Fixed point as P varies

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2376 posts

#1 h Apr 19, 2016, 9:32 PM • 10 Y

Y by AlastorMoody, rart-of-problem-solving, mathematicsy, megarnie, Jc426, HWenslawski, jhu08, Adventure10, Mango247, GeoKing

The isosceles triangle $\triangle ABC$ , with $AB=AC$ , is inscribed in the circle $\omega$ . Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$ , and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$ , respectively.

Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point.

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v_Enhance

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v_Enhance

#2 h Apr 19, 2016, 9:33 PM • 30 Y

Y by mathtastic, InCtrl, SHIKHARS, Heisenberg09, kk108, JasperL, Ultroid999OCPN, Wizard_32, AlastorMoody, CALCMAN, Kanep, tree_3, myh2910, HamstPan38825, math31415926535, 554183, Jc426, HWenslawski, jhu08, megarnie, rayfish, EpicBird08, Adventure10, Mango247, Sedro, Mogmog8, gvccimac_08, NicoN9, MS_asdfgzxcvb, Rounak_iitr

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.

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This post has been edited 3 times. Last edited by v_Enhance, Mar 16, 2024, 7:17 PM
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MathSlayer4444

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MathSlayer4444

#3 h Apr 19, 2016, 9:35 PM • 2 Y

Y by Adventure10, Mango247

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

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mathtastic

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mathtastic

#4 h Apr 19, 2016, 9:35 PM • 2 Y

Y by Adventure10, Mango247

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

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Phie11

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Phie11

#5 h Apr 19, 2016, 9:36 PM • 2 Y

Y by Adventure10, Mango247

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

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sunny2000

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sunny2000

#6 h Apr 19, 2016, 9:36 PM • 2 Y

Y by chezbgone, Adventure10

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.

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mathwizard888

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mathwizard888

#7 h Apr 19, 2016, 9:37 PM • 1 Y

Y by Adventure10

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.

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azmath333

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azmath333

#8 h Apr 19, 2016, 9:47 PM • 2 Y

Y by Adventure10, Mango247

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

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dtxiong

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dtxiong

#9 h Apr 19, 2016, 9:49 PM • 3 Y

Y by Frestho, Adventure10, Mango247

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

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mathwizard888

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mathwizard888

#10 h Apr 19, 2016, 9:50 PM • 2 Y

Y by Adventure10, Mango247

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.

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MathSlayer4444

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MathSlayer4444

#11 h Apr 19, 2016, 9:57 PM • 2 Y

Y by Adventure10, Mango247

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!

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jam10307

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jam10307

#12 h Apr 19, 2016, 10:00 PM • 2 Y

Y by Adventure10, Mango247

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

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mathwizard888

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mathwizard888

#13 h Apr 19, 2016, 10:00 PM • 2 Y

Y by Adventure10, Mango247

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.

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mathgenius64

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mathgenius64

#14 h Apr 19, 2016, 10:01 PM • 1 Y

Y by Adventure10

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

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mathtastic

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mathtastic

#15 h Apr 19, 2016, 10:04 PM • 2 Y

Y by Adventure10, Mango247

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

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HamstPan38825

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HamstPan38825

#75 h Jan 23, 2022, 7:13 PM

Y by

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.

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Brudder

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Brudder

#76 h Aug 9, 2022, 9:24 PM

Y by

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.

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mannshah1211

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mannshah1211

#77 h Jan 17, 2024, 12:57 PM

Y by

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.

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shendrew7

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shendrew7

#78 h Feb 11, 2024, 8:16 PM

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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
$\angle I_BMI_C = \angle M_BMM_C = \angle I_BPI_C \implies I_BI_CPM \text{ cyclic}. \quad \blacksquare$

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Shreyasharma

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#79 h Feb 23, 2024, 4:08 AM • 1 Y

Y by dolphinday

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.

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HamstPan38825

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HamstPan38825

#80 h Mar 11, 2024, 8:47 PM

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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.

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eibc

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eibc

#81 h Mar 16, 2024, 2:24 AM

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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.

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chenghaohu

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chenghaohu

#82 h Jun 5, 2024, 2:39 AM

Y by

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.

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bachkieu

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#83 h Jun 5, 2024, 2:46 AM

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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.

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dolphinday

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dolphinday

#84 h Jul 12, 2024, 3:55 AM

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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.

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ihatemath123

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#86 h Aug 15, 2024, 4:04 PM

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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 $M$ .

