bump $ $

Bump!! Bump

Joint solution with many (Anant Mudgal, Robu Vlad, Anushka Aggarwal, Misheel, and about 10 others): Let $O$, $H$ be as usual. Let $D$ be the refection of $O$ over $A$, and define $E$, $F$ similarly. Let $A'$, $B'$, $C'$ denote the reflections of $\triangle ABC$ over the opposite sides. We will show $I$ lies on $(DEF)$. [asy][asy] size(12cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = 2*A; pair E = 2*B; pair F = 2*C; pair Ap = B+C-B*C/A; pair Bp = A+C-A*C/B; pair Cp = A+B-A*B/C; pair O = origin; pair H = A+B+C; pair M = midpoint(B--Cp); pair N = midpoint(C--Bp); pair P = midpoint(B--C); draw(unitcircle, red); filldraw(circumcircle(D, E, F), invisible, red); draw(A--B--C--cycle, red); draw(B--Bp, orange); draw(C--Cp, orange); filldraw(CP(D, H), invisible, deepgreen); pair I = 2*dir(192); pair X = -D+2*foot(circumcenter(D, Bp, Cp), D, I); pair Y = extension(X, Cp, I, E); pair Z = extension(X, Bp, Y, Ap); draw(A--Ap, orange); draw(Bp--X--Cp, deepcyan); draw(Cp--Y--Ap, deepcyan); draw(Ap--Z--Bp, deepcyan); filldraw(X--Y--Z--cycle, invisible, deepcyan); draw(X--D, blue+dashed); draw(Y--E, blue+dashed); draw(I--F, blue+dashed); draw(D--E--F--cycle, red); draw(B--Cp, brown); draw(C--Bp, brown); draw(M--A--N, brown); draw(A--P, brown); dot("$A$", A, dir(A)); dot("$B$", B, dir(180)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$A'$", Ap, dir(Ap)); dot("$B'$", Bp, dir(Bp)); dot("$C'$", Cp, dir(Cp)); dot("$O$", O, dir(315)); dot("$H$", H, dir(H)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$I$", I, dir(I)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(250)); /* TSQ Source: !size(12cm); A = dir 110 B = dir 210 R180 C = dir 330 D = 2*A E= 2*B F = 2*C A' = B+C-B*C/A B' = A+C-A*C/B C' = A+B-A*B/C O = origin R315 H = A+B+C M = midpoint B--Cp N = midpoint C--Bp P = midpoint B--C unitcircle red circumcircle D E F 0.1 lightred / red A--B--C--cycle red B--Bp orange C--Cp orange CP D H 0.1 yellow / deepgreen I = 2*dir(192) X = -D+2*foot circumcenter D Bp Cp D I Y = extension X Cp I E Z = extension X Bp Y Ap R250 A--Ap orange Bp--X--Cp deepcyan Cp--Y--Ap deepcyan Ap--Z--Bp deepcyan X--Y--Z--cycle 0.1 lightcyan / deepcyan X--D blue dashed Y--E blue dashed I--F blue dashed D--E--F--cycle red B--Cp brown C--Bp brown M--A--N brown A--P brown */ [/asy][/asy] Claim: Point $D$ is the circumcenter of $\triangle HB'C'$. (Similarly for $E$ and $F$.) Proof. First, we show that $\triangle C'DB' \sim \triangle BOC$. Let $M$, $N$, $P$ denote the midpoints of $\overline{BC'}$, $\overline{CB'}$, $\overline{BC}$. I claim that $\triangle MAN \sim \triangle BOC$. This is because $\measuredangle MAN = \measuredangle MAP + \measuredangle PAN = 2 \measuredangle BAP + 2 \measuredangle PAC = 2\measuredangle BAC$, and $MA = PA = NA$. Thus, by mean geometry $ \triangle C'DB' \sim \triangle MAN \sim \triangle BOC $ and since $\measuredangle C'HB' = 180^{\circ} - \measuredangle A$, this implies result. $\blacksquare$ Claim: Line $\overline{XD}$ is the angle bisector of $\angle YXZ$. Similarly, for $\overline{YE}$ and $\overline{ZF}$. Proof. We can compute $\measuredangle YXZ = \measuredangle C'XB' = 2 \measuredangle A$ by the reflections, and thus it follows that $X$, $C'$, $D$, $B'$ are concyclic. $\blacksquare$ If we let $I = \overline{XD} \cap \overline{ZF}$, Finally, \[ \angle DIF = \angle 180^{\circ} - \angle XIZ = 180^{\circ} - \left( 90^{\circ} + \angle XYZ / 2 \right) = \angle ABC = \angle DEF \]gleefully ignoring configuration issues; this implies $I$ lies on $(DEF)$.

