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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 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
IMO Genre Predictions
ohiorizzler1434   57
N 11 minutes ago by EVKV
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
57 replies
ohiorizzler1434
May 3, 2025
EVKV
11 minutes ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   2
N 16 minutes ago by Tkn
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
2 replies
BR1F1SZ
Monday at 9:45 PM
Tkn
16 minutes ago
3-var inequality
sqing   1
N 16 minutes ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^3-ab+b^3=1  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{1}{3}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{1}{3}$$Let $ a,b\geq  0 ,a^3+ab+b^3=3  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{2\sqrt[3]{9}}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{\sqrt[3]{3}}{5}$$
1 reply
sqing
34 minutes ago
sqing
16 minutes ago
IMO ShortList 2001, combinatorics problem 3
orl   37
N 27 minutes ago by deduck
Source: IMO ShortList 2001, combinatorics problem 3, HK 2009 TST 2 Q.2
Define a $ k$-clique to be a set of $ k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.
37 replies
orl
Sep 30, 2004
deduck
27 minutes ago
area of O_1O_2O_3O_4 <=1, incenters of right triangles outside a square
parmenides51   2
N 31 minutes ago by Solilin
Source: Thailand Mathematical Olympiad 2012 p4
Let $ABCD$ be a unit square. Points $E, F, G, H$ are chosen outside $ABCD$ so that $\angle AEB =\angle BF C = \angle CGD = \angle DHA = 90^o$ . Let $O_1, O_2, O_3, O_4$, respectively, be the incenters of $\vartriangle ABE, \vartriangle BCF, \vartriangle CDG, \vartriangle DAH$. Show that the area of $O_1O_2O_3O_4$ is at most $1$.
2 replies
parmenides51
Aug 17, 2020
Solilin
31 minutes ago
Geo metry
TUAN2k8   3
N 32 minutes ago by TUAN2k8
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
3 replies
TUAN2k8
Yesterday at 10:33 AM
TUAN2k8
32 minutes ago
Functional equation of nonzero reals
proglote   8
N an hour ago by jasperE3
Source: Brazil MO 2013, problem #3
Find all injective functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that \[f(x+y) \left(f(x) + f(y)\right) = f(xy)\] for all non-zero reals $x, y$ such that $x+y \neq 0$.
8 replies
proglote
Oct 24, 2013
jasperE3
an hour ago
Interesting inequalities
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^2-ab+b^2+a+b=3  $. Prove that
$$  \frac{39+\sqrt{13}}{78}\geq  \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq  \frac{1}{2}$$Let $ a,b\geq  0 ,a^2+ab+b^2+a+b=3  $. Prove that
$$  \frac{19+\sqrt{10}}{39}\geq  \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq    \frac{39+\sqrt{13}}{78}$$Let $ a,b\geq  0 ,a^2+ab+b^2+a+b=5  $. Prove that
$$  \frac{3}{5}> \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq     \frac{185+3\sqrt{21}}{402}$$
1 reply
sqing
2 hours ago
sqing
an hour ago
4-var inequality
sqing   2
N 2 hours ago by sqing
Source: SXTB (4)2025 Q2837
Let $ a,b,c,d> 0  $. Prove that
$$   \frac{1}{(3a+1)^4}+ \frac{1}{(3b+1)^4}+\frac{1}{(3c+1)^4}+\frac{1}{(3d+1)^4} \geq \frac{1}{16(3abcd+1)}$$
2 replies
sqing
Yesterday at 2:59 PM
sqing
2 hours ago
Inspired by Bet667
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b $ be a real numbers such that $a^2+kab+b^2\ge a^3+b^3.$Prove that$$a+b\leq k+2$$Where $ k\geq 0. $
2 replies
sqing
Yesterday at 2:46 PM
sqing
2 hours ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   14
N 2 hours ago by MathLuis
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
14 replies
parmenides51
Jul 25, 2018
MathLuis
2 hours ago
GCD of terms in a sequence
BBNoDollar   1
N 3 hours ago by mashumaro
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
1 reply
BBNoDollar
Yesterday at 10:29 PM
mashumaro
3 hours ago
Diodes and usamons
v_Enhance   46
N 3 hours ago by HamstPan38825
Source: USA December TST for the 56th IMO, by Linus Hamilton
A physicist encounters $2015$ atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is sure are currently in the same state. Is there any series of diode usage that makes this possible?

