Proving a geometry problem with "if, suppose" is not my strength, but let I try this hard and nice problem. The center of $(A_1B_1C_1)$ lies on $(O)$, which means it lies outside triangle $A_1B_1C_1$.It implies that this triangle is obtuse, and WLOG, assumethag $\widehat{C_1A_1B_1}>90$. Let $(I)$ be the incircle of triangle $ABC$, and its tangent points on $AB,AC$ are $F,E$ respectively. Note that $CC_1,BB_1$ cut each other at the Nagel point, so $BC_1=AF=AE=B_1C,BA_1=AB_1,A_1C=AC_1$. It means if we let $S$ be the midpoint of arc $BAC$,then $S$ is indeed the spiral similarity mapping $B\rightarrow C,C_1\rightarrow B_1$, which means $SC_1=SB_1,SC=SB$.Combining with the constraint, it follows that $S$ is actually the center of $(A_1B_1C_1)$. Let $I_a,I_b,I_c$ be respectively the centers of $A$-excircle, $B$-excircle and $C$-excircle. Note that $I_aA_1,I_bB_1,I_cC_1$ are concurrent at a point $U$. Also note that $U$ is a intersection of $(BC_1A_1)$ and $(CB_1A_1)$. According to that, by angle chasing, we get: $\widehat{BUC}=\widehat{BUA_1}+\widehat{CUA_1}=360-\widehat{BAC}-\widehat{C_1A_1B_1}= 180-\frac{\widehat{C_1SB_1}}{2}$. Combining with $SC_1=SB_1$, we conclude that $S$ is also the center of $(BUC)$. Now by symmetry, $C_1U=BA_1=AB_1,B_1U=CA_1=AC_1$.Hence $AC_1B_1U$ is a parallelogram.But note that $\widehat{AC_1U}=90$ so it is actually a rectangle, and the conclusion follows.

[asy][asy] unitsize(2.5cm); void b() { pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pen dd=linetype("4 8"); /* Define the excenter */ pair excenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)) { return extension(A,bisectorpoint(C,A,B),B,rotate(90,B)*bisectorpoint(A,B,C)); } /* Draw points */ pair A=D("A",dir(115),N), B=D("B",(-1,0),W), C=D("C",(1,0),E), I=D(incenter(A,B,C)), Ia=D("I_a",excenter(A,B,C),S), Ib=D("I_b",excenter(B,C,A),E), Ic=D("I_c",excenter(C,A,B),W), A1=D("A_1",foot(Ia,B,C),SSW), B1=D("B_1",foot(Ib,C,A),E), C1=D("C_1",foot(Ic,A,B),N), Ma=D("M_a",midpoint(Ib--Ic),N), Mb=D("M_b",midpoint(Ic--Ia),SW), Mc=D("M_c",midpoint(Ia--Ib),SE), V=D("V",circumcenter(Ia,Ib,Ic),SE); /* Draw paths */ D(unitcircle,heavyblue); D(circumcircle(B,C,V),linetype("2 2")+rgb(0.6,0,1)); D(circumcircle(B,C1,V),linetype("2 2")+rgb(0.6,0,1)); D(A--B--C--cycle); D(Ia--Ib--Ic--cycle,gray(0.3)+linewidth(1)); D(A1--B1--C1--cycle); D(I--Ia,dd+red); D(I--Ib,dd+red); D(I--Ic,dd+red); D(V--Ia,dd+heavygreen); D(V--Ib,dd+heavygreen); D(V--Ic,dd+heavygreen); } b(); pathflag=false; b(); [/asy][/asy] Let the excenters opposite $A,B,C$ be $I_a,I_b,I_c$. Let the midpoint of $\overline{I_bI_c}$ be $M_a$, which lies on $(ABC)$, the nine-point circle of $\triangle I_aI_bI_c$; analogously define $M_b,M_c$. $M_aB=M_aC$ and $BC_1=s-a=B_1C$, so $\triangle M_aBC_1\cong\triangle M_aCB_1$ (SAS), thus $M_a$ is equidistant from $B_1,C_1$, with analogous results for $M_b,M_c$. It follows that the circumcentre of $\triangle A_1B_1C_1$ is one of $M_a,M_b,M_c$; WLOG, suppose it is $M_a$. By isogonal conjugacy, $I_aA_1,I_bB_1I_cC_1$ concur at the Bevan point $V$ of $\triangle ABC$. $M_aM_b$ is the common perpendicular bisector of $\overline{C_1A_1}$ and $\overline{I_cC}$, so $C_1A_1\parallel I_cC$. $(A_1C_1M_b)$ is the circle on diameter $\overline{VB}$, so by Reim's theorem, $V \in (I_bI_cBC)$. Hence $\angle I_cI_aI_b=\tfrac{1}{2}\angle I_cVI_b=45^{\circ}\implies\angle CAB=180^{\circ}-2\angle I_cI_aI_b=90^{\circ}$, as required.

Fairly easy for a #3. Anyway, here is my solution [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(4.22cm); 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 = 2.24, xmax = 6.46, ymin = -0.88, ymax = 3.1; /* image dimensions */ /* draw figures */ draw((3.5,-0.32)--(3.5,2.62)); draw((3.5,-0.32)--(5.3,-0.32)); draw((3.5,2.62)--(5.3,-0.32)); draw(circle((4.4,1.15), 1.723629890666787)); draw((3.5,1.9736298906667877)--(4.102372300006038,1.6361252433234716)); draw((4.102372300006038,1.6361252433234716)--(4.653629890666787,-0.32)); draw((4.653629890666787,-0.32)--(3.5,1.9736298906667877)); draw((2.93,0.25)--(4.653629890666787,-0.32)); draw((2.93,0.25)--(3.5,1.9736298906667877)); draw((2.93,0.25)--(4.4,1.15)); draw((2.93,0.25)--(3.5,-0.32)); draw((2.93,0.25)--(5.3,-0.32)); draw((2.93,0.25)--(3.5,2.62)); draw((2.93,0.25)--(2.93,-0.32)); draw((2.93,-0.32)--(3.5,-0.32)); draw((2.93,0.25)--(3.5,0.25)); /* dots and labels */ dot((3.5,-0.32),linewidth(3.pt) + dotstyle); label("$A$", (3.22,-0.62), NE * labelscalefactor); dot((3.5,2.62),linewidth(3.pt) + dotstyle); label("$B$", (3.22,2.78), NE * labelscalefactor); dot((5.3,-0.32),linewidth(3.pt) + dotstyle); label("$C$", (5.42,-0.62), NE * labelscalefactor); dot((4.4,1.15),linewidth(3.pt) + dotstyle); label("$O$", (4.48,1.26), NE * labelscalefactor); dot((4.102372300006038,1.6361252433234716),linewidth(3.pt) + dotstyle); label("$A_1$", (4.18,1.76), NE * labelscalefactor); dot((4.653629890666787,-0.32),linewidth(3.pt) + dotstyle); label("$B_1$", (4.58,-0.78), NE * labelscalefactor); dot((3.5,1.9736298906667877),linewidth(3.pt) + dotstyle); label("$C_1$", (3.08,1.84), NE * labelscalefactor); dot((2.93,0.25),linewidth(3.pt) + dotstyle); label("$I$", (2.78,-0.08), NE * labelscalefactor); dot((3.5,0.25),linewidth(3.pt) + dotstyle); label("$H$", (3.58,0.36), NE * labelscalefactor); dot((2.93,-0.32),linewidth(3.pt) + dotstyle); label("$F$", (2.76,-0.64), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $I$ be the circumcenter of $A_1B_1C_1$. By the assumption, $I$ lies on $(O)$, WLOG, assume $I$ is on the arc $BC$ that contains $A$. We have $IB_1=IC_1$ and $CB_1=BC_1=p-a$ (set $BC=a,CA=b,AB=c,p=\frac{a+b+c}{2}$) Since $\angle{ABI}=\angle{ACI}$ hence $IB_1C=IC_1B$ (s.a.s), thus $IB=IC$, meaning $I$ is the midpoint of the arc $BC$ that contains $A$ of $(O)$. Using Stewart's theorem for $IBC$ w.r.t $IB_1$, we have \[ IC^2\cdot BA_1+IB^2\cdot CA_1=BC(IA_1^2+BA_1\cdot CA_1) \]or $(BA_1+CA_1)IB^2=BC(IA-1^2+BA-1\cdot CA_1)$ (since $IB=IC$). Notice that $BA_1+CA_1=BC$, we have \[ IB^2-IA_1^2=\frac{a^2-(b-c)^2}{4}\quad (1) \]Let $H,F$ be the perpendicular projection of $I$ on $AB$ and $AC$. We have $IBH=ICF$, thus $AF=AH$, which gives us $BH=CF=\frac{b+c}{2}$. On the other hand \begin{align*} IB^2-IC_1^2&=HB^2-HC_1^2\\ &=\left(\frac{b+c}{2}\right)^2-\left(\frac{b+c}{2}-\frac{b+c-a}{2}\right)^2\quad (2) \end{align*}Since $IA_1=IC_1$, from $(1)$ and $(2)$, we have \begin{align*} \frac{a^2-(b-c)^2}{4}&=\left(\frac{b+c}{2}\right)^2-\left(\frac{b+c}{2}-\frac{b+c-a}{2}\right)^2\\ \Leftrightarrow a^2&=b^2+c^2 \end{align*}Done. You can have a look at here