Wow this is so visual and fun. Let $l_a, l_b, l_c$ be the corresponding reflections of the lines $d_a, d_b, d_c$. Let $A', B', C'$ be the reflections of $A, B, C$ in $BC, CA, AB$ respectively.Note $A', B', C'$ always lie on $l_a, l_b, l_c$ respectively. Note that if $d_a, d_b, d_c$ rotate with a constant angular velocity $\omega$, then $l_a, l_b, l_c$ rotate with a constant angular velocity $-\omega$. Then, the angles of triangle $XYZ$ remain the same as the lines rotate, and $X, Y, Z$ move on circles having $B'C', C'A', A'B'$ as chords, respectively. Then internal angle bisector of $\angle X$ always passes through a fixed midpoint of arc $B'C'$, similar stuff for $Y, Z$. Thus note that a internal angle bisector of the triangle $XYZ$ has a fixed point, and rotates with constant angular velocity $-\omega$. The intersection of two angle bisectors thus traces a circle, and we're done.

Interesting problem! Here's my solution: Let $A',B',C'$ be defined as in the above solutions. Also, let $d_a \cap BC=A_1$, and define $B_1$ and $C_1$ analogously. Then \begin{align*} \measuredangle B'XC' &=\measuredangle B'B_1A+\measuredangle CAB+\measuredangle AC_1C' \\ &=\measuredangle AB_1B+\measuredangle CAB+\measuredangle CC_1A \\ &= \measuredangle ACC_1+\measuredangle CC_1A+\measuredangle CAB \\ &=2 \measuredangle CAB \\ \end{align*}Thus, as $\measuredangle B'XC'$ is fixed, so $X$ must lie on a fixed circle $\omega_A$ through $B',C'$. Taking $I$ as the incenter of $\triangle XYZ$, we get that, using Fact 5, $XI$ passes through the midpoint $P_A$ of arc $\widehat{BC}$ in $\omega_A$. Similarly, $P_B \in YI$ and $P_C \in ZI$, where $P_B,P_C$ are defined similarly. Then $$2\measuredangle P_BIP_C=2\measuredangle YIZ=\measuredangle YXZ=2\measuredangle CAB$$which gives that $\measuredangle P_BIP_C$ is also fixed. Similarly, $\measuredangle P_AIP_B$ and $\measuredangle P_AIP_C$ must also be fixed, and so $I \in \odot (P_AP_BP_C)$. Thus, the locus of $I$ is a circle. $\blacksquare$ NOTE: There's a slight thing that we have overlooked here. It might be possible that $P_A$ becomes the other arc midpoint, if $X$ goes on the other side of $B'C'$. But it's not hard to see that if $X$ goes on the other side, then the internal angle bisector of $\angle YXZ$ will in fact become the external angle bisector of $\angle B'XC'$. So our solution is not wrong (duh!). P.S. Here's a much faster proof of the $1^{\text{st}}$ Lemma mentioned in v_Enhance's proof: Let $O_a,O_b,O_c$ be the reflections of $O$ in $BC,CA,AB$. Since $O_c$ is the center of $\odot (AHBC')$, and $DO_c$ is parallel to $AB$, and hence perpendicular to $HC'$, so $D$ lies on the perpendicular bisector of $HC'$. Similarly, $D$ lies on the perpendicular bisector of $HB'$, giving the desired result.