Proposed by Linus Hamilton
46 replies
v_Enhance
Dec 17, 2014
HamstPan38825
3 hours ago
China South East Mathematical Olympiad 2013 problem 2
s372102   3
N 3 hours ago by AGCN
$\triangle ABC$, $AB>AC$. the incircle $I$ of $\triangle ABC$ meet $BC$ at point $D$, $AD$ meet $I$ again at $E$. $EP$ is a tangent of $I$, and $EP$ meet the extension line of $BC$ at $P$. $CF\parallel PE$, $CF\cap AD=F$. the line $BF$ meet $I$ at $M,N$, point $M$ is on the line segment $BF$, the line segment $PM$ meet $I$ again at $Q$. Show that $\angle ENP=\angle ENQ$
3 replies
s372102
Aug 10, 2013
AGCN
3 hours ago
angles in triangle
AndrewTom   33
N Apr 30, 2025 by zuat.e
Source: BrMO 2012/13 Round 2
The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$.
33 replies
AndrewTom
Feb 1, 2013
zuat.e
Apr 30, 2025
angles in triangle
G H J
Source: BrMO 2012/13 Round 2
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AndrewTom
12750 posts
#1 • 4 Y
Y by Adventure10, Mango247, Rounak_iitr, ItsBesi
The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$.
Z K Y
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Nilashis
132 posts
#2 • 3 Y
Y by AndrewTom, Adventure10, Mango247
Use sin rule on $\triangle ABQ$, $\triangle ACQ$, $\triangle BAP$, $\triangle CAP$ and comparing them we get that
$\frac{sin\angle BAQ}{sin\angle QAC}=\frac{sin\angle CAP}{sin\angle PAB}$. Now take $\angle BAQ=x$ and $\angle PAC=y$ then the equation reduces to $\frac{sinx}{sin(A-x)}=\frac{siny}{sin(A-y)}$
$2sin(A-x)siny=2sin(A-y)sinx$
$cos(A-x-y)-cos(A-x+y)=cos(A-y-x)-cos(A-y+x)$
$cos(A-x+y)=cos(A-y+x)$
$x-y=y-x$
$x=y$
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nsato
15654 posts
#3 • 2 Y
Y by Adventure10, ehuseyinyigit
This appears as an exercise in Geometry Revisited (Section 1.9, Exercise 3).
Z K Y
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sunken rock
4392 posts
#4 • 3 Y
Y by AndrewTom, Adventure10, Mango247
It has been posted here around few years ago, with a very nice synthetic solution!

Best regards,
sunken rock
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MMEEvN
252 posts
#5 • 12 Y
Y by sunken rock, AndrewTom, kprepaf, jlammy, 93051, v_Enhance, Med_Sqrt, hakN, Adventure10, Mango247, ohiorizzler1434, ehuseyinyigit
Let $R$ be the point such that $APBR$ is a parallelogram . Hence $AR || BP ||QC$ and $AR=BP=CQ$ Hence $ARQC$ is a parallelogram.$\angle ACQ = \angle ARQ$ . But $ \angle ACQ = \angle ABQ$ . Hence $ARBQ$ is cyclic.
.$\angle PAB=\angle ABR =\angle AQR= \angle QAC$. $ \Longrightarrow \angle QAB=\angle PAC$
Z K Y
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jlammy
1099 posts
#6 • 2 Y
Y by Adventure10, Mango247
sunken rock wrote:
It has been posted here around few years ago, with a very nice synthetic solution!

Best regards,
sunken rock

Can you specify the details of this "very nice synthetic solution"?
Z K Y
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sunken rock
4392 posts
#7 • 2 Y
Y by Adventure10, Mango247
@jlammy: Like MMEEvN did!

