Let $I, C, \omega$ and $\Omega$ be the incenter, circumcenter, incircle and circumcircle, respectively, of the scalene triangle $XYZ$ with $XZ > YZ > XY$. The incircle $\omega$ is tangent to the sides $YZ, XZ$ and $XY$ at the points $D, E$ and $F$. Let $S$ be the point on $\Omega$ such that $XS, CI$ and $YZ$ are concurrent. Let $(XEF) \cap \Omega = R$, $(RSD) \cap (XEF) = U$, $SU \cap CI = N$, $EF \cap YZ = A$, $EF \cap CI = T$ and $XU \cap YZ = O$. Prove that $NARUTO$ is cyclic.
Problem
Source: Olympic Revenge 2021 #3
Tags: geometry, incenter, anime, Euler Line, Spiral Similarity, auyesl
MatteD
18.11.2021 19:06
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(27.18979429233048cm); 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 = -3.521348802441401, xmax = 23.668445489889077, ymin = -3.5442873152837056, ymax = 10.083849188207118; /* image dimensions */ /* draw figures */ draw((6.1991390220619795,6.0475094661374875)--(2.8956654033830653,0.), linewidth(1.) + blue); draw((12.,0.)--(6.1991390220619795,6.0475094661374875), linewidth(1.) + red); draw((6.1991390220619795,6.0475094661374875)--(5.,0.), linewidth(1.) + red); draw((5.824976277734518,5.3625478168596015)--(7.392685796240347,0.), linewidth(1.) + blue); draw((6.1991390220619795,6.0475094661374875)--(7.967093076198495,0.), linewidth(1.) + blue); draw((8.5,2.4486387453094984)--(2.8956654033830653,0.), linewidth(1.) + blue); 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draw((12.,0.)--(22.719044380727606,0.), linewidth(1.) + blue); draw(shift((10.078120191541926,-12.615226757365107))*xscale(17.85880484513116)*yscale(17.85880484513116)*arc((0,0),1,44.94170315648353,135.05829684351647), linewidth(1.) + dotted + blue); draw((8.5,2.4486387453094984)--(13.933233165821118,4.822520071053535), linewidth(1.) + blue); draw(shift((0.1664307028696561,3.701527122418355))*xscale(4.598915642679828)*yscale(4.598915642679828)*arc((0,0),1,-126.40234193478837,8.955914948697231), linewidth(1.) + blue); draw((6.1991390220619795,6.0475094661374875)--(22.719044380727606,0.), linewidth(1.) + blue); /* dots and labels */ dot((6.1991390220619795,6.0475094661374875),linewidth(4.pt) + dotstyle); label("$X$", (6.267641929944082,6.178224702450601), NE * labelscalefactor); dot((5.,0.),linewidth(4.pt) + dotstyle); label("$Y$", (5.071025066222937,0.12866166919369929), NE * labelscalefactor); dot((12.,0.),linewidth(4.pt) + dotstyle); label("$Z$", (12.067909783259081,0.12866166919369929), NE * labelscalefactor); 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dot((6.883337853525148,1.742288776069052),linewidth(4.pt) + dotstyle); label("$H$", (6.949048755118624,1.8737279287870363), NE * labelscalefactor); dot((7.967093076198495,0.),linewidth(4.pt) + dotstyle); label("$L$", (8.029327868200214,0.12866166919369929), NE * labelscalefactor); dot((4.2284948613843785,2.4399464607297885),linewidth(4.pt) + dotstyle); label("$S$", (4.289900169071633,2.5717544326243713), NE * labelscalefactor); dot((4.709278267581001,4.417460843010857),linewidth(4.pt) + dotstyle); label("$R$", (4.77187085029265,4.54949619349682), NE * labelscalefactor); dot((8.569584174749293,5.179750893155314),linewidth(4.pt) + dotstyle); label("$U$", (8.64425597872358,5.31400141198533), NE * labelscalefactor); dot((-2.562803997643753,0.),linewidth(4.pt) + dotstyle); label("$A$", (-2.490928725348192,0.12866166919369929), NE * labelscalefactor); dot((13.933233165821118,4.822520071053535),linewidth(4.pt) + dotstyle); label("$T$", (13.995792508143149,4.948368481403868), NE * labelscalefactor); dot((22.719044380727606,0.),linewidth(4.pt) + dotstyle); label("$O$", (22.78760252076101,0.12866166919369929), NE * labelscalefactor); dot((4.755792517838705,3.4052447912049963),linewidth(4.pt) + dotstyle); label("$P$", (4.821729886281031,3.535695795066405), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] CLAIM 1: If $D'$ is the reflection of $D$ in $EF$, then $XD', CI, YZ$ concur. Let $H$ be the orthocenter of $\triangle DEF$, which lies on $CI$, and $L=XI \cap YZ$. Clearly it suffices to show $\frac{DD'}{DH}=\frac{LX}{LI}$, since $DD'$ and $LX$ are parallel. Let $\alpha = \angle FDE, \beta = \angle DEF, \gamma = \angle EFD$ and $r=ID$. Then $\frac{DD'}{DH}=\frac{4r \sin{\beta} \sin{\gamma}}{2r \cos{\alpha}}= \frac{\sin{\beta} \sin{\gamma}}{\cos \alpha}$, while $\frac{LX}{LI}=1+\frac{IX}{LI}=1+\frac{r/\cos\alpha}{r/\cos(\beta - \gamma)}=1+\frac{\cos(\beta - \gamma)}{\cos \alpha}$. It is trivial to verify these two expressions are equal. $\square$ As a consequence, since $XI$ and $RD$ intersect on $(XYZ)$ and $XI$ is parallel to $DD'$, Reim gives $D' \in (DRS)$. Now let $\Phi$ be the composition of an inversion of center $R$ and power $RE \cdot RY$ and a reflection over the bisector of $\angle YER$, and notice it is an involution. Moreover, observe that $\triangle RQ_1Q_2 \sim \triangle R\Phi(Q_2)\Phi(Q_1)$ for all $Q_1, Q_2$. Let $K$ be the concurrence point of $CI$ and $YZ$ and $P$ be the second intersection of $(AEF)$ and $XS$. First, it's well-known that $\Phi(F)=Z$, $\Phi(E)=Y$, $\Phi(I)=D$ and $\Phi(X)=A$. Now, notice that by CLAIM 1 the center of $(DRS)$ lies on $EF$, so $\Phi(C)$, which is the reflection of $R$ in $EF$, lies on $(DRS)$ so $\Phi(CI)=(DRS)$. $U$ lies on $\Phi(YZ)$ and $\Phi(CI)$, so $\Phi(U)=K$. $T=CI \cap EF$, so $\Phi(T)= \Phi(CI) \cap \Phi(EF) = S$. Now notice that $\triangle RAY \sim \triangle REX$, so $\angle KPR=\angle XPR=\angle XER = \angle YAR= \angle KAR$, so $PRAK$ is cyclic. Then $\Phi(O)= \Phi(XU) \cap \Phi(YZ)= (KAR) \cap (REF) = P$. Since $\Phi(A), \Phi(U), \Phi(T), \Phi(O)$ all lie on the same line $SX$, pentagon $ARUTO$ is cyclic, and to finish observe that $\triangle RSK \sim \triangle RUT$, so $R$ is the Miquel point of $TUSK \implies N= SU \cap TK$ lies on $(ARUTO) \implies NARUTO$ is cyclic. $\blacksquare$
jpedrorvc1
19.11.2021 04:56
Naruto can be a little harsh sometimes
This problem is based on Taiwan TST 2020/6: https://artofproblemsolving.com/community/c6h2073612p14866139
There are some other interesting features of this 6-point-circle configuration (for example: $GO \cap \Omega \in (NARUTO)$)
No one managed to fully solve it during the test, although sacrificing the $\triangle ABC$ for Naruto was pretty funny for the teachers
Let: $K = IC \cap YZ$, $B = (XEF) \cap XS$, $J = IC \cap (AEF)$, $Q = UI \cap EF$, $P = SM \cap YZ$, $L = DI \cap (XEF)$, $H = DI \cap XU $, $V = EF \cap DI$. Let $X'$ be the reflection of $X$ in $IC$. Let $M$ be the midpoint of the smaller arc $YZ$ , and $G$ be the midpoint of arc $YXZ$.