v_Enhance wrote:

Someone must ask this, what's the motivation to use $(DEF) ? $

Let $O,H,N$ be the circumcenter, orthocenter, nine-point center of $\triangle ABC$, respectively ; $D,E,F$ be the reflection of point $O$ over points $A,B,C$, respectively ; $A',B',C'$ be the reflections of $A,B,C$ in $\overline{BC},\overline{CA}, \overline{AB}$, respectively. We claim that the locus of incenters of all such triangles is $\odot(DEF)$. Let $N_A,N_B,N_C$ be the reflection of $N$ in $\overline{BC},\overline{CA}, \overline{AB}$, respectively. The following claim is well known: Claim 1: $N_A$ is the midpoint of segment $OA'$, and similar for vertices $B,C$. Proof: Since $AH = OO_A$ (well known) and $\overline{AH} \parallel \overline{OO_A}$, it directly follows that $N$ is also the midpoint of segment $\overline{AO_A}$. Now if we reflect the three points $A,N,O_A$ in $\overline{BC}$, our desired result directly follows. $\square$ Now, let $\ell$ be the line passing through $N$ parallel to lines $d_a,d_b,d_c$ ; $\ell_A,\ell_B,\ell_C$ be the reflections of $\ell$ in $\overline{BC},\overline{CA}, \overline{AB}$, respectively ; $X' = \ell_B \cap \ell_.C , Y' = \ell_C \cap \ell_A , Z' = \ell_A \cap \ell_B$. So using claim 1, we see that under the dilation at $O$ with factor $1/2$, the reflection of line $d_a$ in $\overline{BC}$ gets mapped to $\ell_A$ and similar for vertices $B,C$ ; $\odot(DEF)$ clearly gets mapped to $\odot(ABC)$. So our problem has reduced to showing that locus of incenters of $\triangle X'Y'Z'$ is $\odot(ABC)$. Claim 2: Line $\overline{BY'}$ is the internal angle bisector of $\angle X'Y'Z'$ and similar bisection at vertices $Z',X'$. proof: Set $R = \ell \cap \ell_A, S = \ell \cap \ell_C$. By symmetry, $R \in \overline{BC}, S \in \overline{AB}$. Note that $\angle N_CY'N_A = 180^\circ - (\angle N_CSN + \angle NRN_A) = 180^\circ - (2 \angle BSR + 2 \angle BRS) = 180^\circ - 2 \angle ABC$ and $\angle N_CB = \angle N_CBN + \angle N_ABN = 2 \angle ABN + 2 \angle CBN = 2 \angle ABC$, so it follows that quadrilateral $Y'N_CBN_A$ is cyclic. Also, $BN_C = BN = BN_B$, hence $\overline{BY'}$ bisects $\angle N_CY'N_A$, which is same as $\angle X'Y'Z'$, so we are done. $\square$ So by claim 2, lines $\overline{BY'},\overline{CZ'},\overline{AX'}$ concur at $I$, the incenter of $\triangle X'Y'Z'$. Then, $$ \angle AIC = 180^\circ - \angle XIZ' = 180^\circ - \left(90^\circ + \frac{\angle X'Y'Z'}{2} \right) = 90^\circ - \frac{180^\circ - 2 \angle ABC}{2} = \angle ABC $$so point $I$ lies on $\odot(ABC)$. Conversely, we see that if we pick any point $I$ on $\odot(ABC)$ and set $X' = \overline{AI} \cap \odot(AN_BN_C), Y' = \overline{BI} \cap \odot(BN_CN_A), Z' = \overline{CI} \cap \odot(CN_AN_B)$, and then define $\ell_A,\ell_B,\ell_C$ as the lines passing through the pair of points $\{Y',Z'\},\{Z',X'\},\{X',Z'\}$, then one can prove using a similar proof as above that $I$ will be the incenter of $X'Y'Z'$ and the reflection of lines $\ell_A,\ell_B,\ell_C$ in $\overline{BC},\overline{CA}, \overline{AB}$, respectively coincide, which completes the proof of this problem. $\blacksquare$ As a remark, I want to mention below a special degenerate case: the three points $X',Y',Z'$ will coincide if and only if any two of the points $X',Y',Z'$ coincide and this happens if and only the line $\ell$ passes through the through the orthocenter of $\triangle ABC$. The proof of the above fact can be seen here.