Set $\omega$ as the circumcircle of $\triangle ABC$, $\omega'$ as the circumcircle of $\triangle A_1B_1C_1$. As the circumcircle of $\triangle A_1B_1C_1$ is outside $\triangle ABC$, or outside $\triangle A_1B_1C_1$, one of the angles is obtuse. WLOG $\angle B_1A_1C_1 > 90$. We will show $\angle A = 90$. Set $M_a$ as the midpoint of arc $BAC$ on $\omega$. Set $I_b$ as the $B$-excenter. First, I claim that $M_a$ must be the center of $\omega'$. Trivially $M_aB=M_aC$, $\angle C_1BM_a=\angle B_1CM_a$ and $BC_1=B_1C$, so $\triangle M_aBC_1 \equiv \triangle M_aCB_1$. Now this gives $M_aB_1=M_aC_1$. Meanwhile, the center of $\omega'$ clearly lies on arc $BAC$. Also, there are two intersections between $\omega$ and the perpendicular bisector of $B_1C_1$. However, only one of the two lies on arc $BAC$. This forces $M_a$ to be the center of $\omega'$. Let us consider the intersections of $\omega'$ and the $B$-excircle. Since $AM_a$ is the external angle bisector of $A$, we have $A, M_a, I_b$ colinear. Now note that $B$-excircle is symmetric wrt $\overline{AM_aI_b}$. One intersection of $\omega'$ and the $B$-excircle is $B_1$, so the other must be $B_1$ reflected over $AI_b$. This point is the tangency point of $BA$ and the $B$-excircle. Call this point $B'_1$. By symmetry, note that the $A'_1$ lies on $\omega'$, where $A'_1$ is $A_1$ reflected over the midpoint of $BC$. We are ready to finish. By Power of a Point with point $B$ and circle $\omega'$, we have $$BC_1 \cdot BB'_1 = BA'_1 \cdot BA_1 \implies s(s-a)=(s-b)(s-c) \implies a^2=b^2+c^2$$so $\angle BAC= 90$ as desired.

Slightly different! Let $ M_1,M_2 $ and $M_3$ the midpoints of the sides $I_bI_c,I_aI_c, I_aI_b $ resp. of the excentral triangle $ I_aI_bI_c$ which are in the same time the midpoints of the arcs $\overarc{BAC},\overarc{ABC},\overarc{BCA}$ of the circumcircle of $ABC$ it's known $BC_1=B_1C$ then the similarity that send $B\to C_1,C\to B_1$ is rotation with center ,the second intersection of $\cal{C}$$ (ABC)$ and $ \cal{C}$$(AB_1C_1)$ but the center is on $\cal{C} (ABC)$ and on the bisector of $BC$ that it 's $M_1$ so $M_1B_1=M_1C_1 $ similarly we get $M_2A_1=M_2C_1,M_3B_1=M_3A_1$ hence ,since the center of the circumcircle of $A_1B_1C_1 $ is on the circumcircle of $ABC$ ,the bisectors of the sides of $A_1B_1C_1$ meet at $ M_1,M_2 $ or $M_3$ WLG suppose it 's $M_1$ then $M_1M_2 \perp A_1C_1$ but $M_1M_2\parallel I_aI_b$ thus $ A_1C_1\perp I_aI_b$ similarly $ A_1B_1\perp I_aI_c \implies \widehat{(A_1C1,A1B_1)}=\widehat{(I_aI_b,I_aI_c)}=\frac{1}{2}\widehat{(AC,AB)}$ thus $\widehat {(AB,AC)}=\widehat{(M_1B,M_1C)}=\widehat{(M_1C_1,M_1B_1)}=2\widehat{(A_1C_1,A_1B_1)}=\widehat{(AC,AB)}$ so $\widehat {(AB,AC)}=\frac{\pi}{2}$. R.HAS

Not hard for an IMO 3 Indeed, we see that if $M,N,P$ are the midpoints of arcs $BC,CA,AB$ respectively and $I_a,I_b,I_c$ are the corresponding excentres then, A rotation about $M$ sends $B_1C$ to $C_1B$ and thus, points $A,M,B_1,C_1$ are concyclic. Similarly for the other two vertices. This gives that in fact $M$ lies on the perpendicular bisector of $B_1C_1$ and so one of since the circumcentre of $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$, then it must actually be one of $M,N,P$. Let it be $M$ say. We prove that $\angle A=90^{\circ}$ Indeed, we notice that $MN$ is the perpendicular bisector of $A_1C_1$ and $MP$ of $A_1B_1$. Since, $MN \parallel I_aI_b$, $A_1C_1 \perp I_aI_b$ and similarly $A_1B_1 \perp I_aI_c$. Now, this means that $\angle B_1A_1C_1=180^{\circ}-(90^{\circ}+\frac{\angle A}{2})$ and so $\angle B_1MC_1=180^{\circ}-\angle A=\angle B_1AC_1=\angle A$ and so, $\angle A=90^{\circ}$ Lol, IMO 3 QED

Let $X,Y,Z$ be the intersection points of $(AB_1C_1),(BC_1A_1),(CA_1B_1)$ with $(ABC)$. As $X$ is the center of the spiral similarity that takes $\overrightarrow{C_1B}$ to $\overrightarrow{B_1C}$ and $BC_1=CB_1$, we get that $XB=TC$ and $XB_1=XC_1$. Therefore, $X$ lies on the perpendicular bisector of $B_1C_1$. The analogous relations are true for $Y$ and $Z$. As the circumcenter of $(A_1B_1C_1)$ is both on the perpendicular bisectors of $A_1B_1,B_1C_1,C_1A_1$ and on $(ABC)$, it has to be one of the points $X,Y,Z$. Suppose wlog that it is $X$. In this case, $XY$ is the perpendicular bisector of $C_1A_1$ and $XZ$ is the perpendicular bisector of $A_1B_1$ so $\widehat{YAZ}=\widehat{YXZ}=\dfrac{\widehat{C_1XB_1}}{2}=\dfrac{\hat{A}}{2}$. We are now done: $$\hat{A}=\widehat{BAC}=\widehat{YAB}+\widehat{YAZ}+\widehat{ZAC}=\left ( \hat{A}+\dfrac{\hat{B}}{2}-90^\circ \right ) +\dfrac{\hat{A}}{2}+\left ( \hat{A}+\dfrac{\hat{C}}{2}-90^\circ \right ) $$which implies $\hat{A}=90^\circ$.

will use a mixture of angle chasing and complex numbers (which is my favorite ). first of all according to the problem $\triangle A_1B_1C_1$ has an angle which more than 90 and let $\angle A$ be the largest of all thereupon we can easily see that $B_1C_1$ would the largest side so $\angle B_1A_1C_1$ is the one which is more than 90. afterward we claim the mid point of arc $BAC$ (we call $N$ is the center of $\odot A_1B_1C_1$. initially it's well-known that $N$ is the miquel point of $BCB_1C_1$ and $\overline{NB_1}=\overline{NC_1}$, with the fact center of $\odot A_1B_1C_1$ is on the circumcircle of $\triangle ABC$ the cliam will be proved. as mentioned before $N$ is the miquel point of $BCB_1C_1$ which implies $AB_1C_1N$ is concyclic and $\angle B_1NC_1=\angle A \implies $ $\angle B_1A_1C_1=180-\frac{A}{2}$ and this the beginning of complex. I just explain it briefly and would go into details. take $\odot ABC$ the unit circle. we put $A=a^2, B=b^2$ and $C=c^2$ and then we can calculate $A_1=\frac {c^2+b^2+\sum ab-(\frac{(cb)(a+b+c)}{a})}{2}$ and then we have that $180- \angle AIB=\angle B_1A_1C_1 \implies \frac{\frac{a^2-I}{a^2-b^2}}{\frac{a_1-b_1}{a_1-c_1}}$ has to be real $\frac{\frac{a^2-I}{a^2-b^2}}{\frac{a_1-b_1}{a_1-c_1}}=\frac{\frac{(a+b)(a+c)}{a^2-b^2}}{\frac{(b^2-a^2)(1-\frac{c(\sum a)}{ab}}{(c^2-a^2)(1+\frac{b(\sum a)}{ac}}}$ is real and I think the rest is straightforward to give us $b^2=c^2$