Property 1: $S, D, X'$ are collinear.
Let $B_1 = YI \cap \Omega$, $C_1 = ZI \cap \Omega$ and $W = X'I \cup \Omega$.
$$ (X',X;Z,Y) \stackrel{I}{=} (W, M; C_1, B_1)$$Verify that $MM\perp ID, MW\perp IC, MB_{1}\perp IZ, X_{1}Z_{1}\perp IY$.
$$\Rightarrow M (M, W; B_1, C_1) = I (D, C;Z, Y )\ , $$$$\Rightarrow S (D, K; Z, Y) = (X', X; Z,Y)\ , $$Then $X' \in SD$.
Property 2: ($D, J, U$), ($P , I, U$) and ($S, P, M$) are collinear.
$S, P$ and $M$ are collinear by the inversion of center $M$ and radius $MI$.
Using Property 1:
$$ \angle JUR = \angle JXR = \angle DSR = \angle DUR $$$\Rightarrow$ $D$, $J$ and $U$ are collinear.
We also have:
$$ \angle JUI = \angle JXI = \angle DSP = \angle DUP$$$\Rightarrow$ Then $P, I$ and $U$ are collinear.
Property 3: $(H,I;V,D) = -1$.
This occurs if and only if, $H$ is the ortocenter of $\triangle YIZ$. We know that $\angle PUH = \angle PDH \Rightarrow H \in (RSD)$. It's enough to prove that the ortocenter of $\triangle YIZ$ lies on ($RSD$).
Let $I'$ be the reflection of $I$ in $YZ$. Perform an inversion of center $D$ that fixed $\Omega$. The image of $(RSD)$ under inversion is $MX'$, and $I'$ is the image of $H$. We only have to prove that $M, X', I'$ are collinear, but see that $ \triangle XIC \sim \triangle II'M $ .
$$\Rightarrow \angle I'MI = \angle ICX = \angle X'MX. $$
Property 4: $(A, B, U)$ are collinear.
$(A, B, U)$ are collinear $\Leftrightarrow$ $BU, EF$ and $YZ$ are concurrent. See that:
$$\angle YDB = \angle KIB = \angle JUB = \angle DUB$$$\Rightarrow YZ$ is tangent to $(DUB)$ at point $D$, so, $XY$ is the radical axis of $(DUB)$ and $(DEF)$.
By radical axis on $(DEF), (DUB), (FEBU)$, we are done.
Using $\angle IQR = 90$ with Property 3, we have:
$$\angle DUI = \angle IUV$$.
$$\Rightarrow \angle IUV = \angle DUI = \angle JXI = \angle ITV. $$$\Rightarrow I, T, U,V$ is cyclic.
$$\Rightarrow \angle LRU = \angle LIU = \angle ATU $$As $G, L, R, A$ are collinear (Well known fact). $\Rightarrow ARUT$ is cyclic.
See that:
$$\angle DOU = \angle LIU = \angle LRU$$$\Rightarrow ARUTO$ is cyclic.
And finally, we have:
$$ \angle IUA = \angle IUB = \angle MXS = \angle KPS = \angle DUS $$$$ \Rightarrow \angle SUA = \angle DUI = \angle KTA$$$\Rightarrow N \in (AUT)$. $\Rightarrow NARUTO$ is cyclic.
