Let $I_A$ be the $A-$excenter, $K$ is mindpoint of the $II_a$ by the nine point circle, $\triangle I_BII_A\sim \triangle I_BBC$ and $\triangle I_CII_A \sim \triangle I_CBC \Rightarrow \angle I_BKA=\angle I_BPC$ and $\angle I_CKA=\angle I_CPB$ so $PQI_BI_C$ is cyclic.

Nice config geo! Let $D=AK \cap BC$, $I_A$ be the $A$-excenter, $M$ be the midpoint of the major arc $BC$ and $I_BI_C \cap BC=X$. We have that $AMPD$ and $BCI_BI_C$ are cyclic, so spamming PoP gives $XD.XP=XA.XM=XB.XC=XI_C.XI_B$, so $I_BI_CDP$ is cyclic. We want to prove that $Q$ lies on that circle. Indeed, we want $AI_C.AI_B=AQ.AD=AD.AK \iff$ $D$ is orthocenter for $\triangle I_BI_CK$, which follows by Brokard for the cyclic quadrilateral $IBI_AC$ (since $K$ is its center).

Not that hard but yeah... Nice problem for a NUB like me. Anyways moving onto the solution,

Let $D = AK \cap BC$, $E = I_B I_C$. Also, trivial to notice that $I_BI_CBC$ is cyclic. Now,$$AQ \cdot AD = AK \cdot AD \stackrel{\dagger}{=} (\sqrt{bc})^2 = AB \cdot AC \stackrel{\ddagger}{=} AI_B \cdot AI_C \implies I_B I_C DQ\text{ is cyclic}.$$

. And, $$EI_B \cdot EI_C = EB \cdot EC \stackrel{\star}{=} ED \cdot EP \implies I_B I_C PD \text{ is cyclic}.$$ $\star$ is just EGMO Lemma 9.17 as it's well known $(E, D; B, C) = -1$. Both these together imply that $I_B I_C PDQ$ is cyclic.$\blacksquare$ $\textbf{Remark:}$ This might've been well known for $\sqrt{bc}$ Inversion that $I_B \xleftrightarrow{} I_C$, and due to not being in touch with geo for so long I had forgotten the result and had to prove it myself .

Rafinha wrote: Let $I_A$ be the $A-$excenter, $K$ is mindpoint of the $II_a$ by the nine point circle, $\triangle I_BII_A\sim \triangle I_BBC$ and $\triangle I_CII_A \sim \triangle I_CBC \Rightarrow \angle I_BKA=\angle I_BPC$ and $\angle I_CKA=\angle I_CPB$ so $PQI_BI_C$ is cyclic. nice solution!

My solution is also posted above I just wanted to post this since I think it is a very cool one! 2022 Brazilian National Mathematical Olympiad - Problem 2 wrote: Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic. By some angle chasing we get: $$\angle I_AII_B=\angle BCI_B\land\angle II_AI_B=\angle CBIB_B\implies \Delta II_AI_B\cup\{ K\}\sim\Delta CBI_B\cup\{P\}\implies\angle IKI_B=\angle CPI_B$$By a similar argument, we can get $\angle IKI_C=\angle BPI_C$.Thus, we have $$\angle I_CQI_B=\angle I_CKI_B=\angle CPI_B+\angle BPI_C=180^\circ-\angle I_CPI_B\implies P,Q,I_B,I_C \text{ are concyclic.}\blacksquare$$

Let $L$ be the midpoint of arc $BAC$ and $X=I_BI_C\cap BC$. Let $K'$ be the reflection of $K$ over $L$. Then, $\angle DAL=\angle DPL=90^\circ$, so $ADLP$ is cyclic. Thus, by say Power of a Point at $K$, $PDQK'$ is also cyclic. In addition, Power of a Point gives $XD\cdot XP=XA\cdot XL=XB\cdot XC=XI_C\cdot XI_B$ so $I_BI_CPD$ is cyclic. Note $D$ is the orthocenter of $I_BI_CK$. Note clearly $KA\perp I_BI_C$ (note $I_B,I_C,A,L$ are collinear) and $\measuredangle DI_CI_B=\measuredangle DPI_B=\measuredangle CPI_B$. In addition, if $I_A$ is the $A$-excenter, $II_AI_B\sim CBI_B$, so the midpoints are corresponding. In particular, $\measuredangle CPI_B=\measuredangle I_BKI=90^\circ-\measuredangle I_CI_BK$. Thus, $I_CD\perp I_BK$ and vice versa, so $D$ is the orthocenter. This implies that the reflection of $(I_BI_CK)$ is $(I_BI_CD)$, so the reflection of $K$ over $I_BI_C$ lies on $(I_BI_CD)$, as desired.