Restate the problem in terms of the excentral triangle as follows- Restated problem wrote: Let $O$ be the circumcenter and $\triangle DEF$ be the orthic triangle of $\triangle ABC$. Let $AO \cap EF = A_1$. Define $B_1$ and $C_1$ analogously. Suppose that the circumcenter of $\triangle A_1B_1C_1$ lie on $\odot DEF$. Show that $\triangle DEF$ is right-angled. My solution: Let $\triangle M_AM_BM_C$ be the medial triangle of $\triangle ABC$. Note that $\measuredangle OM_AD = \measuredangle OB_1D = \measuredangle OC_1D = 90^{\circ} \Rightarrow M_A,D,B_1,C_1,O$ lie on a circle. All other such results hold cyclically. Also, $M_A$ lies on $\odot (DEF)$, as it is the nine point circle of $\triangle ABC$. So $M_A$ is the center of the spiral similarity that takes $B_1F$ to $C_1E$ $\Rightarrow \triangle M_AC_1B_1 \sim \triangle M_AEF \Rightarrow$ As $M_AE = M_AF$, we get that $M_AC_1 = M_AB_1$. All other such results hold cyclically. As the circumcenter of $\triangle A_1B_1C_1$ lies on $\odot (M_AM_BM_C)$, WLOG we can assume that $M_A$ is the circumcenter of $\triangle A_1B_1C_1$. Thus, $M_AM_B$ is the perpendicular bisector of $\overline{A_1C_1}$. But $M_AM_B \perp CF$ $\Rightarrow A_1C_1 \parallel CF$ $\Rightarrow \measuredangle OCF = \measuredangle OC_1A_1 = \measuredangle OEA_1 = \measuredangle OEF$ $\Rightarrow O$ lies on $\odot (BCEF)$. Hence, $\measuredangle EDF = \measuredangle C_1DB_1 = \measuredangle C_1OB_1 = \measuredangle COB = \measuredangle CEB = 90^{\circ}$. $\blacksquare$

We ignore for now the given condition and prove the following important lemma. Lemma: Let $(AB_1C_1)$ meet $(ABC)$ again at $X$. From $BC_1 = B_1C$ follows $XC_1 = XB_1$, and $X$ is the midpoint of major arc $\widehat{BC}$. Proof. This follows from the fact that we have a spiral similarity $\triangle XBC_1 \sim \triangle XCB_1$ which must actually be a spiral congruence since $BC_1 = B_1C$. $\blacksquare$ We define the arc midpoints $Y$ and $Z$ similarly, which lie on the perpendicular bisectors of $\overline{A_1 C_1}$, $\overline{A_1 B_1}$. [asy][asy] pair A = dir(110); pair B = dir(180); pair C = dir(0); pair X = dir(90); pair Y = dir(235); pair Z = dir(325); pair A_1 = foot(Y+Z-X, B, C); pair B_1 = foot(Z+X-Y, C, A); pair C_1 = foot(X+Y-Z, A, B); draw(unitcircle, grey); draw(C_1--X--B_1, red); draw(A--B--C--cycle, grey); draw(A_1--B_1--C_1--cycle, heavycyan); draw(Y--X--Z, orange); draw(circumcircle(X, B_1, C_1), dotted+grey); draw(X--A_1, red+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$A_1$", A_1, dir(270)); dot("$B_1$", B_1, dir(B_1)); dot("$C_1$", C_1, dir(C_1)); /* TSQ Source: A = dir 110 B = dir 180 C = dir 0 X = dir 90 Y = dir 235 Z = dir 325 A_1 = foot Y+Z-X B C R270 B_1 = foot Z+X-Y C A C_1 = foot X+Y-Z A B unitcircle 0.1 grey / grey C_1--X--B_1 red A--B--C--cycle grey A_1--B_1--C_1--cycle 0.1 cyan / heavycyan Y--X--Z orange circumcircle X B_1 C_1 dotted grey X--A_1 red dashed */ [/asy][/asy] We now turn to the problem condition which asserts the circumcenter $W$ of $\triangle A_1B_1C_1$ lies on $(ABC)$. Claim: We may assume WLOG that $W = X$. Proof. This is just configuration issues since we already knew that the arc midpoints both lie on $(ABC)$ and the relevant perpendicular bisectors. Suppose (WLOG) that $\angle B_1 A_1 C_1 > 90^{\circ}$ (since $W$ lies $(ABC)$ and hence outside $\triangle ABC$, hence outside $\triangle A_1 B_1 C_1$). Then $A$ and $X$ lie on the same side of line $\overline{B_1 C_1}$, and since $W$ is supposed to lie both on $(ABC)$ and the perpendicular bisector of $\overline{B_1C_1}$ it follows $W = X$. $\blacksquare$ Consequently, $\overline{XY}$ and $\overline{XZ}$ are exactly the perpendicular bisectors of $\overline{A_1 C_1}$, $\overline{A_1 B_1}$. The rest is angle chase, the fastest one is \begin{align*} \angle A &= \angle C_1 X B_1 = \angle C_1 X A_1 + \angle A_1 X B_1 = 2 \angle YXA_1 + 2 \angle A_1 X Z \\ &= 2 \angle YXZ = 180^{\circ} - \angle A \end{align*}which solves the problem. Remark: Angle chasing is also possible even without the points $Y$ and $Z$, though it takes much longer. Introduce the Bevan point $V$ and use the fact that $VA_1B_1C$ is cyclic (with diameter $\overline{VC}$) and similarly $VA_1C_1B$ is cyclic; a calculation then gives $\angle CVB = 180^{\circ} - \frac{1}{2} \angle A$. Thus $V$ lies on the circle with diameter $\overline{I_b I_c}$.

WLOG let $\angle A$ be the largest angle in $\triangle ABC$, and define $K_A$ as the midpoint of $\widehat{BAC}$ on $(ABC)$. Furthermore, let $K_B$ and $M$ be the midpoints of $\widehat{ABC}$ and $\overline{A_1C_1}$ respectively. Lemma: Suppose the circumcenter $K$ of $\triangle A_1B_1C_1$ lies on $(ABC)$, then $K=K_A$. Proof: Already done with basically all the solutions above. [asy][asy] size(9cm); defaultpen(fontsize(9pt)); pair A=(32.5,53.3), B=(0,0), C=(120,0), D=(80,0), E=(100.8,11.7), F=(11.7,19.2), K=(60,60), L=(28.8,-51.2), M=(45.9,9.6); dot(M); draw(A--B--C--cycle, linewidth(0.5)); draw(circle((60,0),60), linewidth(0.5)); draw(F--K--B, linewidth(0.5)); draw(E--K--C, linewidth(0.5)); draw(E--F, linewidth(0.4)+grey); draw(F--D, linewidth(0.4)+grey); draw(circumcircle(D,E,F), linewidth(0.8)+dashed); draw(F--L--A, linewidth(0.5)); draw(D--L--C, linewidth(0.5)); real xmin=-100, ymin=-110, xmax=200, ymax=80; clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw(K--L, linewidth(0.7)+dotted); dot("$A$", A, (0,1)); dot("$B$", B, (-1,0)); dot("$C$", C, (1,0)); dot("$A_1$", D, (0.5,-1)); dot("$B_1$", E, (0,-1)); dot("$C_1$", F, (-0.5,0)); dot("$K_A$", K, (0,1)); dot("$K_B$", L, (-0.5,-1)); [/asy][/asy] Using the above lemma, the condition we are trying to prove is equivalent to showing that $K_B$, $M$, and $K_A$ are collinear. We will proceed with barycentric coordinates wrt $\triangle ABC$ (and using the standard notation). It is well known that $A_1=(0:s-b:s-c)$ and $C_1=(s-a:s-b:0)$, which have coordinate sums of $c$ and $a$ respectively, so $$M=(a(s-a):(a+c)(s-b):c(s-c)).$$Now, since $K_A$ lies on the exterior bisector of angle $A$, $K_A$ must be in the form of $(t:-b:c)$ for some value $t$. Intersecting this with $(ABC)$ gives us $$a^2bc-b^2ct+c^2bt=0\implies t=\frac{a^2}{b-c},$$or $K_A=(a^2:b(c-b):c(b-c))$ (and $K_B=(a(c-a):b^2:c(a-c))$ by symmetry). Finally, the desired collinearity occurs iff $$ 0=ac\cdot\text{det}\left( \begin{array}{ccc} a & b(c-b) & b-c \\ c-a & b^2 & a-c \\ s-a & (a+c)(s-c) & s-c \end{array}\right), $$which can be shown to be equivalent to $b^2+c^2=a^2$ with a few minutes of omitted expansion and row/column operations.

