Quite nice but hard for a J1, in my opinion. Here is a solution with Danielle Wang, which repeatedly uses https://web.evanchen.cc/handouts/Fact5/Fact5.pdf. Let $M$ be the midpoint of arc $BC$ not containing $A$. We claim $M$ is the desired fixed point. Since $\angle MPA = 90^\circ$ and ray $PA$ bisects $\angle I_BPI_C$, it suffices to show that $MI_B = MI_C$ (by external version of Fact 5). Let $M_B$, $M_C$ be the second intersections of $PI_B$ and $PI_C$ with circumcircle. Now $M_BI_B = M_BB = M_CC = M_CI_C$ by Fact 5 again twice, also $MM_B = MM_C$, and $\angle I_BM_BM = \angle I_CM_CM$, so triangles $\triangle I_BM_BM \cong \triangle I_CM_CM$, done.

Do you think putting down that it passes through the midpoint of arc BC not containing A (and then scrawling down some other nonsense) might be worth a 1? Probably not, but I just want to check

Did anyone get this one? I was 3 hours in and I didn't get this one so I went for a walk

Wait I put that it passed through midpoint of arc, drew the Japanese rectangle, mentioned Fact 5, and that's it.

I solved #2 in about 30 minutes, then spent 4 hours working on this question while knowing that the midpoint of the arc BC is the point.

Extending $MI_B$ and $MI_C$ gives a nice complex solution. That's what I did during the test after spending a half hour trying to bash it with messy 20-term expressions.

How many points would I get if I did a bunch of cyclic quad stuff to determine we just need to prove $\angle I_BMI_C = \angle ABC$?

I thought the point was some other point inside the triangle for the whole time...

azmath333 wrote: How many points would I get if I did a bunch of cyclic quad stuff to determine we just need to prove $\angle I_BMI_C = \angle ABC$? 1 at best, since proving that is the bulk of the problem.

Me during test wrote: ok so as P gets closer and closer to B and C, the circumcircle that would be there intersects at the incenter and some other random point [which turns out to be the midpoint of the arc BC not including A], this random point is CLEARLY not where they intersect, therefore, the answer is the incenter!

wait same actually. spent 3 hours trying to complex bash a wrong statement. rip

Hm I thought that for a while but then I proved $\angle I_BII_C$ is right so then I went back to midpoint of arc.

That's what I thought when I looked at it in the beginning, but if you take the point where P=M, the only other time the circumcircle of $PI_BI_C$ intersects $AM$ (say at point X) is at a point that is not the incenter. Note that $I_BMI_C$ and $BIC$ where I is the incenter of ABC are supplementary, so I cannot intersect the circle

Also, interestingly the result seems to be true for any triangle $\triangle ABC$, not necessarily isosceles.

mathwizard888 wrote: Hm I thought that for a while but then I proved $\angle I_BII_C$ is right so then I went back to midpoint of arc. Ah, rats, I proved that $II_BBA$ and $II_CCA$ were cyclic (and then, obviously, $I_BII_C$ is right), but I denied it because I was so convinced that $I$ was the point the problem was looking for

Okay here are configuration things 1. Let $\triangle ABC$ be a triangle with circumcircle $\omega$. Let $P$ be a variable point on the arc $BC$ of $\omega$ not containing $A$. Call the incenters of $ABP$ and $ACP$ to be $I_1$ and $I_2$, respectively. Then (a) The circumcircle of $\triangle I_1PI_2$ passes through a fixed point. (b) Let $Q$ be the second intersection of $AP$ with the circumcircle of $\triangle I_1PI_2$. Then $AQ$ has a constant length. 2. Let $\triangle ABC$ be a triangle. Let $P$ be a variable point on side $BC$. Call the incenters of $ABP$ and $ACP$ to be $I_1$ and $I_2$, respectively. Then

Anyone know why this doesn't work: $A=1$ $M=-1$ $B=x^2$ $C=1/x^2$ $P=y^2$ $I_B=-xy-y-x$ $I_C=-1/x-y/x-y$ then http://bit.ly/1U5VcpW, which is not equal to its conjugate

You have sign issues on $I_B$ and $I_C$ depending on which $x$ and $y$ you use, so it's important to specify that.

