Let $X$ be the reflection of $C_1$ over $\overline{CI}$. We will show that $X$ is the desired concurrency point. [asy][asy] defaultpen(fontsize(10pt)); size(300); pair A, B, C, I, A1, B1, C1, D, E, F, K, L, X; A = dir(180); B = dir(0); C = dir(125); D = foot(C, A, B); I = incenter(A, B, C); A1 = foot(I, B, C); B1 = foot(I, C, A); C1 = foot(I, A, B); E = 2*foot(C, C1, A1)-C; F = 2*foot(C, C1, B1)-C; K = 2*foot(D, C1, A1)-D; L = 2*foot(D, C1, B1)-D; X = 2*foot(C1, C, I)-C1; draw(A--B--C--cycle, orange); draw(incircle(A, B, C), red); draw(circumcircle(A1, E, I), lightblue); draw(arc(circumcenter(B1, F, I), circumradius(B1, F, I), 0, 120), lightblue); draw(C--D, deepgreen+dashed); draw(B1--I--A1^^C1--I, heavygreen); draw(L--C1--K--X--cycle, magenta); dot("$A$", A, dir(200)); dot("$B$", B, dir(340)); dot("$C$", C, dir(90)); dot("$D$", D, dir(270)); dot("$A_1$", A1, dir(90)); dot("$B_1$", B1, dir(110)); dot("$C_1$", C1, dir(260)); dot("$E$", E, dir(70)); dot("$F$", F, dir(100)); dot("$K$", K, dir(270)); dot("$L$", L, dir(160)); dot("$X$", X, dir(80)); dot("$I$", I, dir(0)); [/asy][/asy] The proof proceeds in two main steps. Claim: $X$ lies on $(A_1EI)$ and $(B_1FI)$. Proof. Let $r$ be the inradius of $\triangle ABC$. Then $CA_1IB_1$ is a square, and we deduce $$A_1C = B_1C = A_1E = B_1F = C_1I = XI = r.$$Now we have \begin{align*} \angle IA_1E &= 90^\circ+\angle BA_1E = 90^\circ+(180^\circ - \angle CA_1E) \\ &= 270^\circ - 2\angle C_1A_1B = 90^\circ+\angle B, \\ \angle XIA_1 &= \angle C_1IB_1 = 180^\circ - \angle A = 90^\circ + \angle B. \end{align*}Combining these two observations, we see that $XIA_1E$ is an isosceles trapezoid, and hence it is cyclic. Similarly, $XIB_1F$ is cyclic, as desired. $\blacksquare$ To finish, we prove that $X$ lies on $(C_1KL)$. Observe that $$\angle DC_1L = 2\angle AC_1B_1 = 180^\circ - \angle A = \angle C_1IB_1,$$and similarly, $\angle DC_1K = \angle C_1IA_1$. Noting that $DC_1 = C_1L = C_1K$, we see that figures $DC_1LK$ and $C_1IB_1A_1$ are similar. Claim: We have $\triangle DC_1X\sim \triangle C_1IC$. Proof. First we do the length computation. Note that $$DC_1 = CI\cos \angle(CI, AB) = CI\cos(\angle B+45^\circ) = \sqrt{2}r\cos(\angle B+45^\circ),$$and so \begin{align*} C_1X &= 2r\sin\angle C_1B_1X = 2r\sin(\angle C_1B_1A - \angle C_1A_1B_1) \\ &= 2r\sin(A/2-B/2) = 2r\sin(45^\circ - \angle B) = 2r\cos(\angle B+45^\circ) = \sqrt{2}DC_1. \end{align*}Furthermore, \begin{align*} \angle DC_1X &= 180^\circ - \angle XC_1B = 180^\circ - \angle C_1B_1X \\ &= 180^\circ - (45^\circ - \angle B) = 180^\circ - \angle DCI = \angle CIC_1, \end{align*}which proves the claim by SAS similarity. $\blacksquare$ Combined with the previous similarity, this claim shows that $DC_1LKX\sim C_1IB_1A_1C$; in particular, $XKC_1L$ is a square. This proves the desired result.

(soln with Anant) WLOG, $CA<CB$. Let $P$ be such that $C_1PA_1B_1$ is an isosceles trapezoid with $C_1A_1=B_1P$. $\textbf{Claim:}$ $B_1D=A_1L$ and $A_1D=B_1K$. $\textbf{Proof:}$ Define $X=CI \cap B_1C_1$. Note that $\angle DXC_1=\angle B/2$. Indeed, let $Q=C_1B_1 \cap BI$, note that $CDXQB$ is well-known to be a cyclic pentagon. Then $\angle DXC_1=\angle DBI$ and we're done. Now this shows that $\triangle XDL \sim \triangle XB_1A_1 \implies \triangle XLA_1 \cong \triangle XDB_1$. (Since $\triangle XDL, \triangle XA_1B_1$ are both isosceles.) $\blacksquare$ $\textbf{Claim:}$ $FCEI$ is a parallelogram $\textbf{Proof:}$ By Angle Chasing, $FB_1 \parallel AB$, and $EA_1 \parallel AB$. So, we have that $FB_1EA_1$ is a parallelogram. Since $A_1IB_1C$ is a square, the conclusion follows. $\blacksquare$ $\textbf{Claim:}$ $C, I, K, L$ collinear. $\textbf{Proof:}$ Note that $B_1L=B_1D=A_1L$ hence $L$ on $CI$. Likewise $K$ on $CI$. $\blacksquare$ $\textbf{Claim:}$ $FB_1IP$ and $EA_1IP$ are cyclic. $\textbf{Proof:}$ Note that $\angle IFB_1=\angle IFC-\angle CFB_1=45^{\circ}-\angle A/2=\angle B/2=\angle ICA_1=\angle IPB_1$, proving the claim. Likewise $EA_1IP$ is cyclic. $\blacksquare$ Finally, note that $\angle KLP=\angle KC_1P=90^{\circ}$ by symmetry along $CI$, so $P$ lies on $\odot(C_1KL)$ as well. $\textbf{Remark.}$ The solution was motivated by the natural guess that $A_1L=B_1D$.

[asy][asy] size(9cm); defaultpen(fontsize(10pt)); pen pri=heavycyan; pen sec=lightblue; pen tri=royalblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,I,D,A1,B1,C1,EE,F,K,L,T; O=(0,0); A=dir(150); B=-A; C=reflect(O,(1,0))*A; I=incenter(A,B,C); D=foot(C,A,B); A1=foot(I,B,C); B1=foot(I,C,A); C1=foot(I,A,B); EE=reflect(C1,A1)*C; F=reflect(C1,B1)*C; K=reflect(C1,A1)*D; L=reflect(C1,B1)*D; T=reflect(circumcenter(A1,EE,I),circumcenter(B1,F,I))*I; draw(EE--C--F,tri+dashed); draw(K--D--L,tri+dashed); draw(K--C1--L,sec+dashed); draw(K--T,sec); filldraw(A1--I--T--EE--cycle,sfil,sec+opacity(1)); filldraw(B1--I--T--F--cycle,sfil,sec+opacity(1)); filldraw(A1--B1--C1--cycle,fil,pri); filldraw(incircle(A,B,C),fil,pri); filldraw(A--B--C--cycle,fil,pri); dot("$A$",A,N); dot("$B$",B,SE); dot("$C$",C,SW); dot("$I$",I,SW); dot("$D$",D,N); dot("$A_1$",A1,S); dot("$B_1$",B1,SW); dot("$C_1$",C1,dir(75)); dot("$E$",EE,S); dot("$F$",F,W); dot("$K$",K,NE); dot("$L$",L,SW); dot("$T$",T,E); [/asy][/asy] Since $\overline{A_1C_1}$ is parallel to the external angle bisector of $\angle ABC$, reflection across $\overline{A_1C_1}$ swaps lines parallel to $\overline{AB}$ with lines parallel to $\overline{BC}$. In particular, $\overline{CA_1}\parallel\overline{BC}\implies\overline{EA_1}\parallel\overline{AB}$ (and similarly $\overline{FB_1}\parallel\overline{AB}$). $\overline{DC_1}\parallel\overline{AB}\implies\overline{KC_1}\parallel\overline{BC}$ (and similarly $\overline{LC_1}\parallel\overline{AC}$). $\overline{CD}\perp\overline{AB}\implies\overline{EK}\perp\overline{BC}$ (and similarly $\overline{FL}\perp\overline{AC}$). Consider the point $T$ such that $A_1ITE$ is an isosceles trapezoid with $\overline{A_1I}\parallel\overline{ET}$. I claim $T$ is the common point. Since $IT=A_1E=A_1C=IA_1$, we have $T$ lies on the incircle. Angle chasing gives \begin{align*} \measuredangle B_1IT&=90^\circ+\measuredangle A_1IT=90^\circ+\measuredangle EA_1I=\measuredangle EA_1B=\measuredangle ABC\\ &=90^\circ+\measuredangle BAC=90^\circ+\measuredangle FB_1I=\measuredangle FB_1I, \end{align*}and since $IT=IB_1=B_1C=B_1F$, we have $B_1ITF$ is also an isosceles trapezoid. It follows that $T$ lies on $(A_1EI)$ and $(B_1FI)$, so it remains to show $C_1KTL$ is cyclic. Recall that $\overline{EK}\perp\overline{BC}$. Since $\overline{A_1I}\parallel\overline{ET}$, we have $E$, $T$, $K$ collinear, so $\overline{TK}\parallel\overline{AC}$. We've seen that $\overline{LC_1}\parallel\overline{AC}$, so $C_1KTL$ is a rectangle. This completes the proof.

