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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
H not needed
dchenmathcounts   44
N 17 minutes ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
17 minutes ago
IZHO 2017 Functional equations
user01   51
N 37 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
37 minutes ago
chat gpt
fuv870   2
N 39 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
an hour ago
fuv870
39 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 41 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
41 minutes ago
No more topics!
Easy geometry
rcorreaa   19
N Today at 4:30 AM by ihategeo_1969
Source: 2022 Brazilian National Mathematical Olympiad - Problem 2
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.
19 replies
rcorreaa
Nov 22, 2022
ihategeo_1969
Today at 4:30 AM
Easy geometry
G H J
Source: 2022 Brazilian National Mathematical Olympiad - Problem 2
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rcorreaa
238 posts
#1 • 4 Y
Y by VicKmath7, itslumi, Mango247, Rounak_iitr
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.
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MrOreoJuice
594 posts
#2
Y by
Solution
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Rafinha
51 posts
#3 • 3 Y
Y by BrodyGong1027, Ubfo, Rounak_iitr
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.
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VicKmath7
1385 posts
#4
Y by
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).
This post has been edited 2 times. Last edited by VicKmath7, Nov 22, 2022, 10:00 AM
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kamatadu
466 posts
#5 • 2 Y
Y by teomihai, HoripodoKrishno
Not that hard but yeah... Nice problem for a NUB like me.

Anyways moving onto the solution,
[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 = -28, xmax = 6, ymin = -5, ymax = 19;  /* image dimensions */

 /* draw figures */
draw((-8.132112681822244,6.103664258932486)--(-8.99,-0.77), linewidth(0.4)); 
draw((-8.99,-0.77)--(-0.31,-0.77), linewidth(0.4)); 
draw((-0.31,-0.77)--(-8.132112681822244,6.103664258932486), linewidth(0.4)); 
draw(circle((-4.65,2.1787015619620003), 5.24694586416871), linewidth(0.4)); 
draw((-0.31,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); 
draw((-13.3200477963969,4.134066052134884)--(4.020047796396906,10.71722880012654), linewidth(0.4)); 
draw(circle((-5.0862639235495,8.574769577234072), 9.354947603412338), linewidth(0.4) + linetype("4 4")); 
draw((-8.99,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); 
draw((-0.31,-0.77)--(4.020047796396906,10.71722880012654), linewidth(0.4)); 
draw((-8.99,-0.77)--(-26.237391236422933,-0.77), linewidth(0.4)); 
draw((-26.237391236422933,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); 
draw((-11.614225363644488,15.275572820071684)--(-4.65,-3.0682443022067103), linewidth(0.4)); 
draw((-8.99,-0.77)--(4.020047796396906,10.71722880012654), linewidth(0.4)); 
 /* dots and labels */
dot((-8.132112681822244,6.103664258932486),linewidth(2pt) + dotstyle); 
label("$A$", (-8.516662214016402,6.952946630403522), NE * labelscalefactor); 
dot((-8.99,-0.77),linewidth(2pt) + dotstyle); 
label("$B$", (-10.215226956958455,-2.259379272983231), NE * labelscalefactor); 
dot((-0.31,-0.77),linewidth(2pt) + dotstyle); 
label("$C$", (-0.02429676701291338,-0.9572815308149143), NE * labelscalefactor); 
dot((-4.65,-3.0682443022067103),linewidth(2pt) + dotstyle); 
label("$K$", (-4.669273367310101,-4.62494533814528), NE * labelscalefactor); 
dot((-4.65,-0.77),linewidth(2pt) + dotstyle); 
label("$P$", (-4.696326799454727,-2.2229913594429492), NE * labelscalefactor); 
dot((-13.3200477963969,4.134066052134884),linewidth(2pt) + dotstyle); 
label("$I_C$", (-15.751989828677138,4.626351465965678), NE * labelscalefactor); 
dot((4.020047796396906,10.71722880012654),linewidth(2pt) + dotstyle); 
label("$I_B$", (4.068985215404134,10.91917518067594), NE * labelscalefactor); 
dot((-11.614225363644488,15.275572820071684),linewidth(2pt) + dotstyle); 
label("$Q$", (-13.001730467854944,15.653525805985506), NE * labelscalefactor); 
dot((-5.5225278470990045,-0.77),linewidth(2pt) + dotstyle); 
label("$D$", (-6.941761667458694,-2.1959379272983231), NE * labelscalefactor); 
dot((-26.237391236422933,-0.77),linewidth(2pt) + dotstyle); 
label("$E$", (-27.09656861520515,-0.5244266165008968), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Let $D = AK \cap BC$, $E = I_B I_C$. Also, trivial to notice that $I_BI_CBC$ is cyclic.