[asy][asy] import geometry; size(8cm,0); defaultpen(fontsize(9pt)); pair B = (-25,0), C = (25,0), N = (0,25), A = (-7,24); triangle t = triangle(A,B,C); drawline(t); circle c1 = excircle(t.BC); circle c2 = excircle(t.AC); circle c3 = excircle(t.AB); clipdraw(c1,bp+green); clipdraw(c2,bp+green); clipdraw(c3,bp+green); dot("$B$", B, (-1,0)); dot("$C$", C, (1,0)); dot("$N_A$", N, (0,1)); dot("$A$", A, (0,1)); draw(circumcircle(A,B,C)); draw(box((-60,-20), (60,50)), invisible); draw(extouch(t)); dot("$A_1$",extouch(t).A); dot("$B_1$",extouch(t).B); dot("$C_1$",extouch(t).C); point V = circumcenter(excenter(t.AB),excenter(t.BC),excenter(t.CA)); dot("$V$", V); point NN = (0,25); line l = perpendicular(NN, line(V,NN)); point VV = reflect(l)*V; dot("$V'$", VV); point BB = (-25,0); point CC = (25,0); point BBB = reflect(l)*BB; dot("$B'$", BBB); draw(circumcircle(BB,CC,V)); dot("$I_C$",excenter(t.AB)); dot("$I_B$",excenter(t.AC)); [/asy][/asy] Let $V$ be the circumcenter of $\triangle ABC$ and $N_A$ be the midpoint of major arc $BC$ in $\odot(ABC)$. Then as $\triangle N_AC_1B\cong \triangle N_AB_1C$ by SAS, it follows that $N_AC_1=N_AB_1$ and so $N_A$ must be the circumcenter of $\triangle A_1B_1C_1$ (it lies on $\odot (ABC)$ and on the perpendicular bisector of $B_1C_1$). Now since $A_1B_1C_1$ is the pedal triangle of $V$ WRT $\triangle ABC$ (isogonal lines in $I_AI_BI_C$,) we are motivated to reflect $V$ over $N_A$ to $V'$. Then from homothety at $V$, $V'$ is the circumcenter of the triangle formed by reflecting $V$ across the sidelines of $\triangle ABC$, and so by a well known lemma, $V,V'$ are isogonal conjugates WRT $\triangle ABC$. So if $B'$ is the reflection of $B$ across $AN_A$, $\angle V'BA=\angle VBC$. But using the fact that $V',V$ are actually reflections across $AN_A$ ($\angle AN_AV=90^{\circ}$ since $N_A$ is the midpoint of $I_BI_C$), we get \[\angle VBC=\angle V'BA=\angle VB'A=\angle VB'C\]so $B'BVC$ is cyclic. This means $V$ lies on the circle with diameter $I_BI_C$, so $\angle C_1VB_1=\angle I_CVI_B=90^{\circ}$. Now $AC_1VB_1$ has three right angles so it must be a rectangle, done.

What a nice problem [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11cm); 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 = -21.409127223555547, xmax = 28.190838144140958, ymin = -17.305058049471487, ymax = 12.257551066742169; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); /* draw figures */ draw((-4.209230897322035,2.2998395501682687)--(-7.486483568083494,-2.376689541817193), linewidth(2) + wrwrwr); draw((-7.486483568083494,-2.376689541817193)--(2.492566699066566,-2.376689541817193), linewidth(2) + wrwrwr); draw((2.492566699066566,-2.376689541817193)--(-4.209230897322035,2.2998395501682687), linewidth(2) + wrwrwr); draw(circle((-1.2661559069471517,-4.352433851809797), 9.979070321608386), linewidth(2) + sexdts); draw(circle((-2.496958434508464,-2.3866926575511727), 4.989535160804191), linewidth(2) + wrwrwr); draw((-3.7228357632022506,9.632367459134334)--(-1.2661559069471517,-4.352433851809797), linewidth(2) + wrwrwr); draw((-3.7277609620697754,-0.4209514632925489)--(-1.2661559069471517,-4.352433851809797), linewidth(2) + wrwrwr); draw(circle((-3.7277609620697754,-0.4209514632925489), 9.979070321608388), linewidth(2) + sexdts); draw((-9.4383046679002,1.3745050436612252)--(4.444387798883274,3.8311799628448138), linewidth(2) + dtsfsf); draw((4.444387798883274,3.8311799628448138)--(-1.2661559069471517,-14.331504173418184), linewidth(2) + dtsfsf); draw((-1.2661559069471517,-14.331504173418184)--(-9.4383046679002,1.3745050436612252), linewidth(2) + dtsfsf); draw((6.25128930508029,-0.4009452318245863)--(-9.402657538962494,-8.629324829872353), linewidth(2) + wvvxds); draw((4.433843205457115,-6.162907741096572)--(-13.70681122921984,-0.4009452318245872), linewidth(2) + wvvxds); draw((-7.1523058876969205,8.952112952146338)--(-1.2661559069471517,-4.352433851809797), linewidth(2) + wvvxds); draw((-9.4383046679002,1.3745050436612252)--(2.492566699066566,-2.376689541817193), linewidth(2) + wrwrwr); draw((4.444387798883274,3.8311799628448138)--(-7.486483568083494,-2.376689541817193), linewidth(2) + wrwrwr); draw((-4.209230897322035,2.2998395501682687)--(-1.2661559069471517,-14.331504173418184), linewidth(2) + wrwrwr); /* dots and labels */ dot((-4.209230897322035,2.2998395501682687),dotstyle); label("$A$", (-4.031976727087481,2.732298202011498), NE * labelscalefactor); dot((-7.486483568083494,-2.376689541817193),dotstyle); label("$B$", (-8.236818081788396,-3.145898385682654), NE * labelscalefactor); dot((2.492566699066566,-2.376689541817193),dotstyle); label("$C$", (2.875976927064022,-2.974272207939759), NE * labelscalefactor); dot((-1.2661559069471517,-14.331504173418184),linewidth(4pt) + dotstyle); label("$I_{A}$", (-2.058275683044194,-15.159730827685301), NE * labelscalefactor); dot((4.444387798883274,3.8311799628448138),linewidth(4pt) + dotstyle); label("$I_{B}$", (4.635145248928691,4.191120712826105), NE * labelscalefactor); dot((-9.4383046679002,1.3745050436612252),linewidth(4pt) + dotstyle); label("$I_{C}$", (-10.382145303574577,1.917073857732747), NE * labelscalefactor); dot((-3.7277609620697754,-0.4209514632925489),linewidth(4pt) + dotstyle); label("$I$", (-3.560004738294521,-0.05662718631054505), NE * labelscalefactor); dot((-2.496958434508464,-2.3866926575511727),linewidth(4pt) + dotstyle); label("$O$", (-3.002219660630114,-3.145898385682654), NE * labelscalefactor); dot((-1.2661559069471517,-4.352433851809797),linewidth(4pt) + dotstyle); label("$P$", (-1.1143317054582746,-4.004029274397129), NE * labelscalefactor); dot((-3.7228357632022506,9.632367459134334),linewidth(4pt) + dotstyle); label("$Q$", (-4.5039487158804405,8.782120967448545), NE * labelscalefactor); dot((-10.723305103352601,-7.537425776530027),linewidth(4pt) + dotstyle); label("$D$", (-10.55377148131747,-7.179113562640686), NE * labelscalefactor); dot((-2.494495835074701,2.6399668036622685),linewidth(4pt) + dotstyle); label("$O'$", (-2.315714949658536,2.9897374686258407), NE * labelscalefactor); dot((-7.1523058876969205,8.952112952146338),linewidth(4pt) + dotstyle); label("$Q_A$", (-7.936472270738331,9.64025185616302), NE * labelscalefactor); dot((-13.70681122921984,-0.4009452318245872),linewidth(4pt) + dotstyle); label("$B_1$", (-13.514323047382401,-0.05662718631054505), NE * labelscalefactor); dot((-9.402657538962494,-8.629324829872353),linewidth(4pt) + dotstyle); label("$Q_C$", (-9.609827503731552,-9.581880051041216), NE * labelscalefactor); dot((4.433843205457115,-6.162907741096572),linewidth(4pt) + dotstyle); label("$Q_B$", (4.763864882235862,-6.53551539610483), NE * labelscalefactor); dot((6.25128930508029,-0.4009452318245863),linewidth(4pt) + dotstyle); label("$C_1$", (6.437220115229083,-0.05662718631054505), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $O, I$ be the circumcenter, incenter of triangle $ABC$, and let $I_A, I_B, I_C$ be its excenters. Let $P$ be the circumcentre of $\triangle I_AI_BI_C$, then $P$ is the reflection of $I$ over $O$. Denote the isogonal conjugate of $P$ WRT $\triangle ABC$ by $Q$. Since $\triangle A_1B_1C_1$ is the pedal triangle of $P$ WRT $\triangle ABC$, so the circumcenter of $\triangle A_1B_1C_1$, $O'$, is the midpoint of $PQ$. Hence, $|\overrightarrow{IQ}|=2|\overrightarrow{OO'}|=PI_A$ holds. Denote $Q_A, Q_B, Q_C$ by the image of $Q$ by reflecting on lines $AI, BI, CI$. Since $P$ and $Q$ are isogonal conjugates, $P$, $A$, $Q_A$ are collinear, and similar facts hold for $B$ and $C$. Also, $IQ=IQ_A=IQ_B=IQ_C$, so the quadrilateral $QQ_AQ_BQ_C$ is concyclic with center $I$. Note that $\angle Q_C Q_B Q_A=\angle Q_C Q Q_A=\angle (AI, CI)=\angle I_C I_B I_A$, so $\triangle I_AI_BI_C \sim \triangle Q_AQ_BQ_C$. WLOG assume that $P$ and $Q$ are on opposite sides of $I_BI_C$. Let $BP$ and $CP$ meet $\odot(Q_AQ_BQ_C)$ again at $B_1, C_1$. Now, consider a homothety centered on $P$ sending $\odot(ABC)$ to $\odot(Q_AQ_BQ_C)$, then this homothety sends $\triangle ABC$ to $\triangle Q_AB_1C_1$. Now we chase angles to finish. Note that $$\angle IBP=\angle ABP-\angle ABI=\angle Q_AB_1P- \angle ABI= \angle Q_A Q_C Q_B - \angle ABI=\angle I_AI_CI_B-\angle ABI=\angle BI_CC.$$Similarly $\angle ICP=\angle BI_BC=\angle BI_CC$, so $\angle IBP=\angle ICP=\frac{\pi}{2}-\angle I_BI_AI_C$. Hence, $\angle BPC=2\pi -\angle BIC-\angle IBP-\angle ICP=\angle BIC$, so quadrilateral $BICP$ is a parallelogram. This implies that $O$ is the midpoint of $BC$, so we're done! $\square$