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OronSH

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OronSH

#87 h Dec 26, 2024, 5:05 AM • 1 Y

Y by dolphinday

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.

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bjump

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bjump

#88 h Mar 3, 2025, 10:59 PM • 1 Y

Y by imagien_bad

I am banned. : BUT THIS PROBLEM IS MOR BANNE CUZ TRIV COMPLECZK) )) )

I claim that the arc midpoint call it $D$ of $BC$ lies on this circle. Now toss onto the complex plane with $(ABC)$ as the unit circle, $P=p^2$ , $A=1$ , $B=a^2$ , $C=a^{-2}$ , Then $I_B = -ap -a -p$ , $I_C = -pa^{-1}-a^{-1}-p$ . It suffices to czech the following is real:
$\frac{(p^2+ap+a+p)(pa^{-1}+a^{-1}+p+1)}{(p^2+pa^{-1}+a^{-1}+p)(ap+a+p+1)}= \frac{(p+1)(p+a)(p+a^{-1})(a^{-1}+1)}{(p+1)(p+a^{-1})(p+1)(a+1)} = \frac{(a+p)(a^{-1}+1)}{(a+1)(a^{-1}+p)}$ Which is real bc $a^{-1}$ , $-1$ , $a$ , and $p$ all lie on the unit circle. Done.

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imagien_bad

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imagien_bad

#89 h Mar 10, 2025, 5:04 PM

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bjump wrote:

I am banned. : BUT THIS PROBLEM IS MOR BANNE CUZ TRIV COMPLECZK) )) )

I claim that the arc midpoint call it $D$ of $BC$ lies on this circle. Now toss onto the complex plane with $(ABC)$ as the unit circle, $P=p^2$ , $A=1$ , $B=a^2$ , $C=a^{-2}$ , Then $I_B = -ap -a -p$ , $I_C = -pa^{-1}-a^{-1}-p$ . It suffices to czech the following is real:
$\frac{(p^2+ap+a+p)(pa^{-1}+a^{-1}+p+1)}{(p^2+pa^{-1}+a^{-1}+p)(ap+a+p+1)}= \frac{(p+1)(p+a)(p+a^{-1})(a^{-1}+1)}{(p+1)(p+a^{-1})(p+1)(a+1)} = \frac{(a+p)(a^{-1}+1)}{(a+1)(a^{-1}+p)}$ Which is real bc $a^{-1}$ , $-1$ , $a$ , and $p$ all lie on the unit circle. Done.

YOUR BANNED. No BASH ALLOWED, Im TELLING YOUR MOMMY

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ErTeeEs06

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ErTeeEs06

#90 h Mar 30, 2025, 7:38 PM • 1 Y

Y by Funcshun840

The condition $AB=AC$ doesn't matter. I'll prove the general version with $ABC$ just a random triangle. Let $T$ be the point on $(ABC)$ where the $A$ -mixtilinear incircle is tangent. I claim that $T$ is the desired fixed point. We see $\angle I_bPI_c=\frac{1}{2}\angle BPC=\frac{1}{2}\angle BTC$ , so we want to show $\angle I_bTI_c=\frac{1}{2}\angle BTC$ . It is a well-known fact that $\triangle TBI\sim \triangle TIC$ so there is a spiral similarity with center $T$ taking $BI$ to $IC$ . We have $\angle BI_bA=90^\circ+\frac{1}{2}BPA=90^\circ+\frac{1}{2}BCA=\angle BIA$ , so $I_b$ is on $(ABI)$ and similarly $I_c$ is on $(ACI)$ . I claim that triangles $BII_b$ and $ICI_c$ are similar. Angle chase $\angle BII_b=\angle BAI_b=\frac{1}{2}\angle BAP=\frac{1}{2}\angle BAC-\frac{1}{2}\angle PAC=\angle IAC-\angle I_cAC=\angle IAI_c=\angle ICI_c$ and the symmetrical equality implies $BII_b$ and $ICI_c$ are similar. Now this means that the spiral similarity at $T$ sending $BI$ to $IC$ sends $I_b$ to $I_c$ . Therefore $\angle BTI_b=\angle ITI_c$ and $\angle I_bTI_c=\angle BTI=\frac{1}{2}\angle BTC$ and we are done.

This post has been edited 1 time. Last edited by ErTeeEs06, Mar 30, 2025, 7:39 PM
Reason: typo

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