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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.

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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
combi/nt
blug   0
6 minutes ago
Prove that every positive integer $n$ can be written in the form
$$n=2^{a_1}3^{b_1}+2^{a_2}3^{b_2}+..., $$where $a_i, b_j$ are non negative integers, such that
$$2^x3^y\nmid 2^z3^t$$for every $x, y, z, t$.
0 replies
blug
6 minutes ago
0 replies
Interesting inequalities
sqing   2
N 13 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 , a(b+c)=k.$ Prove that
$$\frac{1}{a+1}+\frac{2}{b+1}+\frac{1}{c+1}\geq  \frac{4\sqrt{k}-6}{ k-2}$$Where $5\leq  k\in N^+.$
Let $ a,b,c\geq 0 , a(b+c)=9.$ Prove that
$$\frac{1}{a+1}+\frac{2}{b+1}+\frac{1}{c+1}\geq \frac{6}{7}$$
2 replies
1 viewing
sqing
2 hours ago
sqing
13 minutes ago
Find smallest value of (x^2 + y^2 + z^2)/(xyz)
orl   7
N 22 minutes ago by Bryan0224
Source: CWMO 2001, Problem 4
Let $ x, y, z$ be real numbers such that $ x + y + z \geq xyz$. Find the smallest possible value of $ \frac {x^2 + y^2 + z^2}{xyz}$.
7 replies
orl
Dec 27, 2008
Bryan0224
22 minutes ago
easy substitutions for a functional in reals
Circumcircle   9
N 44 minutes ago by Bardia7003
Source: Kosovo Math Olympiad 2025, Grade 11, Problem 2
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ with the property that for every real numbers $x$ and $y$ it holds that
$$f(x+yf(x+y))=f(x)+f(xy)+y^2.$$
9 replies
Circumcircle
Nov 16, 2024
Bardia7003
44 minutes ago
writing words around circle, two letters
jasperE3   1
N an hour ago by pi_quadrat_sechstel
Source: VJIMC 2000 2.2
If we write the sequence $\text{AAABABBB}$ along the perimeter of a circle, then every word of the length $3$ consisting of letters $A$ and $B$ (i.e. $\text{AAA}$, $\text{AAB}$, $\text{ABA}$, $\text{BAB}$, $\text{ABB}$, $\text{BBB}$, $\text{BBA}$, $\text{BAA}$) occurs exactly once on the perimeter. Decide whether it is possible to write a sequence of letters from a $k$-element alphabet along the perimeter of a circle in such a way that every word of the length $l$ (i.e. an ordered $l$-tuple of letters) occurs exactly once on the perimeter.
1 reply
jasperE3
Jul 27, 2021
pi_quadrat_sechstel
an hour ago
Interesting inequality
imnotgoodatmathsorry   0
an hour ago
Let $x,y,z > \frac{1}{2}$ and $x+y+z=3$.Prove that:
$\sqrt{x^3+y^3+3xy-1}+\sqrt{y^3+z^3+3yz-1}+\sqrt{z^3+x^3+3zx-1}+\frac{1}{4}(x+5)(y+5)(z+5) \le 60$
0 replies
imnotgoodatmathsorry
an hour ago
0 replies
Arithmetic Sequence of Products
GrantStar   19
N an hour ago by OronSH
Source: IMO Shortlist 2023 N4
Let $a_1, \dots, a_n, b_1, \dots, b_n$ be $2n$ positive integers such that the $n+1$ products
\[a_1 a_2 a_3 \cdots a_n, b_1 a_2 a_3 \cdots a_n, b_1 b_2 a_3 \cdots a_n, \dots, b_1 b_2 b_3 \cdots b_n\]form a strictly increasing arithmetic progression in that order. Determine the smallest possible integer that could be the common difference of such an arithmetic progression.
19 replies
GrantStar
Jul 17, 2024
OronSH
an hour ago
Inequality Involving Complex Numbers with Modulus Less Than 1
tom-nowy   0
an hour ago
Let $x,y,z$ be complex numbers such that $|x|<1, |y|<1,$ and $|z|<1$.
Prove that $$ |x+y+z|^2 +3>|xy+yz+zx|^2+3|xyz|^2 .$$
0 replies
tom-nowy
an hour ago
0 replies
Inequality
nguyentlauv   2
N an hour ago by nguyentlauv
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
2 replies
nguyentlauv
May 6, 2025
nguyentlauv
an hour ago
japan 2021 mo
parkjungmin   0
an hour ago

The square box question

Is there anyone who can release it
0 replies
parkjungmin
an hour ago
0 replies
easy sequence
Seungjun_Lee   17
N an hour ago by GreekIdiot
Source: KMO 2023 P1
A sequence of positive reals $\{ a_n \}$ is defined below. $$a_0 = 1, a_1 = 3, a_{n+2} = \frac{a_{n+1}^2+2}{a_n}$$Show that for all nonnegative integer $n$, $a_n$ is a positive integer.
17 replies
Seungjun_Lee
Nov 4, 2023
GreekIdiot
an hour ago
Japan MO Finals 2023
parkjungmin   0
an hour ago
It's hard. Help me
0 replies
parkjungmin
an hour ago
0 replies
I Brazilian TST 2007 - Problem 4
e.lopes   77
N an hour ago by alexanderhamilton124
Source: 2007 Brazil TST, Russia TST, and AIMO; also SL 2006 N5
Find all integer solutions of the equation \[\frac {x^{7} - 1}{x - 1} = y^{5} - 1.\]
77 replies
e.lopes
Mar 11, 2007
alexanderhamilton124
an hour ago
Japan MO Finals 2024
parkjungmin   0
an hour ago
Source: Please tell me the question
Please tell me the question
0 replies
parkjungmin
an hour ago
0 replies
Equality of angles [Iran TST 2010]
Omid Hatami   32
N Jan 25, 2024 by math_comb01
Circles $W_1,W_2$ intersect at $P,K$. $XY$ is common tangent of two circles which is nearer to $P$ and $X$ is on $W_1$ and $Y$ is on $W_2$. $XP$ intersects $W_2$ for the second time in $C$ and $YP$ intersects $W_1$ in $B$. Let $A$ be intersection point of $BX$ and $CY$. Prove that if $Q$ is the second intersection point of circumcircles of $ABC$ and $AXY$
\[\angle QXA=\angle QKP\]
32 replies
Omid Hatami
Jun 9, 2010
math_comb01
Jan 25, 2024
Equality of angles [Iran TST 2010]
G H J
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Omid Hatami
1275 posts
#1 • 11 Y
Y by tenplusten, Abdollahpour, RedFlame2112, Adventure10, Mango247, and 6 other users
Circles $W_1,W_2$ intersect at $P,K$. $XY$ is common tangent of two circles which is nearer to $P$ and $X$ is on $W_1$ and $Y$ is on $W_2$. $XP$ intersects $W_2$ for the second time in $C$ and $YP$ intersects $W_1$ in $B$. Let $A$ be intersection point of $BX$ and $CY$. Prove that if $Q$ is the second intersection point of circumcircles of $ABC$ and $AXY$
\[\angle QXA=\angle QKP\]
This post has been edited 1 time. Last edited by Omid Hatami, Jun 9, 2010, 9:10 PM
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frenchy
150 posts
#2 • 3 Y
Y by RedFlame2112, Adventure10, Mango247
do you mean the intersection of the circumscribed cercles of $ABC$ and $AXY$??
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goodar2006
1347 posts
#3 • 5 Y
Y by RedFlame2112, Adventure10, Mango247, and 2 other users
up to this point I have reached that $AXKC$ and $AYKB$ are concyclic . is it useful ?
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Luis González
4148 posts
#4 • 17 Y
Y by bigbang195, Zeref, mad, ilovemath121, HQN, wiseman, canhhoang30011999, Hermitianism, darkeagle, lahmacun, RedFlame2112, kamatadu, Adventure10, Mango247, and 3 other users
Let $D$ be the second intersection of $\odot(ABC)$ with $\omega_1$ and $E$ the second intersection of $DC$ with $\omega_1.$ For convenience, denote $\angle AXY=\angle AYX=\theta.$