The following solution sketch is written in the order the solution was found: (i) $AB_1IC_1$ is a square, so we obtain $r=AB_1=B_1I=IC_1=C_1A=FB_1=EA_1$. (ii) Angle-chasing using the reflections yields $AB \parallel FB_1 \parallel EA_1$, $AC \parallel LC_1$, and $BC \parallel KC_1$. (iii) Thus $KC_1 \perp LC_1$. (iii) Let the circumcircles of $A_1EI$ and $B_1FI$ intersect at $P$. (iv) Similar angle-chasing yields $\angle A_1PI=45-\frac{\angle B}{2}$ and $\angle B_1PI=45-\frac{\angle A}{2}$, thus $\angle EPF=45^\circ=\frac{1}{2} \angle B_1PA_1$. So $P$ lies on the incircle of $ABC$. (v) $A_1EIP$ and $B_1FIP$ are isosceles trapezoids because of cyclicity and equal lengths. So, since $A_1I \perp BC$, we have that $EP \perp BC$. Similarly $FP \perp AC$. (vi) By reflection since $CD \perp DC_1$, $FL \perp LC_1$, so $FL \perp AC$. Thus, $F,L,P$ collinear. Similarly $E,K,P$ collinear. (vii) $KP \perp LP$ by (vi) and (ii). (viii) $KC_1LP$ is cyclic with diameter $KL$ (in fact it is a square). This concludes the proof.

Here's a solution found with amar_04 and mueller.25. 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draw((2.9367042904010443,-2.0185298305110715)--(7.83163440613279,-5.072018354875753), linewidth(0.4)); draw((6.900536522780633,-3.1402771367106057)--(8.035518531166327,-1.3208312751962268), linewidth(0.4)); draw(shift((4.0317414835085055,-5.713299232377903))*xscale(3.853625226622336)*yscale(3.853625226622336)*arc((0,0),1,9.57914051601262,106.50850548755722), linewidth(0.8) + linetype("4 4")); draw(shift((9.119725962980144,-3.26089428172582))*xscale(2.2224649079714722)*yscale(2.2224649079714722)*arc((0,0),1,61.5085014267715,234.5791397507545), linewidth(0.8) + linetype("4 4")); /* dots and labels */ dot((6.216072669651946,-0.18584926681053188),dotstyle); label("$C$", (6.252707482304711,-0.08075266786739245), NE * labelscalefactor); dot((3.020410473726549,-5.30869202045405),dotstyle); label("$B$", (3.0569446187083895,-5.202167513374285), NE * labelscalefactor); dot((14.31654455898904,-5.238975632537797),dotstyle); label("A", (14.354785767896667,-5.140710535228202), NE * labelscalefactor); 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Now notice that $(CB_1IA_1)$ is a square $\implies$ $\overline{EA_1}=\overline{A_1I}=\overline{A_1C}=\overline{B_1C}=\overline{B_1I}=\overline{B_1F}$. Now let $G$ be the reflection of $C_1$ over $\overline{AI}$. We claim that $G$ is the required point. Now firstly we'll show that $EA_1FB_1$ is a parallelogram. Notice that by the lengths we obtained earlier we have that $A_1$ is the circumcentre of $\odot(ECI)$ and similiarly $B_1$ is the centre of $\odot(CIF)$. Also we have that $\Delta IEC \cong \Delta FCI$ $\implies EFCI$ is a parallelogram $\implies$ $EF$ passes through the midpoint of $A_1B_1$ which quickly implies the claim. Notice that the this claim quickly implies $\overline{EA_1} \parallel \overline{B_1F} \parallel \overline{BC}$ $\square$. Now we'll show that $\overline{KC_1} \parallel \overline{AB}$. For this notice that $\angle KC_1D=2\angle A_1C_1B=180^\circ-\angle A_1BC_1$. Similiarly obtain that $\overline{LC_1} \parallel \overline{AC}$ $\square$. Now we claim that $\angle KC_1L=90^\circ$. But notice that $\angle KC_1A=\angle CBA$ and $\angle LC_1A=\angle C_1AC$ which implies the result $\square$. Notice that from the previous claim we have by definition that $(KC_1LG)$ is a square. But since we have $\overline{CA_1} \parallel \overline{KC_1}$ and similiarly for the other sides we have that there exists a dilation such that $(CA_1IB_1) \mapsto (KGLC_1)$. Note that this implies $\overline{A_1G},\overline{B_1C_1},\overline{CK}$ concurrent. Call this concurrency point $J$. Now from the dialation we have that $\angle JA_1I=\angle JGE \implies$ $\overline{JA_1} \parallel \overline{EG}$ which along with the previous length equalities implies that $(A_1IGE)$ is a cyclic trapezoid. Similiarly obtain that $(B_1FIG)$ is a cyclic trapezoid with which we are done $\blacksquare$.

Really good problem.I think many people (this is just my guess)would have been stuck on how to bring D into the picture: Anyways here's my solution: Claim $1$: $\triangle C_1A_1B_1$ is similar to $\triangle DKL$ is similar to $CFI\equiv CEI$ stating that $\quad CDIE$ is a parallelogram Proof:It is clear that $CB_1IA_1$ is a square.So $B_1C=B_1I=B_1F$ So $B_1$ is circumcenter of $FCI$.Since we have $\angle CB_1I=90\Rightarrow \angle CFI=45$.Also $\angle FB_1C=180-A\Rightarrow \angle FIC=90-\frac {A}{2}$.So $\triangle CFI$ is similar to $B_1C_1A_1$.Similar angle chasing would yield $CFIE$ is a parallelogram and $\triangle KDL$ is similar to $\triangle A_1C_1B_1$. (It may also be noted that $F-I-B$ and $E-I-A$ though this did not come in use) Claim $2$:(crux of the problem):$C$,$I$,$L$ and $K$ are collinear. Proof:Let $LK$ intersect $AB$ in $P$.Observe that $C_1$ is circumcenter of $KDL$ and $LC_1K$ is a $45-90-45$ triangle. Using sine rule we get that $$\frac{C_1D}{C_1P}=\sqrt2\times cos(45-A)=cosA+cosB$$.Also if $CI$ intersects $BC$ in $P^*$ then $$\frac {P^*I}{IC}=\frac {1}{cosA+cosB}=\frac{P^*C_1}{C_1D}\Rightarrow P=P^*$$Now from the similar triangles we found in the first claim it is easy to deduce that $CI$ is parallel to $KL$.So we have proved Claim $2$. Claim $3$:Let $FL\cap EK$=$W$;Then $W$ is the concurrency point. Proof:First of all it is direct to notice that $C_1$ is circumcenter of $FCE$ and by claim 1$\angle FCE=\angle FIE=135$ so $\angle FC_1E=90$. Thus there is a spiral similarity about $C_1$ sending $LK$ to $FE$.So $M$ lies on circumcircle of $C_1KL$ and $C_1FE$.Since center of $(F_1CE)$ (midpoint of $FE$) lies on $CI$ we have $M$ lies on the incircle as well ($IC_1=IW$). $\angle FMI=\angle LMI=\angle LC_1I=B$ and since $\angle FB_1I=180-B$ we have that $FB_1IM$ is cyclic.Similarly $EA_1IM$ is cyclic and we are done.$\blacksquare$

Where are the problems located at?

did no one use iran lemma??????? you can use that to prove $C_1KPL$ cyclic where $P$ is the ref of $C_1$ over $CI$

Cute one We first have $CE \perp C_1A_1 \perp IB \implies CE \parallel IB$. Hence $\angle EA_1B=2\angle ECA_1 =\angle CBA \implies A_1E \parallel AB$. Similarly $B_1F \parallel AB$. Now we have $B_1F \parallel A_1E$, and by isosceles triangles $B_1F=CB_1=CA_1=A_1E$, so $EF$ passes through the midpoint of $A_1B_1$. But from $\angle C= 90^{\circ}$, $CA_1IB_1$ is a square, hence $EF$ passes through its center. Denote $J$ by the center of the square. Now, by reflection at $J$, we have $\angle IEA_1=\angle CFB_1=\angle IAB$, so $A, I, E$ are collinear. Construct $T=\odot(IA_1E) \cap \odot(IB_1F)$, then we have $\angle ITA_1=\angle IEA_1=\angle IC_1B_1$, and $\angle ITB_1=\angle IC_1A_1$, so $T$ is the reflection of $C_1$ on $CI$. Now it suffices to prove that $T$ lies on $\odot(C_1LK)$, but we have $\angle LC_1T=\angle LC_1B-\angle TC_1B=\frac{\angle A+\angle B}{2}=45^{\circ}$ and $\angle KC_1T=45^{\circ}$ simlarly, and $C_1T=2r\sin(\angle A -45^{\circ})=\sqrt{2} C_1D$, so $C_1LTK$ is a square. Thus we are done.