Solution from Twitch Solves ISL: Let $I_A$ be the $A$-excenter, and let $H = \overline{AIK} \cap \overline{BC}$. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-3.81664,2.88806); pair B = (-3.5,-0.5); pair C = (1.5,-0.5); pair I = (-2.45079,0.65183); pair K = (-1.,-1.72347); pair P = (-1.,-0.5); pair Q = (-6.63328,7.49959); pair I_B = (3.85362,7.57292); pair I_C = (-5.85362,1.64391); pair I_A = (0.45079,-4.09877); pair H = (-1.74727,-0.5); pair T = (-9.36373,-0.5); import graph; size(9cm); pen zzttqq = rgb(0.6,0.2,0.); pen qqffff = rgb(0.,1.,1.); pen qqwuqq = rgb(0.,0.39215,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw(I_A--I_B--I_C--cycle, linewidth(0.6) + qqffff); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw(circle((-1.,1.44247), 3.16594), linewidth(0.6)); draw(circle((-1.37363,5.22015), 5.73234), linewidth(0.6)); draw(I_A--I_B, linewidth(0.6) + qqffff); draw(I_B--I_C, linewidth(0.6) + qqffff); draw(I_C--I_A, linewidth(0.6) + qqffff); draw(Q--I_A, linewidth(0.6)); draw(B--I_B, linewidth(0.6) + blue); draw(C--I_C, linewidth(0.6) + blue); draw(I_B--T, linewidth(0.6) + qqwuqq); draw(T--B, linewidth(0.6) + qqwuqq); dot("$A$", A, dir((-14.232, 52.186))); dot("$B$", B, dir((-41.152, -54.958))); dot("$C$", C, dir((13.913, -43.888))); dot("$I$", I, dir((5.740, 13.299))); dot("$K$", K, dir((6.148, 12.041))); dot("$P$", P, dir(110)); dot("$Q$", Q, dir((-50.427, 29.076))); dot("$I_B$", I_B, dir((6.270, 12.255))); dot("$I_C$", I_C, dir((-9.789, 50.090))); dot("$I_A$", I_A, dir((-12.421, -42.985))); dot("$H$", H, dir(225)); dot("$T$", T, dir((-60.449, -26.493))); [/asy][/asy] Then by Brokard's theorem on cyclic quadrilateral $BICI_A$, it follows that $\triangle I_B I_C H$ is self-polar to this circle. In particular $K$ is the orthocenter. Equivalently, $H$ is the orthocenter of $\triangle I_B K I_C$. Define $T = \overline{I_B I_C} \cap \overline{BC}$. Then $(TH;BC) = -1$. I claim that $I_B I_C P H$ is cyclic. Indeed, $TH \cdot TP = TB \cdot TC = TI_B \cdot I_C$. The reflection of the orthocenter $K$ of $\triangle I_BI_CH$ over a side $\overline{I_B I_C}$ then coincides with $Q$, which now lies on $(PHI_BI_C)$.