By Reim's theorem for $\odot(ABC)$ and $\omega_1$ with common chord $BD,$ it follows that $XE \parallel AC$ $\Longrightarrow$ $\angle BXE=\pi -2\theta.$ Since $\angle BDX=\theta,$ then $\angle XDE=\angle XYA=\theta$ $\Longrightarrow$ $X,Y,C,D$ are concyclic. Therefore, $AQ,XY,DC$ concur at the radical center $O$ of $\odot(AXY) ,$ $ \odot(ABC)$ and $\odot(XYC),$ which becomes the exsimilicenter of $\omega_1,\omega_2,$ since $O$ is also center of the positive inversion taking $\omega_1,\omega_2$ into each other. Thus, the tangent of $\odot(XYK) \equiv \omega_3$ at $K$ passes through $O.$ Further, $AB,AC$ are tangent lines of $\omega_3$ at points $X,Y.$ Then $KA$ is the polar of $O$ WRT $\omega_3$ cutting $\omega_3$ again at $R$ such that $OR$ is tangent to $\omega_3$ at $R.$ If $P \equiv AK \cap XY,$ then the cross ratio $(X,Y,P,O)$ is harmonic.

Consequently, if $M \equiv PK \cap XY$ (midpoint of XY), it follows that $OR^2=OK^2=OQ \cdot OA=OP \cdot OM.$ Since $K,P,R,A$ are collinear, being K,R double points, $K,$ $M,$ $R,$ $Q,$ $O$ are concyclic, yielding $\angle QKP=\angle QOY.$ But because of $AY^2=AR \cdot AK= AQ \cdot AO,$ we deduce that $\angle QOY=\angle QYA=\angle QXA$ $\Longrightarrow$ $\angle QXA=\angle QKP.$
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RaleD
118 posts
#5 • 6 Y
Y by me9hanics, RedFlame2112, Adventure10, Mango247, and 2 other users
goodar2006 wrote:
up to this point I have reached that $AXKC$ and $AYKB$ are concyclic . is it useful ?

This is useful in my solution. After noticing this and that $AX=AY$ we make an inversion with center in $A$. After some angle chasing we actually got following problem:
Let tangents in $K, X, Y$ on circumcircle of $KXY$ intersect sides $XY, KY, KX$ in $Q, C, B$ respectively. Show that $Q, C, B$ are collinear
(in problem we have that $Q, B, C$ are collinear and need to get that $QK$ is tangent but this is same
I don't have nice explanation for this, however it follows from Ceva's theorem.
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goodar2006
1347 posts
#6 • 3 Y
Y by RedFlame2112, Adventure10, and 1 other user
RaleD wrote:
goodar2006 wrote:
up to this point I have reached that $AXKC$ and $AYKB$ are concyclic . is it useful ?

This is useful in my solution. After noticing this and that $AX=AY$ we make an inversion with center in $A$. After some angle chasing we actually got following problem:
Let tangents in $K, X, Y$ on circumcircle of $KXY$ intersect sides $XY, KY, KX$ in $Q, C, B$ respectively. Show that $Q, C, B$ are collinear
(in problem we have that $Q, B, C$ are collinear and need to get that $QK$ is tangent but this is same
I don't have nice explanation for this, however it follows from Ceva's theorem.

beautiful solution! thanks. :)
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paul1703
222 posts
#7 • 3 Y
Y by RedFlame2112, Adventure10, Mango247
frenchy wrote:
do you mean the intersection of the circumscribed cercles of $ABC$ and $AXY$??
actualy the point you have reached with the cyclic quadrilaterals folows from the fact that k is the miquel point of the quadrilateral AXPY
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vladimir92
212 posts
#8 • 11 Y
Y by anantmudgal09, fjm30, RedFlame2112, Adventure10, Mango247, and 6 other users
As always, Iran have nice problems to propose for contestants.
My solution
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wsjradha
48 posts
#9 • 24 Y
Y by MMEEvN, hal9v4ik, mad, meho96, Olemissmath, wiseman, DenisO1998, Abubakir, harapan57, Hermitianism, TRYTOSOLVE, gemcl, leru007, microsoft_office_word, RedFlame2112, Adventure10, Mango247, and 7 other users
Here is a solution with spiral similarity.


Note that $K$ is the center of spiral similarity that takes $BX$ to $YC$. In particular, $BXK \sim YCK$. But, we know that
$\angle KYC =  \angle KBX = \angle KXY$
since $XY$ is tangent to $\omega_1$. Similarly, $\angle KXB = \angle KCY = \angle KYX$. Thus, we have that
$BXK \sim XYK \sim YCK$,
and so there is a spiral similarity at $K$ that takes $BX$ to $XY$ to $YC$. Let $\angle BKX = \angle XKY = \angle YKZ = \theta$. Then, we have
$\angle BAC = 180 - \angle AXY - \angle AYX = 180 - \theta - \theta = 180 - 2\theta$.
Let $M$, $N$ be the midpoints of $BX$, $CY$, respectively. Then, the spiral similarity at $K$ taking $BX$ to $YC$ takes $M$ to $N$, or
$\angle MKN = \angle BKY = 2\theta$.
Since $\angle MAN = \angle BAC = 180-2\theta$, we have that $AMKN$ is a cyclic quadrilateral.