crazyeyemoody907
12.12.2022 02:00
Solved with CyclicISLscelesTrapezoid, Eyed, and v4913 We do a massive refactoring and simplification; consider the following equivalent problem, a breakdown of the given, despite being longer: naruto simplified wrote: In triangle $ABC$ with circumcircle $\Omega$ centered at $O$, the incircle $\omega$ centered at $I$ touches the sides at $D,E,F$. Let $I',I_a'$ be the respective reflections of $I$ and the orthocenter of $\triangle BIC$ in $\overline{BC}$, and $M$ the midpoint of arc $BC$ on $\Omega$. Further define: $S$ as the intersection of the Euler lines $\overline{OI}$ of $\triangle DEF$, $\overline{MI'}$ of $\triangle I_a'BC$; $T=\overline{EF}\cap\overline{BC}, U=\overline{EF}\cap\overline{OI}, V=\overline{MI'}\cap\overline{BC},R=\overline{AV}\cap (AI)$; \item $K=\overline{OI}\cap\overline{BC}$; Prove that (a) $Q,R,S,T,U,V$ are concyclic, and (b) $\overline{AK},\Omega,(QRD), \overline{RS}$ concurrent; (a) The concyclicity Let the spiral similarity $s$ at $Q$ with (directed) angle $\theta$ map $E,F\to C,B$ and thus $D,I$ and the orthocenter of $\triangle DEF$ to $I',M,I_a'$ respectively. Clearly, $S$ is the intersection of the Euler lines of two triangles related by $s$: $DEF,I_a'CB$. By design, we have $U\overset s\to V$, so \[\measuredangle VQU=\theta=\measuredangle(\overline{BC},\overline{EF})\overset s=\measuredangle(\overline{MI'},\overline{OI})=\measuredangle VSU,\]whence $Q,S,T,U,V$ concyclic. To see that the last point is also concyclic with the other five, let $N$ be the midpoint of $\widehat{BAC}$, so that $\overline{NA}$ touches $(AI)$. Indeed, then \[\measuredangle QRV=\measuredangle QRA=\measuredangle QAN\overset s=\measuredangle QUV\]as needed.
In fact, by design, $S$ is the exsimilicenter of the incircle and the circle at $O$ with radius half that of $\Omega$, so it's actually the inverse of $I$ wrt $\Omega$.
(b) The concurrence Let $D'$ be the reflection of $D$ in $\overline{EF}$, and $G$ the orthocenter of $\triangle BIC$, so that $D'\overset s\to G$. We easily have $DD'GQ$ cyclic. As $\measuredangle(\overline{AD'},\overline{NG})=\theta$, the point $X=\overline{AD'}\cap\overline{NG}$ lies on both $(DD'GQ), \Omega$. We require the following result(s): Theorem: weird concurrences Let scalene triangle $ABC$ have circumcenter $O$, circumcircle $\Omega$, and orthocenter $H$. let $K$ be the polar of $\overline{BC}$ wrt $\Omega$, and $A'$ be the reflection of $A$ in $\overline{BC}$. Then $\overline{OH},\overline{A'K}$ and the tangent to $\Omega$ at $A$ are concurrent. Let $E$ be the reflection of the point $E_0$ (such that $A$ is the incenter or excenter of $\triangle E_0BC$) in the perpendi-cular bisector of $\overline{BC}$. Then $\overline{OH},\overline{BC},\overline{EA'}$ are also concurrent. (parentheses used above for easier grammatical parsing) [asy][asy] //thm(a) //setup; size(9cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(226,235,255); blu2=RGB(196,216,255); lightpurple=RGB(237,186,255); // blu1 lighter //defn pair A,B,C,O,H,A1,Ha; A=(3.71,9.95); B=(0,0); C=(14,0); O=circumcenter(A,B,C); H=orthocenter(A,B,C); A1=2*foot(A,B,C)-A; Ha=orthocenter(A1,B,C); pair K,J,X; K=2*circumcenter(O,B,C)-O; J=extension(O,(B+C)/2,A,A+rotate(90)*(A-O)); X=extension(O,H,K,A1); pair