the projection of $A$ on $ID$ , then it is easy to show that $\Delta LEF$ and $\Delta ABC$ are oppositely similar, $Z$ is the circumcenter of $\Delta LEF$ and $S$ is the incenter of $\Delta LEF$ by Fact 5. So by similarity we have, in formal sum,
$\begin{align*} ZS + IO &= EF + BC \\ AP + IO &= (AI + 90^\circ) + BC \\ (AI + BC - XI) + IO &= (AI + 90^\circ) + BC \\ IO - XI = 90^\circ \end{align*}$ so $\angle XIO = 90^\circ$ .

Finally, let $BI$ and $CI$ meet $\Omega$ again at $S_b, S_c$ , then by Butterfly Theorem we have that the reflection $X'$ of $X$ across $I$ lies on $S_bS_c$ . So
$\measuredangle AIX = \measuredangle AIX' = \measuredangle X'AI = \measuredangle IYX$ which finishes the problem.

https://i.imgur.com/YWeWmLU.png

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#14 h Mar 2, 2025, 12:48 PM

Y by

Let $A'$ be the reflection of $A$ over $BC$ . Note that $P=JA'\cap BC$ . Apply $\sqrt{bc}$ inversion.

New Problem Statement: $ABC$ is a triangle with incenter, $A-$ excenter $I,J$ . $Y\in BC$ such that $YA=YJ$ . $(AIO)$ meets $(ABC)$ at $P$ and $JP$ intersects $(ABC)$ at $X$ . Prove that $(JXY)$ and $AJ$ are tangent.
Apply $\sqrt{bc}$ inversion on $\triangle JBC$ .

New Problem Statement: $ABC$ is a triangle with orthocenter $H$ and $AH$ meets $BC$ at $D$ . Let $A'$ be the reflection of $A$ over $BC$ . Point $P$ lies on $(BHC)$ such that $PA'$ is the exterior angle bisector of $\measuredangle DPA$ . Let $AP$ meet $(BHC)$ at $X$ and $Y\in (AC)$ such that $HA=HY$ . Prove that $XY\perp BC$ .
$AH\cap (ABC)=H'$ . Let $Y'\in (BHC)$ such that $H'A'=H'Y'$ . Note that $YY'\perp BC$ . Set $\triangle A'BC$ the main triangle.

New Problem Statement: $ABC$ is a triangle with orthocenter $H$ , $\ AH$ meets $BC,(ABC)$ at $D,E$ respectively. $A'$ is the reflection of $A$ over $BC$ . Point $P$ lies on $(ABC)$ such that $PA$ is exterior angle bisector of $\measuredangle PDA'$ . Let $PA'$ intersect $(ABC)$ at $X$ and $Z$ lies on $(ABC)$ such that $HA=HZ$ . Prove that $XZ\perp BC$ .
First we will show that $W,H,Z$ are collinear. It sufficies to show that $HW=HE$ . Let $K$ be the antipode of $A$ on $(ABC)$ . Note that $-1=(PK,PA;PD,PA')=(PK\cap AD,A;D,A')$ hence $P$ is unique. Note that $W,H,P,A'$ are concyclic since $DQ.DP=DB.DC=DH.DA'$ .
$\measuredangle HWE=\measuredangle HWP+\measuredangle PWE=\measuredangle AA'P+\measuredangle PAA'=180-\measuredangle A'PA=\measuredangle WPA=\measuredangle WEA$ Thus, $W,H,Z$ are collinear. $ZK\cap AH=Q$ . Let $X'\in (ABC)$ where $X'Z\perp BC$ . Since $KZ\perp AZ$ and $HA=HZ$ , we have $HA=HQ$ .
$(A,K;W,X)=(PA,PK;PW,PA')=-1=(A,Q;H,AH_{\infty})=(ZA,ZK;ZH,ZX')=(A,K;W,X')$ So $X=X'$ as desired. $\blacksquare$

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