I have a solution using complex numbers: Let $A_1=b$, $B_1=a$, with $\frac{b}{a}=i$ ($\omega$ is the unit circle). Then, $C=\frac{2ab}{a+b}$, and $F=a+c-ac \cdot \frac2{a+b}=\frac{a^2+ab+bc-ac}{a+b}=\frac{a+ai+ci-c}{1+i}=a+ci$. Similarly, $E=b-ci$. Let $\omega$ intersect $(FB_1I)$ at $P$, we have: $\frac{p-0}{p-a} \cdot \frac{a+ci-a}{a+ci-0} = \frac{p}{p-a} \cdot \frac{ci}{a-ci} = \frac{\frac1{p}}{\frac{a-p}{ap}} \cdot \frac{\frac{-i}{c}}{\frac{c-ai}{ac}} \Leftrightarrow \frac{pc}{a+ci} = \frac{a^2}{c-ai} \Leftrightarrow p=\frac{a+ci}{a-ci} \cdot \frac{a^2}{c}=\frac{a}{c} \cdot \frac{a^2+aci}{c-ai} = \frac{a}{c} \cdot \frac{-b^2+bc}{c-b} = \frac{ab}{c}$ Since $p$ is symmetric for $a$ and $b$, it turns out that $(EA_1I)$ also meets $\omega$ for the second time at $p$. Now, we calculate $d$: $d=\frac12 \left( \frac{2ab}{a+b}+c+c-c^2 \cdot \frac2{a+b} \right)=\frac{ab+ac+bc-c^2}{a+b}$ From here, we get: $c \overline{d}=\frac{c^2+bc+ac-ab}{(a+b)c} \Rightarrow l= a+c-ac \overline{d}=\frac{a^2c+abc+ac^2+bc^2-ac^2-abc-a^2c+a^2b}{(a+b)c}=\frac{(a^2+c^2)b}{(a+b)c}$ Similarly, $k=\frac{(b^2+c^2)a}{(a+b)c}$. Since $C_1$ is the circumcenter of $DLK$ and $\angle LDK=45^{\circ}$, it suffices to prove that $LP$ and $KP$ are perpendicular, because $\angle LC_1K=90^{\circ}$. $\frac{k-p}{l-p}=\frac{ab^2+ac^2-a^2b-ab^2}{a^2b+bc^2-a^2b-ab^2}=\frac{a}{b} \cdot \frac{c^2-ab}{c^2-ab} \in i\mathbb{R}$, thus we are done.

Claim: $K, L \in CI$. Proof. Define $A'$ as the reflection of $A$ over $B_1C_1$ (so $C_1, K, A'$ collinear) and let $I' \in C_1I$ such that $A'I \parallel CI$. Observe that $AC_1A'B_1$ is a rhombus and $\angle I'A'C_1 = 180^\circ - \angle ACI = 135^\circ$. If WLOG the inradius is $1$, then sine law in $\triangle C_1I'A'$ yields \[\frac{C_1I'}{C_1I} = \frac{\sin 135^\circ}{\sin \angle C_1I'A}C_1A' = \frac{C_1A'}{\sqrt{2} \sin \angle DCI } = \frac{C_1A'}{C_1D} = \frac{C_1A'}{C_1K}, \]whence $\triangle C_1IK \sim \triangle C_1I'A'$ and thus $IK \parallel I'A \parallel CI$. Similar work shows that $C, I, L$ collinear, so the claim is proven. $\Box$ Observe that $C_1$ is the circumcenter of $\triangle DKL$. Moreover, since $\angle KDL = 180^\circ - \angle AIB = 45^\circ$, we have $\angle KC_1L = 90^\circ$. Define $C'$ as the antipode of $C_1$ on $(C_1KL)$, which by our previous observations mean that $C'$ and $C_1$ are symmetric about $CI$. Some angle chase (lazily omitted ) yields $\angle C_1IA_1 = \angle IA_1E$, and since $IC'=IA_1=A_1E$, we have that $C'IA_1E$ is an isosceles trapezoid, i.e. it is cyclic. Similarly, $C'IB_1F$ is cyclic, and it follows that $C'$ is the desired concurrency point.

Here is a terrible solution. [asy][asy] size(8cm); pair A1 = dir(270), B1 = dir(180), C1 = dir(53), C = (-1, -1), A = extension(C, B1, C1, C1+C1*dir(90)), B = extension(C, A1, A, C1), D = foot(C,A,B), H = orthocenter(A1, B1, C1), X = dir(90)/C1, I = 0, E = 2*foot(C,C1,A1)-C, Y = A1*C1; D(C--A--B--C--D); D(unitcircle, purple); D(A1--I--B1); D(C1--I--X); D(A1--B1--C1--A1, purple); D(X--I--A1--E--X, heavygreen); D(circumcircle(I,A1,E), heavygreen+dashed); D(B1--Y, heavygreen); D("A", A, A); D("B", B, B); D("C",C,C); D("D",D,D); D("E",E,E); D("A_1", A1, A1); D("B_1",B1,B1); D("X",X,X); D("C_1",C1,C1); D("Y",Y,Y); D("I",I,dir(116)); D("H",H,dir(105)); label("$\omega$", dir(116), 0.5*dir(116)); [/asy][/asy] We use directed angles. Note that $\measuredangle BA_1E = 2\measuredangle BA_1C_1 = \measuredangle A_1BC_1$, so $A_1E\parallel AB$. If $X$ is the reflection of $C_1$ over $CI$, then $X,E$ lie on the same side of $IA_1$, $\measuredangle XIA_1 = \measuredangle B_1IC_1 = \measuredangle (AC, AB) = \measuredangle (IA_1, A_1E) = \measuredangle IA_1E$, and $IX = IA_1 = CA_1 = A_1E$. It follows that $IA_1EX$ is an isosceles trapezoid. Thus, $X\in (A_1EI)$. Similarly, $X\in (B_1FI)$. Now, note that (by complex numbers, say), the orthocenter $H$ of triangle $A_1B_1C_1$ forms parallelogram $CHC_1I$, so $CH\perp AB$, and so $H$ lies on $CD$. Furthermore, $CB_1 = IC_1 = CH$. Now, if $BH$ meets $\omega$ at $Y$, we see that $YE$ is the reflection of $HC$ over $A_1C_1$, so \[ \measuredangle B_1YE = \measuredangle CHY = \measuredangle HB_1C, \]so $B_1C\parallel YE$ and $X,Y,E$ are collinear. Thus, $X$ is the anti-Steiner point of $CD$. However, it is well known (see EGMO 2017/6, say) that $(CKL)$ passes through the anti-Steiner point of $CD$, so $X\in (CKL)$ as well. $\blacksquare$

Here is a sketch of another proof. Consider $X = (A_1EI) \cap (B_1FI)$ 1. Notice the tons of tons of parallel lines, which cry for us to angle chase. In fact, $A,I,E$ and $B, F, I$ are collinear. 2. Show that $C,K,L,I$ are collinear. (Sine Law) 3. Let $T=B_1C_1 \cap (A_1BI).$ Show that $T \in CI$ and that $T \in XA_1$ (by angle chase) 4. An Angle chase shows that $IX=IA_1$, hence $X$ is on the incircle. 5. Another angle chase shows that $K,X, E$ are collinear. 6. Lastly, a simple angle chase shows that $\angle{TXK}=\angle{A_1IE}= \frac{\angle{A}}{2}$ and $\angle{TXF}= 90 +\frac{\angle{A}}{2} $. This leaves us with $\angle{KXL}=\angle{KC_1L}=90,$ which proves $X$ belons to the circumcircle of $C_1KL.$

By drawing the diagram, we can guess that the three circles intersect on the incircle of $\triangle ABC$. Let $C'$ be the point on the incircle such that $FB_1IC'$ form an isosceles trapezoid, then by angle chasing it's easy to see that $EA_1IC'$ is also an isosceles trapezoid, so the first two given circles indeed intersect on the incircle. Next, we will prove that $(C_1KL)$ also passes through the point $C'$. Observe that $\angle KC_1L = \frac{\pi}{2}$ and $KC_1 = C_1L$, so we will use a phantom point argument: suppose $P$ is a point such that $PLC_1K$, in that order, is a square. Extend $PL$ and $PK$ to meet $CD$ at points $A'$ and $B'$. We claim that $\triangle PA'B'$ is similar to $\triangle CAB$. Indeed, by angle chasing we get $\triangle LKD \sim \triangle B_1A_1C_1$, and since $(LKD)$ is the incircle of $\triangle PA'B'$, the claim follows. Now, we have $\angle DA'C_1 = \frac{1}{2} \angle PA'B' = \frac{1}{2} \angle CAB = \frac{\pi}{2} - \angle B_1C_1A$, so $A'$ lies on $B_1C_1$; since line $B_1C_1$ bisects both $\angle CA'F$ and $\angle PA'B'$, we see that $P,A',L,F$ lie on a line. By a similar argument, $P,B',K,E$ also lie on a line, so $P = FL\cap EK$. But $\measuredangle (CA, FL) = \measuredangle (CA, CD) + \measuredangle (CD, FL) = \frac{\pi}{2}$, so $FP\parallel IB_1$. This is enough to imply that $P = C'$.

Inversion comes handy. The square and intersection on incircle become obvious.