We know that $Q$ is the center of spiral similarity that takes $BX$ to $CY$, so $\angle MQN = \angle XQY = \angle XAY = 180 - 2\theta$. Thus, the five points $A$, $M$, $K$, $N$, $Q$ are concyclic.

Let $PK \cap XY = Z$. Then, by radical-axis, we know that $Z$ is the midpoint of segment $XY$. By the spiral similarity at $K$, we know that $\angle ZKN = \theta$, or
$\angle QKP = \theta - \angle QKN$.
But, we have
$\angle QKN = \angle QAN = \angle QAY = \angle QXY = \angle QAX - \angle QXA = \theta - \angle QXA$.
It follows that $\angle QXA = \angle QKP$, as desired.
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vntbqpqh234
286 posts
#10 • 9 Y
Y by hqdung, mad, RedFlame2112, Adventure10, Mango247, and 4 other users
That is my proof.
Let $(O),(O_{4})$ be the circumcircle of $\triangle KXY, \triangle AXY$
And $M$ is the intersection of $QO$ with $XY$
$KP$ meets $XY$ at $N$.
Easy to see:
$\angle AXY=\angle AYX=\angle XKY$
then $AX,AY$ touch with $(O$ then
$O$ lies on $O_{4}$ and $O,A,N$ conllinear.
And $KA$ is symmedian of $\triangle XKY$
On the other hand: $\triangle KBX, KXY,KYC$ are similar
th­en:
$\frac{XM}{MY}=\frac{QX}{QY}=\frac{BX}{BY}=\frac{KX^{2}}{KY^{2}}$
Hence $K,M,A$ is conllinear.
Easy to see $M,N,Q,A$ lie on circle.(*)
Have $OK^{2}=OX^{2}=ON.OA$ (**)
when (*) and (**) have
$\angle OQN=\angle KAO=\angle OKN$ th­en $K,O,N,Q$ lie on a circle
Hence $\angle QKA=\angle QOA=\angle QXA$
QED
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dragon96
3212 posts
#11 • 3 Y
Y by RedFlame2112, Adventure10, Mango247
wsjradha wrote:
We know that $Q$ is the center of spiral similarity that takes $BX$ to $CY$, so $\angle MQN = \angle XQY = \angle XAY = 180 - 2\theta$. Thus, the five points $A$, $M$, $K$, $N$, $Q$ are concyclic.

Sorry to revive, but I'm not following why Q is the center of spiral similarity. Also, how does the fact that $\angle MQN = \angle XQY$ follow from that? Thanks.

EDIT: Nevermind.
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antimonyarsenide
875 posts
#12 • 5 Y
Y by bigbang195, me9hanics, RedFlame2112, Adventure10, Mango247
It's amazing how many geometry problems this paper helps you solve:

http://web.mit.edu/yufeiz/www/olympiad/cyclic_quad.pdf

Unfortunately I'd forgotten almost everything in there, so it was hard getting this problem to work D:
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Pedram-Safaei
132 posts
#13 • 2 Y
Y by RedFlame2112, Adventure10
use of the two centers of rotational homothety and two midpoints of two side...
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sayantanchakraborty
505 posts
#14 • 3 Y
Y by RedFlame2112, Adventure10, Mango247
There is also a nice solution using inversion.Just invert with center $A$ and arbitrary radius.Let $X'$ be the image of $X$ under the inversion.Then $X'P'KB'$ and $KPY'C'$ are cyclic.So
$\angle{X'P'Y'}=\angle{X'P'K'}+\angle{Y'PK}=\angle{K'B'A}+\angle{K'C'A}=\angle{BKA}+\angle{CKA}=180-\angle{BAC}$

so $AX'PY'$ is also cyclic.Now some trivial steps yeild you $\angle{AQ'X'}=\angle{AQ'K'}+\angle{AP'K'}$ which means $\angle{QXA}=\angle{QKP}$ as desired.
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stephcurry
2095 posts
#15 • 3 Y
Y by TanMath, RedFlame2112, Adventure10
dang this question took me forever, but it was nice :)

First, WLOG let $Q$ be closer to $Y$ than $X$. Let $\angle BKX = \angle 1, \angle XBY = \angle 2,$ and $\angle XCY = \angle 3$. Since $K$ is the second intersection of the circumcircles of $\triangle PXB$ and $\triangle PYC$, there is a spiral similarity centered at $K$ that takes $\triangle KBX$ to $\triangle KYC$. Notice that this spiral similarity also takes $\triangle KBY$ to $\triangle KXC$, so those triangles are similar as well. Thus, $\angle KYC = \angle 1$ as well. By angle chasing, we find that $\angle CXY = \angle 2$, $\angle BYX = \angle 3$, and $\angle AXY = \angle AYX = \angle 1$. By exterior angles of $\triangle BXY$, we see that $\angle 2 + \angle 3 = \angle 1$. Thus, $\angle XKY = \angle XKP + \angle PKY = \angle 2 + \angle 3 = \angle 1$. Also, we have $\angle KXY = \angle KXC + \angle CXY = \angle KBY + \angle 2 = \angle KBX$, so $\triangle KBX \sim \triangle KXY$, and the spiral similarity that sends $\triangle KBY$ to $\triangle KXC$ also sends $\triangle KBX$ to $\triangle KXY$. It also sends $\triangle KXY$ to $\triangle KYC$ by similar reasoning.

Since $Q$ is the second intersection of the circumcircles of $\triangle AXY$ and $\triangle ABC$, $Q$ is the center of spiral similarity that sends $\triangle QXY$ to $\triangle QBC$, which implies that $\triangle QXY \sim \triangle QBC$ are similar. Since quadrilateral $AQYX$ is cyclic, $\angle XQY = \angle XAY = 180 - 2\angle 1$.

Let M and N be midpoints of BX and CY, respectively.

Lemma 1: Quadrilateral $QMKN$ is cyclic
Proof: By MGT, spirally similar triangles $QXY$ and $QBC$ are similar to $\triangle QMN$, so $\angle MQN = \angle $XQY$ = 180 - 2\angle 1$. By MGT again, spirally similar triangles $KBY$ and $KXC$ are similar to $\triangle KMN$ are similar, so $\angle MKN = \angle BKY = \angle BKX + \angle XKY = 2\angle 1$. This implies that quad $QMKN$ is cyclic because $\angle MQN + \angle MKN = 180$.

Thus by Lemma 1, we have $\angle QKN = \angle QMN = \angle QXY$. By angle chasing, $\angle QKP = \angle PKN - \angle QKN = \angle PKN - \angle QXY$. Now we extend $KP$ to intersect $XY$ at a point $Z$.