Best regards,
sunken rock
Z K Y
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IDMasterz
1412 posts
#8 • 1 Y
Y by Adventure10
no angle chasing: Let $P'$ be the $P$ isogonal conjugate and $P''$ be its reflection over $BC$. The angle bisectors of $BPC$ and $BAC$ are obviously parallel. $AP, PP''$ are antiparallel wrt $BPC$ so $AP' \parallel PP'' \parallel QP'$ since $PP'QP''$ form a parallelogram, so done.
Z K Y
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DottedCaculator
7348 posts
#9
Y by
Solution
This post has been edited 1 time. Last edited by DottedCaculator, Dec 10, 2021, 10:45 PM
Z K Y
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Project_Donkey_into_M4
148 posts
#11
Y by
AndrewTom wrote:
The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$.

For a non complicated solution unlike above here's a hint
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guptaamitu1
656 posts
#12 • 1 Y
Y by Rounak_iitr
Here's a different proof with similar triangles and homothety (plus reflection in angle bisector)
Let $E = \overline{BP} \cap \overline{AC}, F = \overline{CP} \cap \overline{AB}$. Then points $B,C,E,F$ are concyclic. Using $BPCQ$ is a parallelogram we get
$$ \angle QCB = \angle PBC = \angle EBC = \angle EFC = \angle EFP $$Similarly $\angle QBC = \angle FEP$. Hence,
$$ \triangle QBC \sim \triangle PEF $$[asy]
size(200);
pair B=dir(-160),C=dir(-20),E=dir(70),F=dir(135),A=extension(B,F,C,E),P=extension(B,E,C,F),Q=B+C-P;
draw(unitcircle,cyan);
fill(P--E--F--P--cycle,purple+grey+grey);
fill(B--Q--C--B--cycle,purple+grey+grey);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$P$",P,dir(-90));
dot("$Q$",Q,dir(Q));
draw(A--B--C--A,royalblue);
draw(B--E^^C--F,red);
draw(P--A--Q,green);
draw(B--Q--C^^E--F,brown);
[/asy]
Let $\mathbb H$ denote homothety at $A$ with scale $\frac{AF}{AC} = \frac{AE}{AB}$ followed by reflection in internal angle bisector of $\angle BAC$. Note $\mathbb H(F) = C$ and $\mathbb H(E) = B$. Thus $\mathbb H(P) = P'$ is a point such that $$\triangle PEF \sim \triangle P' BC$$Hence $P' \equiv Q$. As $\mathbb H$ also consists of reflection in internal angle bisector of $\angle BAC$, so $\angle BAP = \angle CAQ$ follows. $\blacksquare$
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TechnoLenzer
55 posts
#13 • 2 Y
Y by ike.chen, allin27x
Let $\infty_1 = BP \cap CQ$ and $\infty_2 = CP \cap BQ$. Since these are parallel pairs of lines, $\infty_1, \infty_2$ are the points at infinity for those pencils of parallel lines respectively. Note that $\measuredangle \infty_1AB = \measuredangle PBA = \measuredangle ACP = \measuredangle CA\infty_2$ by $A\infty_1 \; || \; CP$ and $A\infty_2 \; || \; BP$. Thus, $A\infty_1$ is isogonal to $A\infty_2$ wrt. $\triangle ABC$. Hence by DDIT on complete quadrilateral $P, B, Q, C, \infty_1, \infty_2$, there exists a projective involution swapping $(AB, AC)$, $(A\infty_1, A\infty_2)$, $(AP, AQ)$. This is taking the isogonal, and so $AP, AQ$ are isogonal. $\blacksquare$
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samrocksnature
8791 posts
#14
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Any complex sols?
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AwesomeYRY
579 posts
#15
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Consider the following two series of laws of sines:
\begin{align*}
    \frac{BQ}{\sin(\angle BAQ)} = \frac{AQ}{\sin(\angle ABP + \angle PBQ)} &= \frac{AQ}{\sin(\angle PCA + \angle ACP)} = \frac{CQ}{\sin (\angle QAC)},\\
    \frac{BP}{\sin(\angle BAP)} = \frac{AP}{\sin(\angle ABP)} &= \frac{AP}{\sin(\angle PCA)} = \frac{PC}{\sin(\angle PAC)}.
\end{align*}Putting them together, we get
\[\frac{\sin(\angle QAC)}{\sin(\angle BAQ)} = \frac{CQ}{BQ}= \frac{BP}{PC} = \frac{\sin(\angle BAP)}{\sin(\angle PAC)}\]and since $\angle WAC + \angle BAQ = \angle BAC = \angle BAP + \angle PAC$, we have that $\angle BAP = \angle QAC$ and we're done. $\blacksquare$.
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anurag27826
93 posts
#16
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Amazing problem, although my solution is the same as of guptaamitu1's solution, but still im posting it for the sake of storage.