Weird problem . Especially coz the last part kind of forces you to do trig . Anyway, here's my solution: WLOG take $AC \leq BC$, and let $\odot (B_1IF) \cap \omega=T$. I'll use the diagram from post #2. The main synthetic claim is the following:- CLAIM Lines $C_1T$ and $A_1B_1$ are parallel, or equivalently, $A_1B_1C_1T$ is an isosceles trapezoid. Proof of Claim Note that $B_1IA_1C$ is a rectangle with $CB_1=CA_1$, and so it must in fact be a square. This gives $B_1F=B_1C=B_1I=IT$, which translates to the fact that $FB_1IT$ is an isosceles trapezoid with $FT \parallel B_1I$. Thus, $$\angle B_1IT=\angle IB_1F=\angle IB_1C_1+\angle C_1B_1F=\angle IAC_1+\angle C_1B_1C=\angle IAC_1+(180^{\circ}-\angle C_1IA)=90^{\circ}+A=180^{\circ}-B$$where we continuously use the fact that $AB_1IC_1$ is a cyclic quadrilateral. Since $\angle A_1IC_1=180^{\circ}-B$, so we directly get that $T$ is simply the reflection of $C_1$ in line $A_1B_1$, as desired. Note that, by the symmetry of our Claim, $T$ must also lie on $\odot (A_1IE)$, and so it suffices to show that $T \in \odot (C_1KL)$. Now, $$\angle TC_1L=\angle TC_1B_1-\angle B_1C_1L=(180^{\circ}-\angle C_1B_1A_1)-\angle DC_1B_1=180^{\circ}-\angle C_1B_1A_1-\angle C_1A_1B_1=\angle A_1C_1B_1=\frac{\angle A_1IB_1}{2}=45^{\circ}$$Similarly, we have $\angle TC_1K=45^{\circ}$. Combined with $C_1K=C_1D=C_1L$, we get that $C_1T$ is the perpendicular bisector of $KL$, and $\angle KC_1L=90^{\circ}$. Thus, proving $T \in \odot (C_KL)$ is equivalent to showing that $C_KTL$ is a square, or in other words, showing that $C_1T=\sqrt{2}C_1D$. Now, since $T$ is the reflection of $C_1$ in $CI$, so $$C_1T=2d(C_1,CI)=2C_1I \sin \angle C_1IC=2r \sin (180^{\circ}-A+45^{\circ})=2r \sin (A-45^{\circ})=\sqrt{2}r (\sin A-\cos A)$$where $r$ is the radius of $\omega$. Thus, it suffices to prove $C_1D=r (\sin A-\cos A)$. Now we use $$C_1D=C_1A-AD=(s-a)-b \cos A=\frac{b+c-a}{2}-\frac{b^2}{c}=\frac{bc+c^2-ac-2b^2}{2c}=\frac{bc+a^2-ac-b^2}{2c}=\frac{(a-b)(a+b-c)}{2c}$$But, we also have $$r=s-c=\frac{a+b-c}{2}, \sin A=\frac{a}{c}, \cos A=\frac{b}{c} \Rightarrow r (\sin A-\cos A)=\frac{(a+b-c)(a-b)}{2c}=C_1D$$Hence, done. $\blacksquare$

Here's the solution I submitted for the contest. First, easy angle chasing gives $A_1 I B_1 C$ is a square, which gives us $\triangle EA_1 I$ and $\triangle FB_1 I$ both being isosceles. Furthermore, this gives us $A_1$ is the circumcenter of $\triangle CEI$, $B_1$ is the circumcenter of $\triangle FCI$. Now, suppose $(A_1 EI) \cap (B_1 IF) = N$. We will prove that $C_1KNL$ is a square, which is cyclic. Claim 01. $N$ lies on the incircle. Proof. Notice that $\angle A_1 N B_1 = \angle A_1 NI + \angle B_1 NI = \angle A_1 EI + \angle B_1 FI = 180^{\circ} - \frac{1}{2} ( \angle EA_1 I + \angle FB_1 I) = 180^{\circ} - \angle ECF = 45^{\circ} = \angle A_1 C_1 B_1 $ where $\angle ECF = 135^{\circ}$ is easy to proved by simple angle chasing. Claim 02. $\angle KC_1 L = 90^{\circ}$. Proof. Just notice that $C_1 K \parallel BC$ and $C_1 L \parallel AC$ from the reflection hypothesis. Claim 03. $\angle KC_1 N = 45^{\circ}$. Proof. Notice that $A_1 E = A_1 I = IN$ and they are cyclic, hence \[ \angle KC_1 N = \angle KC_1 A - \angle NC_1 A = \angle CBA - \angle NA_1 C_1= \angle CBA - \angle C_1 A_1 I + \angle NA_1 I = \frac{B}{2} + \angle A_1 EI = \frac{A + B}{2} = 45^{\circ} \] It suffices to prove that $C_1 N = DC_1 \sqrt{2}$, which immediately follows by simple trigonometry length bash on $\triangle A_1 C_1 N$.

Pretty sure no one defined the point of intersection as $T=EK \cap FL$. Let us define $T$ as above. Lines $EK$ and $FL$ are the reflections of line $CD$ across $A_1C_1$ and $B_1C_1$, respectively. It follows that $\angle ETF=2 \angle A_1C_1B_1=90^{\circ}$. $C_1$ is the circumcircle of $\triangle DKL$, and we can therefore easily establish that $\angle KC_1L=2 \angle KDL= \dots =90^{\circ}$. Hence $C_1KTL$ is a cyclic quadrilateral. An easy angle chase yields $FT \perp AC$ and $TE \perp BC$. Some more angle chasing yields $FB_1 \parallel AB \parallel A_1E$. The configuration we're interested in is the purple part of @TheUltimate123 's sketch. We want to establish $IT=A_1E=FB_1$. Let us draw a line $l$ through point $I$, parallel to $FB_1$ and $A_1E$. Let's label $B_2=FT \cap l$ and $A_2=ET \cap l$. $FB_1IB_2$ and $EA_2IA_1$ are parallelograms. It follows that $IT$ is the median of the hypotenuse in right triangle $\triangle TA_2B_2$, and its lenght is half of the hypotenuse's. Hence quadrilaterals $TFB_1I$ and $TIA_1E$ are isosceles trapezoids, and are therefore cyclic. $\blacksquare$

khina wrote: did no one use iran lemma??????? you can use that to prove $C_1KPL$ cyclic where $P$ is the ref of $C_1$ over $CI$ What is Iran lemma?

Solved with nprime06 [asy][asy] defaultpen(fontsize(8pt)); size(10cm); dotfactor *= 0.75; pair A = dir(155); pair C = conj(A); pair B = -A; pair I = incenter(A,B,C); pair D = foot(C,A,B); pair AA = foot(I,B,C); pair BB = foot(I,C,A); pair CC = foot(I,A,B); pair E = reflect(CC,AA)*C; pair F = reflect(CC,BB)*C; pair K = reflect(CC,AA)*D; pair L = reflect(CC,BB)*D; pair X = intersectionpoints(circumcircle(AA,E,I),circumcircle(BB,F,I))[0]; pen bfil=purple+pink+white+white+white+white+white; pen ufil=purple+pink+white+white+white+white; pen tfil=purple+white+white+white; pen qfil=purple+purple+purple+cyan+white+white+white; pen qfil2=purple+purple+purple+cyan+white+white; fill(circumcircle(I,C,F)^^circumcircle(I,C,E)^^circumcircle(D,L,K)^^circumcircle(I,X,E)^^circumcircle(F,I,X)^^circumcircle(X,K,L), bfil); fill(A--B--C--cycle, ufil); fill(circumcircle(AA,BB,CC),tfil); fill(AA--BB--CC--X--cycle,qfil2); draw(A--B--C--cycle); draw(circumcircle(AA,BB,CC)); draw(F--BB^^AA--E^^A--B,fuchsia); draw(circumcircle(I,C,F)^^circumcircle(I,C,E)^^circumcircle(D,L,K),linewidth(0.1)); draw(circumcircle(I,X,E)^^circumcircle(F,I,X)^^circumcircle(X,K,L),linewidth(0.4)); draw(AA--BB--CC--X--AA,linewidth(0.4)); draw(A--E^^B--F^^C--K,darkcyan+linewidth(0.1)); draw(E--K^^F--X,linewidth(0.1)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(220)); dot("$I$",I,dir(320)); dot("$D$",D,dir(100)); dot("$A_1$",AA,dir(240)); dot("$B_1$",BB,dir(210)); dot("$C_1$",CC,dir(100)); dot("$E$",E,dir(300)); dot("$F$",F,dir(200)); dot("$K$",K,dir(60)); dot("$L$",L,dir(L)); dot("$X$",X,dir(315)); [/asy][/asy] Lemma 1. $EA_1\parallel FB_1 \parallel AB$. Proof. Note that \[\angle AC_1B_1=\angle C_1B_1A=180^\circ-\angle C_1B_1C=180^\circ-\angle FB_1C_1\implies AB\parallel FB_1.\]Similarly, we can show that $AB\parallel EA_1$. $\square$ Lemma 2. $ADIF$ and $BDIE$ are cyclic. Proof. Note that $B_1I=B_1C=B_1F$, so $(FCI)$ has center $B_1$. In particular, because $BC\perp B_1C$, we must have $BI\cdot BF=BC^2=BD\cdot BA$, which implies the lemma. $\square$ Lemma 3. $BIF$ and $AIE$ are collinear. Proof. We have \[\angle CIB=90^\circ+\frac12 \angle CAB\]and \[\angle CIF=\frac12 \angle CB_1F=\frac12 (180^\circ-\angle AB_1F)=90^\circ -\frac12\angle CAB\]so $\angle CIB=180^\circ-\angle CIF$ which implies $BIF$ collinear. Similarly, $AIE$ is collinear. $\square$ Lemma 4. $CILK$ are collinear. Proof. It suffices to show that $CIL$ collinear, and $CIK$ collinear follows by symmetry. We have \[\angle LCB_1=\angle DFB_1=\angle ADF=\angle AIF=45^\circ = \angle ICB_1,\]as desired. $\square$ Let $X$ be the point such that $A_1B_1C_1X$ is an isosceles trapezoid. Claim 1. $X\in (FB_1I)$ and $(EA_1I)$. Proof. Note that $IX=IB_1=B_1F$ and \[\angle FB_1I=90^\circ + \angle FB_1A=90^\circ + \angle CAB =180^\circ-\angle ABC = \angle A_1IC_1 = \angle B_1IX,\]so $FB_1IX$ is an isosceles trapezoid, hence $X\in (FB_1I)$. A similar argument applies to $EA_1IX$. $\square$ Claim 2. $X\in(KC_1L)$. Proof. Because $C_1K=C_1D=C_1L$, $C_1$ is the center of $(KDL)$. it follows that $CD$ and hence $EK$ are tangent to $(KDL)$; thus $\angle EKC_1=90^\circ$. But by reflection in $A_1C_1$ we see that $EK\parallel AC\iff C_1K\parallel CA_1\iff C_1D\parallel A_1E$, which is just Lemma 1. However, we also note that $EX\parallel IA_1\parallel AC$, so $EXK$ are collinear. It follows that $\angle XKC_1=90^\circ$ and similarly $\angle XLC_1=90^\circ$, and we are done. $\blacksquare$

