We will prove that points $A$, $I_{1}$, $I_{2}$ and $B$ lie on the same circle. Assume that incircles of $AEF$ and $BDF$ touch $AC$ and $BC$ at $P$ and $Q$ respectively. Triangles $AEF$ and $BDF$ are of course similar (with triangle $ABC$), so if $PI_{1} \cap AB=M$ then \[\frac{QD}{BQ}=\frac{AP}{PE}=\frac{AM}{MB},\] thus points $M$, $Q$, $I_{2}$ are collinear. Easy to see that $FMED$ is cyclic. So, \[\angle AI_{1}I_{2}+\angle ABI_{2}=\angle AI_{1}F+\angle FI_{1}I_{2}+\frac{1}{2}\angle CBA=\frac{\pi}{2}+\frac{1}{2}\angle CBA+\angle BMI_{2}+\frac{1}{2}\angle CBA=\angle MI_{2}B+\angle BMI_{2}+\frac{1}{2}\angle CBA=\pi.\] Thus quadrangle $AI_{1}I_{2}B$ is cyclic, so circumcircles of triangles $ACI_{1}$ and $BCI_{1}$ intersect at incenter $I_{3}$ of triangle $CED$. Now, if point $I$ is incenter of triangle $ABC$, by angle chasing we have $\angle II_{1}I_{3}=\angle II_{2}I_{3}$, $\angle II_{3}I=\angle II_{2}I_{1}$ and $\angle II_{3}I_{2}=\angle II_{1}I_{2}$ $\Longrightarrow I$ is orthocenter of triangle $I_{1}I_{2}I_{3}\Longrightarrow O_{1}O_{2}\perp CI_{3}$ and $CI_{3} \perp I_{1}I_{2}\Longrightarrow O_{1}O_{2}\parallel I_{1}I_{2} .$

my solution: let $I$ be the incenter of $\triangle ABC$ then we need to prove that $AI_{1}I_{2}B$ is cyclic so we need to prove that $II_{1}.IA=II_{2}.IB $ now let the incircle of $ABC$ intersect $AB,BC,AC$ at $F',D',E'$ respectively then we will prove that $F'E'$ is perpendicular bisector of $II_{1}$ ,if we prove this then $II_{1}.IA=2.r^2$ so $AI_{1}I_{2}B$ is cyclic now we have $\triangle AEI_{1}$ is similar to$ \triangle AIB $ so $ \frac{AI_{1}}{AI}=\frac{AE}{AB}=cos(A) $ so now let $J$ be a point on $AI$ such that $E'F'$ is perpendicular bisector of $IJ$ then we have $\frac{AJ}{AI}=\frac{sin(AE'J)}{sin(AE'I)}.\frac{E'J}{E'I}=sin(AE'J)=sin(90-A)=cos(A)$ so $I_{1}=J$ then we are done (the rest is like Burii

A very short way to show that A, B, I_1, I_2 are cyclic: Triangles FAE and FDB are similar, with spiral similarity centered at F. Since I_1 is the incenter of AEF and I_2 is the incenter of DBF, this spiral similarity centered at F also takes BI_2 to EI_1. Then a spiral similarity centered at F also takes BE to I_2I_1. Because of the spiral similarity, if we let the intersection of BE and I_2I_1 be P, we have that PFBI_2 is cyclic. This implies that 1/2 * <C = <BFI_2 = <BPI_2, so I_1I_2 and BE intersect at an angle of C/2.from here it is easy to chase out that I_1I_2 intersects with BI_2 at an angle of 180 - A/2, so AI_1I_2B is cyclic. (sorry for no latex)

Nice problem. New solution: Call $X$ the intersection point of the angle bisector of $\angle ACB$ and the circumcircle of $AI_1C$. Note that $I \in CX$. Say $AI=a$. By homothety $AI_1=a\times cosA$. So $II_1(IA)=a^2(1-cosA)$. Similarly $II_2(IB)=b^2(1-cosB)$. But $\left(\frac{a}{b}\right)^2 = \left(\frac{sin(B/2)^2}{sin(A/2)^2}\right) = \frac{1-cosB}{1-cosA}$ by the double angle formulas, since $1-cosB=2sin(B/2)^2$ and the analogous for $A$. From this, $IX(IC)=II_1(IA)=II_2(IB)$, and so $ABI_1I_2$ is cyclic and $I$ is the radical center of the 3 circles: circumcircle of $ACI_1$, of $BCI_2$ and of $I_1I_2AB$. Now, consider the similarity between $BFD$ and $BCA$. We get that $FI_2 = CI(cosB)$. Analogously, $FI_1=CI(cosA)$. So $\frac{FI_1}{FI_2} = \frac{cosA}{cosB} = \frac{sin\angle HDE}{sin\angle HED} = \frac{HE}{HD}$ and since $\angle DHE = \angle I_2FI_1$ we see that $\triangle DHE \equiv \triangle I_2FI_1$, they are similar (KEY FACT of this solution). Therefore $\angle FI_2I_1 = 90-A$. And therefore if $I_1I_2 \cap BC = Y$, we see $\angle I_2YC = 180-(90-A)-(A+C/2) = 90-C/2$. Similarly if $I_1I_2 \cap AC = Z$, $\angle I_1ZC=90-C/2$ and therefore $YZC$ is isosceles, so that $I_1I_2 \perp CI$. Finally, note that the radical axis of the circumcircles of $AI_1C$ and $BI_2C$ is $CI$, therefore $O_1O_2 \perp CI$. Therefore $O_1O_2 \parallel I_1I_2$ and we're done. Comment: $X$ is actually the incenter of triangle $CDE$. So if $X=I_3$, $\triangle I_1I_2I_3$ is homothetic to $\triangle O_2O_1O_3$. For symmetry, we can rename $O_1=Q_2$, $O_2=Q_1$ and $O_3=Q_3$, so that $\triangle I_1I_2I_3 \equiv \triangle Q_1Q_2Q_3$. An interesting question would be: Where is the center of homothety? I have tried locating it , maybe it's the circumcenter, centroid, symedian point, incenter, nagel point, georgenne point, orthocenter, or something, of a triangle, but I have had no luck so far.