Lemma 2: $Z$ is the midpoint of $XY$
Proof: As $KP$ is the radical axis of circles $W_1$ and $Q_2$, $Z$ lies on the radical axis. This implies that it has equal power with respect to both circles, so $ZX^2 = ZY^2$, which implies that $ZX = ZY$, so $Z$ is the midpoint of $XY$. Finally we have $\triangle KXZ \sim \triangle KYN$, where Z is midpoint of XY.

By Lemma 2, the spiral similarity that takes $\triangle KXY$ to $\triangle KYC$ also takes $Z$ to $N$, as both are corresponding midpoints, so $\triangle KZN \sim \triangle KYC$ by spiral similarity. Thus, $\angle ZKN = \angle YKC = \angle 1$, so $\angle QKP = \angle 1 - \angle QXY = \angle QXA$, as desired. $QED$
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tranquanghuy7198
253 posts
#16 • 8 Y
Y by DenisO1998, kapilpavase, lucaslcmlmo, gemcl, Tanloc05, RedFlame2112, Adventure10, Mango247
My solution:
Notice that $\triangle{QXB} \sim \triangle{QYC}$ (same direction)
We construct $\triangle{QKL} \sim \triangle{QXB} \sim \triangle{QYC}$ (same direction)
$\Rightarrow \exists S_{(Q, \alpha, k)}$ which is a spiral similarity that maps:
$X\mapsto{B}, Y\mapsto{C}, K\mapsto{L}$
$\Rightarrow \triangle{BCL} \sim \triangle{XYK}$
$\Rightarrow \angle{BLC} = \angle{XKY} = 180-\angle{XPY} = 180-\angle{BPC}$
$\Rightarrow BPCL$ is cyclic
$\Rightarrow \angle{CPL} = \angle{CBL} = \angle{YXK} = \angle{CPK}$ (because $XY$ is tangent to $\omega_1$)
$\Rightarrow \overline{P, K, L}$
$\Rightarrow \angle{QKP} = 180-\angle{QKL} = 180-\angle{QXB} = \angle{QXA}$
Q.E.D
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gavrilos
233 posts
#17 • 6 Y
Y by USHdog, ayan_mathematics_king, TanMath, noslorn, RedFlame2112, Adventure10
My solution:

Let $T$ be the second intersection point of $W_2$ and the circumcircle of $\triangle{ABC}$.Also,suppose that $R\equiv TY\cap AB$.

We will first show that $\triangle{AXY}$ is isosceles.

Indeed,$\angle{AXY}=180^{\circ}-\angle{BXY}=180^{\circ}-\angle{PXY}-\angle{BXP}=180^{\circ}-\angle{PBX}-\angle{BXP}=\angle{BPX}=\angle{CPY}=$

$=180^{\circ}-\angle{PCY}-\angle{PYC}=180^{\circ}-\angle{PYX}-\angle{PYC}=180^{\circ}-\angle{XYC}=\angle{AYX}$ q.e.d.

We will go on with showing that $XYTB$ is cyclic.Indeed,$\angle{CTY}=180^{\circ}-\angle{YPC}=180^{\circ}-\angle{AXY}$

and $\angle{CTB}=\angle{CAB}=\angle{YAX}=180^{\circ}-2\angle{AXY}$.

Thus,$\angle{YTB}=\angle{CTY}-\angle{CTB}=180^{\circ}-\angle{AXY}-\left(180^{\circ}-2\angle{AXY}\right)=\angle{AXY}$ which is what we wanted.

Hence,$RY\cdot RT=RX\cdot RB$ implying that $R$ has equal powers wrt $W_1,W_2$.Thus,it lies on their radical axis which is $PK$.

Thus,$P,K,R$ are collinear.We have $\angle{QTR}=\angle{QTY}=\angle{QTC}-\angle{CTY}=$

$=180^{\circ}-\angle{QAY}-\left(180^{\circ}-\angle{YPC}\right)=\angle{YPC}-\angle{QAY}=\angle{AXY}-\angle{QAY}=\angle{QXA}=\angle{QXR}$

whence we get that $QRXT$ is cyclic $(1)$.

Also,$\angle{QTK}=\angle{QTY}+\angle{YTK}=\angle{QXA}+\angle{KPB}=$

$=\angle{QXA}+\angle{KXB}=180^{\circ}-\angle{QXK}$ which gives that $QXKT$ is also cyclic $(2)$.

From $(1),(2)$ we conclude that $Q,R,X,K,T$ all lie on the same circle.

Thus,$\angle{QXA}+\angle{QXR}=\angle{QKR}=\angle{QKP}$ which ends the proof.
Attachments:
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SmartClown
82 posts
#18 • 2 Y
Y by RedFlame2112, Adventure10
By easy angle chasing we get $BKYA$ and $CKXA$ are cyclic. Now let $AQ \cap XY= L $ and $PK \cap XY =R$ and $R$ is the midpoint of $XY$. Notice that by inversion at $A$ and radius $\sqrt{XY}$ point $Q$ goes to $L$ so we have $\angle QXA=\angle QLR$ so we need to prove that $QLKR$ is cyclic so it is enough to prove that $\angle KRY=\angle AQK$. Let $\angle XYP=\angle YKR=x$ and $\angle XYP=\angle XKR=y$. Because $AX$ and $AY$ are tangents to the circumcircle of $KXY$ we have $KA$ is the symmedian in $\triangle KXY$. So we have $\angle AKY=y$. Now let the tangent at $K$ of the circumcircle $KXY$ cut $AY$ at $S$. We get $\angle SKA=\angle YRK$ so we need to prove that $\angle SKA=\angle AQK$ which is equivalent to $SK$ being the tangent to the circle $AKQ$ so we need to prove that circumcircles $KXY$ and $KQA$ are tangent to each other. Now we invert at $A$. Now we let the picture of some point $X$ be $X_1$. Now $K_1$ is the point inside the isosceles triangle $AXY$ such that $\angle AY_1K_1=\angle K_1X_1Y_1$ and the equivalent when we switch $X_1$ and $Y_1$. Now because $BKYA$ and $XKCA$ are cyclic we have that $ B_1= AX_1 \cap K_1Y_1$ and $C_1=AY_1 \cap X_1K_1$ and because $AQBC$ and $AQXY$ are cyclic we have $Q_1= X_1Y_1 \cap B_1C_1$. Now we only need to prove that $Q_1K_1$ is tangent to circumcircle $XKY$. Now because $B$ is the intersection of one side of the triangle and the tangent and the same holds for $C_1$ and we need to prove that the same holds for $Q_1$ it is enough to prove that intersections of tangents at $K,X,Y$ and $XY,KY,KX$ respcetively in any triangle $KXY$ are collinear and it follows from easy Menelaus so we are finished.
These angles are like this in my configuration but the idea is the same in any.
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Abubakir
68 posts
#19 • 4 Y
Y by me9hanics, RedFlame2112, Adventure10, Mango247
RaleD wrote:
Let tangents in $K, X, Y$ on circumcircle of $KXY$ intersect sides $XY, KY, KX$ in $Q, C, B$ respectively. Show that $Q, C, B$ are collinear
(in problem we have that $Q, B, C$ are collinear and need to get that $QK$ is tangent but this is same
I don't have nice explanation for this, however it follows from Ceva's theorem.
Actually, this called Pascal's theorem for triangle. Beautiful solution!
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ABCDE
1963 posts
#20 • 4 Y
Y by TanMath, RedFlame2112, Adventure10, Mango247
//cdn.artofproblemsolving.com/images/e/6/b/e6b5363124f5d2731c83dd12eb7e9e04931c0d37.png