Really nice problem, but how is this a 1... Let $X$ be the reflection of $C_1$ over $\overline{CI}.$ $\textbf{Claim: }$ $\overline{XE}\perp\overline{BC}$ and $\overline{XF}\perp\overline{AC}.$ $\emph{Proof: }$ Let $\overline{CI}$ intersect $\overline{BC}$ at $T.$ Write \begin{align*} \angle A_{1}IX &= \angle A_{1}IC_{1} - \angle XIC_{1}\\ &= (180^\circ-B) - 2\angle TIC_{1}\\ &= 180^\circ-B - 2(90^\circ - \angle ATC)\\ &= 2\angle ATC - B\\ &= 2(180^\circ - \frac{1}{2}C - \angle A)-B\\ &= 270^\circ-2A-B\\ &= 90^\circ+B \end{align*}Moreover, note that \begin{align*} \angle EA_{1}I &= \angle EA_{1}C_{1}+\angle C_{1}A_{1}A\\ &= \angle C_{1}A_{1}C+\frac{1}{2}B\\ &= (180^\circ-\angle BA_{1}C_{1})+\frac{1}{2}B\\ &= 90^\circ+B. \end{align*}Finally, remark that $IX=IA_{1}=A_{1}C=A_{1}E.$ This combined with the fact that $\angle A_{1}IX=\angle EA_{1}I$ implies that $XIA_{1}E$ is an isosceles trapezoid, which yields $\angle(\overline{XE},\overline{BC})=\angle(\overline{IA_{1}},\overline{BC}) = 90^\circ.$ We can similarly show that $\overline{XF}\perp\overline{AC},$ so we are done. $\blacksquare$ $\textbf{Claim: }$ $C_{1},K,L,X$ are concylic. $\emph{Proof: }$ Since the reflection of $\overline{CD}$ across $\overline{C_{1}A_{1}}$ is the line through $E$ perpendicular to $\overline{BC},$ which is $\overline{EX}$ by our first claim, we know $K$ lies on $\overline{EX}.$ Similarly, $L$ lies on $\overline{FX},$ so $\angle KXL=\angle FXE = 90^\circ.$ Next, let $U=\overline{C_{1}A_{1}}\cap\overline{DK}.$ Compute that \begin{align*} \angle BC_{1}K &= 180^\circ - \angle KC_{1}D\\ &= 180^\circ-2\angle UC_{1}D\\ &= 180^\circ-2\angle A_{1}C_{1}B\\ &= B. \end{align*}By symmetry, $\angle LC_{1}B = A,$ so we have $\angle LC_{1}K = A+B = 90^\circ.$ Thus, $\angle LC_{1}K+\angle KXL=180^\circ,$ so quadrilateral $C_{1}KLX$ is cyclic. $\blacksquare$ We already showed that $XIA_{1}E$ and $XIB_{1}F$ are isosceles trapezoids (and are thus cyclic), so $X$ is the desired point.