My Solution: With some cyclic angle chasing, it's easy to prove that $\triangle AFE\sim \triangle DFB\sim \triangle ACB$, thus from the two triangles' spiral similarity transformation we also have $\triangle FI_1F_2\sim \triangle FEB$ $\Rightarrow$ $\angle (CB,I_1I_2)=\angle (CB,BE)+\angle (BE,I_1I_2)=90-\frac {\angle C}{2}$ $\angle AI_1I_2=\angle FI_1I_2+\angle AI_1F=\angle BEF+90+\frac {\angle AEF}{2}$ $=180-\frac {\angle AEF}{2}=180-\angle FBI_2$, which means that $A,I_1,I_2,B$ are concyclic. Now let $(O_1),(O_2)$ meet at $C,X$, since $OX\perp O_1O_2$. So now $I_1I_2\parallel O_1O_2$ $\iff$ $CX\perp I_1I_2$ $\iff$ $CX$ bisects $\angle ACB$(deducted from (*)). Let $AI_1, BI_2$ meet at incenter $I$, so now we just have to prove $C,X,I$ are collinear which is immediate from the radical axis theorem regarding $(O_1),(O_2),(AI_1I_2B)$ hence we are done. Q.E.D

my solution also similar to other posts. It is easy problem. But, other solutions? Please, we can proof with use since complex numbers?

Easy to get that $AI_1=AIcosA$.So $II_1=2AIsin^2\frac{A}{2}$Hence $II_1*AI=2AI^2sin^2\frac{A}{2}$.Analogously we have $II_2*IB=2IB^2sin^2\frac{B}{2}$.It is easy to check that both of these are equal.Hence points $A,I_1,I_2,B$ are concyclic.We also see that powers of $I$ with respect to $\odot{AI_1C}$ and $\odot{BI_2C}$ are equal.Hence $CI$ is the radical axis of the two circles $\Rightarrow O_1O_2 \perp CI$.Let $CI \cap I_1I_2=X$.Then $\angle{XI_1I}=\angle{ABI}=\frac{B}{2}$.Also $\angle{XII_1}=90^{\circ}-\frac{B}{2}$.Hence $CI \perp I_1I_2$ and the result follows....

Hello!Another solution. Lemma: The bisector of $\hat{ACB}$ is perpendicular to $I_1I_2$ Proof: Let $M\equiv FI_1\cap AC$ and $N\equiv FI_2\cap BC$ and $P\equiv l\cap MN$ where $l$ the bisector of $\hat{ACB}$ The triangles $BDF,AFE,ABC$ are pairwise similar. Thus $\hat{MFN}=180-\hat{BFN}-\hat{AFM}=180-BFI_2-AFI_1=$ $=180-\frac{\hat{ACB}}{2}-\frac{\hat{ACB}}{2}=180-\hat{ACB}$ thus $CNFM$ is cyclic. That is,$\hat{CNM}=\hat{CFM}=90-\frac{\hat{ACB}}{2}$. We now have $\hat{CPN}=180-\hat{CNP}-\hat{NCP}=180-\left(90-\frac{\hat{ACB}}{2}\right)-\frac{\hat{ACB}}{2}=90\Leftrightarrow l\perp MN$. It suffices now to show that $MN\parallel I_1I_2$ The theorem of bisector gives $\frac{FI_2}{I_2N}=\frac{FD}{ND}$ and $\frac{FI_1}{I_1M}=\frac{AF}{MF}$ However,the similarity of the triangles$FDN,AFM$ gives $\frac{FD}{ND}=\frac{AF}{MF}$ thus $\frac{FI_2}{I_2N}=\frac{FI_1}{I_1M}$. Now the converse of Thales's theorem gives $I_1I_2\parallel MN$ q.e.d. Back to our problem now,let $G$ be the incenter of $ABC$ and $T$ the intersection of $I_1I_2$ with the bisector of $\hat{ACB}$ According to the lemma $\hat{GTI_1}=90^{\circ}$.Also $TGI_1=\frac{\hat{BAC}+\hat{ACB}}{2}=90-\frac{\hat{ABC}}{2}$. Thus,$\hat{TI_1G}=\frac{\hat{ABC}}{2}=\hat{ABI_2}$ which means that $ABI_2I_1$ is cyclic. Now,is $S$ is the second intersection point of the circles $(O_1),(O_2)$,then $AI_1,BI_2,CS$ are the common chords of the circles $(O_1),(O_2)$ and $(A,B,I_1,I_2)$. Thus they concur at the radical centre of the circles.Since $AI_1,BI_2$ are bisectors they intersect at $G$. Since $CS$ passes from $G$ it is the bisector of $\hat{ACB}$. Thus,according to the lemma,$CS\perp I_1I_2$.Also $CS\perp O_1O_2$ (obvious) and we conclude that $I_1I_2\parallel O_1O_2$ q.e.d. The diagram on the top is for the lemma. Edit:I just noticed that my solution has many things in common with JuanOrtiz's one.