First, note that $\angle XKY=\angle XKP+\angle PKY=\angle XBP+\angle YCP=\angle XBP+\angle XYP=\angle AXY$ and similarly $\angle XKY=\angle AYX$, so the circumcircle of $XKY$ is tangent to $AX$ and $AY$. Let $I$ be the incenter of $ABC$. Note that $\angle BPC=180^\circ-\angle XPB=180^\circ-\angle AXY=90^\circ+\angle XAY=\angle BIC$, so $BPIC$ is cyclic. This means that the circumcircle of $BPC$ is symmetric with respect to the angle bisector of $\angle BAC$. Let $M$ be the midpoint of $XY$, $D$ be the harmonic conjugate of $K$ with respect to $X$ and $Y$ (well-known that $D$ is reflection of $P$ across $XY$), $P'$ be the reflection of $P$ over $M$, and $D'$ be the reflection of $D$ over $M$. Note that $DPD'P'$ is a rectangle that is symmetric with respect to $BC$ and the perpendicular bisector of $BC$.

Now, perform a $\sqrt{bc}$ inversion with respect to $KXY$. Note that the circumcircle of $KPX$ maps to $AY$ and the circumcircle of $KQY$ maps to $AX$. This means that $A$ maps to $P$ and $B$ maps to $C$. Now, $Q$ maps to the intersection of the circumcircles of $PBC$ and $PXY$. Note that both circles are symmetric with respect to the angle bisector of $\angle BAC$, so $Q$ maps to $D'$. Now, we have $\angle QKP=\angle AKD'$ and $\angle QXA=\angle QXK-\angle AXK=\angle YD'K-\angle YPK=\angle PYD'+\angle PKD'$. Hence, it suffices to show that $\angle AKP=\angle PYD'$. But $\angle AKP=\angle DKP'$, and the circumcircles of $XYK$ and $XYP$ are reflections of each other across $XY$. But since $P'D$ is the reflection of $PD'$ across $XY$, the corresponding arcs and thus angles are equal as desired.
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anantmudgal09
1980 posts
#21 • 5 Y
Y by TanMath, RedFlame2112, Adventure10, Mango247, everythingpi3141592
WLOG, let $KX<KY$. Note that $$\angle KXB=\angle KPB=180^{\circ}-\angle KPY=\angle KYX$$from which it follows that $AX$ is tangent to $(KXY)$. Similar argument shows that line $AY$ is tangent to $(KXY)$.

Define $R \ne P$ as the other intersection point of circles $(BPC)$ and $(XPY)$. We will show that $XY \parallel PR$. Denote by $T$ the circumcenter of triangle $BPC$ and observe that $$XP\cdot XC=XY^2=YP\cdot YB \Longrightarrow TX=TY \Longrightarrow XY \parallel PR.$$
Apply inversion about $K$ of radius $r^2=KX\cdot KY$ followed by reflection in the bisector of angle $XKY$. Observe that the pairs $\{X,Y\},$ $\{B,C\},$ and $\{P,A\}$ are interchanged under this map. As $(ABC) \rightarrow (PBC)$ and $(AXY) \rightarrow (PXY),$ the points $Q$ and $R$ are also interchanged.

Finally, note that $\angle QKP=\angle RKA=\angle RKP+\angle PKA$ and $$\angle QXA=\angle KXQ-\angle KXA=\angle KRY-\angle KPY=\angle RKP+\angle RYP.$$Thus, we have $$\angle QXA=\angle QKP \iff \angle RYP=\angle PKA \iff \angle PXY-\angle PYX=\angle PKA$$which holds as $\angle PXY=\angle PKX$ and $\angle PYX=\angle PKY$ proving the result. $\square$
This post has been edited 2 times. Last edited by anantmudgal09, Dec 11, 2016, 9:34 AM
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TheUltimate123
1740 posts
#23 • 4 Y
Y by mijail, RedFlame2112, Adventure10, Mango247
[asy]
        size(8cm);
        defaultpen(fontsize(9pt));

        pen pri=deepgreen+linewidth(0.5);
        pen sec=royalblue+linewidth(0.5);
        pen tri=rgb(41, 207, 255)+linewidth(0.5);
        pen fil=invisible;
        pen sfil=invisible;
        pen tfil=invisible;

        pair X, Y, O1, O2, P, K, B, C, A, Q, L;
        X=(-1, 0);
        Y=(1, 0);
        O1=(-1, -0.9);
        O2=(1, -1.3);
        P=intersectionpoints(circle(O1, length(X-O1)), circle(O2, length(Y-O2)))[0];
        K=intersectionpoints(circle(O1, length(X-O1)), circle(O2, length(Y-O2)))[1];
        B=intersectionpoint(circle(O1, length(X-O1)), (P+(P-Y)*0.01) -- (P+(P-Y)*100));
        C=intersectionpoint(circle(O2, length(Y-O2)), (P+(P-X)*0.01) -- (P+(P-X)*100));
        A=extension(B, X, C, Y);
        Q=intersectionpoints(circumcircle(A, B, C), circumcircle(A, X, Y))[1];
        L=intersectionpoint(K -- (K+(K-P)*100), circumcircle(B, C, P));