First of all we claim that $\triangle PEF \sim \triangle BQC$. Note that $\angle EPF = \angle BPC = \angle BQC$. Also note that $\angle PFE = \angle PBC = \angle QCB$. Both angle equalities are there since $BPCQ$ is a parallelogram. So, consider homothety $\psi$ under $A$ follow by reflection along the angle bisector of $\angle BAC$ with scale $\frac{AF}{AC}$. Note that $\psi$ sends $F$ to $C$ and $E$ to $B$. Then $\psi$ sends $P$ to $P'$ such that $\triangle CP'B \sim FPE \implies P' = Q$. So, it also implies that the line $AQ$ is a reflection of the line $AP$ along the angle bisector of $\angle BAC$, which implies $\angle BAP = \angle CAQ$. So, we're done.
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OronSH
1730 posts
#17
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Great problem.

We use directed angles. Let $P'$ be the isogonal conjugate of $P.$ Consider the lines through $P'$ parallel to $PB,PC.$ These lines intersect $AB$ at points $D,F$ with $D$ between $A$ and $F,$ and they intersect $AC$ at points $E,G$ with $E$ between $A$ and $G.$ Since $\angle DFE=\angle ABP=\angle PCA=\angle DGE,$ we have that $DEGF$ is cyclic. Also, since $\angle BFP'=\angle ABP=\angle PCA=\angle BCP',$ we have $BFCP'$ is cyclic, and similarly $BCGP'$ is cyclic, so $BFCG$ is cyclic, and by Reim's theorem we have $DE,BC$ parallel. Now, take a homothety at $A$ sending $DE$ to $BC.$ It is not hard to see that this takes $P'$ to $Q,$ so $A,P',Q$ are collinear, and we have $\angle QAB=\angle CAP.$
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huashiliao2020
1292 posts
#18
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It states that for a point P inside a triangle ABC with ABP=ACP, the point Q such that BPCQ is a parallelogram, AQ-AP are pairwise isogonal lines.

Here's the first one, quoted from my sl2012g2 (a good application for anyone who's reading!)
Sketch. If we let D be the point s.t. APBD is a parallelogram, we have that BDQ is congruent to APC by a translation of vector AD. (This is just done by length equalities and parallel lines.) Now, from DQB=ACP=ABP=DAB, ADBQ is cyclic. Then BAQ=BRQ=PAC, as desired. $\square$

Second solution I just came up with: Let BP,CP intersect AC,AB at E,F, respectively. It's obvious that $$F\in(BCE)\implies EFP=EBC=QCB,FEP=FCB=CBQ\implies EFP\sim CBQ,AEF\sim ABC\implies AEPF\sim ABQC\implies BAP=FAP=CAQ,$$as desired. $\blacksquare$
This post has been edited 2 times. Last edited by huashiliao2020, Aug 30, 2023, 11:01 PM
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Ianis
411 posts
#19
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Let $P'$ be the isogonal conjugate of $P$ in $ABC$, the angle condition implies that $P'$ is on the perpendicular bisector of $BC$. Use barycentric coordinates with respect to $ABC$ and let $P=(x,y,z)$. Then $P'=\left (\frac{a^2}{x}:\frac{b^2}{y}:\frac{c^2}{z}\right )$ and $Q=B+C-P=(-x,1-y,1-z)=(-x,z+x,x+y)$. We have\begin{align*}\begin{vmatrix}1&0&0\\\dfrac{a^2}{x}&\dfrac{b^2}{y}&\dfrac{c^2}{z}\\-x&z+x&x+y\end{vmatrix} & =\begin{vmatrix}\dfrac{b^2}{y}&\dfrac{c^2}{z}\\z+x&x+y\end{vmatrix} \\
& =b^2\frac{x+y}{y}-c^2\frac{z+x}{z} \\
& =\frac{b^2yz-c^2yz+b^2zx-c^2xy}{yz} \\
& =\frac{x}{a^2}\left ((b^2-c^2)\frac{a^2}{x}+a^2\left (\frac{b^2}{y}-\frac{c^2}{z}\right )\right ) \\
& =0,
\end{align*}where the last equality holds because $P'$ is on the perpendicular bisector of $BC$. Hence $A,P',Q$ are collinear, so $\angle QAB=\angle P'AB=\angle CAP$. Done.
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asdf334
7585 posts
#20
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hello;;; ddit with $A$ and $BPCQ$
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shendrew7
795 posts
#21
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Trivial by first isogonality lemma.