Let $I, I_3$ be the incenters of $\triangle ABC$, $\triangle CDE$, respectively, and let $M$ be the midpoint of $\widehat{AB}$ on $\odot(ABC).$ Since $B, C, E, F$ are inscribed in the circle of diameter $\overline{BC}$, we have $\triangle AEF \sim \triangle ABC.$ Therefore, $\tfrac{AI_1}{AI} = \tfrac{AE}{AB} = \cos A.$ It follows that \[IA \cdot II_1 = IA(IA - I_1A) = IA^2(1 - \cos A) = 2 \cdot IA^2 \cdot \sin^2 \frac{A}{2}.\]Then if $Z$ is the projection of $I$ onto $AB$, it follows that $IA \cdot II_1 = 2 \cdot IZ^2.$ Similarly, $IB \cdot II_2 = 2 \cdot IZ^2$, so by Power of a Point, $A, B, I_1, I_2$ are concyclic. Analagously, $B, C, I_2, I_3$ are concyclic and $C, A, I_3, I_1$ are concyclic. Then since $CI_3$ is the common chord of $\odot(ACI_1), \odot(BCI_2)$, we have $O_1O_2 \perp CI_3.$ In particular, it is sufficient to show that $I_1I_2 \perp CI_3.$ Recall that $M$ is the circumcenter of $\triangle AIB.$ Then since $I_1I_2$ is antiparallel to $AB$ WRT $\angle AIB$, it is well-known that $I_1I_2 \perp IM \equiv CI_3$, as desired. $\square$

Let the radius of the incircle of triangle $AEF=r_a$ and the radius of the incircle of triangle $BDF=r_b$. First, I claim that $ABI_1I_2$ is a cyclic quadrilateral. By standard formula about the orthic triangle and some trig, \[ r_a (b \cos A + c \cos A + 2R \cos A \sin A) = \sin A \cdot \cos^2 A \cdot bc \]and obviously \[ r(a+b+c) = \sin A \cdot bc \implies \frac{AI_1}{AI}=\frac{r_a}{r} = \cos A \]Evidently $AI_1 \cap BI_2 \equiv I$, thus by PoP it is enough to show \[ AI (AI - AI_1) = BI (BI - BI_2) \]With our work above this rearranges to \[ AI^2 (1-\cos A) = BI^2 (1-\cos B) \]true as $\frac{AI}{BI}=\frac{\sin \frac{B}{2}}{\sin \frac{A}{2}}$. Applying radical axis theorem on the circumcircles of $AI_1I_2B, AI_1C, BI_2C$, we have $CI \perp O_1O_2$. Now I claim $CI \perp I_1I_2$. We want \[ \angle AIH + \angle I_2I_1I = 90 \]Using cyclic $AI_1I_2B$ and standard angle formulas \[ 180-\angle AIC + \angle IBA = 90 \]\[ 180-(90+\frac{B}{2})+\frac{B}{2} = 90 \]We are done.

[asy][asy] unitsize(3cm); pair A, B, C, D, E, F, I, I1, I2, I3, O1, O2, K, P; A = dir(70); B = dir(200); C = dir(-30); D = foot(A, B, C); E = foot(B, A, C); F = foot(C, A, B); I = incenter(A, B, C); I1 = incenter(A, E, F); I2 = incenter(B, D, F); I3 = incenter(C, D, E); O1 = circumcenter(A, C, I1); O2 = circumcenter(B, C, I2); P = IP(Line(C, I, 10), Line(A, B, 10)); K = IP(Line(C, I, 10), Line(I1, I2, 10)); draw(A--B--C--A); draw(D--E--F--D); draw(A--I, blue); draw(B--I, blue); draw(C--P, green); draw(I1--I2--I3--I1, orange + linewidth(1)); draw(circumcircle(A, B, I1), red + linewidth(0.8)); draw(circumcircle(B, C, I2), red + linewidth(0.8)); draw(circumcircle(C, A, I3), red + linewidth(0.8)); draw(O1--O2); clip(CR((0,0),1.75)); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, S); dot("$D$", D, S); dot("$E$", E, E); dot("$F$", F, W); dot("$I$", I, E); dot("$I_1$", I1, E); dot("$I_2$", I2, S); dot("$I_3$", I3, S); dot("$O_1$", O1, E); dot("$O_2$", O2, S); dot("$P$", P, W); dot("$K$", K, W); [/asy][/asy] Let $I$ and $I_3$ be the incenters of triangles $ABC$ and $CDE$ respectively, let $K = \overline{CI} \cap \overline{I_1I_2}$ and let $P = \overline{CI} \cap \overline{AB}$. Note that $\overline{AI_1}, \overline{BI_2},$ and $\overline{CI_3}$ are the bisectors of $A, B,$ and $C$, respectively, so they concur at $I$. We claim that $ABI_2I_1$ is cyclic. By the converse of power of a point, it suffices to show that \begin{align*} \frac{II_1}{BI} &= \frac{II_2}{AI} \\ \iff \frac{AI - AI_1}{AI} &= \frac{BI - BI_2}{AI}. \end{align*}Due to properties of the orthic triangle, \begin{align*} \triangle AEF \sim \triangle DBF \sim \triangle ABC. \end{align*}Hence, \begin{align*} AI_1 &= \frac{AE}{AB}AI \\ BI_2 &= \frac{BD}{AB}BI. \end{align*}We have, \begin{align*} \frac{AI - AI_1}{BI} &= \frac{BI - BI_2}{AI} \\ \iff \frac{AI - \frac{AE}{AB}\cdot AI}{BI} &= \frac{BI - \frac{BD}{AB}\cdot BI}{AI} \\ \iff \frac{AI}{BI}(AB - AE) &= \frac{BI}{AI}(AB - BD) \\ \iff \left(\frac{AI}{BI}\right)^2 &= \frac{AB - BD}{AB - AE}. \end{align*}By the Law of Sines, we find that \begin{align*} \left(\frac{AI}{BI}\right)^2 &= \left(\frac{\sin\angle ABI}{\sin\angle BAI}\right)^2 \\ &= \left(\frac{\sin\left(\frac{B}{2}\right)}{\sin\left(\frac{A}{2}\right)}\right)^2 \\ &= \frac{1 - \cos{B}}{1 - \cos{A}}, \end{align*}with the last equality following from the half-angle formula. Again, by the Law of Sines, \begin{align*} \frac{AB - BD}{AB - AE} &= \frac{AB - AB\cos{B}}{AB - AB\cos{A}} \\ &= \frac{1 - \cos{B}}{1 - \cos{A}}. \end{align*}Therefore, \begin{align*} \left(\frac{AI}{BI}\right)^2 &= \frac{AB - BD}{AB - AE}. \end{align*}Hence, $ABI_2I_1$ is cyclic, as claimed. By identical arguments, we find that $BCI_3I_2$ and $CAI_1I_3$ are also cyclic. Note that the radical axis of $(BCI_3I_2)$ and $(CAI_1I_3)$ is $\overline{CI_3} \equiv \overline{CI}$. Hence, $\overline{CI} \perp \overline{O_1O_2}$. To finish, we will show that $\overline{CI} \perp \overline{I_1I_2}$. Note that \begin{align*} \angle BIP &= 180^\circ - \angle BIC \\ &= 180^\circ - \left(90^\circ + \frac{A}{2}\right) \\ &= 90^\circ - \frac{A}{2}. \end{align*}Also, \begin{align*} \angle II_2I_1 &= 180^\circ - \angle BI_2I \\ &= \angle BAI_1 \\ &= \frac{A}{2}. \end{align*}Hence, \begin{align*} \angle IKI_2 &= 180^\circ - \angle BIP - \angle II_2I_1 \\ &= 90^\circ. \end{align*}Thus, $\overline{CI} \perp \overline{I_1I_2}$, and $\overline{CI} \perp \overline{O_1O_2}$, which implies $\overline{I_1I_2} \parallel \overline{O_1O_2}$. This completes the proof.