The answer is the circle centered at $O$ with twice the radius of $(ABC)$. Let $P,Q,$ and $R$ be the reflections of $O$ over $A,B,$ and $C$. Then define $D,E,$ and $F$ to be the reflection of $A,B,$ and $C$ across $BC,AC,$ and $AB$. Note that $D,E,$ and $F$ lie on the reflections of $d_a,d_b,$ and $d_c$. [asy][asy] size(12cm); defaultpen(fontsize(6pt)); dotfactor*=0.75; pen bfil=purple+pink+white+white+white+white+white; pen ufil=purple+pink+white+white+white+white; pen tfil=0.3*purple+white+white; pen qfil=purple+cyan+white+white+white; pen qfil2=purple+cyan+white+white; pen qfil3=rgb(184, 205, 245); pen qbfil=0.6*bfil+0.8*qfil2; pen qtfil=pink+cyan+cyan+purple+white+white; pen darkpurple=cyan+purple+purple; pair A=dir(110); pair B=dir(200); pair C=dir(340); pair J=dir(90); pair K=dir(0); pair L=dir(-140); pair D=2*foot(A,B,C)-A; pair E=2*foot(B,A,C)-B; pair F=2*foot(C,A,B)-C; pair P=2*A; pair Q=2*B; pair R=2*C; pair O=0; pair H=A+B+C; pair AA=extension(A,J,B,C); pair BB=extension(B,K,A,C); pair CC=extension(C,L,A,B); pair X=extension(E,BB,F,CC); pair Y=extension(F,CC,D,AA); pair Z=extension(E,BB,D,AA); pair I=extension(Q,Y,Z,R); draw(unitcircle,qtfil); draw(circumcircle(P,Q,R),qtfil); draw(circumcircle(F,E,H),darkpurple); draw(3*P--(-6*P)^^3*Q--(-6*Q)^^3*R--(-6*R),linewidth(0.4)+red); draw((50*J-49*A)--(50*A-49*J),blue+dashed); draw((50*K-49*B)--(50*B-49*K),darkgreen+dashed); draw((50*L-49*C)--(50*C-49*L),purple+dashed); draw(A--(3*D-2*A)^^B--(3*E-2*B)^^C--(3*F-2*C)); draw((5*X-4*Z)--(100*Z-99*X),darkgreen); draw((100*Y-99*Z)--(100*Z-99*Y),blue); draw((100*Y-99*X)--(100*X-99*Y),purple); draw((100*A-99*C)--(100*C-99*A)^^(100*B-99*A)--(100*A-99*B)^^(100*B-99*C)--(100*C-99*B),fuchsia); draw(circumcircle(P,F,E),orange); draw((100*P-99*X)--(100*X-99*P)^^(100*Q-99*Y)--(100*Y-99*Q)^^(100*Z-99*R)--(100*R-99*Z),orange); draw(circumcircle(foot(I,X,Z),foot(I,Z,Y),foot(I,X,Y)),fuchsia); dot("$A$",A,dir(105)*2); dot("$B$",B,dir(110)*1.5); dot("$C$",C,dir(-110)*1.6); dot("$D$",D,dir(50)*1.2); dot("$E$",E,dir(5)*1.9); dot("$F$",F,dir(75)*2.4); dot("$R$",R,dir(-80)*2); dot("$P$",P,dir(70)*2); dot("$Q$",Q,dir(-90)*1.5); dot("$H$",H,dir(120)*2.3); dot("$O$",O,dir(60)*3); dot("$X$",X,dir(-20)*2); dot("$Y$",Y,dir(Y)*1.6); dot("$Z$",Z,dir(-60)*1.5); dot("$I$",I,dir(-110)*2); clip((2.6,3.7)--(-3.1,3.7)--(-3.1,-3.3)--(2.6,-3.3)--cycle); [/asy][/asy] Claim: The center of $(FEH)$ is $P$. Proof. We prove this using complex numbers. Rotating the diagram so that $a=1$, we have $e=-\frac cb+c+1,h=1+b+c,$ and $p=2$, so \[PH=PE\iff (1-b-c)\left(1-\frac1b-\frac1c\right)=\left(1+\frac cb-c\right)\left(1+\frac bc-\frac1c\right)\]which can be easily verified by expansion. Similarly, $PF=PH$, which implies the desired. Claim: $FPEX$ is cyclic. Proof. We have \[\angle FPE=2(180^\circ-\angle BHC)=2\angle BAC=\angle(XY,d_c)+\angle(XZ,d_b)=180^\circ-\angle FXE.\] Since $P$ is the midpoint of arc $FE$ in $(FPEX)$, line $PX$ bisects $\angle YXZ$. Thus $I=XP\cap YQ\cap ZR$, so \[\angle QIR=90^\circ+\frac12\angle YXZ=180^\circ-\angle BAC=180^\circ-\angle QPR,\]as needed.