Using $A$-indexing, we reword the problem to the following: Quote: Let $ABC$ be a triangle with a right angle at $A$. Let $I$ be the incentre of triangle $ABC$, and let $D$ be the foot of the altitude from $A$ to $BC$. The incircle $\omega$ of triangle $ABC$ is tangent to sides $BC$, $CA$, and $AB$ at $A_1$, $B_1$, and $C_1$, respectively. Let $E$ and $F$ be the reflections of $A$ in lines $A_1C_1$ and $A_1B_1$, respectively. Let $K$ and $L$ be the reflections of $D$ in lines $A_1C_1$ and $A_1B_1$, respectively. Prove that the circumcircles of triangles $B_1FI$, $C_1EI$, and $A_1KL$ have a common point. Let $G$ be the second intersection of $(IC_1E)$ with $\omega$ and let $G'$ be the second intersection of $(IB_1F)$ with $\omega.$ Since $AB_1IC_1$ is a square, it follows that $GI=C_1I=C_1A=C_1E$ and $G'I=B_1I=B_1A=B_1F,$ so $IC_1EG$ and $IB_1FG'$ are both isosceles trapezoids. Then since \[\measuredangle C_1IG = \measuredangle EC_1I = \measuredangle EC_1B + \measuredangle BC_1I = -\measuredangle AC_1E + 90^\circ = -2\measuredangle BC_1A_1 + 90^\circ = \measuredangle CBA + 90^\circ\]and similarly $\measuredangle G'IB_1 = \measuredangle ACB + 90^\circ,$ this means that \[\measuredangle C_1IG + \measuredangle B_1IC_1 + \measuredangle G'IB_1 = (\measuredangle CBA + 90^\circ) + 90^\circ + (\measuredangle ACB + 90^\circ) = 0^\circ,\]so in fact $G = G'.$ Futhermore, also note that $\measuredangle EC_1B=\measuredangle CBA$ and $\measuredangle CB_1F=\measuredangle ACB$ implies that $C_1E\parallel BC\parallel B_1F.$ Next, observe that due to the cyclic quadrilateral formed by $A_1B_1,A_1C_1,DK,$ and $DL,$ we have $\measuredangle LA_1K = 2\measuredangle LDK = 2\measuredangle B_1A_1C_1 = 90^\circ$ since $A_1$ and $I$ are the circumcenters of $\triangle DKL$ and $\triangle A_1B_1C_1,$ respectively. Finally, since $\{A,E\}$ and $\{D,K\}$ are reflected across $A_1C_1$ and $AD\perp BC\parallel C_1E,$ it follows that $EK\perp AB\perp C_1I\parallel EG,$ so in fact $E,K,G$ are collinear. By the same logic $F,G,L$ are also collinear, which means that $\measuredangle LGK = \measuredangle FGE=90^\circ,$ whence $A_1KGL$ is cyclic as desired. $\blacksquare$ Remark: Another way to finish is to observe that $AD$ and $EK$ being reflections across $A_1C_1$ implies that $\measuredangle A_1KG = \measuredangle ADA_1 = 90^\circ$ and similarly $\measuredangle GLA_1 = \measuredangle ADA_1 = 90^\circ$ so we get that $A_1KGL$ is cyclic again (and in fact is a square by homothety). [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -10.2, xmax = 12.2, ymin = -7.33, ymax = 7.33; /* image dimensions */ /* draw figures */ draw(circle((-2.2145742352477003,1.0574334108658943), 3.148917446193928), linewidth(0.7)); draw(circle((-6.5296683537847,-3.358964905239265), 6.174513044631867), linewidth(0.7)); draw(circle((0.9127128823761499,1.093716705432946), 3.127497592888079), linewidth(0.7)); draw(circle((-1.406656419803343,-1.868132343196593), 0.8390559606420185), linewidth(0.7)); draw((-3.4,5.35)--(-7.62,-2.09), linewidth(0.7)); draw((-7.62,-2.09)--(9.725363221344413,-2.0947624723216984), linewidth(0.7)); draw((9.725363221344413,-2.0947624723216984)--(-3.4,5.35), linewidth(0.7)); draw((-8.102487739690662,2.6118684132890193)--(-4.953570412190903,2.6110038230567962), linewidth(0.7)); draw((-0.6610038230567974,3.7964295878090963)--(2.4879135044429637,3.7955649975768724), linewidth(0.7)); draw((-2.2145742352477003,1.0574334108658943)--(-4.953570412190903,2.6110038230567962), linewidth(0.7)); draw((-2.2145742352477003,1.0574334108658943)--(-0.6610038230567974,3.7964295878090963), linewidth(0.7)); draw((-2.2145742352477003,1.0574334108658943)--(-0.5978740141267617,-1.6447807697593197), linewidth(0.7)); draw((-0.5978740141267617,-1.6447807697593197)--(-8.102487739690662,2.6118684132890193), linewidth(0.7)); draw((-3.4,5.35)--(-3.403278555687832,-6.590802046824329), linewidth(0.7)); draw((-3.403278555687832,-6.590802046824329)--(2.4879135044429637,3.7955649975768724), linewidth(0.7)); draw((-0.6610038230567974,3.7964295878090963)--(-3.403278555687832,-6.590802046824329), linewidth(0.7)); draw((-1.183304846366073,-2.6769147488731733)--(-3.4020430998824125,-2.091158113712269), linewidth(0.7)); draw((-3.4020430998824125,-2.091158113712269)--(-1.630007993240614,-1.0593499375200122), linewidth(0.7)); draw((-4.953570412190903,2.6110038230567962)--(-1.183304846366073,-2.6769147488731733), linewidth(0.7)); draw((-2.2154388254799233,-2.091483916633864)--(-4.953570412190903,2.6110038230567962), linewidth(0.7)); /* dots and labels */ dot((-3.4,5.35),dotstyle); label("$A$", (-3.32,5.55), NE * labelscalefactor); dot((-7.62,-2.09),dotstyle); label("$B$", (-8.4,-2.33), NE * labelscalefactor); dot((9.725363221344413,-2.0947624723216984),dotstyle); label("$C$", (10,-2.33), NE * labelscalefactor); dot((-2.2145742352477003,1.0574334108658943),linewidth(4pt) + dotstyle); label("$I$", (-2.58,1.45), NE * labelscalefactor); dot((-3.4020430998824125,-2.091158113712269),linewidth(4pt) + dotstyle); label("$D$", (-4.1,-2.71), NE * labelscalefactor); dot((-2.2154388254799233,-2.091483916633864),linewidth(4pt) + dotstyle); label("$A_{1}$", (-3.11,-2.88), NE * labelscalefactor); dot((-0.6610038230567974,3.7964295878090963),linewidth(4pt) + dotstyle); label("$B_{1}$", (-1,4.02), NE * labelscalefactor); dot((-4.953570412190903,2.6110038230567962),linewidth(4pt) + dotstyle); label("$C_{1}$", (-5.58,2.85), NE * labelscalefactor); dot((-8.102487739690662,2.6118684132890193),dotstyle); label("$E$", (-8.44,2.81), NE * labelscalefactor); dot((2.4879135044429637,3.7955649975768724),dotstyle); label("$F$", (2.56,3.99), NE * labelscalefactor); dot((-1.630007993240614,-1.0593499375200122),linewidth(4pt) + dotstyle); label("$K$", (-2.45,-1.07), NE * labelscalefactor); dot((-1.183304846366073,-2.6769147488731733),linewidth(4pt) + dotstyle); label("$L$", (-1.1,-3.25), NE * labelscalefactor); dot((-0.5978740141267617,-1.6447807697593197),linewidth(4pt) + dotstyle); label("$G$", (-0.3,-1.83), NE * labelscalefactor); dot((-3.403278555687832,-6.590802046824329),linewidth(4pt) + dotstyle); dot((-3.4014839166338655,-0.05456117452007657),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Neat question! Note that $CB_1IA_1$ is a square since $BCA=CB_1I=CA_1I=90$, $CB_1=CA_1$ Hence we get the relation $$CB_1=B_1F=B_1I=IA_1=A_1E=A_1C$$ Note that $FCB_1= 90- AB_1C_1= \frac{a}{2}$ Similarly, $ECA_1=\frac{b}{2}$ Note $BCI=45$ , $B_1FI=\frac{b}{2}$, $A_1EI=\frac{a}{2}$ Claim 1: $FB_1I$ intersect $EA_1I$ lies on $w$ Proof: Let $EA_1I$ intersect the incircle at $P$ $A_1PI=IEA_1=\frac{a}{2}$ $B_1PA_1=BC_1A_1=\frac{a}{2} + \frac{b}{2}$ $IPB_1=\frac{b}{2}= B_1FI$ Hence, $P,B_1,F,I$ concyclic Claim 2: $K,P,E$ and $F,P,E$ collinear Note $F$ and $L$ are both reflections of $C$ and $D$ across $B_1C_1$ Hence, $FCD=CFL=CFB_1+90-a=CFB_1+ IPB_2+IB_1P= CFP$ Which implies that $F,L,P$ collinear, and by symmetry, $K,P,E$ also collinear Claim 3: $LK$ is the diameter of $C_1LK$ $LDC_1= 90-B_1C_1A= \frac{a}{2}$ $C_1DK=90-A_1C_1B= \frac{b}{2}$ $LC_1K=2(\frac{a}{2}+\frac{b}{2})=90$ Claim 4: $P,L,C_1,K$ concyclic Proof: Note that $FPE= B_1FP+A_1EP=a+b=90$ $LPK=180-90=90=180-LC_1K$ Hence, $P,L,C_1,K$ concyclic

[asy][asy] size(9cm); defaultpen(fontsize(10pt)); pair A, B, C, D; A = dir(180); B = dir(0); C = dir(130); D = foot(C, A, B); pair I, A1, B1, C1; I = incenter(A, B, C); A1 = foot(I, B, C); B1 = foot(I, C, A); C1 = foot(I, A, B); pair E, F, K, L; E = 2 * foot(C, C1, A1) - C; F = 2 * foot(C, C1, B1) - C; K = 2*foot(D, C1, A1) - D; L = 2*foot(D, C1, B1) - D; pair X; X = K + L - C1; draw(A--B--C--A, royalblue); draw(incircle(A, B, C), royalblue); draw(B1--C1--A1, deepgreen); draw(B1--F^^A1--E, purple); draw(L--C1--K, magenta); draw(K--X--L, magenta+dashed); draw(C--D^^I--C1, royalblue); draw(C--K, blue + dashed); draw(A1--I--X--E^^B1--I^^F--X, purple+dotted); dot("$A$", A, dir(200)); dot("$B$", B, dir(340)); dot("$C$", C, dir(90)); dot("$D$", D, dir(270)); dot("$A_1$", A1, dir(80)); dot("$B_1$", B1, dir(110)); dot("$C_1$", C1, dir(260)); dot("$I$", I, dir(180)); dot("$E$", E, dir(60)); dot("$F$", F, dir(100)); dot("$K$", K, dir(270)); dot("$L$", L, dir(110)); dot("$X$", X, dir(80)); [/asy][/asy] Beautiful. Three steps. Observation: We can determine the directions of many lines. Explanation: Reflection with respect to $A_1C_1$ takes lines with direction $BC$ to lines with direction $AB$. Reflection with respect to $B_1C_1$ takes lines with direction $AB$ to lines with direction $AC$. Hence, reflection $C_1K$ of $C_1D$ over $A_1C_1$ is parallel to $BC$, and reflection $C_1L$ of $C_1D$ over $B_1C_1$ is parallel to $AC$. Similarly, $A_1E$ and $B_1F$ are both parallel to $AB$. $\square$ Proposition: $C, I, L, K$ are collinear. Proof: $LK$ is in the direction of the external angle bisector of $\angle LC_1K$, which is parallel to the internal angle bisector of $\angle ACB$. Hence, $LK$ and $CI$ are parallel. Let $d(X,l)$ denote a point-line distance. Find \[\frac{1}{2}(d(C_1, CI) \cdot CI) = [C_1IC] = \frac{1}{2}(C_1D \cdot C_1I),\]from which we conclude $d(C_1, CI) = \frac{1}{\sqrt{2}}C_1D = d(C_1, KL)$. Hence, $C, I, L, K$ are collinear, as desired. $\square$ Claim: Construct $X$ so that $C_1LXK$ is a square. $A_1EXI$ and $B_1FXI$ are isosceles trapezoids. Proof: Since $C_1$ and $X$ are symmetric with respect to $CI$, $X$ lies on the incircle. By reflection, $A_1E = A_1C = A_1I = IX$. Without loss of generality, assume $BC > AC$. Evaluate \begin{align*} \angle EA_1I & = 90^\circ + \angle ABC = (180^\circ - \angle ABC) - 2(45^\circ - \angle ABC) \\ & = \angle A_1IC_1 - 2\angle C_1IL = \angle A_1IX. \end{align*}Hence, we conclude that $A_1EXI$ is an isosceles trapezoid. The rest is analogous. $\square$ $(A_1EI)$, $(B_1FI)$, and $(C_1KL)$ concur at $X$.