niyu thank you for your solution. I like it.

math_science wrote: niyu thank you for your solution. I like it. Glad to hear it!

Key Lemma: $I_1I_2BA$ is cyclic.

Now, let $\omega_1=\odot(AI_1C)$ and $\omega_2=\odot(BI_2C)$ intersect at $C$ and $I_3$. Then our lemma forces $I$ to lie on the radical axis of $\omega_1, \omega_2$. Hence, $O_1O_2 \perp CI$. If $AI, BI$ meet $\odot(ABC)$ again at $M, N$ respectively, then $\angle II_1I_2=\angle IBA=\angle NMA$ and so $I_1I_2 \parallel MN$, but it is well known that $MN \perp CI$. Hence $I_1I_2 \perp CI$ and we are done. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -64.12480498146493, xmax = 69.86062085974757, ymin = -25.843937807832337, ymax = 74.74002210129348; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-28.082321837039625,53.172232012915444)--(-45.29239001234524,-0.23797272940859554)--(22.43052577761703,-1.2345365879987495)--cycle, linewidth(0.5)); /* draw figures */ draw((-28.082321837039625,53.172232012915444)--(-45.29239001234524,-0.23797272940859554), linewidth(0.5)); draw((-45.29239001234524,-0.23797272940859554)--(22.43052577761703,-1.2345365879987495), linewidth(0.5)); draw((22.43052577761703,-1.2345365879987495)--(-28.082321837039625,53.172232012915444), linewidth(0.5)); draw((-39.21313645390692,18.628552023967433)--(-9.417874581181733,33.068984802224), linewidth(0.5) + wrwrwr); draw((-9.417874581181733,33.068984802224)--(-28.871825398011143,-0.4796064665477631), linewidth(0.5) + wrwrwr); draw((-28.871825398011143,-0.4796064665477631)--(-39.21313645390692,18.628552023967433), linewidth(0.5) + wrwrwr); draw(circle((-11.151706469718487,18.238921752624023), 38.81986479185563), linewidth(0.5) + wrwrwr); draw((-28.082321837039625,53.172232012915444)--(-28.871825398011143,-0.4796064665477631), linewidth(0.5) + wrwrwr); draw((-45.29239001234524,-0.23797272940859554)--(-9.417874581181733,33.068984802224), linewidth(0.5) + wrwrwr); draw((-39.21313645390692,18.628552023967433)--(22.43052577761703,-1.2345365879987495), linewidth(0.5) + wrwrwr); draw((-28.082321837039625,53.172232012915444)--(-11.722891066294,-20.57674068684218), linewidth(0.5) + wrwrwr); draw((-45.29239001234524,-0.23797272940859554)--(17.29725139481586,44.65177277026128), linewidth(0.5) + wrwrwr); draw((-56.394010042291846,33.834471391601326)--(22.43052577761703,-1.2345365879987495), linewidth(0.5) + wrwrwr); draw(shift((-65.5634541942046,35.77171145814406))*xscale(41.32327913358693)*yscale(41.32327913358693)*arc((0,0),1,-86.79902453242678,41.01039812399654), linewidth(0.5) + linetype("2 2") + wrwrwr); draw((-37.955631807891805,5.024003552705843)--(-24.240253119742448,35.8520315173558), linewidth(0.5) + wrwrwr); draw((-30.71222600548734,21.304991610933694)--(-30.057573572733677,19.975157785035076), linewidth(0.5) + wrwrwr); draw((-30.71222600548734,21.304991610933694)--(-32.13831135490057,20.90087728502656), linewidth(0.5) + wrwrwr); draw((-11.722891066294,-20.57674068684218)--(17.29725139481586,44.65177277026128), linewidth(0.5) + wrwrwr); draw((3.1728966225907085,12.90449011761243)--(3.827549055344367,11.574656291713813), linewidth(0.5) + wrwrwr); draw((3.1728966225907085,12.90449011761243)--(1.746811273177476,12.500375791705295), linewidth(0.5) + wrwrwr); draw(shift((30.742679201319177,57.134904251302))*xscale(58.958320179867684)*yscale(58.958320179867684)*arc((0,0),1,163.03856623055563,280.985130126059), linewidth(0.5) + wrwrwr); draw(shift((-11.99284194794007,-38.92163717045144))*xscale(51.04200031229999)*yscale(51.04200031229999)*arc((0,0),1,10.809638497587533,154.18170827143484), linewidth(0.5) + wrwrwr); /* dots and labels */ dot((-28.082321837039625,53.172232012915444),dotstyle); label("$A$", (-26.6746765121553,55.635395753773984), NE * labelscalefactor); dot((-45.29239001234524,-0.23797272940859554),dotstyle); label("$B$", (-47.80363412828607,-4.714980191701505), NE * labelscalefactor); dot((22.43052577761703,-1.2345365879987495),dotstyle); label("$C$", (22.28883604738128,-5.474104417430756), NE * labelscalefactor); dot((-28.640773132330857,15.221879190260053),linewidth(4pt) + dotstyle); label("$H$", (-31.229421866530796,17.17310165015858), NE * labelscalefactor); dot((-28.871825398011143,-0.4796064665477631),linewidth(4pt) + dotstyle); label("$D$", (-28.82552848505484,-5.221063008854339), NE * labelscalefactor); dot((-9.417874581181733,33.068984802224),linewidth(4pt) + dotstyle); label("$E$", (-8.961777911806148,34.12687602477853), NE * labelscalefactor); dot((-39.21313645390692,18.628552023967433),linewidth(4pt) + dotstyle); label("$F$", (-42.48976454818133,17.04658094587037), NE * labelscalefactor); dot((17.29725139481586,44.65177277026128),linewidth(4pt) + dotstyle); label("$N$", (17.860611397293994,45.64026011500551), NE * labelscalefactor); dot((-11.722891066294,-20.57674068684218),linewidth(4pt) + dotstyle); label("$M$", (-12.883919744740604,-24.325689356373836), NE * labelscalefactor); dot((-24.240253119742448,35.8520315173558),linewidth(4pt) + dotstyle); label("$I_1$", (-23.76470031352651,36.91033151911912), NE * labelscalefactor); dot((-37.955631807891805,5.024003552705843),linewidth(4pt) + dotstyle); label("$I_2$", (-37.04937426378837,3.255824178455636), NE * labelscalefactor); dot((-20.222977651400768,17.74199193260845),linewidth(4pt) + dotstyle); label("$I$", (-19.716037776303846,18.691350101617083), NE * labelscalefactor); dot((-56.394010042291846,33.834471391601326),dotstyle); dot((-7.047500726543539,11.880226930039877),linewidth(4pt) + dotstyle); label("$I_3$", (-6.557884530330193,14.136604747241574), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] P.S. Our lemma gives us that $I_3$ is, in fact, the incenter of $\triangle CED$.