        X=(X-B)/(C-B);
        Y=(Y-B)/(C-B);
        O1=(O1-B)/(C-B);
        O2=(O2-B)/(C-B);
        P=(P-B)/(C-B);
        K=(K-B)/(C-B);
        A=(A-B)/(C-B);
        Q=(Q-B)/(C-B);
        L=(L-B)/(C-B);
        B=(0,0);
        C=(1,0);

        filldraw(circle(O1, length(X-O1)), fil, pri);
        filldraw(circle(O2, length(Y-O2)), fil, pri);
        draw(B -- Y -- X -- C, pri);
        draw(B -- A -- C -- B, sec);
        filldraw(circumcircle(A, B, C), sfil, sec);
        filldraw(circumcircle(A, X, Y), sfil, sec);
        draw(P -- L, tri);
        draw(B -- L -- C, tri);
        filldraw(circumcircle(B, P, C), tfil, tri);

        dot("$X$", X, dir(100));
        dot("$Y$", Y, dir(80));
        dot("$P$", P, dir(100));
        dot("$K$", K, dir(-30));
        dot("$B$", B, dir(210));
        dot("$C$", C, dir(-30));
        dot("$A$", A, N);
        dot("$Q$", Q, dir(120));
        dot("$L$", L, S);
[/asy]
Note that $K$ is the Miquel point of $BXCY$ and $Q$ is the Miquel point of $BXYC$. Let the spiral similarity at $Q$ sending $\overline{XY}$ to $\overline{BC}$ send $K$ to $L$, so $\triangle BLC\sim\triangle XKY$. Note that
\begin{align*}
    \measuredangle BLC&=\measuredangle XKY=\measuredangle XKP+\measuredangle XKP+\measuredangle PKY\\
    &=\measuredangle YXP+\measuredangle PYX=\measuredangle YPX=\measuredangle BPC,
\end{align*}so $BPCL$ is cyclic. Moreover, \[\measuredangle BPL=\measuredangle BCL=\measuredangle XYK=\measuredangle YCK=\measuredangle BXK=\measuredangle BPK,\]whence $L$ lies on $\overline{PK}$. It follows that $\measuredangle QKP=\measuredangle QKL=\measuredangle QXB=\measuredangle QXA$, as required.
This post has been edited 1 time. Last edited by TheUltimate123, Nov 3, 2019, 3:19 AM
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v_Enhance
6877 posts
#24 • 7 Y
Y by Physicsknight, EulersTurban, AlastorMoody, v4913, myh2910, RedFlame2112, Adventure10
With Aatman Supkar, Alan Chen, Alexandru Girban, Anushka Aggarwal, Daniel Sheremeta, Derek Liu, Dhrubajyoti Ghosh, Gene Yang, Jeff Lin, Paul Hamrick, Raymond Feng, Robin Son, and Sean Li (plus Eric Shen cheerleading :P looks like we even ended up with the same letters!):

Let the spiral similarity at $Q$ mapping $\overline{XY}$ to $\overline{BC}$ take $K$ to $L$.

Claim: The point $L$ lies on $(PBC)$ and $\overline{KP}$.
Proof. This is only place where the fact $\overline{XY}$ is tangent gets used. Angle chase: \begin{align*} 		\measuredangle BPK &= \measuredangle YPK = \measuredangle XYK = \measuredangle BCL \\ 		\measuredangle CPK &= \measuredangle XPK = \measuredangle YXK = \measuredangle CBL. 	\end{align*}Adding the two gives $BPCL$ cyclic. Then $\measuredangle BPK = \measuredangle BPL$, done. $\blacksquare$

Now, by spiral similarity, $\measuredangle QXA = \measuredangle QXB = \measuredangle QKL = \measuredangle QKP$, as needed.


A few behind-the-scenes quotes while we were working on this:
  • ms paint > ggb confirmed
  • if we were in ms paint, "we would [unfortunately] be trying to do spiral similarities at Q and K over and over"
  • evan during the solving: something's wrong i can feel it
  • wow you can turn the random properties we found into a problem now :)
  • i thought it would be spiral sim the whole way through but then you guys found all of these nice properties
This post has been edited 6 times. Last edited by v_Enhance, Aug 25, 2021, 2:22 AM
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TheUltimate123
1740 posts
#25 • 7 Y
Y by Th3Numb3rThr33, HolyMath, AlastorMoody, RedFlame2112, YaoAOPS, Adventure10, Mango247
Here are proofs to a bunch of observations made on Evan's wild goose chase. Evan's diagram

Let $\overline{AK}$ and $\overline{XY}$ intersect at $D$, and let $O$ be the circumcenter of $\triangle KXY$. Let $M$ be the midpoint of $\overline{XY}$, $E$ the second intersection of $\overline{AK}$ and $(KXY)$, and $T$ the intersection of the tangent to $(KXY)$ at $K$ and line $XY$. Denote by $B_1$ and $C_1$ the midpoints of $\overline{BX}$ and $\overline{CY}$ respectively, and $Z$ the image of $Q$ wrt.\ inversion at $K$ with radius $\sqrt{KX\cdot KY}$ followed by a reflection over the angle bisector of $\angle XKY$.

By construction, $P$ is an HM-point of $\triangle KXY$. Thus $\overline{KP}$ bisects $\overline{XY}$ (also by radical axes), and $(XPY)$ passes through the orthocenter of $\triangle KXY$.

Claim 1. $\overline{AX}$ and $\overline{AY}$ are tangent to $(KXY)$. Proof

Claim 2. $O$, $D$, $Q$ are collinear. Proof

Claim 3. $A$, $Q$, $T$ are collinear. Proof

Claim 4. $A$, $Q$, $B_1$, $K$, $C_1$ lie on a circle tangent to $(KXY)$. Proof

Claim 5. $P$ and $Z$ are reflections across $\overline{AO}$. Proof
This post has been edited 2 times. Last edited by TheUltimate123, Nov 4, 2019, 1:15 AM
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FISHMJ25
293 posts
#30 • 4 Y
Y by fiste, RedFlame2112, Adventure10, Mango247
solution
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sriraamster
1492 posts
#31 • 3 Y
Y by jj_ca888, RedFlame2112, Adventure10
Note that $K$ is the center of a spiral similarity sending $BX \mapsto YC$ and $Q$ is the center of a spiral similarity sending $XY \mapsto BC$. Also, observe that $PQ \cap XY  = M$, the midpoint of $XY$ from a well known lemma. We also have $\measuredangle KYX = \measuredangle KCY$ and $\measuredangle KBX = \measuredangle YXK$ whence $\triangle KXY \sim \triangle KBX \sim \triangle KYC$. Due to this, $\measuredangle BXY = \measuredangle XYC$ whence $\measuredangle AXY = \measuredangle XYA$ so $AXY$ is isosceles. Combined with the angle statements, this implies that $AX$ and $AY$ are tangents to $(KXY)$. Let $AK \cap XY = L$. Observe that $KL$ is a symmedian in $\triangle KXY$. We have $$\frac{QX}{QY} = \frac{BX}{YC} = \frac{BK}{KY} = \frac{KX^2}{KY^2}$$which follows from $\frac{BK}{KX} = \frac{KX}{KY}$. Likewise due to symmedian ratios, $\frac{KX^2}{KY^2} = \frac{XL}{LY}$, so $QL$ bisects $\angle XQY$. Observe that $\measuredangle LQY = \measuredangle MAY$, so $LQ$ and $AM$ intersect on $(AXY)$, say at point $O$. As $\triangle AXY$ is isosceles and $\measuredangle AXY = \measuredangle XYA = \measuredangle XKY$, we have $\measuredangle YAX = 180 - 2 \measuredangle XKY$ so $\measuredangle XOY = 2 \measuredangle XKY$ which, combined with $AM \perp XY$, implies that $O$ is the circumcenter of $\triangle XKY$.