Let $X = BP \cap CA$ and $Y = CP \cap AB$. Then $BCXY$ is cyclic and $BPCQ$ is a parallelogram, so
\[\angle PXY = \angle BCP = \angle CBQ, \quad \angle PXY = \angle CBP = \angle BCQ.\]
Hence we have $\triangle PXY \sim \triangle QBC$ as well as $\triangle AXY \sim \triangle ABC$, so the quadrilaterals $AXPY$ and $ABQC$ are also similar with opposite orientation, which implies the desired. $\blacksquare$
This post has been edited 1 time. Last edited by shendrew7, Dec 31, 2023, 7:14 AM
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joshualiu315
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#22
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Let $R$ be the reflection of $P$ across the midpoint of $\overline{AB}$. It is clear that $ARBP$ is a parallelogram. Then, notice that $\overline{AR} \parallel \overline{BP} \parallel \overline{QC}$, and $AR=BP=QC$. Hence, $ARQC$ is a parallelogram, meaning that $\angle ARQ = \angle ACQ$.

Also, we have

\[\angle ABQ = \angle PBQ+\angle ABP =\angle PCQ+\angle PCA = \angle ACQ.\]
This means that $\angle ABQ = \angle ARQ$, so $ARBQ$ is cyclic. Thus,

\[\angle BAP = \angle ABR = \angle AQR = \angle CAQ. \ \square\]
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EpicBird08
1751 posts
#23
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Let $H$ be the point such that $APCH$ is a parallelogram. First note that $AH = CP = BQ$ since $BPCQ$ is a parallelogram as well. Additionally, this gives $AH \parallel PC \parallel BQ,$ so $AHQB$ is also a parallelogram. Then $$\measuredangle AHQ = \measuredangle QBA = \measuredangle QBP + \measuredangle PBA = \measuredangle PCQ + \measuredangle ACP = \measuredangle ACQ,$$so $AHCQ$ is cyclic.

Next, note from our three parallelograms that $QC = BP, CH = AP,$ and $QH = AB,$ so $\triangle ABP \cong \triangle HQC.$ Finally, $\angle BAP = \angle QHC = \angle QAC,$ as desired.
This post has been edited 1 time. Last edited by EpicBird08, Jan 13, 2024, 5:09 PM
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dolphinday
1324 posts
#24
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Construct parallelograms $APCR$, $APBS$, and $BPCQ$.
And it follows that $ARQB$ is a parallelogram since $BQ \parallel PC \parallel AR$ and $AP \parallel RC$. Similarly, $ACQS$ is a parallelogram. Then we have $\angle SBP = 180^{\circ} = \angle SAP = \angle QAR = \angle BQA$ so $ASBQ$ is cyclic. Then $\angle CAQ = \angle{AQS} = \angle SBA = \angle{PAB}$ so we are done.
This post has been edited 2 times. Last edited by dolphinday, Feb 4, 2024, 3:53 PM
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Assassino9931
1324 posts
#25
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Denote $\angle BAC = \alpha$, $\angle ABP = \angle ACP = x$, $\angle BAP = y$, $\angle CAQ = z$. The Sine Law in $ABP$ and $ACP$ gives $\frac{BP}{\sin y} = \frac{AP}{\sin x} = \frac{CP}{\sin(\alpha - y)}$, i.e. $\frac{CP}{BP} = \frac{\sin(\alpha - y)}{\sin y}$. Analogously $\frac{BQ}{CQ} = \frac{\sin(\alpha - z)}{\sin z}$. However, $CP = BQ$ and $BP = CQ$ from the parallelogram $BPCQ$, thus $\frac{\sin(\alpha - y)}{\sin y} = \frac{\sin(\alpha - z)}{\sin z}$. Hence $\sin\alpha \cot y - \cos \alpha = \sin\alpha \cot z - \cos \alpha$, equivalent to $\cot y = \cot z$, i.e. $y=z$, as desired.
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dolphinday
1324 posts
#26
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Alternate solution involving DDIT;