Dear Mathlinkers, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=572225 Jean-Louis

My solution: Let $I$ be the incenter of $\triangle ABC$. We start off with the following Lemma. Lemma $I$ lies on the radical axis of $\odot (ACI_1)$ and $\odot (BCI_2)$. PROOF: Note that it suffices to show $IA \cdot II_1=IB \cdot II_2$. Now, an easy angle chase yields that $\triangle AI_1E \sim \triangle DI_2B \sim \triangle AIB$. Using $AF \cdot AB=AE \cdot AC$, we get that $$\frac{AI_1}{AI}=\frac{AE}{AB}=\frac{AF}{AC}=\cos A \Rightarrow AI_1=AI \cos A \Rightarrow IA \cdot II_1=IA(IA-AI_1)=IA^2(1-\cos A)=2IA^2 \sin ^2 \frac{A}{2}=2r^2$$where $r$ is the inradius of $\triangle ABC$. Similarly, we have $IB \cdot II_2=2r^2$, giving $IA \cdot II_1=IB \cdot II_2$. $\Box$ Return to the problem at hand. As $CI$ is the radical axis of $\odot (ACI_1)$ and $\odot (BCI_2)$, we have $O_1O_2 \perp CI$. Let $P$ and $Q$ be the midpoints of $\overarc{BC}$ and $\overarc{CA}$. Then by Fact 5, we have that $PQ$ is the perpendicular bisector of $CI$, and so $PQ$ is parallel to $O_1O_2$. But, Using $IA \cdot II_1=IB \cdot II_2$ and $IA \cdot IP=IB \cdot IQ$, we get that $I_1I_2 \parallel PQ$. Thus, $I_1I_2$ is parallel to $O_1O_2$. $\blacksquare$

ISL 2012 G3 wrote: In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel. Solution: Let $I_3$ be Incenter WRT $\Delta CDE$. Let $\odot (DI_3E)$ $\cap$ $\odot (BI_2D)=L$ $\implies$ $L-I_2-I_3$ and Since, $\angle BLD$ $=$ $90^{\circ}+$ $\tfrac{C}{2}$ $=$ $\angle EI_3D$ $\implies$ $BL \perp AC$ $\implies$ $\angle I_3CB$ $=$ $\tfrac{C}{2}$ $=$ $\angle LI_2B$ $\implies$ $BI_2I_3C$ is cyclic & Similarly, $AI_1I_2B$ and $AI_1I_3C$ are cyclic $\implies$ $O_1O_2 \perp CI$. Let $I_1I_2$ $\cap$ $AB$ $=$ $X_1$ & $CI$ $\cap$ $I_1I_2$ $=$ $X_2$ $\implies$ $X_1FX_2C$ is cyclic $\implies$ $I_1I_2||O_1O_2$ $\qquad \blacksquare$

Note that there is a spiral similarity at $F$ from $AI_1E$ to $DI_2B$, so $AFD\sim I_1FI_2$. Then, $\angle AI_1I_2=\angle FI_1I_2+\angle AI_1F=180-\frac{\angle B}{2}\implies (AI_1I_2B)$. Also, we also have that $\angle\widehat{I_1I_2,AD}=\angle DFI_2=\angle C/2$, so $\angle\widehat{I_1I_2,CI}=\angle\widehat{I_1I_2,AD}+\angle\widehat{AD,CI}=\frac{\angle C}{2}+90-\frac{\angle C}{2}=90\implies I_1I_2\perp CI$. Radical axis on $(AI_1I_2B),(AI_1C),(BI_2C)$ gives that the radical axis of the latter two is $CI$, so $CI\perp O_1O_2$, and thus $O_1O_2\parallel I_1I_2$ as desired.