Now,$$AQ \cdot AD = AK \cdot AD \stackrel{\dagger}{=} (\sqrt{bc})^2 = AB \cdot AC \stackrel{\ddagger}{=} AI_B \cdot AI_C \implies I_B I_C DQ\text{ is cyclic}.$$
$\dagger$ follows straight from $\sqrt{bc}$ Inversion, and $\ddagger$ follows from here.

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 :).
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teomihai
2944 posts
#7
Y by
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!
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geometry6
304 posts
#8
Y by
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 $$\angle I_CQI_B=\angle I_CKI_B=\angle CPI_B+\angle BPI_C=180^\circ-\angle I_CPI_B\implies P,Q,I_B,I_C \text{ are concyclic.}\blacksquare$$
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bruckner
106 posts
#9
Y by
Straightforward by complex numbers
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naman12
1358 posts
#10 • 3 Y
Y by Mango247, Mango247, Mango247
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.
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v_Enhance
6857 posts
#13 • 4 Y
Y by HamstPan38825, mod_x, Rounak_iitr, ehuseyinyigit
Solution from Twitch Solves ISL:

Let $I_A$ be the $A$-excenter, and let $H = \overline{AIK} \cap \overline{BC}$.
[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]
  • 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 $(PHI_BI_C)$.
This post has been edited 1 time. Last edited by v_Enhance, Nov 26, 2022, 3:53 AM
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Quidditch
815 posts
#14 • 1 Y
Y by teomihai
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. :D
This post has been edited 1 time. Last edited by Quidditch, Nov 27, 2022, 9:30 AM
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IAmTheHazard
5000 posts
#15 • 1 Y
Y by samrocksnature
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. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Mar 8, 2023, 3:00 AM
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math_comb01
659 posts
#16 • 1 Y
Y by teomihai
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
Attachments:
This post has been edited 3 times. Last edited by math_comb01, May 20, 2023, 7:39 PM
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L567
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#17
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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. $\blacksquare$
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IAmTheHazard
5000 posts
#18 • 1 Y
Y by L13832
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. $\blacksquare$
This post has been edited 2 times. Last edited by IAmTheHazard, Oct 26, 2023, 1:54 PM
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Thapakazi
53 posts
#19 • 1 Y
Y by surpidism.
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.
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L13832
248 posts
#20
Y by
[asy]
        /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
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 [/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.
This post has been edited 1 time. Last edited by L13832, Jun 28, 2024, 5:10 AM
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anantmudgal09
1979 posts
#21
Y by
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.
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bin_sherlo
661 posts
#22 • 2 Y
Y by ehuseyinyigit, teomihai
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.$\blacksquare$
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ihategeo_1969
161 posts
#23
Y by
One of my quickest headsolves.

If we let $D=\overline{AI} \cap \overline{BC}$ then see that $(I_BI_CQD)$ is cyclic since $AI_B \cdot AI_C=AD \cdot AK=AD \cdot AQ$ (first two quantities are both equal to $AB \cdot AC$ by say $\sqrt{bc}$ inversion).

Now let $X$ be $I_A$-Ex point of $\triangle I_AI_BI_C$ And now $(XD;BC)=-1$ (by Cevs Menelaus or whatever) and hence by Mclaurin's we get $XI_C \cdot XI_B=XB \cdot XC=XD \cdot XP$ and done by PoP.
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