horror show of a solution Let $X$ be the reflection of $C_1$ over $\overline{CI}$. I claim $X$ is the desired concurrency point. We first need the following claim. Claim 1: $K,L$ lie on $\overline{CI}$. Proof: By symmetry suffices to show that $L$ lies on $\overline{CI}$. We use coordinates, with the following unusual setup: let the midpoint of $\overline{B_1C_1}$ be the origin, and let $B_1=(-1,0),C_1=(1,0)$. Let $I=(0,a)$, so that $A=(0,-\tfrac{1}{a})$ by power of a point, since $AEFI$ is cyclic. Because $B_1IA_1C$ is a square, it is clear that $C=(a-1,-1)$. Then $\overline{AC_1}$ has equation $y=\tfrac{x-1}{a}$ while $\overline{CD}$ has equation $y+1=-a(x-a+1)$. Thus $D=(\tfrac{a^3-a^2-a+1}{a^2+1},\tfrac{a^2-2a-1}{a^2+1})$, so $L=(\tfrac{a^3-a^2-a+1}{a^2+1},-\tfrac{a^2-2a-1}{a^2+1})$. Since $\overline{CI}$ and $\overline{IL}$ both have slope $\tfrac{a+1}{1-a}$, we are done. $\blacksquare$ We now prove a second quick claim. Claim 2: $\overline{C_1L} \parallel \overline{AC}$. Likewise, $\overline{C_1K} \parallel \overline{BC}$. Proof: We only do the first case: $\angle BC_1L=180^\circ-\angle AC_1L=180^\circ-2\angle AC_1B_1=\angle BAC$. $\blacksquare$ This implies that $C_1KXL$ is a square, hence cyclic. We now show that $X$ lies on $(A_1EI)$ and $(B_1FI)$ as well. For symmetry reasons, $\overline{CI},\overline{A_1C_1},\overline{B_1X}$ concur at some point $P$. I claim both $P$ and $X$ lie on $(A_1EI)$. This follows because \begin{align*} \angle PEA_1=\angle PCA_1&=45^\circ\\ \angle PIA_1=180^\circ-\angle CIA_1&=135^\circ\\ \angle PXA_1=\angle B_1XA_1=\frac{\angle B_1IA_1}{2}&=45^\circ. \end{align*}Likewise, $X$ also lies on $(B_1FI)$, hence we are done. $\blacksquare$ Remark: It is also true that $K,X,E$ and $L,X,F$ are collinear. One way to show this is to prove that $\overline{KE} \parallel \overline{AC}$, which can be proved by an argument of lines chase. Set $\overline{BC}$ as the "real axis"; then because $CDKE$ is an isosceles trapezoid, we have (modulo $180^\circ$) $$\arg(\overline{KE})=2\arg(\overline{C_1A_1})-\arg(\overline{CD})=2\left(90^\circ-\frac{\angle B}{2}\right)-(90^\circ-\angle B)=90^\circ,$$where $\angle B$ is taken to be the value in $(0,90^\circ)$. Remark: the following well-organized diagram was produced

why is asy so hard [asy][asy] unitsize(3cm); pair A, B, C, D, I, A1, B1, C1, E, F, K, L, X; A1=dir(-135); B1=dir(-45); C1=dir(105); X=A1*B1/C1; I=circumcenter(A1,B1,C1); A=2*B1*C1/(B1+C1); B=2*C1*A1/(C1+A1); C=2*A1*B1/(A1+B1); D=foot(C,A,B); E=2*foot(C,C1,A1)-C; F=2*foot(C,C1,B1)-C; K=2*foot(D,C1,A1)-D; L=2*foot(D,C1,B1)-D; draw(circumcircle(A1,B1,C1)); draw(circumcircle(A1,E,I)); draw(circumcircle(B1,F,I)); draw(circumcircle(C1,K,L)); draw(A1--B1--C1--A1); draw(C--A1--I--B1--C--D--C1); draw(A1--E--C--F--B1); draw(L--C1--K--X--L); label("$C$", C, S); label("$D$", D, NW); label("$E$", E, S); label("$F$", F, SE); label("$I$", I, W); label("$A_1$", A1, S); label("$B_1$", B1, S); label("$C_1$", C1, NW); label("$K$", K, N); label("$L$", L, N); label("$X$", X, SE); [/asy][/asy] Since $\angle ACB=90^{\circ}$, we get $\angle A_1C_1B_1=45^{\circ}$. Let $X$ be the point on $\omega$ such that $XC_1A_1B_1$ is an isosceles trapezoid. Then, since $A_1CB_1I$ is a square, we get that $A_1$ is the circumcenter of $CEI$, so we have $$\angle IEA_1=90^{\circ}-\angle ICE=45^{\circ}-\angle A_1CE=90^{\circ}-\angle C_1A_1B_1=\angle XIA_1.$$Therefore, $A_1EIX$ is cyclic. Similarly, $B_1FIX$ is also cyclic. Now, we claim $C_1KLX$ is cyclic. We will show $C_1KXL$ is a square. Since $\angle KC_1X=180^{\circ}-\angle C_1A_1B_1-\angle DC_1A_1=45^{\circ}$, it suffices to show $C_1X=\sqrt2C_1D$. This follows from $\angle(C_1D,CI)=\angle(C_1I,C_1X)$ and $\frac{CI}{C_1I}=\sqrt2$. Therefore, the circumcircles of $A_1EI$, $B_1FI$, and $C_1KL$ all pass through $X$.

Solved with kingu. This was a pretty tough problem for the Problem 1 position and we were significantly questioning our skills. I think the main obstacle here is that there are so many nice properties in the picture it's hard to decide which of them actually matter and lead towards the solution. Before starting, we make the preliminary observation that $B_1$ and $A_1$ are the centers of $(CEI)$ and $(CFI)$ respectively since $CB_1=B_1E$ and $CA_1=A_1F$ due to reflection and $CB_1=B_1I=A_1I=CA_1$ since quadrilateral $CB_1IA_1$ is most clearly a rectangle. We first make some observations in order to locate $E$ and $F$. Claim : Points $E$ and $F$ lie on $\overline{BI}$ and $\overline{CI}$ respectively. Proof : Since $E$ is the reflection of $C$ across $\overline{B_1C_1}$, we have that $\overline{CE} \parallel \overline{AI}$ (both these lines are perpendicular to $\overline{B_1C_1}$). Thus, \[2\measuredangle EIB_1 = \measuredangle EB_1I = 2\measuredangle ECI = 2(\measuredangle ECA + \measuredangle ACI) = 90 + \measuredangle BAC\]Also, \[2\measuredangle IBC = \measuredangle ABC = 90 + \measuredangle BAC\]from which it clearly follows that $2\measuredangle EIB_1 = 2\measuredangle IB_1C$ and thus, $\measuredangle EIB_1 = \measuredangle IBC$. Further, since it is clear that $IB_1 \parallel BC$ (since both these lines are perpendicular to $AC$) we conclude that points $E$ , $I$ and $B$ are indeed collinear as claimed. The proof that $F$ lies on $\overline{CI}$ is entirely similar. Now, we can deal with the two easy circles, with the following characterization of our concurrency point. Claim : Circles $(B_1EI)$ and $(C_1FI)$ concur on $\omega$, say at $X$. Proof : Let $X$ be the second intersection point of $(B_1EI)$ and $(C_1FI)$. Now, as a result of our previous observation, we have $\measuredangle EIB_1 = \measuredangle IBC$ and $\measuredangle A_1IF = \measuredangle CAI$. Thus, \[\measuredangle B_1XA_1 = 45^\circ\]from which the claim is clear. It is easy to see that $EDLC$ and $FKDC$ are isoceles trapezoids since $CL$ and $CK$ are the reflections of $ED$ and $FD$ across $\overline{B_1C_1}$ and $\overline{A_1C_1}$ respectively. We then proceed to observe that, \[2\measuredangle CEI = \measuredangle CB_1I =90^\circ\]and so $\measuredangle CEI = 45^\circ$. Now, this implies that $\triangle EBC \sim \triangle CBI$. Thus, \[\frac{BI}{BC} = \frac{BC}{BE}\]This means, $BI \cdot BE = BC^2 = BD \cdot BA$ (since we know $\triangle BCD \sim \triangle BAC$ the last equality follows) and thus, quadrilateral $EADI$ must be cyclic. Similarly, $FBDI$ is also cyclic. This leas us to the following key result. Claim : Lines $\overline{EL}$ and $\overline{FK}$ intersect at $X$. Proof : Note that, since $X$ lies on $\omega$, $B_1I=IX$ and thus, \[\measuredangle IEX = \measuredangle IB_1X = \measuredangle B_1XI = \measuredangle B_1EI = \measuredangle IBC\]And also, due to our previous concyclicity observation we have, \[\measuredangle IEL = \measuredangle IED + \measuredangle DEL = \measuredangle IAB + \measuredangle DCL = \measuredangle IAB + (45 + 90 + \measuredangle BAC)\]\[ = \measuredangle IAB + \measuredangle BAC - 45 = 45 - \measuredangle CAI = \measuredangle IBC\]from which it is clear that $\measuredangle IEX = \measuredangle IEL$ and thus, points $E$ , $L$ and $X$ have to be collinear. Similarly we can show that poitns $F$ , $K$ and $X$ are collinear, proving the claim. Now we are as good as done. Since $EL$ is the reflection of $CD$ across $\overline{B_1C_1}$, we have that $C_1L \perp EL$ and thus, $\measuredangle XLC_1 = 90^\circ$. Similarly, $\measuredangle C_1KX = 90^\circ$ implying that $(C_1LK)$ also passes through $X$, finishing the problem.

RMM 2020 p1 Here is a Sketch of the solution. Let $Z$ be a point such that $ZA_1B_1C_1$ is an isosceles trapezoid. Claim 1: $A_1E \parallel B_1F \parallel AC$. Pf: Easy angle chase. Claim 2: $B-I-F$ are collinear, and similarly $A-I-F$. Pf: Easy angle chase. Claim 3: $C-I-L-K$ are colinear. Pf: Another easy angle chase. Claim 4: $BDIE$ and $ADIF$ are cyclic. Pf: PoP Claim 5: $Z$ lies on $(FB_1I)$ and $(EA_1I)$ Pf: More angle chase. Finally we need to show. Claim 6: $Z$ Lies on $(KC_1L)$ And we are done.