Not too bad for G3. We will prove the result by showing that $\overline{CI} \perp \overline{I_1I_2}$ and that $I$ lies on the radical axis of $(CBI_2)$ and $(CAI_1)$. To show the latter fact, notice that $$II_2 \cdot IB = IB^2 (1-\cos B) = 2 IB^2 \sin^2(B/2) = 2IA^2 \sin^2 (A/2) = II_1 \cdot IA,$$hence $I_1ABI_2$ is cyclic. In particular, this also implies the latter result. To show the former result, it suffices to do an angle chase: $$\measuredangle(\overline{I_1I_2}, \overline{CI}) = \measuredangle(\overline{I_1I_2}, \overline{BC}) + \measuredangle(\overline{BC}, \overline{CI}) = \frac B2 - \frac A2 + A+\frac C2 = 90^\circ.$$

Note that $AI_1 \cap BI_2 = I$. Using orthic similarity ratios, we can determine $AI_1I_2B$ is cyclic using POP, as \[\left(\frac{IA}{IB}\right)^2 = \frac{II_2/IB}{II_1/IA} \iff \frac{\sin^2 \frac B2}{\sin^2 \frac C2} = \frac{1-\cos B}{1-\cos C}\] is true from sine half angle formula. If we define $I_3 \in CI$ as the incenter of $\triangle CDE$, we know $(O_1) \cap (O_2) = C, I_3$, so $O_1O_2 \perp CI$. Focusing on $\triangle AIB$, noting that $CI$ passes through the center of $(AIB)$ and $I_1I_2$ is antiparallel to $AB$, we have $CI \perp I_1I_2$, as desired. $\blacksquare$

Denote $I_3$ as the incenter of $\triangle CED$. Claim: $AI_1I_2B$, $AI_1I_3C$, $BI_2I_3C$ are cyclic quadrilaterals. Proof: First it i enough to prove one of the quadrilaterals is cyclic. Lets show that $AI_1I_2B$ is cyclic. Let $AI_1 \cap BI_2 = I$ $\Rightarrow$ we want to prove that $II_1.IA = II_2.IB$. Now let $\angle AFI_1 = \angle I_1FE = \alpha$, also $\angle EFH = \angle HFD = 90 - 2\alpha$ $\Rightarrow$ $\angle DFI_2 = \angle I_2FB = \alpha$. Let $\angle FAI_1 = \angle I_1AE = \beta$, from FAEH being cyclic $\angle FHE = 180 - 2\beta$ $\Rightarrow$ $\angle FHB = \angle FDB = 2\beta$ $\Rightarrow$ $\angle FDI_2 = \angle I_2DB = \beta$ and now $\angle ABI_2 = \angle I_2BC = 90 - \alpha - \beta$ and $\angle ACI_3 = \angle I_3CB = \alpha$. Now we have that $\triangle AEF \sim \triangle ABC$ $\Rightarrow$ $\frac{AI_1}{AI} = \frac{AE}{AB}$. Also $\triangle BFD \sim \triangle BCA$ $\Rightarrow$ $\frac{BI_2}{BI} = \frac{BD}{BA}$. We want to prove $II_1.IA = II_2.IB$ $\Leftrightarrow$ $AI.(AI - AI_1) = BI(BI - BI_2)$ $\Leftrightarrow$ $AI^2 - AI.AI_1 = BI^2 - BI.BI_2$ $\Leftrightarrow$ $AI^2(1 - \frac{AI_1}{AI}) = BI^2(1 - \frac{BI_2}{BI})$ $\Leftrightarrow$ $AI^2(1 - \frac{AE}{AB}) = BI^2(1 - \frac{BD}{BA})$ $\Leftrightarrow$ $(\frac{AI}{BI})^2 = \frac{(1 - \frac{BD}{BA})}{(1 - \frac{AE}{AB})}$. We know that $\frac{AI}{BI} = \frac{\sin(\frac{B}{2})}{\sin(\frac{A}{2})}$. Also $\frac{BD}{BA} = \cos B$ and $\frac{AE}{AB} = \cos A$ $\Rightarrow$ $(\frac{AI}{BI})^2 = \frac{(1 - \frac{BD}{BA})}{(1 - \frac{AE}{AB})}$ $\Leftrightarrow$ $(\frac{\sin(\frac{B}{2})}{\sin(\frac{A}{2})})^2 = \frac{(1 - \cos B)}{(1 - \cos A)}$, which is obvious from the trigonometry formulas $\Rightarrow$ with that, our claim is proven. Now we have that $BI_2I_3C$ is cyclic $\Rightarrow$ $I_3O_2 = CO_2$ $\Rightarrow$ $O_2 \in S_{I_3C}$, also from $AI_1I_3C$ cyclic it follows that $I_3O_1 = O_1C$ $\Rightarrow$ $O_1 \in S_{I_3C}$ $\Rightarrow$ $O_1O_2 \equiv S_{I_3C}$ $\Rightarrow$ $O_1O_2 \perp I_3C$. Let $CI_3 \cap I_1I_2 = P$. Now $\angle PI_1I_3 = \angle PI_1I + \angle II_1I_3 = \angle I_2BA + \angle I_3CA = 90 - \alpha - \beta + \alpha = 90 - \beta$. Also $\angle I_1I_3P = \angle I_1AC = \beta$ $\Rightarrow$ $\angle I_1PI_3 = 180 - \angle PI_1I_3 - \angle I_1I_3P = 180 - (90 - \beta) - \beta = 90^{\circ}$ $\Rightarrow$ $I_1I_2 \perp I_3C$ $\Rightarrow$ we know that $O_1O_2 \perp I_3C$ and $I_1I_2 \perp I_3C$ $\Rightarrow$ it follows that $I_1I_2 \parallel O_1O_2$ which is what we wanted to prove. We are ready.