[asy][asy] size(7cm); defaultpen(fontsize(9pt)); pair A,B,C,MA,MB,MC,I,IA,IB,IC,A1,B1,C1; real r=110; A=dir(r); B=dir(180); C=dir(0); MA=dir(90); MB=-dir(r/2); MC=-dir(90+r/2); I=incenter(A,B,C); IA=-2*MA-I; IB=-2*MB-I; IC=-2*MC-I; A1=foot(IA,B,C); B1=foot(IB,C,A); C1=foot(IC,A,B); draw(circumcircle(A,B1,C1),gray); draw(circumcircle(B,C1,A1),gray); draw(circumcircle(C,A1,B1),gray); draw(MB--MA--MC,gray); draw(MB--IC--IB--MC,gray); draw(circumcircle(A,B,C)); draw(A--B--C--A); draw(A1--B1--C1--A1); dot("$A$",A,A); dot("$B$",B,dir(195)); dot("$C$",C,dir(-15)); dot("$M_A$",MA,MA); dot("$M_B$",MB,MB); dot("$M_C$",MC,MC); dot("$I_B$",IB,NE); dot("$I_C$",IC,dir(150)); dot("$A_1$",A1,SW); dot("$B_1$",B1,NE); dot("$C_1$",C1,dir(290)); [/asy][/asy] Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $CAB$, $ABC$, and $BCA$, respectively, and let $I_A$, $I_B$, $I_C$ be the $A$-, $B$-, and $C$-excenters. Furthermore let $X$ be the circumcenter of $\triangle A_1B_1C_1$. Lemma. For any triangle $ABC$, $M_A$ lies on $(AB_1C_1)$, and $M_AB_1=M_AC_1$. Proof. Since $M_AB=M_AC$, $BC_1=CB_1$, and $\measuredangle M_ABC_1=\measuredangle M_ABA=\measuredangle M_ACA=\measuredangle M_ACB_1$, by SAS $\triangle M_ABC_1\cong\triangle M_ACB_1$. It follows that $M_A$ is the Miquel point of $BCB_1C_1$, so the lemma is clearly true. $\blacksquare$ Claim. $X\in\{M_A,M_B,M_C\}$. Proof. Assume otherwise. Then by our lemma, $\overline{XM_A}\perp\overline{B_1C_1}$, $\overline{XM_B}\perp\overline{C_1A_1}$, and $\overline{XM_C}\perp\overline{A_1B_1}$, so $-\measuredangle M_CM_AM_B=\measuredangle M_BXM_C=\measuredangle C_1A_1B_1$. Hence $\triangle A_1B_1C_1\sim\triangle M_AM_BM_C$, which is acute. This implies that $X$ lies strictly within $\triangle A_1B_1C_1$, and therefore cannot lie on $(ABC)$, contradiction. $\blacksquare$ Without loss of generality $X=M_A$. Then $\overline{M_AM_B}\perp\overline{A_1C_1}$ and $\overline{M_AM_C}\perp\overline{A_1B_1}$, so \[\angle A=\angle B_1M_AC_1=360^{\circ}-2\angle B_1A_1C_1=2\angle M_BM_AM_C=180^{\circ}-\angle A,\]id est $\angle A=90^{\circ}$, as desired.

Let $M_A, M_B, M_C$ denote the midpoints of arcs $BC$, $AC$, and $AB.$, and $O$ the circumcenter of $A_1B_1C_1.$ Claim: $O$ is one of $M_A, M_B, M_C$ Proof: Since $A_1$, $B_1$, and $C_1$ are excircle tangency points, it is well known that they are reflections of the incircle tangency points over the corresponding midpoints. Therefore, $AC_1 = CA_1$, and we can derive similar length conditions for the other points. Now, suppose $X = (BC_1A_1) \cap (ABC)$. $X$ is the spiral center sending $CA_1 \mapsto AC_1$, but since $CA_1 = AC_1$, it also follows that $XCA_1 \cong XC_1A$, meaning $XA = XC \implies X = M_B$. We can similarly derive that $M_A = (AC_1B_1) \cap (ABC)$ and $M_C = (CA_1B_1) \cap (ABC)$. Now, $M_A, M_B$, and $M_C$ satisfy $M_BC_1 = M_BA_1$, $M_CA_1 = M_CB_1$, and $M_AC_1 = M_AB_1$. Since $M_A, M_B, M_C \in (ABC)$, the claim is proven, as the circumcenter $O$ satisfies $OA_1 = OB_1 = OC_1$. WLOG let $O = M_B.$ Since $M_B$ and $M_A$ both lie on the perpendicular bisector of $C_1$ and $B_1$, and $M_B$ and $M_C$ lie on the perpendicular bisector of $A_1B_1$, we have $$\measuredangle ABC = \measuredangle C_1M_BA_1 = \measuredangle C_1M_BB_1 + \measuredangle B_1 M_BA_1 = 2 \measuredangle M_AM_BM_C = 180 - \measuredangle ABC,$$meaning $\angle ABC = 90^{\circ}$ as desired. $\blacksquare$

IMO 2013 Problem 3 and Shortlist G6 wrote: Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled. Proposed by Alexander A. Polyansky, Russia Let $O$ be the center of the circumcircle of triangle $\vartriangle A_1B_1C_1$. Since $O$ lies outside of triangle $\vartriangle A_1B_1C_1$, we have that this triangle is obtuse. WLOG, assume that $\angle C_1A_1B_1>\pi/2$. Let now $I_A,I_B,I_C$ be the three excentres. We proceed with a Claim. Claim 1: Point $O$ coincides with the midpoint of $I_BI_C$. Proof: It is well-known that the midpoint of $I_BI_C$ lies on the circumcircle of $\vartriangle ABC$ (indeed, this midpoing lies on the nine point circle of $\vartriangle I_AI_BI_C$, which is the circle that passes through $A,B$ and $C$) Let the midpoint of $I_BI_C$ be $M$. From the previous remark, we have that $M \in (A,B,C)$, and since $I_CBCI_B$ is inscribed in a circle with center $M$, we have that $MB=MC$. So, triangles $C_1MB$ and $B_1MC$ are equal (they have $BM=CM$, $BC_1=B_1C$ and $\angle C_1BM=\angle MB_1C$). So, it follows that $MC_1=MB_1$. So, point $M$ can be defined as the intersection point of the perpedicular bisector of $B_1C_1$ with the circumcircle of $\vartriangle ABC$ (with this poing lying on the same arc of $BC$ as $A$). So, we must have $O \equiv M$ $\blacksquare$. Back in the problem, we have that $O$ is the midpoint of $I_BI_C$, by the claim. Let now $R$ be the circumcenter of $\vartriangle I_AI_BI_C$. We present now another Claim. Claim 2: $I_C,C_1,R$ are collinear, as do $I_B,B_1,R$ and $I_A,R,A_1$. Proof:By Nagel's theorem, we have $I_CR \perp AB$, and since $I_CC_1 \perp AB$, we have that $I_C,C_1,R$ are collinear. Similarly, we obtain the other two collinearities. $\blacksquare$ Therefore, by the Claim above, we have $\angle BC_1R=\angle BA_1R$, so quadrilateral $BC_1A_1R$ is inscribed in a circle. Similarly we can obtain that $A_1B_1CR$ is also inscribed in a circle. Hence we have that: $$\angle BRC=\angle BRA_1+\angle A_1RC=\angle AC_1A_1+\angle AB_1A_1=360^\circ-\angle A-\angle A_1B_1C_1=360^\circ - \angle A-(180^\circ-\angle C_1OB_1/2)=360^\circ-\angle A+\angle C_1OB_1/2=180^\circ -\angle A+\angle A/2=180^\circ-\angle A/2$$where the latter equality follows from $$\angle C_1OB_1=\angle C_1OB+\angle BOB_1=\angle B_1OC+\angle BOB_1=\angle BOC=\angle A$$ We have thus obtained that $\angle BRC=180^\circ-\angle A/2$. This, combined with $BO=CO$ and $\angle BOC=\angle A$, yield that $O$ is the circumcenter of triangle $\vartriangle BRC$, so $$OB=OR=OC=OB_1=OC_1$$This gives that in triangle $\vartriangle I_CRI_B$ that $I_CO=OI_B=OR$, which implies that $\angle I_CRI_B=90^\circ$. Lastly, $$\angle A=2\pi - \angle AC_1R-\angle AB_1R-\angle C_1RB_1=90^\circ,$$and the proof is complete.