Solved with Max Lu. Define $A',B',C'$ as the reflections of $A,B,C$ over $BC,AC,AB$ respectively. Let $O,H$ be the circumcenter and orthocenter of $\triangle ABC$, and let $A_1, B_1, C_1$ be the reflections of $O$ over $A,B,C$ respectively. Let $\ell_a, \ell_b, \ell_c$ be the reflections of $d_a, d_b, d_c$. Finally, let $I$ be the incenter of $XYZ$. By trivial angle chasing, $\angle YXZ = 180 - 2\angle A$ so $\angle YIZ = 180 - \angle A$. Next, since $A'$ is the reflection of $A$ over $BC$, and $\ell_a$ is $d_a$ reflected over $BC$, we have $A'\in \ell_a$. Similarly, $B'\in \ell_b, C'\in \ell_c$, which means $A'\in YZ$. I claim $A_1$ is the circumcenter of $HB'C'$. If we let $A_2$ be the reflection of $A_1$ over $AB$, and $H_C$ the reflection of $H$ over $AB$, then \[A_1C' = A_2 C = A_2 H_C = A_1H\]Similarly, we get $A_1B' = A_1H$. Now, this means $\angle B'A_1C' = 360 - 2\angle C'HB' = 2\angle A$. This means \[\angle C'A_1B' = 2\angle A = 180 - (180 - 2\angle A) = 180 - \angle YXZ = 180 - \angle C'XB'\]Therefore, $(XC'A_1B')$ is cyclic. Furthermore, since $A_1C' = A_1B'$, we have $XA_1$ bisects $\angle C'XB'$ so $XA_1$ passes through $I$. Similarly, $YB_1$ and $ZC_1$ pass through $I$. Finally, $\angle B_1IC_1 = 180 - \angle A$, so $\angle B_1IC_1$ is fixed. Since $B_1, C_1$ are also fixed, we conclude $I$ lies on a circle. Furthermore, since this circle goes through $A_1,B_1,C_1$, this circle has center $O$ and has twice the circumradii, so this circle is the locus.