Solved with Aadish. Switch the indexing to $A-$ index. Claim. $I_1BCI_2$ are cyclic. Proof. Note that $A,F,D,C$ and $A,E,D,B$ are cyclic. So, $\angle A=\angle BAC=\angle FAC=\angle FDB$ and, similarly, $\angle BAC=\angle CDE$. So, $\angle I_2DI_1=180^\circ-\angle I_2DB-\angle I_1DC=180^\circ-\angle A$. We show that $\angle DI_1I_2=90^\circ-\angle C$ and $\angle I_1I_2D=90^\circ-\angle B$. It suffices to show that $\frac{DI_1}{DI_2}=\frac{\sin(90^\circ-\angle B)}{\sin(90^\circ-\angle C)}=\frac{\cos(\angle B)}{\cos(\angle C)}$. But that is true, because $\angle FDB=\angle CDE$ which means $\frac{DI_1}{DI_2}=\frac{\triangle FDB}{\triangle EDC}=\frac{\cos(\angle B)}{\cos(\angle C}$. Finally note $\angle BI_1D=90^\circ+\frac{\angle BFD}2=90^\circ+\frac{\angle C}2$ which means $\angle DI_1I=90^\circ-\frac{\angle C}{2}$ but $\angle DI_1I_2=90^\circ-\angle C$, which means $\angle I_2I_1D=\frac{\angle C}2=\angle I_2CB=\angle ICB$. So, we're done. [This also means $I_1I_2\perp AI$.] So, we have $II_1\cdot IB=II_2\cdot IC$ which means $I$ has same power from both circles. So, $AI$ is the radical axis and $AI\perp O_1O_2$ which means $O_1O_2\parallel I_1I_2$. $\blacksquare$

first, let $I_1I_2$ intersect $AD$ and $BE$ at $P_1$ and $P_2$, and $AC$ and $BC$ at $Q_1$ and $Q_2$. let the orthocenter of the triangle be $H$, and let the incenter of the triangle be $I$ some simple angle chasing yields that $AP_1I_1F$ and $BP_2I_2F$ are both cyclic quadrilaterals then, angle chasing yields that $AP_1I_1=BP_2I_2$ and $HP_1=HP_2$, as well as $CQ_1=CQ_2$ also, even more angle chasing yields $I_1II_2$, $I_1Q_1A$, and $BQ_2I_2$ are similar, so $AI_1I_2B$ are cyclic then, by radical axis and stuff we get that $O_1O_2$ is perpendicular to $CI$ however, $CI$ is also perpendicular to $I_1I_2$ since $CQ_1=CQ_2$, so $I_1I_2$ is parallel to $O_1O_2$

Claim: Line $I_1 I_2$ is perpendicular to the bisector of $\angle C$. Proof: Extend $FI_1$ and $FI_2$ to meet $\overline{AC}$ and $\overline{BC}$ at $D_1$ and $D_2$, respectively. Since $\triangle AFE \sim \triangle DFB$, it follows that $\overline{I_1 I_2} \parallel \overline{D_1 D_2}$. Since $\angle D_1 F D_2 = 180^{\circ} - \angle C$, it follows that $CD_1 FD_2$ is cyclic. Furthermore, since $\angle D_1 F C = \angle D_2 FC$, it follows that $D_1 C = D_2 C$. So, $\overline{D_1 D_2}$ is perpendicular to the bisector of $\angle C$, which proves our claim. $\square$ Claim: $AI_1 I_2 B$ is cyclic. Proof: Let $\ell$ be the line through $I$ tangent to $(AIB)$. It's well known that $\ell$ is perpendicular to the bisector of $\angle C$, so our claim follows by Reim's theorem. $\square$ Claim: Line $O_1 O_2$ is perpendicular to the bisector of $\angle C$. Proof: It is clear that $I$ is the radical center of $(AI_1 I_2 B)$, $(CI_1 A)$ and $(CI_2 B)$, so it follows that the bisector of $\angle C$ is the radical axis of $(CI_1 A)$ and $(CI_2 B)$. Therefore, the line connecting these two circles' centers, line $O_1 O_2$, is perpendicular to the radical axis, the bisector of $\angle C$. $\square$ Since lines $I_1 I_2$ and $O_1 O_2$ are both perpendicular to the bisector of $\angle C$, they are parallel, as desired.