Without loss of generality let $\angle \max(A, B, C) = \angle A$. Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $\widehat{BAC}$, $\widehat{ABC}$, and $\widehat{ACB}$. Finally let $D$ be the midpoint of $A_1C_1$. Claim: $M$ is the center of $(A_1B_1C_1)$. Proof. Note that $BC_1 = B_1C = s - a$. Then by say, 14 ARMO 11/4, there exists a spiral similarity centered at $M$ mapping $B_1C_1 \mapsto CB$. Then as $MB = MC$, we find $MB_1 = MC_1$, which is sufficient, as $\angle A$ is the largest angle in $\triangle A BC$. $\blacksquare$ From similar statements we find $MK$ and $MN$ are the perpendicular bisectors of $A_1B_1$ and $A_1C_1$ respectively. Let $P$ be the midpoint of $A_1B_1$. Now we proceed with barycentric coordinates, with reference $\triangle A BC$. Now noting that $AM_A$ we can parametrize $M_A$ as $(t:-b:c)$. Then intersecting with the circumcircle we easily find $$M_A = (a^2 : b(c-b) :c(b-c))$$Similarly, $$M_B = (a(c-a): b^2: c(a-c))$$Now we finally calculate $$D = (a(s-a): (a+c)(s-b):c(s-c))$$Now we encode the condition $D$, $M_A$ and $M_B$ collinear to find, \begin{align*} 0&= \begin{vmatrix} a^2 & b(c-b) & c(b-c) \\ a(c-a) & b^2 & c(a-c) \\ a(s-a) & (a+c)(s-b) & c(s-c) \end{vmatrix}\\ \end{align*}which upon expansion can be shown to be equivalent to $a^2 = b^2 + c^2$. Indeed upon computation we find the determinant is exactly, \begin{align*} 0 &= -\frac{ac}{2}(a^4 - 2 a^2 (b^2 - b c + c^2) + (b - c)^2 (b^2 + c^2))\\ 0 &= a^4 - 2a^2(b^2-bc+c^2) + (b^2-2bc+c^2)(b^2+c^2)\\ 0 &= -bc(b^2+c^2) + a^4 - a^2(b^2-bc+c^2) - (b^2-bc+c^2)(b^2 + c^2 - a^2)\\ 0 &= bc(a^2-b^2 -c^2) + a^2(a^2-b^2-c^2) + (b^2-bc+c^2)(a^2-b^2-c^2)\\ 0 &= (a^2-b^2-c^2)(a^2 + b^2 + c^2) \end{align*}

oof configuration issues

Solved with a small hint. Pure synthetic eludes me. WLOG let $\angle A$ be the largest angle of $\triangle ABC$ and let $O$ be the center of $(A_1B_1C_1)$. Let $M$ be the midpoint of arc $BAC$. Then $MB=MC$, $BC_1=CB_1$, and $\angle C_1BM=\angle B_1CM$, so $\triangle MC_1B \cong \triangle MB_1C \implies MB_1=MC_1$. Thus $M$ is the intersection between the perpendicular bisector of $\overline{B_1C_1}$ and $(ABC)$ that lies on the same side of $\overline{BC}$ as $A$. $O$ is also an intersection between this perpendicular bisector and $(ABC)$, and it's clear that $A$ and $O$ lie on the same side of $\overline{BC}$, hence $O=M$. WLOG let $\angle B>\angle C$ $\angle OBA_1=90^\circ-\tfrac{1}{2}\angle A$, so $\angle OBC_1=\tfrac{1}{2}\angle A+\angle B-90^\circ$. Furthermore, $OB=\tfrac{a}{2\cos(90^\circ-\frac{1}{2}\angle A)}=\tfrac{a}{2\sin \frac{1}{2}\angle A}$. Thus by the law of cosines, since $OC_1=OA_1$ we should have \begin{align*} \left(\frac{a}{2\sin \frac{1}{2}\angle A}\right)^2+(s-a)^2-2(s-a)\left(\frac{a}{2\sin \frac{1}{2}\angle A}\right)\cos\left(\frac{1}{2}\angle A+\angle B-90^\circ\right)&=\left(\frac{a}{2\sin \frac{1}{2}\angle A}\right)^2+(s-c)^2-2(s-c)\left(\frac{a}{2\sin \frac{1}{2}\angle A}\right)\cos\left(90^\circ-\frac{1}{2}\angle A\right)\\ (s-a)^2-a(s-a)\frac{\sin(\frac{1}{2}\angle A+\angle B)}{\sin \frac{1}{2}\angle A}&=(s-c)^2-a(s-c)=-(s-c)(s-b)\\ \frac{\sin(\frac{1}{2}\angle A+\angle B)}{\sin \frac{1}{2}\angle A}&=\frac{(s-a)^2+(s-b)(s-c)}{a(s-a)}. \end{align*}Let $D$ be the intersection of the (interior) $\angle BAC$-bisector with $\overline{BC}$. Then by law of sines $\tfrac{\sin(\frac{1}{2}\angle A+\angle B)}{\sin \frac{1}{2}\angle A}=\tfrac{AB}{BD}$, and by angle bisector theorem this equals $\tfrac{AC}{CD}$, hence it equals $\tfrac{b+c}{a}$. Then \begin{align*} \frac{b+c}{a}&=\frac{(s-a)^2+(s-b)(s-c)}{a(s-a)}\\ 2(b+c)(b+c-a)&=(b+c-a)^2+(a-b+c)(a+b-c)\\ 2(b+c)^2-2a(b+c)&=(b+c)^2-2a(b+c)+a^2+a^2-(b-c)^2\\ (b+c)^2+(b-c)^2&=2a^2\\ b^2+c^2&=a^2, \end{align*}or $\angle BAC=90^\circ$, as desired. $\blacksquare$

WLOG let $\angle ABC$ be the largest angle in $\triangle ABC$. Denote the midpoint of arc $ABC$ as $O_1$. It is clear that $O_1A = O_1C$, $A_1C=AC_1$, and $\angle A_1CO_1 = \angle C_1AO_1$. This implies that $\triangle OAC_1 \cong \triangle O_1CA_1$; in particular, $O_1A_1=O_1C_1$, making $O_1$ the desired circumcenter. Define $O_2$ and $O_3$ similarly to $O_1$ except with arcs $BAC$ and $BCA$ instead. Note that $\triangle O_1B_1O_2 \cong \triangle O_1C_1O_2$ and $\triangle O_1A_1O_3 \cong \triangle O_1B_1O_3$. Thus, \[\angle B = \angle A_1O_1C_1 = 2 \angle O_2O_1O_3.\] Convert this into arc lengths as follows: Denote $m (XY)$ denote the measure of arc $XY$. The factor of $2$ caused by the inscribed angle theorem will be omitted. \[2 \angle O_2O_1O_3 = 2 m(O_1O_2) = 2( m(AC) - m(AO_2) - m(CO_3)).\] Realize that arc $AO_2$ has measure $\frac{\angle A}{2}$ because the angle bisector of $\angle BAC$ would intersect $(ABC)$ at the antipode of $O_2$. Analogously, arc $CO_3$ has measure $\frac{\angle C}{2}$. Plugging this in, we get \[2 \angle O_2O_1O_3 = 2 \left( \angle B - \frac{\angle A}{2} - \frac{\angle C}{2} \right) = \angle B \]\[\implies \angle A + \angle C = \angle B.\] We are done. $\square$