Mine was quite convoluted, but I hope it's right...

Let $M$ be the midpoint of arc $BC$. Let $PI_B \cap \odot{ABC}=D, PI_C \cap \odot{ABC}=E$. By Fact 5, we know that $BD=DA$ and $AE=EC$. Furthermore, $\triangle{DAB} \cong \triangle{EAC}$ so we have $DB=EC=DA=AE$. We also have $\angle{DBM}=\angle{ECM}$ so $MD=ME$. $\angle{I_BDM}=\angle{PDM}=\angle{PEM}=\angle{I_CEM}$ so by congruency $\angle{I_BMD}=\angle{I_CME}$. To finish, notice that $\angle{DPE}=\angle{I_BDI_C}=\angle{DME}$. But we also have $\angle{I_BMD}=\angle{I_CME}$ so $\angle{DME}=\angle{I_BMI_C}$. This implies that $I_BI_CMP$ is cyclic, so we are done. Remarks: Extending $PI_C, PI_B$ was fairly natural for me since I consider not constructing $AI \cap \odot{ABC}$ a sin

Let $M$ be the midpoint of the arc $AC$ not containing $A$ and $M_B,M_C$ be the intersection of $PI_B,PI_C$ with $(ABC)$. Claim. We have $\triangle I_BM_BM\cong \triangle I_CM_CM$ because \begin{eqnarray*} MM_B&=&MM_C\\ \measuredangle MM_BP&=&\measuredangle MM_CP\\ M_BI_B&=&M_CI_C \hspace{5mm}\text{[fact 5]} \end{eqnarray*}So we have $$\measuredangle MI_BM_B=\measuredangle MI_CM_C\Longrightarrow \measuredangle MI_BP=\measuredangle MI_CP$$And we are done.$\square$

Let $M,M_B,$ and $M_C$ be the midpoints of arcs $BC,AC,$ and $AB,$ respectively. Notice $I_BM_C=I_CM_B$ by the Incenter-Excenter lemma, $MM_B=MM_C$ by symmetry, and $\angle MM_CP=\angle MM_BP,$ so $\triangle MI_BMC\cong\triangle MI_CM_B.$ Hence, $M$ is the Miquel point of $M_BM_CI_BI_C.$ $\square$

An inefficient solution. Slightly hard for JMO1, in my opinion. The circumcircle passes through the fixed arc midpoint $M$ of minor arc $\widehat{BC}$. Let $D, E$ denote the midpoints of minor $\widehat{BP}$, $\widehat{CP}$ respectively. Claim. The triangles $I_BDM$ and $MEI_C$ are congruent. Proof. $\overline{DI_B}$ and $\overline{EI_C}$ pass through $A$ by Fact 5. Yet $M$ is the $A$-antipode, so $\angle ADM = \angle AEM = 90^\circ$. Next, by an easy ``arc chase" we obtain $\widehat{DP} = \widehat{ME}$ and $\widehat{DM} = \widehat{EC}$. Thus $I_BD=EM$ and $DM=I_CE$ by Fact 5, and the claim is proven by SAS. $\blacksquare$ As a result, $I_BMI_C$ is isosceles, and by the converse of Fact 5 it suffices to show that $\overline{PM}$ bisects the external $\angle I_BPI_C$. This is just angle chasing: $\angle I_BPI_C = \angle B$, while $$\angle I_CPM = \angle APM - \angle I_CPA = 90^\circ - \frac 12 \angle B.$$So $\angle I_BPI_C + 2\angle I_CPM = 180^\circ$, and we are done.

Sorry for the bump, but I've got a quick question, I've never really done too much olympiad geo even though I'm decent at AIME Geo, but how do you actually draw the figure in order to even hypothesize that the intersection point is the middle of the arc? I thought the fixed point was the incenter and spent ~30 minutes trying to prove it via angle chasing to no avail (Tried to prove $II_CI_BP$ was cyclic but ended up with opposite angles adding to more than 180). Just curious if there's some insight on how someone can draw figures accurately enough to deduce that the midpoint of the arc is the correct locus.