This problem dies if you know spiral similarity. Notice that there is a spiral similarity at $F$ sending $\triangle FAE$ to $\triangle FDB.$ (This is a well-known property of the orthic triangle.) This spiral similarity sends $I_1$ to $I_2,$ so $F$ is the spiral center sending $AI_1$ to $DI_2.$ This implies that $F$ is the spiral center sending $AD$ to $I_1 I_2.$ Hence $\boxed{\triangle FAD \sim \triangle FI_1 I_2}.$ We will repeatedly use this fact, which we call (*) for convenience. Let $I = AI_1 \cap BI_2$ be the incenter of the triangle. Claim: $A,B,I_1,I_2$ are concyclic. Proof: By (*), $\angle FI_2 I_1 = \angle FDA = \angle FCA = \angle FCA = 90^\circ - \angle BAC$ since $AFDC$ is cyclic. Also, $\angle BDF = \angle BAC,$ so $\angle BIF = 90^\circ + \frac{\angle BDF}{2} = 90^\circ + \frac{\angle BAC}{2}.$ Therefore, $$\angle BI_2 I_1 = 90^\circ - \angle BAC + 90^\circ + \frac{\angle BAC}{2} = 180^\circ - \frac{\angle BAC}{2}.$$However, $\angle BAI_1 = \frac{\angle BAC}{2},$ so $\angle BAI_1 + \angle BI_2 I_1 = 180^\circ,$ concluding our proof of the claim. Claim: $CI \perp I_1 I_2.$ Proof: We note that $AD \perp BC,$ and the spiral similarity in (*) sends $AD$ to $I_1 I_2$, so we wish to show that it sends $BC$ to a line parallel to $CI.$ In other words, we wish to show that the spiral similarity rotates things by $\angle BCI = \frac{\angle BCA}{2}.$ Indeed, it rotates $FA$ to $FI_1,$ and the angle between these two lines is $\angle AFI_1 = \frac{\angle AFE}{2} = \frac{\angle ACB}{2}.$ Thus $CI \perp I_1 I_2,$ as claimed. To finish, by applying the radical center theorem on $(ABI_1 I_2), (ACI_1),$ and $(BCI_2),$ we see that $CI$ is the radical axis of $(ACI_1)$ and $(BCI_2).$ Since $CI \perp I_1 I_2,$ we see that the radical axis is perpendicular to $I_1 I_2.$ However, it is clear that this radical axis is perpendicular to $O_1 O_2.$ Therefore, $O_1 O_2$ and $I_1 I_2$ are perpendicular to a common line, and we are done.

Let $I_3$ be defined similarly to $I_1$ and $I_2$, thus proving $ABI_1I_2$, $ACI_1I_3$ and $BCI_2I_3$ are all cyclic suffices as this gives that $I$ is the orthocenter of $I_1I_2I_3$ and also gives that the radical axis of $(ACI_1I_3)$ and $(BCI_2I_3)$ is $II_3$ and thus gives that $O_1O_2$ is perpendicular to $II_3$, thus meaning $I_1I_2 \parallel O_1O_2$. To prove the cyclic consider the spiral similarity taking $FAE$ to $FBD$, this also takes $I_1$ to $I_2$, thus giving us that $FI_1I_2$ is similar to $FBE$, giving that $\angle I_1I_2F = 90-\angle CAB$, we have that $\angle FIB= 90+\frac{\angle CAB}{2}$, which gives the cyclic, by similar arguements we get all the cyclics thus we are done.

$ $

@navi_09220114 orz, I can finally do geo. Define $I_3, I$ in the sensible way. Then the main claims are as follows: 1. $A, I_1, I_2, B$ are cyclic. (Proof: power of point on $I$, trig bash) 2. $O_1O_2 \perp CI$. (Proof: Radical axis, noting that $I$ is the radical centre of three circles $AI_1C, BI_2C, AI_1I_2B$) 3. $I$ is the orthocentre of $\Delta I_1I_2I_3$. (Proof: Angle chase) And tada!! We're done!

hehe

Say Intersections of $AI$1$C$ and $BI$2$C$ be $X$ then intuitively $CX$ is the angle bisector of $C$. Now AI1, BI2 and CX concur at I. So by Pop AI1I2B is concyclic. Then O1O2 is perpendicular to CI and by simple angle chasing I1I2 is also perpendicular to CI so O1O2 is parallel to O1O2.

Let $AI_1$ and $BI_2$ intersect at the incenter $I$. Claim: $A, I_1, I_2, B$ are concyclic. Proof. It is easy to see that $\Delta{DI_2B} \sim \Delta{AIB} \sim \Delta{AI_1E}$ by angle chasing. Hence: $I_1 A = IA \cdot \frac{AE}{AB} = \frac{r}{\sin{\frac{A}{2}}} \cdot \cos{A} \implies II_1 = IA - I_1A = \frac{r}{\sin{\frac{A}{2}}} (1 - \cos{A}) = 2 \frac{r}{\sin{\frac{A}{2}}} \cdot \sin^2{\frac{A}{2}} = 2 r \sin{\frac{A}{2}}$ Similarly, $II_2 = 2 r \sin{\frac{B}{2}}$. Hence $II_1 \cdot IA = 2 r \cancel{\sin{\frac{A}{2}}} \frac{r}{\cancel{\sin{\frac{A}{2}}}} = 2 r^2 = II_2 \cdot IB$ as desired. Now since $II_1 \cdot IA = II_2 \cdot IB$, the point $I$ must be on the radical axis of $(ACI_1)$ and $(BCI_2)$. In fact, this radical axis must be the line $CI$. Hence, since the radical axis is perpendicular to the line joining the centers, $CI \perp O_1O_2$ Further, a trivial angle chase using the Claim yields $CI \perp I_1I_2$ Hence $O_1O_2 \parallel I_1I_2$ as desired.

AshAuktober wrote: @navi_09220114 orz, I can finally do geo. Define $I_3, I$ in the sensible way. Then the main claims are as follows: 1. $A, I_1, I_2, B$ are cyclic. (Proof: power of point on $I$, trig bash) 2. $O_1O_2 \perp CI$. (Proof: Radical axis, noting that $I$ is the radical centre of three circles $AI_1C, BI_2C, AI_1I_2B$) 3. $I$ is the orthocentre of $\Delta I_1I_2I_3$. (Proof: Angle chase) And tada!! We're done! My solution is almost identical to this. Nice!