assume $AB \geq BC \geq AB$. let the excenters be $A_2, B_2,$ and $C_2$ respectively, and let the circumcenter of $A_1B_1C_1$ be $O$, let $AA_2$ intersect $BC$ at $Y$, let $AO$ intersect $CC_2$ at $Z$, let $OD\perp B_1C_1$, and $OE\perp B_1A_1$. note that $AA_2,BB_2,$ and $CC_2$ intersect at $X$ cyclic quadrilaterals give that $B_2BC=B_2A_2X=A_1A_2B=C_1C_2B=AC_2X$, $AB_2C=ABC_2=CBA_2$ we can then prove $OAB=OCB$, $OBC=OAC=AC_2B-CC_2C_1-BAO$, as well as $AYB=OZC=CA_2B$ $XCO=BB_2A_2$, so $AOC=90$ and $ABC=90$

solved with GrantStar (angle chasing too hard grantstar too orz) Rename $A_1,B_1,C_1$ to $D,E,F.$ Notice that $BF=CE$ and symmetric. Now construct $A'$ to be the intersection of perpendicular bisectors of $EF,BC.$ We have $\triangle A'FB\cong\triangle A'EC$ directly congruent by SSS so $A'$ is the Miquel point of $BFEC,$ so it is on $(ABC).$ Thus it must be the midpoint of arc $BAC.$ Define $B',C'$ similarly. Let $A_1$ be the point on $(ABC)$ such that $A'A_1$ is the perpendicular bisector of $EF,$ and define $B_1,C_1$ similarly. If the circumcenter of $DEF$ lies on $(ABC)$ we must have that three of $A',B',C',A_1,B_1,C_1$ coincide. However $A',B',C'$ are distinct, and if $A_1,B_1,C_1$ coincide, it would imply that the circumcenter of $DEF$ lies inside $\triangle DEF,$ impossible. Thus we may assume that $A',B_1,C_1$ coincide. This implies that $A'B',A'C'$ are the perpendicular bisectors of $DF,DE$ respectively. Let $A'D,A'E,A'F$ intersect $(ABC)$ again at $D',E',F'.$ Since $A'B'$ bisects $DF$ we have that $B'$ is the arc midpoint of $D'F',$ and since $AC,D'F'$ share an arc midpoint we have $AC\parallel D'F'.$ Similarly $AB\parallel D'E',$ so $\angle A=\angle E'D'F'=180^\circ-\angle EA'F.$ However from the spiral similarity we have $\angle EA'F=\angle BA'C=\angle A,$ so $\angle A=90^\circ$ as desired.

Suppose $P$ and $P'$ are the midpoints of arc $ACB$ and $AB$, respectively. Note that $AB_1 = A_1B$ as well as $AP = BP$ and $AP' = BP'$. SSA (in)congruency tells us at least one of \[\triangle PAB_1 \cong \triangle PBA_1, \quad \triangle P'AB_1 \cong \triangle P'BA_1\] must hold. We can suppose WLOG the first holds. Then our desired center is $P$, so we can also consider the midpoints $M$ and $N$ of arcs $CAB$ and $ABC$, respectively. Then $MP$ and $NP$ are the perpendicular bisectors of $B_1C_1$ and $C_1A_1$, respectively, so \[\angle C = \angle APB = \angle A_1PB_1 = 2 \angle NPM = 180 - \angle C \implies \angle C = 90. \quad \blacksquare\]

Similar ideas are used in GOTEEM 2020/5.

Let X be the arc midpoint of arc BAC. We have that $BC_1=CB_1$ from equal tangents, $XB=XC$ by X being the midpoint, also $\angle ACX =\angle ABX$ since ABCX is cyclic $\Rightarrow$ $\triangle XC_1B\cong\triangle XB_1C$ $\Rightarrow$ X is on the perpendicular bisector of $B_1C_1$. Denote the arc midpoints of arc ABC and ACB, be Y and Z. Similarly they lie on the perpendicular bisectors of $A_1C_1,A_1B_1$, respectively. Assume WLOG that the circumcenter of $\triangle A_1B_1C_1$ is X (from the three arc midpoints). Now $\angle BAC = \angle BXC = \angle BXB_1 + \angle B_1XC = \angle BXB_1 + \angle BXC_1 = \angle C_1XB_1$ $\Rightarrow$ $\angle C_1AB_1 = \angle C_1XB_1$ $\Rightarrow$ $C_1AXB_1$ is cyclic. Now we have that $\angle YXZ = \angle YXA_1 + \angle A_1XZ = \frac{1}{2}\angle C_1XA_1 + \frac{1}{2}\angle A_1XB_1 = \frac{1}{2}\angle C_1XB_1 = \frac{1}{2}\angle C_1AB_1 = \frac{1}{2}\angle BAC$ $\Rightarrow$ $\angle YXZ = \frac{1}{2}\angle BAC$. Also $\angle XYZ = 90 - \frac{1}{2}\angle BAC$ $\Rightarrow$ $\angle YXZ = \frac{1}{2}\angle BAC = 90 - \frac{1}{2}\angle BAC$ $\Rightarrow$ $\angle BAC = 90^{\circ}$ $\Rightarrow$ $\triangle ABC$ is right-angled as we wanted. We are ready.

wow, there was a mop class (sparrow v cannon) where we were building up to this problem, but it was kind of ruined as we were never able to get to it, possibly due to too much stuff happening during class this is pretty nice though Let $S$ denote the circumcenter of $\triangle A_0B_0C_0$. The perpendicular bisector of $B_0C_0$ meets $(ABC)$ at two points. We claim that one of them is the midpoint of arc $BAC$ (this is the "sparrow lemma"). Note that $BC_0=CB_0$. If $J_a$ is the midpoint of arc $BAC$, then $J_aB=J_aC$, and $\angle C_0BJ_a=\angle B_0CJ_a$ as well, so $\triangle J_aC_0B$ and $\triangle J_aB_0C$ are congruent and thus $J_aC_0=J_aB_0$, showing the claim. Call the second such point the $A$-buh point (since it's annoying) and denote it by $X_a$. We have that $X_a$ is second intersection of $(ABC)$ with the perpendicular to $B_0C_0$ through $J_a$. By the given condition, $S$ is either $J_a$ or $X_a$, and similarly either $J_b$ or $X_b$ and either $J_c$ or $X_c$. Clearly, $X_a$, $X_b$, and $X_c$ do not coincide, so $S$ must be some arc midpoint, WLOG $J_a$. Importantly, this means that $J_a$ is also both $X_b$ and $X_c$, which means that $J_aJ_b$ is perpendicular to $A_0C_0$. Since $J_aJ_b$ is the midline in the excentral triangle, $J_aJ_b$ is parallel to $I_aI_b$, which is in turn perpendicular to the internal $C$-bisector. Therefore, the internal $C$-bisector is parallel to $A_0C_0$. Let $CI$ meet $AB$ at $F$. Then, $$\frac{BA_0}{BC}=\frac{BC_0}{BF}$$$$\frac{s-c}{a}=\frac{s-a}{\frac{ac}{a+b}}$$$$(s-c)c=(s-a)(a+b)$$$$(a+b-c)c=(-a+b+c)(a+b)$$$$a^2=b^2+c^2$$and we are done. Alternative angle chase (after getting that $A_0C_0$ and $A_0B_0$ are parallel to the internal bisectors): we know $AJ_aB_0C_0$ cyclic so $\angle C_0AB_0=\angle C_0J_aB_0=2(180-\angle C_0A_0B_0)=2(180-\angle BIC)=2(90-\angle A/2)$ so $\angle A=90$.