I claim that the desired fixed point is $M$, the midpoint of minor arc $BC$ on $(ABC)$, clearly fixed. Let $M_B, M_C$ be the midpoints of minor arcs $CA$, $AB$ on $(ABC)$. Clearly by symmetry, we have that $MM_B = MM_C$. Also, $P, I_B, M_C$, and $P, I_C, M_B$ are collinear, so we have $\angle MM_CI_B = \angle MM_CP = \angle MM_BP = \angle MM_BI_C$. By Fact 5, we have $M_CI_B = AM_C = BM_C,$ and $M_BI_C = AM_B = CM_B,$ and by symmetry, we have $AM_C = BM_C$, so $M_CI_B = M_BI_C$, so $\triangle MM_CI_B \cong \triangle MM_BI_C$, which gives $\angle MI_BM_C = \angle MI_CI_B$, and thus $\angle MI_BP = \angle MI_CP$, which gives the desired conclusion.

We claim the desired fixed point is $M$, the midpoint of arc $BC$ not containing $A$. Define $M_B$ and $M_C$ as the midpoints of arcs $AB$ and $AC$, respectively. Incenter-Excenter Lemma and our isosceles condition tells us \[M_BB = M_BI_B = M_CC = M_CI_C.\] Then we have $\triangle MM_BI_B \cong \triangle MM_CI_C$ from SAS, so \[\angle I_BMI_C = \angle M_BMM_C = \angle I_BPI_C \implies I_BI_CPM \text{ cyclic}. \quad \blacksquare\]

Kind of irritating. Let $M_A$, $M_B$, $M_C$ be the arc midpoints of $\widehat{BC}$, $\widehat{AB}$ and $\widehat{AC}$. We claim $M_A$ is the fixed point. By Incenter-Excenter Lemma we know that, \begin{align*} M_BI_B = M_BA = M_CA = M_CI_C \end{align*}and paired with the fact that $M_AM_B = M_AM_C$ we know that $\triangle M_AM_BI_B \sim M_AM_CI_C$ with ratio $1$. This forces $M_AI_B = M_AI_C$. Now this means $M_A$ is the Miquel point of $I_BM_BM_CI_C$ and hence $M_A \in (I_BI_CP)$ and so we're done.

Here's the complex solution. Align the real axis with the perpendicular bisector of $\overline{BC}$, and let $(ABC)$ be the unit circle. I claim that there exist complex numbers $y$, $z$, and $t$, such that $b = y^2$, $c = z^2$, $p = t^2$, and the incenters of $ABP$ and $ACP$ are given by $-(y+yt + t)$ and $-(z - zt - t)$ respectively, and the arc midpoint of minor arc $\widehat{BC}$ is given by $-yz$. Suppose that $A, B, C$ appear in counterclockwise order. Then for $y = \exp \theta_1$ and $z = \exp \theta_2$, let $y = \exp \frac{\theta_1}2$ and $z = \exp \left(\frac{\theta_2}2 + \pi\right)$. Then as $a = 1$, we can check that the arc midpoints of triangle $ABC$ are indeed given by $-y, -z$, and $-yz$. Furthermore, for $p = \exp \theta$, setting $t_1 = \exp\left(\frac \theta2 + \pi\right)$ and $t_2 = \exp\left(\frac{\theta_2}2\right)$ yields the arc midpoints $-y, -t_1, -yt_1$ for triangle $ABP$ and similar for triangle $ACP$. This is enough to prove the result by the incenter lemma. Now, it suffices to show that $-1 = -yz, t^2, -(y+yt+t), -(z-zt-t)$ are concyclic, or the expression $$\frac{t^2+y+yt+t}{t^2+z-zt-t} \div \frac{1+y+yt+t}{1+z-zt-t} = \frac{(t+y)(z-1)}{(t-z)(y+1)}$$is real. This is clear as the expression equals its conjugate.