I claim the common point is the point $P$ on the incircle such that $C_1P\parallel A_1B_1$. First, note that $\angle PIB_1=\angle C_1IA_1=\angle ABC/2$. However, as $B_1I=B_1C=B_1F$ and \[\angle IB_1F= 90 ^{\circ} + \left(180 ^{\circ} - 2\angle AB_1C_1\right)=180 ^{\circ} - \angle ABC\]we get that $P$ lies on $(IB_1F)$. It similarly lies on $(IA_1E)$ As a result of the above, as $IP=B_1F$, $IB_1 \parallel PF$ or $PF \perp AC$. However, angle chase to get that $C_1L \perp BC$ and $FL \perp AC$, so $L,P,F$ collinear and $LP \perp AC$. Thus $\angle C_1LP=90^{\circ} $ and similarly $\angle C_1KP=90^{\circ}$, concluding.

but even nicer. [asy][asy] unitsize(0.6cm); pair A = (0,15); pair B = (8,0); pair C = (0,0); pair I = (3,3); pair C1= 5/17*A+12/17*B; pair A1 = (3,0); pair B1 = (0,3); pair D = 64/289*A+225/289*B; pair E = A1+ 3/17*(B-A); pair F = B1 + 3/17*(A-B); pair K = (75/17,75/17); pair L = (96/17,96/17); pair X = (K+L-C1); draw(A--B--C--cycle); draw(B1--F, royalblue+linewidth(1.3)); add(pathticks(B1--F, 1, s=8, royalblue)); draw(B1--C, royalblue+linewidth(1.3)); add(pathticks(B1--C, 1, s=8, royalblue)); draw(B1--I, royalblue+linewidth(1.3)); add(pathticks(B1--I, 1, s=8, royalblue)); draw(A1--E, deepgreen+linewidth(1.3)); add(pathticks(A1--E, 1, s=8, deepgreen)); draw(A1--C, deepgreen+linewidth(1.3)); add(pathticks(A1--C, 1, s=8, deepgreen)); draw(A1--I, deepgreen+linewidth(1.3)); add(pathticks(A1--I, 1, s=8, deepgreen)); draw(C1--L, mediumred+linewidth(1.3)); add(pathticks(C1--L, 2, s=8, mediumred)); draw(C1--K, mediumred+linewidth(1.3)); add(pathticks(C1--K, 2, s=8, mediumred)); draw(C1--D, mediumred+linewidth(1.3)); add(pathticks(C1--D, 2, s=8, mediumred)); draw(I--X, black); add(pathticks(I--X, 1, s=8,black)); draw(F--X--E, dashed); dot("$A$", A, dir(110), black+3+fontsize(8)); dot("$B$", B, dir(330), black+3+fontsize(8)); dot("$C$", C, dir(225), black+3+fontsize(8)); dot("$I$", I, dir(225), black+3+fontsize(8)); dot("$D$", D, dir(30), mediumred+3+fontsize(8)); dot("$X$", X, dir(120), black+3+fontsize(8)); dot("$C_1$", C1, dir(30), mediumred+4+fontsize(8)); dot("$K$", K, 1.5*dir(300), mediumred+3+fontsize(8)); dot("$L$", L, dir(90), mediumred+3+fontsize(8)); dot("$A_1$", A1, 1.5*dir(135), deepgreen+5+fontsize(8)); dot("$F$", F, 1.5*dir(135), royalblue+3+fontsize(8)); dot("$B_1$", B1, 1.5*dir(315), royalblue+5+fontsize(8)); dot("$E$", E, 1.5*dir(315), deepgreen+3+fontsize(8)); [/asy][/asy] It's easy to show by angle chasing that $\overline{A_1E}$ and $\overline{B_1F}$ are parallel to $\overline{AB}$. Therefore, if we define $X$ as the point such that $\overline{FX} \parallel \overline{B_1 I}$ and $\overline{EX} \parallel \overline{A_1 I}$, then $FB_1 I X$ and $EA_1 IX$ are isosceles trapezoids, so this is the point that we need to show lies on $(C_1KL)$. It's easy to show by angle chasing that reflecting $\overline{AB}$ over $\overline{A_1 C_1}$ and $\overline{B_1 C_1}$ results in a horizontal line and vertical line, respectively. Therefore: We have $\overline{C_1 K} \parallel \overline{CB}$. Similarly, we have $\overline{C_1 L} \parallel \overline{AC}$. The reflection of $\overline{EX}$ across $\overline{A_1 C_1}$ is a line through $C$ perpendicular to $\overline{AB}$, which is precisely $\overline{CD}$. Therefore, $K$ lies on line $EX$. Similarly, $L$ lies on line $FX$. Combining this together, it follows that $XLC_1 K$ is a square, so $X$ lies on $(C_1 KL)$ as desired.

the main motivation for the finish was my mom yelling at me to hurry up because i had to go somewhere WLOG $AC > BC$. Now start by observing $FB_1 = A_1 E = r$, $C_1K = C_1L = C_1 D$. Also notice $FL \parallel BC$ because $$\angle(C_1B_1, DC) = 180^\circ - \angle C_1B_1C - \angle DCA = \angle AB_1 C_1 -\angle B = \tfrac{180^\circ -\angle A}{2}- \angle B = \tfrac{\angle A}{2}$$$$\angle(Fl, CD) = 2\angle(C_1B_1, DC) = \angle A = \angle DCB$$Similarly $KE \parallel AC$, also observe $FB_1 \parallel A_1 E \parallel AB$. Let $FL \cap EK =G$, observe $\angle KGL = 90^\circ$, now observe \begin{align*} \angle LC_1K &= 360^\circ- \angle KC_1D - \angle LC_1D \\ &= 360^\circ- 2\angle BC_1A_1 - 2(180^\circ-\angle BC_1B_1) \\ &= 2(\angle BC_1 B_1 -\angle BC_1 A_1) \\ &= 2(90^\circ + \tfrac{\angle A}{2} - (90 - \tfrac{\angle B}{2} ))\\ &= \angle A + \angle B \\ &= 90^\circ \end{align*}Therefore $GKC_1L$ is cyclic. Since $\angle C_1A_1B = \angle A_1 C_1 B = \angle KC_1A_1$ then $KC_1 \parallel GL$. Therefore $GKC_1L$ is a square. Now let $BC = a$, $AC = b$, and $AB =c$ then $$C_1D = C_1B - DB = (s-b) - \tfrac{a^2}{c} = \tfrac{ac-bc+b^2}{2c}$$$$GH = C_1D + \tfrac{b}{c} (s-b) = a-b$$$$IG^2 = A_1H^2+ (GH-r)^2 = (\tfrac{ab}{a+b+c})^2 = r^2$$Now since $B_1F = IG = A_1E$ then $B_1 F G I$, and $GIA_1 E$ are isosceles trapezoids. Therefore $G$ lies on $(A_1 EI )$, $(B_1 F I)$, and $(C_1 KL)$ and we conclude.

Impressive. I claim the desired tangency point is the reflection $X$ of $C_1$ about the $C$-bisector, which lies on the incircle. Claim: $E$ lies on $\overline{AI}$ and $F$ lies on $\overline{BI}$. Proof: $\angle BA_1 E = 2\angle A_1 CE = \angle B$ and $IA_1 = CA_1 = A_1E$, so it follows that $\angle A_1IE = \frac A2 = \angle CAI$. As $\overline{IA_1} \parallel \overline{AC}$, the result follows. $\blacksquare$ Claim: $X= (FB_1I) \cap (IA_1E)$ lies on the incircle. Proof: Note that \[\angle FXE = \angle FXI + \angle IXE = 360^\circ -180^\circ-A-B = 90^\circ\]and \[\angle FXB_1 + \angle A_1XE + \angle IFE - \angle B_1IA_1 = 45^\circ.\]It follows that $\angle B_1XA_1 = 45^\circ = \angle B_1C_1A_1$, implying the result. $\blacksquare$ Claim: $C_1XA_1B_1$ is an isosceles trapezoid. Proof: $IX = IA_1 = A_1E$, so $XIA_1E$ is an isosceles trapezoid. Hence $\angle KIA_1 = \angle IA_1E = 90^\circ + B = \angle B_1IC_1$. $\blacksquare$ Claim: $\overline{KL}$ coincides with $\overline{CI}$. Proof: We will show that the reflection of $D$ about $\overline{B_1C_1}$ lies on the $C$-bisector; the other case will follow similarly. To see this,s let $P = \overline{CI} \cap \overline{B_1C_1}$, which lies on the circle $(BCD)$. It follows that $\angle DPC = \angle DBC = 2\angle B_1PC$, so $\overline{PD}$ and $\overline{CI}$ are symmetric about $\overline{B_1C_1}$. The result follows. $\blacksquare$ Putting everything together, $\angle KDL = \angle B_1C_1A_1 = 45^\circ$, so $\angle KC_1L = 90^\circ$ as $C_1$ isws the circumcenter of triangle $DKL$. Furthermore, $C_1L = C_1K$, so the reflection $X$ of $C$ over $\overline{KL} = \overline{CI}$ forms a square $CKXL$. Hence $X$ lies on $(C_1KL)$, which finishes.