rcorreaa wrote: Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic. Can't believe no one has posted this sol yet. We bary with respective triangle $ABC$ and set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. First, let's find $K$. Note that $K$ lies on $AI$ where $I$ is the incenter so $K=(x:b:c)$. Moreover, it lies on $(ABC)$ so $$a^2bc+b^2cx+c^2xb=0\implies x=-\frac{a^2}{b+c}\implies K=(-\frac{a^2}{b+c}:b:c).$$Now, let's find $Q$. We know $K=(-\frac{a^2}{b+c}:b:c)$ and $A=(b+c-\frac{a^2}{b+c}:0:0)$. Thus, $$Q=2A-K=(2(b+c)-\frac{a^2}{b+c}:-b:-c)=(2(b+c)^2-a^2:-b(b+c):-c(b+c)).$$And lastly, we know $P=(0:1:1),I_B=(a:-b:c),I_C=(a:b:-c)$. Let's find the equation of $(PI_BI_C)$. $$(x,y,z)\mapsto (0:1:1)\implies -a^2+2(v+w)=0\implies v+w=\frac{a^2}{2}$$$$(x,y,z)\mapsto (a:-b:c)\implies a^2bc-b^2ca+c^2ab+(ua-vb+wc)(a-b+c)=0\implies abc+ua-vb+wc=0$$$$(x,y,z)\mapsto (a:b:-c)\implies a^2bc+b^2ca-c^2ab+(ua+vb-wc)(a+b-c)=0\implies abc+ua+vb-wc=0$$The second plus the third equation gives $2abc+2ua=0\implies u=-bc$. Plug this back to the second equation to get $vb=wc$, so $w=\frac{vb}{c}$. Plug this into the first equation, and we'll have $v+\frac{vb}{c}=\frac{a^2}{2}\implies v=\frac{a^2c}{2(b+c)}$, so $w=\frac{a^2b}{2(b+c)}$. $$(PI_BI_C): -a^2yz-b^2zx-c^2xy+\left(-bcx+\frac{a^2c}{2(b+c)}y+\frac{a^2b}{2(b+c)}z\right)(x+y+z)=0.$$Finally, we plug $(x,y,z)\mapsto (2(b+c)^2-a^2:-b(b+c):-c(b+c))=Q$. $$-a^2yz-b^2zx-c^2xy=-a^2bc(b+c)^2+b^2c(b+c)(2(b+c)^2-a^2)+bc^2(b+c)(2(b+c)^2-a^2)=2bc(b+c)^2((b+c)^2-a^2)$$$$-bcx+\frac{a^2c}{2(b+c)}y+\frac{a^2b}{2(b+c)}z=(-2bc(b+c)^2+a^2bc)-\frac{a^2bc}{2}-\frac{a^2bc}{2}=-2bc(b+c)^2$$$$x+y+z=(2(b+c)^2-a^2)-b(b+c)-c(b+c)=(b+c)^2-a^2$$Hence, we're done.

Rename $K$ to $M_A$. Suppose $(ABC)$ is the unit circle, and let $A=x^2$, etc., so $M_A=-yz$, etc., and $I=-xy-yz-zx$. By incenter-excenter we then have $I_B=-xy+xz+zy$ and $I_C=-xz+yx+yz$. Furthermore, $P=\frac{y^2+z^2}{2}$ and $Q=2x^2+yz$. We then wish to prove that $$\frac{-xz+yx+yz-\frac{y^2+z^2}{2}}{-xz+yx+yz-(2x^2+yz)} \div \frac{-xy+xz+zy-\frac{y^2+z^2}{2}}{-xy+xz+zy-(2x^2+zy)} \in \mathbb{R}.$$We can simplify the expression to $$\frac{2x(y-z)-(y-z)^2}{x(-z+y-2x)} \div \frac{-2x(y-z)-(y-z)^2}{x(-y+z-2x)}=\frac{2x-y+z}{-2x+y-z} \div \frac{-2x-y+z}{-2x-y+z}=-1,$$which of course is real, so we are done. $\blacksquare$

Beautiful problem! I liked the configuration Let $M$ be the midpoint of $I_BI_C$, then $M$ is on $(ABC)$ because it is nine-point circle of $I_AI_BI_C$ Let $K'$ be the reflection of $K$ with respect to $A$ Claim 1:$I_CQK'I_B$ is cyclic. Proof: Notice $\angle I_CKI_B= \angle I_CQI_B$ because of the reflection and $KA \perp I_BI_C$, then we observe $I_CKI_BK'$ is parallelogram as diagonals bisect each other, hence $\angle I_CKI_B= \angle I_CK'I_B$, hence the claim. Claim 2:The problem itself Proof: Note that by Claim 1 it suffices to prove $I_CK'I_BP$ is cyclic, $$\angle I_CK'P = \angle I_CI_BP$$$$\angle K'KI_B=I_CI_BP$$$$MP*MK = MI_B^2$$$$MP^2+MP*PK=MI_B^2$$$$MP^2+PC^2=MI_B^2$$$$MC=MI_B$$which is true since $I_BCI_C=90^\circ$ and $M$ is midpoint. To make more readable: here's diagram

Solved with mueller.25, starchan, Siddharth03, AdhityaMV Let $AK$ intersect segment $BC$ at $R$ and let $I_BI_C$ intersect line $BC$ at $T$. Let $N$ be the midpoint of major arc $BC$. Then note that $(N,K;B,C) = -1$ so projecting through $A$, we get that $(T,R;B,C) = -1$ as well. This means that since $P$ is the midpoint of $BC$ and $(BCI_BI_C)$, we have $TR \cdot TP = TB \cdot TC = TI_B \cdot TI_C$ so points $P,R,I_B,I_C$ are concyclic. Let $I$ be the incenter of $\triangle ABC$. Note that $(A,R;I,I_A) = -1$ and $K$ is the midpoint of $II_A$. So, $AQ \cdot AR = AK \cdot AR = AI \cdot AI_A = AI_B \cdot AI_C$, implying that points $R,Q,I_B,I_C$ are concyclic. Together, these imply that points $P,Q,I_B,I_C$ are concyclic, as desired. $\blacksquare$