reflect(pair P,pair A,pair B){return 2*foot(P,A,B)-P;} pair E0,E,L; E0=extension(B,reflect(C,A,B),C,reflect(B,A,C)); E=reflect(E0,O,K); L=extension(O,H,B,C); //draw filldraw(A--B--C--cycle,blu1,blu); draw(circumcircle(A,B,C),blu); draw(A--A1^^J--K^^1.4*B-.4*C--L,blu); draw(J--X--K,purple); draw(L--X,magenta); draw(B--K--C--A1--B,blu+dashdotted); draw(B--E--C,purple+dotted); draw(E--L,red); //label label("$A$",A,dir(120)); label("$B$",B,dir(60)); label("$C$",C,dir(-30)); label("$O$",O,dir(50)); label("$H$",H,dir(50)); label("$A'$",A1,-dir(70)); label("$H_a$",Ha,dir(50)); label("$K$",K,dir(0)); label("$J$",J,dir(90)); label("$E$",E,-dir(90)); [/asy][/asy] Proof. These two parts actually aren't connected at all... Part a, by CyclicISLscelesTrapezoid Let $J$ be the intersection of the tangent to $\Omega$ at $A$ with the perpendicular bisector of $\overline{BC}$, and $H_a\in\Omega$ be the reflection of $H$ in $\overline{BC}$. We contend that the triples $(A,H,A'),(J,O,K)$ are homothetic. Indeed, they lie on parallel lines. To finish, check that (if $R$ denotes the radius of $\Omega$) \[JO=\frac R{\cos(B-C)},OK=\frac R{\cos A}, AH=2R\cos A,HA'=AH_a=2R\cos(B-C)\Rightarrow \frac{JO}{OK}=\frac{AH}{HA'}.\][asy][asy] //thm(b) //setup size(8cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(226,235,255); blu2=RGB(196,216,255); lightpurple=RGB(237,186,255); // blu1 lighter //defn pair A,B,C,H,O,M,D; A=(4,9.4); B=(0,0); C=(14,0); H=orthocenter(A,B,C); O=circumcenter(A,B,C); M=(B+C)/2; D=extension(A,H,B,C); pair A2,Oa,X,L,F; A2=B+C-A; Oa=B+C-O; X=extension(H,A2,B,C); L=extension(O,H,B,C); F=extension(A,L,H,A2); //draw filldraw(A--B--C--cycle,blu1,blu); draw(D--A--A2,blu); draw(O--Oa,blu+dashdotted); draw(H--L--F,magenta); draw(B--F--C^^F--A2,purple); //label void pt(string s,pair P,pair v, pen a) {filldraw(circle(P,0.11),a,linewidth(.3)); label(s,P,v);} pt("$A$",A,-dir(20),blu); pt("$B$",B,-dir(30),blu); pt("$C$",C,dir(-30),blu); pt("$H$",H,-dir(20),blu); pt("$O$",O,dir(60),blu); pt("$M$",M,dir(30),blu); pt("$D$",D,-dir(80),blu); pt("$A_2$",A2,dir(0),blu); pt("$O_a$",Oa,-dir(40),blu); pt("$X$",X,-dir(60),purple); pt("$L$",L,dir(-90),magenta); pt("$F$",F,dir(90),purple); [/asy][/asy] Part b, by me Let $F=B+C-E_0$, and $A_2=B+C-A$, so that $A_2$ is an incenter or excenter of $\triangle FBC$. Since $H$ is the antipode of $A_2$ on $(BA_2C)$, it is another incenter / excenter. To prove that $A,L,F$ collinear where $X=\overline{FHA_2}\cap\overline{BC}, L=\overline{OH}\cap\overline{BC}$, verify that (where $O_a\in\overline{H_aA_2}$ is the reflection of $O$ in $\overline{BC}$) \[(\overline{AF}\cap\overline{BC},X; D,M)\overset A=(FX;HA_2)=-1\text{ while } (DM;XL)\overset H=(\infty_{\perp BC}M; O_aO)=-1.\qquad\qquad\square\] Returning to the problem, applying respective parts of the theorem to $\triangle DEF,I_a'BC$, we obtain $(A,D',K)$ and $(A,G,V)$ collinear. Since $R\in(UVQ),\overline{GV}$, and $Q$ is the Miquel point of $D'GVU$, we must have $R=\overline{D'U}\cap\overline{GV}$-- an intersection of opposite sides. Hence, by definition of Miquel point, $R\in(QD'G)$. It remains to prove that $R,X,S$ collinear. In fact, there is a spiral similarity at $Q$ mapping $D',X\to U,S$ since $Q\in(URS),(D'XR)$, so we're done!