Apply DDIT on quadrilateral $BC\infty_{BP}\infty_{BQ}$ from point $A$.
Then $B\infty_{BP} \cap C\infty_{BQ} = P$, and $B_\infty{BQ} \cap C\infty_{BP} = Q$, so we get that $(AB, AC)$, $(A\infty_{BP}, A\infty_{BQ})$, and $(AP, AQ)$ are involutions.
However $\angle \infty_{BQ}AC = \angle ACP = \angle ABP = \angle \infty_{BP}AB$, so $(AB, AC)$ and $(A\infty_{BP}, A\infty_{BQ})$ are both involutions around the angle bisector of $\angle BAC$. So it follows that $AP$ and $AQ$ are isogonal as desired.
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Martin2001
149 posts
#27
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Let the reflection of $P$ over the midpoint of $AC$ be $K.$ We see that $AKQB$ is a parralelogram. We show $AKCQ$ is cyclic because then $\angle QAC=\angle QKC=\angle BAQ.$ Let $\angle QAC=y, \angle ACP=x.$ Then $a-y=\angle BAQ=\angle KQA.$ Then note that $\angle CAK=x.$ Therefore $$180-a-x=\angle AKQ=\angle ACQ=b+c-x,$$as desired$.\blacksquare$
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ihatemath123
3446 posts
#28 • 1 Y
Y by OronSH
Vertically stretch about the angle bisector of $\angle BAC$ until $P$ is the orthocenter of $\triangle ABC$, then $Q$ is the antipode so it's obvious.
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bachkieu
136 posts
#29
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I think this works?
Let $CQ \cap AB = X, BQ \cap AC = Y$; it's easy to show that $\triangle ABC \sim \triangle AYX$ and that $P$ of $\triangle$ $ABC$ corresponds to $Q$ of $\triangle AYX$.
This post has been edited 1 time. Last edited by bachkieu, Sep 5, 2024, 12:14 AM
Reason: forgot period
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Lleeya
11 posts
#30 • 1 Y
Y by endless_abyss
This is Romanian Lemma, construct $EPFR$ to be parralelogram and trivial by similarity of $AEF$ and $ABC$.
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endless_abyss
45 posts
#31
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Here's my solution :)
Attachments:
This post has been edited 1 time. Last edited by endless_abyss, Nov 25, 2024, 5:31 PM
Reason: typo haha
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AndreiVila
210 posts
#32 • 2 Y
Y by trigadd123, Ciobi_
Lleeya wrote:
This is Romanian Lemma, construct $EPFR$ to be parralelogram and trivial by similarity of $AEF$ and $ABC$.

Now I can finally go to sleep knowing that we've achieved our goal at MOP. We won.
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AshAuktober
1005 posts
#33
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Yay!
The given is equivalent to $A\infty_{BP}$, $A\infty_{CP}$ being isogonal. From here Isogonal Line Lemma (or DDIT if you wish) gives the required.
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Tsikaloudakis
981 posts
#35
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see the figure:
Attachments:
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zuat.e
56 posts
#36
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Let $E,F=CP\cap AB, BP\cap AC$, then $EFBC$ is cyclic and $\triangle PEF\sim \triangle QCB$.
Now, we consider the transformation composed by the homothety of ratio $\frac{AC}{AF}$ and its reflection $w.r.t.$ the $\angle A$ angle bisector.
Clearly $E$ is sent to $B$ and $F$ to $C$ and as this transformation preserves similarity, it sends $P$ to $Q$, hence $AP$ and $AQ$ are isogonal.
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