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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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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

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.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]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
BMN is equilateral iff rectangle ABCD is square
parmenides51   4
N 9 minutes ago by Tsikaloudakis
Source: 2004 Romania NMO SL - Shortlist VII-VIII p8 https://artofproblemsolving.com/community/c3950157_
Consider a point $M$ on the diagonal $BD$ of a given rectangle $ABCD$, such that $\angle AMC = \angle  CMD$. The point $N$ is the intersection point between $AM$ and the parallel line to $CM$ that contains $B$. Prove that the triangle $BMN$ is equilateral if and only if $ABCD$ is a square.

Valentin Vornicu
4 replies
parmenides51
Sep 16, 2024
Tsikaloudakis
9 minutes ago
Loop of Logarithms
scls140511   10
N 10 minutes ago by SomeonecoolLovesMaths
Source: 2024 China Round 1 (Gao Lian)
Round 1

1 Real number $m>1$ satisfies $\log_9 \log_8 m =2024$. Find the value of $\log_3 \log_2 m$.
10 replies
scls140511
Sep 8, 2024
SomeonecoolLovesMaths
10 minutes ago
Cool Number Theory
Fermat_Fanatic108   3
N 25 minutes ago by mpcnotnpc
For an integer with 5 digits $n=abcde$ (where $a, b, c, d, e$ are the digits and $a\neq 0$) we define the \textit{permutation sum} as the value $$bcdea+cdeab+deabc+eabcd$$For example the permutation sum of 20253 is $$02532+25320+53202+32025=113079$$Let $m$ and $n$ be two fivedigit integers with the same permutation sum.
Prove that $m=n$.
3 replies
Fermat_Fanatic108
2 hours ago
mpcnotnpc
25 minutes ago
Proving a kite
Bugi   4
N 28 minutes ago by ali123456
Source: Serbian JBTST 3, Day 2
Let $ ABCD$ be a convex quadrilateral, such that

$ \angle CBD=2\cdot\angle ADB, \angle ABD=2\cdot\angle CDB$ and $ AB=CB$.

Prove that quadrilateral $ ABCD$ is a kite.
4 replies
Bugi
May 31, 2009
ali123456
28 minutes ago
Inequality
Marinchoo   6
N 36 minutes ago by sqing
If $abc=1$ prove that $8(a^3+b^3+c^3) \geq 3(a^2+bc)(b^2+ac)(c^2+ab)$
6 replies
Marinchoo
Apr 28, 2020
sqing
36 minutes ago
Interesting inequality
sqing   4
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$$$$(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$$$$(a^2-1)(b-1)(c^2-1) -2abc\geq -7$$$$(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$$$$(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$$$$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$$
4 replies
sqing
3 hours ago
sqing
an hour ago
Orthocentre is collinear with two tangent points
vladimir92   42
N an hour ago by AshAuktober
Source: Chinese MO 1996
Let $\triangle{ABC}$ be a triangle with orthocentre $H$. The tangent lines from $A$ to the circle with diameter $BC$ touch this circle at $P$ and $Q$. Prove that $H,P$ and $Q$ are collinear.
42 replies
vladimir92
Jul 29, 2010
AshAuktober
an hour ago
Problem 4
den_thewhitelion   3
N an hour ago by DensSv
Source: Second Romanian JBMO TST 2016
We have a 4x4 board.All 1x1 squares are white.A move is changing colours of all squares of a 1x3 rectangle from black to white and from white to black.It is possible to make all the 1x1 squares black after several moves?
3 replies
den_thewhitelion
Jun 15, 2016
DensSv
an hour ago
Find the period
Anto0110   2
N an hour ago by YaoAOPS
Let $a_1, a_2, ..., a_k, ...$ be a sequence that consists of an initial block of $p$ positive distinct integers that then repeat periodically. This means that $\{a_1, a_2, \dots, a_p\}$ are $p$ distinct positive integers and $a_{n+p}=a_n$ for every positive integer $n$. The terms of the sequence are not known and the goal is to find the period $p$. To do this, at each move it possible to reveal the value of a term of the sequence at your choice.
If $p$ is one of the first $k$ prime numbers, find for which values of $k$ there exist a strategy that allows to find $p$ revealing at most $8$ terms of the sequence.
2 replies
Anto0110
Yesterday at 7:37 PM
YaoAOPS
an hour ago
Very interesting inequality
sqing   0
an hour ago
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a-1)(b^2-2)(c^3-3)-  \frac{5}{2}abc\geq -10$$$$(a-\frac{3}{2})(b^2-2)(c^3-3)-  \frac{5}{2}abc\geq -15$$$$(a-\frac{3}{2})(b^2-\frac{3}{2})(c^3-3)-  \frac{25}{8}abc\geq - \frac{155}{8}$$$$(a-\frac{3}{2})(b^2-\frac{3}{2})(c^3-3)- 3abc\geq - \frac{363}{20}$$$$(a-\frac{3}{2})(b^2-\frac{3}{2})(c^3-\frac{5}{2})- \frac{55}{16}abc\geq - \frac{341}{16}$$
0 replies
sqing
an hour ago
0 replies
inequality
senku23   3
N an hour ago by SunnyEvan
Let x,y,z in R+ prove that 8(x^3+y^3+z^3)2≥9(x^2+yz)(y^2+zx)(z^2+xy).
3 replies
senku23
4 hours ago
SunnyEvan
an hour ago
ratio chasing inside a triangle, segment trisecting
parmenides51   10
N 2 hours ago by sangsidhya
Source: CRMO 2012 Region 2 p5
Let $ABC$ be a triangle. Let $D, E$ be a points on the segment $BC$ such that $BD =DE = EC$. Let $F$ be the mid-point of $AC$. Let $BF$ intersect $AD$ in $P$ and $AE$ in $Q$ respectively. Determine $BP:PQ$.
10 replies
parmenides51
Sep 30, 2018
sangsidhya
2 hours ago
Geo: incircle, escircle, isotomic conjugate
XAN4   0
2 hours ago
Source: Own
For $\triangle{ABC}$, Its incircle $\odot I$ and $A-$escircle $\odot I_A$ are tangent to $BC$ at $D$ and $E$ respectively. $AI$ intersects line $BC$ at $J$. Line $AD$ intersects $\odot I$ at $F$, and line $AE$ intersects $\odot I_A$ at $G$. Line $FG$ intersects $BC$ at $H$. Prove that $BJ=CH$.
0 replies
XAN4
2 hours ago
0 replies
f(x)+f(x+y) \leq f(xy)+f(y)
augustin_p   7
N 2 hours ago by MuradSafarli
Source: Estonia TST 2022
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following condition for any real numbers $x{}$ and $y$ $$f(x)+f(x+y) \leq f(xy)+f(y).$$
7 replies
augustin_p
Jul 12, 2023
MuradSafarli
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
662 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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