Claim: $KOMQ$ is cyclic
Proof: Invert about $(KXY)$. Observe that $M \mapsto A$ since they are polar. Now note that as $AO$ is a diameter of $(AXY)$, $\measuredangle  OQA = 90$. Also, $AM \perp XY$, so $AQLM$ is cyclic whence $OL \cdot OQ = OM \cdot OA$, so $L$ and $Q$ are swapped after inversion. Therefore, $(MQK)$ is mapped to a line after inversion about $O$, so $KOMQ$ is cyclic as desired. $\blacksquare$

To finish, $\measuredangle QKM = \measuredangle QOA = \measuredangle QXA$, and we are done. $\blacksquare$
This post has been edited 2 times. Last edited by sriraamster, Nov 25, 2019, 3:19 AM
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AlastorMoody
2125 posts
#32 • 3 Y
Y by gamerrk1004, RedFlame2112, kamatadu
Solution
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Ninjasolver0201
722 posts
#33 • 1 Y
Y by RedFlame2112
[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 = -17.017872221696305, xmax = 27.256889742760723, ymin = -9.900050867795619, ymax = 14.96610115795648;  /* image dimensions */
pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqqqcc = rgb(0,0,0.8); pen qqzzqq = rgb(0,0.6,0); pen ccqqqq = rgb(0.8,0,0); 

draw(circle((-1.8490848403873745,6.896735451806183), 4.883756988528579), linewidth(1.2) + qqzzqq); 
draw((-3.7239545994373704,11.406274863941478)--(-7.019406119164214,-2.623582796016489)--(14.291278423016704,0.3099251759874293)--cycle, linewidth(2) + ccqqqq); 
 /* draw figures */
draw(circle((6.15395941298003,-0.282390454652267), 8.158847864582619), linewidth(2) + yqqqyq); 
draw(circle((-4,0), 4), linewidth(1.6) + qqqqcc); 
draw((-5.53559627430592,3.6935002480489656)--(3.0217853171095594,7.251286198255153), linewidth(2) + linetype("2 2") + blue); 
draw((-5.53559627430592,3.6935002480489656)--(14.291278423016704,0.3099251759874293), linewidth(2)); 
draw((-7.019406119164214,-2.623582796016489)--(3.0217853171095594,7.251286198255153), linewidth(2)); 
draw((-7.019406119164214,-2.623582796016489)--(-3.7239545994373704,11.406274863941478), linewidth(2) + red); 
draw((-3.7239545994373704,11.406274863941478)--(14.291278423016704,0.3099251759874293), linewidth(2) + red); 
draw(circle((3.1476653838199984,2.3902501171896704), 11.336130884404211), linewidth(1.6)); 
draw((-7.019406119164214,-2.623582796016489)--(14.291278423016704,0.3099251759874293), linewidth(2) + red); 
draw((-6.527616930354384,8.297599401713773)--(-5.53559627430592,3.6935002480489656), linewidth(2)); 
draw((-6.527616930354384,8.297599401713773)--(-1.4958442565649777,-3.119167198567814), linewidth(2)); 
draw((-1.4958442565649777,-3.119167198567814)--(-1.3263554851829753,2.9751680638896403), linewidth(2)); 
draw((-3.7239545994373704,11.406274863941478)--(-7.019406119164214,-2.623582796016489), linewidth(2) + ccqqqq); 
draw((-7.019406119164214,-2.623582796016489)--(14.291278423016704,0.3099251759874293), linewidth(2) + ccqqqq); 
draw((14.291278423016704,0.3099251759874293)--(-3.7239545994373704,11.406274863941478), linewidth(2) + ccqqqq); 
 /* dots and labels */
dot((6.15395941298003,-0.282390454652267),dotstyle); 
label("$O_2$", (6.547143942243271,-0.17767456702917347), NE * labelscalefactor); 
dot((-4,0),dotstyle); 
label("$O_1$", (-4.042656266398287,-0.9005278235917346), NE * labelscalefactor); 
dot((-1.3263554851829753,2.9751680638896403),linewidth(4pt) + dotstyle); 
label("$P$", (-1.621097856913698,3.5088770414398875), NE * labelscalefactor); 
dot((-1.4958442565649777,-3.119167198567814),linewidth(4pt) + dotstyle); 
label("$K$", (-1.837953833882467,-3.9365115011544907), NE * labelscalefactor); 
dot((-5.738502857552112,-3.6024446996842476),linewidth(4pt) + dotstyle); 
dot((-5.53559627430592,3.6935002480489656),linewidth(4pt) + dotstyle); 
dot((-5.53559627430592,3.6935002480489656),linewidth(4pt) + dotstyle); 
label("$X$", (-6.283501361742234,3.6534476927524), NE * labelscalefactor); 
dot((3.0217853171095594,7.251286198255153),linewidth(4pt) + dotstyle); 
label("$Y$", (3.149733636399222,7.55685527819023), NE * labelscalefactor); 
dot((-7.019406119164214,-2.623582796016489),linewidth(4pt) + dotstyle); 
label("$B$", (-7.656922549211105,-3.1775155817638017), NE * labelscalefactor); 
dot((14.291278423016704,0.3099251759874293),linewidth(4pt) + dotstyle); 
label("$C$", (14.426244438775216,0.5813213523615156), NE * labelscalefactor); 
dot((-3.7239545994373704,11.406274863941478),linewidth(4pt) + dotstyle); 
label("$A$", (-3.97037094074203,11.82168949190934), NE * labelscalefactor); 
dot((-6.527616930354384,8.297599401713773),linewidth(4pt) + dotstyle); 
label("$Q$", (-7.259353258101695,8.09899522061215), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */

[/asy]

