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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
How many ordered pairs of numbers can we find?
BR1F1SZ   2
N 5 minutes ago by BR1F1SZ
Source: 2024 Argentina TST P6
Given $2024$ non-negative real numbers $x_1, x_2, \dots, x_{2024}$ that satisfy $x_1 + x_2 + \cdots + x_{2024} = 1$, determine the maximum possible number of ordered pairs $(i, j)$ such that
\[
x_i^2 + x_j \geqslant \frac{1}{2023}.
\]
2 replies
BR1F1SZ
Jan 25, 2025
BR1F1SZ
5 minutes ago
IMO 2009, Problem 5
orl   87
N 10 minutes ago by asdf334
Source: IMO 2009, Problem 5
Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths
\[ a, f(b) \text{ and } f(b + f(a) - 1).\]
(A triangle is non-degenerate if its vertices are not collinear.)

Proposed by Bruno Le Floch, France
87 replies
orl
Jul 16, 2009
asdf334
10 minutes ago
1978 USAMO #1
Mrdavid445   54
N 23 minutes ago by Marcus_Zhang
Given that $a,b,c,d,e$ are real numbers such that

$a+b+c+d+e=8$,
$a^2+b^2+c^2+d^2+e^2=16$.

Determine the maximum value of $e$.
54 replies
Mrdavid445
Aug 16, 2011
Marcus_Zhang
23 minutes ago
The return of a legend inequality
giangtruong13   4
N an hour ago by Double07
Source: Legacy
Given that $0<a,b,c,d<1$.Prove that: $$ (1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d $$
4 replies
giangtruong13
3 hours ago
Double07
an hour ago
degree of f=2^k
Sayan   15
N an hour ago by Gejabsk
Source: ISI 2012 #8
Let $S = \{1,2,3,\ldots,n\}$. Consider a function $f\colon S\to S$. A subset $D$ of $S$ is said to be invariant if for all $x\in D$ we have $f(x)\in D$. The empty set and $S$ are also considered as invariant subsets. By $\deg (f)$ we define the number of invariant subsets $D$ of $S$ for the function $f$.

i) Show that there exists a function $f\colon S\to S$ such that $\deg (f)=2$.

ii) Show that for every $1\leq k\leq n$ there exists a function $f\colon S\to S$ such that $\deg (f)=2^{k}$.
15 replies
1 viewing
Sayan
May 13, 2012
Gejabsk
an hour ago
Local-global with Fibonacci numbers
MarkBcc168   26
N an hour ago by pi271828
Source: ELMO 2020 P2
Define the Fibonacci numbers by $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n\geq 3$. Let $k$ be a positive integer. Suppose that for every positive integer $m$ there exists a positive integer $n$ such that $m \mid F_n-k$. Must $k$ be a Fibonacci number?

Proposed by Fedir Yudin.
26 replies
MarkBcc168
Jul 28, 2020
pi271828
an hour ago
Cauchy functional equations
syk0526   10
N an hour ago by imagien_bad
Source: FKMO 2013 #2
Find all functions $ f : \mathbb{R}\to\mathbb{R}$ satisfying following conditions.
(a) $ f(x) \ge 0 $ for all $ x \in \mathbb{R} $.
(b) For $ a, b, c, d \in \mathbb{R} $ with $ ab + bc + cd = 0 $, equality $ f(a-b) + f(c-d) = f(a) + f(b+c) + f(d) $ holds.
10 replies
syk0526
Mar 24, 2013
imagien_bad
an hour ago
Three circles are concurrent
Twoisaprime   21
N an hour ago by L13832
Source: RMM 2025 P5
Let triangle $ABC$ be an acute triangle with $AB<AC$ and let $H$ and $O$ be its orthocenter and circumcenter, respectively. Let $\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centered at $A$ cross $\Gamma$ at $A’$ and $F$, respectively. Prove that $\Gamma$ , the circle on diameter $AA’$ and circle $AFH$ are concurrent.
Proposed by Romania, Radu-Andrew Lecoiu
21 replies
Twoisaprime
Feb 13, 2025
L13832
an hour ago
IMO Shortlist 2011, Algebra 3
orl   45
N an hour ago by Ilikeminecraft
Source: IMO Shortlist 2011, Algebra 3
Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\] for all real numbers $x$ and $y$.

Proposed by Japan
45 replies
orl
Jul 11, 2012
Ilikeminecraft
an hour ago
Hard FE with positive reals
egxa   8
N 2 hours ago by megarnie
Source: Turkey Olympic Revenge 2023 Shortlist A4
Find all functions $f:\mathbb{R^+}\to \mathbb{R^+}$ such that for all $x,y\in \mathbb{R^+}$
$f(xf(y)+y)=f(f(y))+yf(x)$
Proposed by Şevket Onur Yılmaz
8 replies
egxa
Jan 22, 2024
megarnie
2 hours ago
Like Father Like Son... (or Like Grandson?)
AlperenINAN   1
N 2 hours ago by hakN
Source: Turkey TST 2025 P4
Let $a,b,c$ be given pairwise coprime positive integers where $a>bc$. Let $m<n$ be positive integers. We call $m$ to be a grandson of $n$ if and only if, for all possible piles of stones whose total mass adds up to $n$ and consist of stones with masses $a,b,c$, it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of $m$. Find the greatest possible number that doesn't have any grandsons.
1 reply
AlperenINAN
Today at 6:09 AM
hakN
2 hours ago
Crazy number theory
MTA_2024   5
N 2 hours ago by bjump
Find all couple $(p;q)$ of primes (greater than 5) such that : $$pq \mid (5^q-3^q)(5^p-3^p)$$
5 replies
MTA_2024
4 hours ago
bjump
2 hours ago
hard number theory problem
Zavyk09   0
2 hours ago
Source: forgotten
Find all couple $(x, y)$ of positive integers such that:
$$2^n + 3^n \mid x^n + y^n, \forall n \in \mathbb{N}^*$$
0 replies
Zavyk09
2 hours ago
0 replies
Slightly weird points which are not so weird
Pranav1056   9
N 2 hours ago by Retemoeg
Source: India TST 2023 Day 4 P1
Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$. Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$. Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$. Let $M$ be the midpoint of arc $BZC$, and let $V$ be the midpoint of $ID$. Prove that $\angle IML = \angle DVM$
9 replies
Pranav1056
Jul 9, 2023
Retemoeg
2 hours ago
Easy geometry
rcorreaa   19
N Mar 16, 2025 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
Mar 16, 2025
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.
Z K Y
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MrOreoJuice
594 posts
#2
Y by
Solution
Z K Y
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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.
Z K Y
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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
6858 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
1184 posts
#17
Y by
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
Y by
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
250 posts
#20
Y by
[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 */
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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]
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
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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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