I claim that $I$ lies on a circle centered at $O$ with radius $2R$. In particular, let $D, E, F$ be the reflections of $O$ over $A, B, C$ respectively. Furthermore, let $A', B', C'$ be the reflections of $A, B, C$ over their opposite sides, respectively. Claim. $\triangle DB'C' \sim \triangle OBC$. Proof. Let $U$ be the midpoint of $\overline{BC'}$ and $V$ the midpoint of $\overline{B'C}$; furthermore, let $M$ be the midpoint of $\overline{BC}$. By symmetry over perpendicular bisectors, $AU=AM=AV$, and in particular $$\angle UAV = 2(180^\circ - \angle UMV) = 2(\angle UMB + \angle VMC) = 2A = \angle BOC.$$Thus $\triangle UAV \sim \triangle BOC$, and as $\triangle UAV$ is precisely the spiral average of $\triangle DB'C'$ and $\triangle OBC$, we have the result. $\blacksquare$ Now, notice that \begin{align*} \measuredangle(\overline{XY}, \overline{XZ}) &= \measuredangle(\overline{XY}, \overline{AB}) + \measuredangle(\overline{AB}, \overline{AC}) +\measuredangle(\overline{AC}, \overline{XZ}) \\ &= \measuredangle(\overline{AB}, \ell_C) + \measuredangle BAC + \measuredangle(\ell_B, \overline{AC}) \\ &= 2\measuredangle BAC. \end{align*}This implies that $DB'C'X$ is cyclic, so by Fact 5, $X, I, D$ are collinear along the bisector of $\angle YXZ$. To finish, $$\angle XIF = 90^\circ + \frac 12 \angle ZYX = \angle DEF$$by previous computation. This implies the result.

Can someone explain motivation behind (DEF)?

Let $\triangle PQR$ be the image of $\triangle ABC$ under a homothety with ratio 2 at the circumenter $O$. Let $A'$, $B'$, and $C'$ as the reflections of the vertices over the sides, and suppose $H$ is the orthocenter. We proceed with three claims. $A'$, $B'$, $C'$ lie on corresponding sides of $\triangle XYZ$: Follows by construction. $P$ is the center of $(B'C'H)$: There are many ways to prove this, but a complex bash suffices. This implies $\angle B'PC' = 2 \angle A$ and $PB' = PC'$. $XP$ bisects $\angle X$: Lines $d_b$ and $d_c$ are reflections of parallel lines, so we can angle chase to find $\angle X = 180-2 \angle A$. Thus $B'PC'X$ is cyclic, which suffices. The incenter $I$ of $\triangle XYZ$ lies on $(PQR)$, a fixed circle: We can express $\angle RIP$ as \[\angle (RZ, PB) = 90 - \frac{\angle Y}{2} = \angle B = \angle RQP. \quad \blacksquare\]

Sharygin 2009 Here is a sketch of the solution. We claim that it’s the circle Centered at $O$ with radius twice the radius is circle $(ABC)$. Let $A’,B’,C’$ denote the intersection of $O$ wrt $A,B,C$, and let $A_1,B_1,C_1$ denote the reflection of vertices $A,B,C$ wrt opposite side in $\triangle ABC$. We divide the problem into the following claims. Claim 1: $A’$ is the Center of $(B_1C_1H)$. Claim 2: $B_1C_1A’X$ is cyclic. Which is provable by easy angle chase, hence we are done.

Let the reflections of $O,$ the circumcenter of $(ABC)$ across $A,B,C,$ be $A_1, B_1, C_1$ respectively. Then let the reflections of $A,B,C$ across the opposite segment of the triangle be $A', B', C',$ respectively. This problem can be split into two parts. First step is to show that $A_1$ is the circumcenter of $(B'C'H),$ where $H$ is the orthocenter of $\triangle ABC.$ Do this by mean geometry. Above gives us $\angle C'A_1B'=2\angle A.$ Then by extending all the parralel lines and angle chasing with their reflections we see that $\angle ZXY=180-2\angle A.$ Then by a simple angle chase we can show the incenter lies on $(A_1B_1C_1),$ which is the desired fixed circle$.\blacksquare$

Very nice..

I think many people are qurious of the motivation of $(DEF)$. By getting the line that is perpendicular to $AB,BC,CA$ you can find $(DEF)$ I won't write the solution since it is similar from above. (Line perpendicular to $BC$ gives reflection of $O$ to $A$) If you let $I$ be the incenter and $X,Y,Z$ the intersections, $AO^2 = AY \times AZ =AI^2$ proves this through a little trig bash.