Let $M_A$, $M_B$, $M_C$ denote the midpoints of arcs $\widehat{BC}, \widehat{AC}, \widehat{AB}$, respectively, not containing the third point. I claim that $M_A$ is the desired fixed point. Claim: $\triangle M_AI_BM_B \cong \triangle M_AI_CM_C$ Proof: By fact $5$, we have $$M_BI_B = M_BA = M_CA = M_CI_C.$$Also, note that $M_AM_B = M_AM_C$ by symmetry, and $$\angle M_AM_BI_B = \angle M_AM_BP = \angle M_AM_CP = \angle M_AM_CI_C,$$so the claim follows by SAS. Then, note that $\overline{M_AP} \perp \overline{PA}$ and $$\angle I_BPA = \tfrac{1}{2} \angle APB = \tfrac{1}{2}\angle APC = \angle I_CPA,$$so $M_A$ lies on the external bisector of $\angle I_BPI_C$. Combined with $M_AI_B = M_AI_C$, this implies that it must be the midpoint of arc $\widehat{I_BI_C}$ containing $P$ of $(PI_BI_C)$, which finishes.

How would people know that the point is actually the midpoint of arc BC? I thought that it was the center of the big circle at first and spent nearly an hour trying to prove it.

chenghaohu wrote: How would people know that the point is actually the midpoint of arc BC? I thought that it was the center of the big circle at first and spent nearly an hour trying to prove it. good intuition ok just kidding. while i feel that intuition does play a role in these guessing points types of problems, i think that choosing said midpoint is most motivated by looking at the incenters and remembering the incenter-excenter lemma. hopes this helps.

Let $M_1$, $M_2$, $M_3$ be the minor arc midpoints of $BC$, $AC$, and $AB$. We claim that $M_1$ is the desired fixed point on the circumcircle, so it suffices to show that $M_1$ is the Miquel point of quadrilateral $M_3I_BI_CM_2$. We can apply Fact $5$ repeatedly to get $M_3I_B = M_3A = M_2A = M_2I_C$ and clearly $M_1M_3 = M_2$ and $\angle M_1M_3I_B = M_1M_2I_C$. Hence, $M_1$ is the center of the spiral similarity sending $M_3I_B$ to $M_2I_C$ as desired.

Let the arc midpoints of $BC$, $CA$ and $AB$ be $M_A$, $M_B$ and $M_C$. We claim the fixed point is $M_A$. Claim: We have $\triangle M_A I_B M_C \cong \triangle M_A I_C M_B$. Proof: This is by SAS congruence: we have that $MM_C = MM_B$, that $\angle I_B M_C M = \angle I_C M_B M$ and that \[I_B M_C = M_C A = A M_B = I_C M_B.\] So, $\angle PI_B M = \angle PI_C M$, proving that $PI_B I_C$ passes through $M$.

Invert at $A$, so the problem becomes: given isosceles $\triangle ABC$ and variable $P$ on segment $BC$, if $I_B,I_C$ are the $A$-excenters of $\triangle ABP,\triangle ACP$, then $(PI_BI_C)$ passes through a fixed point. We claim it is the midpoint $M$ of $BC$. Note $\angle I_BPI_C=90^\circ$ so it suffices to show $\angle I_BMI_C=90^\circ$. We use moving points, varying $P$ such that line $AI_B$ and $AI_C$ have degree $1$, which is possible since they are rotations by a fixed angle. Then $I_B,I_C$ have degree $1$, so lines $MI_B,MI_C$ have degree $1$, so $e^{2i\measuredangle I_BMI_C}$ has degree $2$. To show it is $-1$ always, we only need to check $3$ cases: If $P=B$ it is clear since $I_B=B$ and $I_C$ lies on the perpendicular bisector of $BC$. Similarly $P=C$ works. Finally, if $P=M$ then $\measuredangle I_BMI_C=\measuredangle I_BPI_C=90^\circ$. We are done.