Moderator note: the original wording of the problem was as follows. Quote: $ \triangle ABC $ is a triangle with incenter $I$. Show that the circumcenters of $ \triangle IAB, \triangle IBC, \triangle ICA $ lie on a circle whose center is the circumcenter of $ \triangle ABC $. Those circumcentres are the meeting points of the angle bisectors of $\triangle ABC$ with the circumcircle of $\triangle ABC$, by a well-known (and easy-to-prove) lemma.

Proof of Lemma: Construct $\triangle ABC$ as shown below. Let $I$ be the incenter. Let the intersection of $AI$ and the circumcircle of $\triangle ABC$ be $P \neq A.$ [asy][asy] import olympiad; unitsize(50); pair A,B,C,O,I; A=origin; B=2*right; C=1.5*dir(70); I=incenter(A,B,C); // olympiad - incenter draw(A--B--C--cycle); dot(I); draw(circumcircle(A,B,C)); // olympiad - circumcircle draw(incircle(A,B,C)); // olympiad - incircle label("$I$",I,S); label("$A$",B,E); label("$B$",A,W); label("$C$",C,N); [/asy][/asy] Lemma: $PC=PI=PB.$ Proof: Since $AI$ bisects $\angle CAB,$ $P$ is the midpoint of arc $BC.$ Now, let $\angle BCP = \angle BAP \angle PAC = \alpha$ and $\angle ICB = \angle ICA = \beta.$ Thus, $\angle PIC = \angle PCI = \alpha + \beta \implies PI=PC.$ Similarly, $PI=PB.$ This means that $P$ is the circumcenter of $\triangle IBC.$

let $P,Q,R$ be circumcentres of triangle $IBA,ICA,IBC$ respectively. than $\angle BPA=360-2\angle AIB = 180-C$ and hence $BPAC$ is cyclic quad. similarly $B,P,A,Q,C,R$ all lie on circumcircle of $ABC$ and hence circumcircle of triangle $PQR$ and triangle $ABC$ coincide. we are done

Dear Mathlinkers, this usefull result comes from Jules Mention, a french mathematician around 1850. Sincerely Jean-Louis

By incenter-excenter lemma (kind of), the circumcenters $A', B', C'$ lie on the circumcircle of $\Delta ABC$, and $\Delta A'B'C'$ is also known as the circum-incentral triangle of $\Delta ABC$ and hence, have the same circumcenter

By the Incenter-Excenter Lemma, $C'A = C'B = C'I$ and $C'$ lies of $(ABC)$. Similarly, $A'$ and $B'$ also lie on $(ABC)$. Thus $O$ is the circumcenter of $\triangle{A'B'C'}$. P.S. Are we allowed to state this lemma directly on contests?

Yes, it's very known result aka trillium theorem

PhysicsMonster_01 wrote: P.S. Are we allowed to state this lemma directly on contests? Yes, you can directly state it as Incenter-Excenter Lemma in a contest.

Consider the circumcircle of $ABC$. By the Incenter-Excenter Lemma, since $A'$ is equidistant from $I, B, C$, it lies on the circumcircle of $ABC$. Similarly, $B'$ and $C'$ lie on the circumcircle of $ABC$ as well, so triangles $ABC$ and $A'B'C'$ have the same circumcenter.

Let $AI, BI,$ and $CI$ hit $(ABC)$ at $L_A, L_B$, and $L_C$ respectively. By the Incenter-Excenter Lemma, $L_A,L_B$, and $L_C$ are the circumcenters of $IBC, ICA$, and $IAB$ respectively so $L_A=A', L_B=B'$ and $L_C=C'$. Hence, because $A', B'$ and $C'$ are all on $(ABC)$, the circumcenter of $A'B'C'$ is $(ABC)$. $\blacksquare$

Storage. Let $\Omega$ denote the circumcircle of $\triangle ABC$. By Fact 5, $\overline{AI}\cap\Omega = A'$. Similarly, $\overline{BI}\cap\Omega = B'$ and $\overline{CI}\cap\Omega =C'$ so $A,B,C,A'B'C'$ are concyclic, implying the result.

Another application of Even Chen’s in center excenter theorem

Incenter excenter lemma to win

Just storage … Let the center of $\odot{AIB}=D$. Note that $\angle{BIA}=90+\frac{C}{2}$, therefore $\angle{BFA}=2(90-\frac{C}{2})=180-C$, so this lies on $\odot{ABC}$. The same holds for the others.

The shortest solution I have written......

Oh! USAMO 1997/2 was the shortest one....... Nevermind......

Storage

By Incenter-Excenter Lemma, the circumcenters of each triangle bisects segment $IE_k$, where $E_k$ is the excenter of the $k$-th vertex. By the lemma it also coincides with the midpoints of minor arcs $AB$, $BC$, and $CA$, so the three points lie on the circumcircle of $ABC$, proving our original assertion $\blacksquare$

Proof: Incenter excenter lemma tells us that the circumcenters of $A',B',C'$ are on $(ABC)$ and $(A'B'C')$ so they have the same center

By Fact 5, the circumcenters of these triangles are the midpoints of their respective arcs, so since they all lie on $(ABC)$, we're done.

first of all we denote the circumcenter of $\triangle{BIC},\triangle{AIC} , \triangle{AIB}$ as $O_{A},O_{B},O_{C}$ respectively. We notice $\angle{BO_{A}C}=180^{\circ}-A$, so $\angle{O_{A}BC}=\angle{O_{A}CB}=\frac{A}{2}$ Hence $O_{A}$ is midpoint of $\widehat{BC}$, so we have $O_{A}, O_{B},O_{C} \in \odot(ABC)$ and hence circumcenter of $\triangle{O_{A}O_{B}O_{C}}$ coincides with that of $\triangle{ABC}$ $\blacksquare$

By the Incenter/Excenter lemma, we have that the circumcenters of $\triangle IBC$, $\triangle ICA$, and $\triangle IAB$ are the midpoints of arcs $BC$, $CA$, and $AB$ on the circumcircle of $\triangle ABC$, respectively. Therefore, we have that $\odot ABC$ and $\odot A'B'C'$ are cocentric, as desired.

We know that $A'$, $B'$, and $C'$ are the midpoints of arcs $BC$, $CA$, and $AB$ through the Incenter-Excenter lemma. Hence the circumcircles of $ABC$ of $A'B'C'$ coincide. $\blacksquare$

By the Incentre/Excentre Lemma, we have that $A'$ is in fact the midpoint of $\widehat{BC}$ in the circumcircle of $\Delta ABC$. As such, $A', B', C'$ all lie on the circumcircle of $\Delta ABC$, proving the result. $\square$

It follows directly from the incenter excenter lemma

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