IAmTheHazard wrote: Rename $K$ to $M_A$. Suppose $(ABC)$ is the unit circle, and let $A=x^2$, etc., so $M_A=-yz$, etc., and $I=-xy-yz-zx$. By incenter-excenter we then have $I_B=-xy+xz+zy$ and $I_C=-xz+yx+yz$. Furthermore, $P=\frac{y^2+z^2}{2}$ and $Q=2x^2+yz$. We then wish to prove that $$\frac{-xz+yx+yz-\frac{y^2+z^2}{2}}{-xz+yx+yz-(2x^2+yz)} \div \frac{-xy+xz+zy-\frac{y^2+z^2}{2}}{-xy+xz+zy-(2x^2+zy)} \in \mathbb{R}.$$We can simplify the expression to $$\frac{2x(y-z)-(y-z)^2}{x(-z+y-2x)} \div \frac{-2x(y-z)-(y-z)^2}{x(-y+z-2x)}=\frac{2x-y+z}{-2x+y-z} \div \frac{-2x-y+z}{-2x-y+z}=-1,$$which of course is real, so we are done. $\blacksquare$ oh Let $D$ and $E$ be the intersections of the interior and exterior $\angle A$-bisectors with $\overline{BC}$ respectively. By $\sqrt{bc}$ inversion we have $$AI_B\cdot AI_C=AB\cdot AC=AD\cdot AK=AD\cdot AQ,$$hence $I_BI_CQD$ is cyclic. On the other hand, it is well-known (extraversion of incenter/excenter) that $I_BI_CBC$ is cyclic, so $EI_B\cdot EI_C=EB\cdot EC$. Furthermore, since $(B,C;D,E)=-1$, we have $EB\cdot EC=ED\cdot EP$, hence $I_BI_CPD$ is cyclic by power of a point, so we're done. $\blacksquare$