Nice one, but quite easy for P3. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.651746085090494cm); 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 = -9.548320615797111, xmax = 11.103425469293384, ymin = -3.6265775477501117, ymax = 5.545945680581804; /* image dimensions */ pen qqffff = rgb(0.,1.,1.); pen zzttqq = rgb(0.6,0.2,0.); pen zzffzz = rgb(0.6,1.,0.6); pen ffccww = rgb(1.,0.8,0.4); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen ffefdv = rgb(1.,0.9372549019607843,0.8352941176470589); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); pen ffzztt = rgb(1.,0.6,0.2); /* draw figures */ draw(circle((0.39170126705322017,1.2593588720578868), 1.857541583711824), linewidth(1.) + qqffff); draw((-1.465734736439853,1.2791636226574639)--(-0.17979008689644715,3.0268028931175093), linewidth(1.) + zzttqq); draw((2.2491372705462935,1.2395541214583097)--(-0.17979008689644715,3.0268028931175093), linewidth(1.) + zzttqq); draw((-1.465734736439853,1.2791636226574639)--(2.2491372705462935,1.2395541214583097), linewidth(1.) + zzttqq); draw((-1.465734736439853,1.2791636226574639)--(3.00856072208783,3.5120521747993507), linewidth(1.) + zzffzz); draw((2.2491372705462935,1.2395541214583097)--(-2.1855486867822345,2.7215375763025706), linewidth(1.) + zzffzz); draw((-0.17979008689644715,3.0268028931175093)--(0.7671538157020328,-3.195131835870218), linewidth(1.) + zzffzz); draw((0.7671538157020328,-3.195131835870218)--(0.8146013590071536,1.2548497364354834), linewidth(1.) + ffccww); draw((0.8067633169011869,0.5197401711159279)--(3.00856072208783,3.5120521747993507), linewidth(1.) + ffccww); draw((0.8067633169011869,0.5197401711159279)--(-2.1855486867822345,2.7215375763025706), linewidth(1.) + ffccww); draw((0.7671538157020328,-3.195131835870218)--(3.00856072208783,3.5120521747993507), linewidth(1.) + qqwuqq); draw((-2.1855486867822345,2.7215375763025706)--(3.00856072208783,3.5120521747993507), linewidth(1.) + qqwuqq); draw((0.7671538157020328,-3.195131835870218)--(-2.1855486867822345,2.7215375763025706), linewidth(1.) + qqwuqq); draw(circle((0.8067633169011869,0.5197401711159279), 3.7150831674236473), linewidth(1.) + afeeee); draw((0.8146013590071536,1.2548497364354834)--(-1.030036728431871,1.871290934767146), linewidth(1.) + afeeee); draw((0.8067633169011869,0.5197401711159279)--(-1.465734736439853,1.2791636226574639), linewidth(1.) + afeeee); draw((0.8067633169011869,0.5197401711159279)--(2.2491372705462935,1.2395541214583097), linewidth(1.) + afeeee); draw((0.8146013590071536,1.2548497364354834)--(1.6570099584366114,1.6752521294662914), linewidth(1.) + afeeee); draw((0.8146013590071536,1.2548497364354834)--(0.4115060176527976,3.116794875550961), linewidth(1.) + ffefdv); draw((0.4115060176527976,3.116794875550961)--(-1.030036728431871,1.871290934767146), linewidth(1.) + ffefdv); draw((0.4115060176527976,3.116794875550961)--(1.6570099584366114,1.6752521294662914), linewidth(1.) + ffefdv); draw(circle((1.5279502937237401,0.8796471462871188), 0.806004768171585), linewidth(1.) + dotted + eqeqeq); draw(circle((-0.32948570976933295,0.899451896886696), 1.1980162124518678), linewidth(1.) + dotted + eqeqeq); draw(circle((0.3134866150023695,1.7732715321167192), 1.3470941977637165), linewidth(1.) + dotted + cqcqcq); draw(circle((0.4115060176527973,3.11679487555096), 2.6269605003572587), linewidth(1.) + dotted + ffzztt); draw((-1.030036728431871,1.871290934767146)--(1.6570099584366114,1.6752521294662914), linewidth(1.) + afeeee); /* dots and labels */ dot((-1.465734736439853,1.2791636226574639),dotstyle); label("$B$", (-1.4070278095853517,1.421023992101179), NE * labelscalefactor); dot((2.2491372705462935,1.2395541214583097),linewidth(4.pt) + dotstyle); label("$C$", (2.2972604172409987,1.3531798853827477), NE * labelscalefactor); dot((-0.17979008689644715,3.0268028931175093),dotstyle); label("$A$", (-0.13155860327884275,3.157833124093021), NE * labelscalefactor); dot((-2.1855486867822345,2.7215375763025706),linewidth(4.pt) + dotstyle); label("$Ic$", (-2.1261753408007236,2.8321814118445503), NE * labelscalefactor); dot((3.00856072208783,3.5120521747993507),linewidth(4.pt) + dotstyle); label("$Ib$", (3.0571144124874294,3.619173049778354), NE * labelscalefactor); dot((0.7671538157020328,-3.195131835870218),linewidth(4.pt) + dotstyle); label("$Ia$", (0.8182588907791958,-3.083824694002661), NE * labelscalefactor); dot((0.8067633169011869,0.5197401711159279),linewidth(4.pt) + dotstyle); label("$V$", (0.8589653548102546,0.6340323541673756), NE * labelscalefactor); dot((-1.030036728431871,1.871290934767146),linewidth(4.pt) + dotstyle); label("$C_{1}$", (-0.9728255265873912,1.9773456671923157), NE * labelscalefactor); dot((1.6570099584366114,1.6752521294662914),linewidth(4.pt) + dotstyle); label("$B_{1}$", (1.7138010994624893,1.787382168380708), NE * labelscalefactor); dot((0.8146013590071536,1.2548497364354834),linewidth(4.pt) + dotstyle); label("$A_{1}$", (0.8725341761539409,1.366748706726434), NE * labelscalefactor); dot((-0.023360782794746484,1.998977572999846),linewidth(4.pt) + dotstyle); label("$I$", (0.031267252845392446,2.1130338806291786), NE * labelscalefactor); dot((0.4115060176527976,3.116794875550961),linewidth(4.pt) + dotstyle); label("$M_{1}$", (0.46546953584335293,3.2256772308114523), NE * labelscalefactor); dot((-0.7091974355401003,-0.2367971297838247),linewidth(4.pt) + dotstyle); label("$M_{3}$", (-0.660742635682607,-0.12582164107905533), NE * labelscalefactor); dot((1.8878572688949316,0.15846016946456634),linewidth(4.pt) + dotstyle); label("$M_{2}$", (1.9444710623051558,0.26767417788784637), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The problem can be divided into two claims: Claim 1: Proving that $O$ the Center of $\triangle A_1B_1C_1$ is the midpoint of $I_cI_b$. Claim 2: Showing that $V$ lies on $(I_cBCI_b)$ The finishing is an easy angle chase.

This problem was mostly just me relearning some $9$ point circle properties. We look at this problem from the excentral triangle, let them be $I_A, I_B, I_C.$ Let $M_A$ be the midpoint of $I_BI_C,$ define $M_B,M_C$ similarly. WLOG let $M_A$ be the center of $(A_1B_1C_1)$ by spiral. Then note that by using the previous spiral on $M_B,$ we have that $\overline{M_AM_B}$ is the perpendicular bisector of both of $I_CC, C_1A_1,$ so $I_CC\| C_1A_1.$ As $V$ lies on $(BC_1A_1)$ because it is the antipode of $B,$ by Reim we see that $(I_CBVCI_B)$ is cyclic. Therefore we get that $\angle I_CI_AI_B=45\deg,$ so we're done$.\blacksquare$

Edit: Line $2$ and $7$ are meant to say "perpendicular bisector of $B_1C_1$".

Posting a solution from old, for storage. Let $C_3$ be the point where the $C-$excircle is tangent to $\overline{AC}$. Consider the case when $O$ lies inside the angle $C_3ACB$. We first note the following claim. Claim : $O$ lies on the circle $(AB_1C_1)$. Proof : Let $O'=(AB_1C_1) \cap (ABC)$. Then, $$\angle BC_1O' = 180 - \angle O'C_1A = 180 - \angle O'B_1A = \angle OB_1C$$and $$\angle O'BC_1 = \angle O'BA = \angle O'CA = \angle O'CB_1$$Let $B_2,C_2$ be the points where the incircle touches $AC$ and $AB$ respectively. Now, we have the following key lemma. Let $ABC$ be a triangle. Suppose its incircle and $A$-excircle are tangent to $BC$ at $X$ and $D$ respectively. Then, $BX = CD$ and $BD = CX$. Then, $$B_1C = AB_2 = AC_2 = BC_1$$Thus, clearly $$\triangle O'C_1B \cong \triangle O'B_1C$$Thus, $O'C_1=O'B_1$ which means $O'$ lies on the perpendicular bisector of $B_1C_1$ and also lies on $(ABC)$ but then, this means that $O'=O$. Now, since $\triangle O'C_1B \cong \triangle O'B_1C$ we have $$\angle C_1OB_1 = \angle C_1OC + \angle COB_1 = \angle C_1OC + \angle BOC_1 = \angle BOC = \angle A$$We then also have, $$\angle OAC_3 = \angle OBC = \angle OCB = \angle OAB = \angle OAC_1$$Then, note that $AC_3=AC_1$ which gives us that $\triangle OC_3A \cong \triangle OAC_1$. Thus, $$OC_3=OC_1=OB_1$$Thus, $C_3$ also lies on $(A_1B_1C_1)$. Thus, \begin{align*} 360 - 2 \angle C_1A_1B_1 &= \angle BAC\\ 360 - 2(180 - \angle C_1C_3A) &= \angle A\\ 360 - 180 - \angle A &= \angle A\\ 2\angle A &= 180 \\ \angle A &= 90^\circ \end{align*}The other cases are entirely similar (with $\angle B$ and $\angle C$ being $90^\circ$). Thus, $\triangle ABC$ is right angled as claimed and we are done.