Let $H$ be the intersection of $AK$ and $BC$. We first show that $QI_CHI_B$ is cyclic. First as $\triangle I_CAC \sim \triangle I_BAB$, we see that \[\frac{I_CA}{AC} = \frac{AB}{I_BA} \implies I_CA \cdot I_BA = AB \cdot AC.\] Secondly, as $\triangle ABK \sim \triangle ACH$, we again see \[\frac{AK}{AC} = \frac{AB}{AH} \implies AK \cdot AH = AB \cdot AC.\] So, $$QA \cdot AH = KA \cdot AH = BA \cdot AC = I_CA \cdot AI_B$$ implying the claim. Now let $J$ be the midpoint of $I_BI_C$. Note that $J$ lies on $(ABC)$ as it is the nine point circle of $\triangle I_AI_BI_C$. Furthermore $J$ is the arc midpoint of arc $BC$ containing $A$. That means $PHAJ$ is cyclic. Now, letting $I_CI_B$ intersect $BC$ at $X$, we see \[HX \cdot XP = AX \cdot XJ = BX \cdot XC = I_CX \cdot XI_B\] finishing the problem.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -5.803255101716253, xmax = 20.94508689023541, ymin = -2.1002703047388525, ymax = 22.319316398917017; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((7.575733531948671,10.863916969703546)--(5,5)--(12,5)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((7.575733531948671,10.863916969703546)--(5,5), linewidth(2)); draw((5,5)--(12,5), linewidth(2)); draw((12,5)--(7.575733531948671,10.863916969703546), linewidth(2)); draw(circle((8.5,6.960275942712022), 4.011568492694037), linewidth(2)); draw((1.6247981902196311,10.169020646842817)--(8.970521844693746,-1.0807354791321093), linewidth(2)); draw((8.970521844693746,-1.0807354791321093)--(15.375201809780371,11.7746682239693), linewidth(2)); draw((1.6247981902196311,10.169020646842817)--(15.375201809780371,11.7746682239693), linewidth(2)); draw(circle((8.380234318788622,11.99749071039707), 6.998515561209342), linewidth(2)); draw((7.575733531948671,10.863916969703546)--(8.970521844693746,-1.0807354791321093), linewidth(2)); draw((1.6247981902196311,10.169020646842817)--(12,5), linewidth(2)); draw((5,5)--(15.375201809780371,11.7746682239693), linewidth(2)); draw((15.375201809780371,11.7746682239693)--(8.5,5), linewidth(2) + linetype("2 2") + qqwuqq); draw((8.5,2.9487074500179844)--(15.375201809780371,11.7746682239693), linewidth(2) + linetype("2 2") + zzttqq); draw((8.5,5)--(1.6247981902196311,10.169020646842817), linewidth(2) + linetype("2 2") + qqwuqq); draw((1.6247981902196311,10.169020646842817)--(8.5,2.9487074500179844), linewidth(2) + linetype("2 2") + zzttqq); /* dots and labels */ dot((7.575733531948671,10.863916969703546),dotstyle); label("$A$", (7.716463099778057,11.193041132615535), NE * labelscalefactor); dot((5,5),dotstyle); label("$B$", (5.128957223893979,5.338809088427837), NE * labelscalefactor); dot((12,5),dotstyle); label("$C$", (12.11522308878099,5.338809088427837), NE * labelscalefactor); dot((8.5,5),linewidth(4pt) + dotstyle); label("$P$", (8.622090156337485,5.274121441530735), NE * labelscalefactor); dot((8.5,2.9487074500179844),linewidth(4pt) + dotstyle); label("$K$", (8.622090156337485,3.2041167408234825), NE * labelscalefactor); dot((1.6247981902196311,10.169020646842817),linewidth(4pt) + dotstyle); label("$I_C$", (1.7651995852446765,10.416789369850315), NE * labelscalefactor); dot((8.970521844693746,-1.0807354791321093),linewidth(4pt) + dotstyle); label("$I_A$", (9.107247508065749,-0.8065173667968194), NE * labelscalefactor); dot((15.375201809780371,11.7746682239693),linewidth(4pt) + dotstyle); label("$I_B$", (15.511324550878843,12.033980542277858), NE * labelscalefactor); dot((6.651467063897343,18.779126489389107),linewidth(4pt) + dotstyle); label("$Q$", (6.7784922197700785,19.052590230613387), NE * labelscalefactor); dot((8.029478155306254,6.978150379168081),linewidth(4pt) + dotstyle); label("$I$", (8.169276628057771,7.247094671892335), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $I_A$ be the $A-$excenter. Note that $\angle I_CKI_B=\angle I_CQI_B=90^{\circ}$ this is because $Q$ is the reflection of $K$ over $A$, so we have to show that $90^{\circ}=\angle I_CDI_B$, so we need to show that $\angle I_BI_CD+\angle I_CI_BD=90^{\circ}$. Now observe that $K$ is midpoint of the $II_a$ because the circumcircle of the orthic triangle is nine point circle of the reference triangle. Note that $II_CBA$ and $II_ABC$ are cyclic, this gives us $\triangle I_BI_AI\sim \triangle I_BCI$ and $\triangle I_CI_AI \sim \triangle I_CCB$ because of this similarity we have $\angle I_BPC=\angle I_BKA$ and $\angle I_CPB=\angle I_CKA$ so $\angle I_BI_CD+\angle I_CI_BD=90^{\circ}$ so $P,Q,I_B,I_C$ are concyclic.

Cute! rcorreaa wrote: Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic. Since $KA \perp I_BI_C$ and $KQ$ bisects $I_BI_C$, we just need to prove $Q$ is the $K$-HM point in $\triangle KI_BI_C$. This is equivalent to showing that $L\overset{\text{def}}{:=} KA \cap BC$ is the orthocentre of $\triangle KI_BI_C$, as $LQ \perp KQ$. Finally, this last part follows as $L$ lies on ray $AK$ and $AL \cdot AK = AB \cdot AC = AI_B \cdot AI_C$, as desired.

If we rephrase the problem according to $\triangle I_AI_BI_C$ where $I_A$ is the $A-$excenter, we get the following problem. New Problem Statement: $ABC$ is a triangle with altitudes $AD,BE,CF$ and $K=AD\cap (DEF), \ M$ is the midpoint of $EF$. Prove that reflection of $K$ with respect to $BC,B,C,M$ are concyclic. Proof: Let $N$ be the midpoint of $BC$ and $Q,K'$ be the reflections of $K$ to $BC,N$. $P=EF\cap AD,R=EF\cap BC$. Note that $K,M,N$ are collinear. $NB.NC=NM.NK=NM.NK'$ hence $B,C,M,K'$ are concyclic. Also $RB.RC=RE.RF=RD.RN=RP.RM$ which implies $B,C,M,P$ are cyclic. Since $P,M,K,Q$ lie on the circle with diameter $PK',$ we get that $P,K'\in (BCM)$ and $Q\in (PMK)$ thus, $B,C,M,P,K',Q$ are concyclic as desired.$\blacksquare$