Let $O_3$ be circumcircle of $\Delta KXY$ and $O_4$ be the circumcircle of $\Delta AXY$. Let $XY \cap QO_3=R$, $KP \cap XY=S$. Notice, that the angles in circle $\angle XKY=\angle AXY= \angle XKY$, then $O_3$ lies on $O_4$ and hence $O_3, A, N$ are collinear.
Hence, by Angle Chasing we can say through AA- similarity that $\Delta KBX, \Delta KXY, \Delta KYC$ are similar triangles. Using the simillarity results we can say that $K, R, A$ are collinear. Hence, points $Q,A, R,S$ form a cyclic quadrilateral with another circle.With the property of power of point $O_3K^2=O_3K\cdot O_3A=O_3X^2$, hence $\angle O_3QS=\angle KAO_3=\angle OKS$, therefore $K, O_3, S, Q$ form a cyclic quadrilateral, hence all the $4$ points lie on a circle.
$\therefore \angle QO_3A=\angle QKA=\angle QKA. \blacksquare$
This post has been edited 1 time. Last edited by Ninjasolver0201, Jan 21, 2021, 9:10 AM
Reason: ADD
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myh2910
41 posts
#34 • 2 Y
Y by v_Enhance, RedFlame2112
v_Enhance wrote:
\begin{align*} \measuredangle BPK &= \measuredangle YPK = \measuredangle XYP = \measuredangle BCL \\ \measuredangle CPK &= \measuredangle XPK = \measuredangle YXP = \measuredangle CBL. \end{align*}
I think it should be \begin{align*} \measuredangle BPK &= \measuredangle YPK = \measuredangle XYK = \measuredangle BCL \\ \measuredangle CPK &= \measuredangle XPK = \measuredangle YXK = \measuredangle CBL. \end{align*}
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kamatadu
480 posts
#35 • 2 Y
Y by RedFlame2112, HoripodoKrishno
Anshu-Droid's solve again :)
IRAN TST 2010 P5 wrote:
Circles $W_1,W_2$ intersect at $P,K$. $XY$ is common tangent of two circles which is nearer to $P$ and $X$ is on $W_1$ and $Y$ is on $W_2$. $XP$ intersects $W_2$ for the second time in $C$ and $YP$ intersects $W_1$ in $B$. Let $A$ be intersection point of $BX$ and $CY$. Prove that if $Q$ is the second intersection point of circumcircles of $ABC$ and $AXY$
\[\angle QXA=\angle QKP\]

https://i.imgur.com/KXO9s31.png

Let $D = W_1 \cap \odot ABC, G = W_2 \cap \odot ABC, F = DX \cap \odot ABC, F' = GY \cap \odot ABC, E = DX \cap AC$ and $Z$ be a point on $AF$ such that $A$ lies between $Z$ and $F$.

Claim: $\overline{F - Y - G}$ are collinear.
Proof: By Reim's on $\{ W_1, \odot ABC \}$ wrt lines $DX$ and $BX$, we get $XX \parallel AF$. Now again by Reim's on $\{ W_2, \odot ABC \}$ wrt lines $GY$ and $CY$, we get $YY \parallel AF'$. But now, $$AF \parallel XX \equiv YY \parallel AF' \implies F' = F.$$
Claim: $DXYC$ is cyclic.
Proof: $\measuredangle YCD = \measuredangle ACD = \measuredangle AFD = \measuredangle AFX = \measuredangle YXF = \measuredangle YXD$.

Now by Radical Center Theorem on $\{ W_1, W_2, \odot DXYC \}$, we get $E = DX \cap PK \cap CY$.

Claim: $AX = AY$.
Proof: $\measuredangle AYX = \measuredangle YAF = \measuredangle CAF = \measuredangle CGF = \measuredangle CGY = \measuredangle CPY = \measuredangle XPB = \measuredangle XDB = \measuredangle FDB = \measuredangle FAB = \measuredangle FAX = \measuredangle YXA$.

So, from our previous claim, we get that $AF$ is tangent to $\odot AXY$.

Claim: $DEYK$ is cyclic.
Proof: $\measuredangle YDE = \measuredangle YDX = \measuredangle YCX = \measuredangle YCP = \measuredangle YKP = \measuredangle YKE$.

Claim: $QEYD$ is cyclic.
Proof: $\measuredangle QYE = \measuredangle QYA = \measuredangle QAZ = \measuredangle QAF = \measuredangle QDF = \measuredangle QDE$.

So, from our previous $2$ claims, we come to the conclusion that $QEYKD$ is cyclic.

Now, $$\angle QXA = \angle QYA = \angle QYE = \angle QKE = \angle QKP$$and we are done :D :blush:.
This post has been edited 2 times. Last edited by kamatadu, Dec 12, 2022, 6:25 PM
Reason: asd
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CyclicISLscelesTrapezoid
372 posts
#36 • 1 Y
Y by Mango247
https://cdn.discordapp.com/attachments/1052853505578246184/1062571873147310100/Screen_Shot_2023-01-10_at_7.20.28_PM.png

wow.

We have $\angle AXY=\angle XBY+\angle XYB=\angle XKP+\angle YKP=\angle XKY$, so $\overline{AX}$ is tangent to the circumcircle of $KXY$. Similarly, $\overline{AY}$ is tangent to the circumcircle of $KXY$. Let $M$ be the midpoint of $\overline{BC}$, let $P'$ be the reflection of $P$ over $M$, and let $R$ be the reflection of $P$ over $\overline{BC}$. It's clear that $M$ and $P'$ lie on $\overline{PK}$. We have \[\angle XP'Y=\angle XPY=360^\circ-\angle KPX-\angle KPY=\angle KXY+\angle KYX=180^\circ-\angle XKY,\]so $XKYP'$ is cyclic. Thus, $KR$ is a symmedian of $\triangle KXY$, so $A$, $R$, and $K$ are collinear.

Let $\overline{AP'}$ and $\overline{AR}$ intersect the circumcircle of $AXY$ again at $T$ and $T'$, respectively. Notice the following similarities:
  • Since $\angle KBX=\angle KXY$ and $\angle KXB=\angle KYX$, we have $\triangle KBX \sim \triangle KXY$. Similarly, $\triangle KYC \sim \triangle KXY$.
  • Since $Q$ is the Miquel point of $BXYC$, $\triangle QBX \sim \triangle QCY$.
  • We have \[\angle KXT'=\angle KXY-\angle YXT'=180^\circ-\angle AYK-\angle KAY=\angle YKT',\]and similarly, $\angle KYT'=\angle XKT'$, so $\triangle T'XK \sim \triangle T'KY$.
Thus, we have \[\frac{T'X}{T'Y}=\frac{T'X}{T'K} \cdot \frac{T'K}{T'Y}=\left(\frac{KX}{KY}\right)^2\]and \[\frac{QX}{QY}=\frac{BX}{CY}=\frac{BX}{XY} \cdot \frac{XY}{YC}=\left(\frac{KX}{KY}\right)^2,\]so $QXT'Y$ is harmonic.

Since $\overline{QT'}$ is a symmedian of $QXY$, $\overline{QT}$ is a median. We have $MK \cdot MP'=MX \cdot MY=MQ \cdot MT$, so $KQP'T$ is cyclic, which means $\angle QKP=\angle QTP'=\angle QXA$.
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Apr 12, 2023, 4:25 AM
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YaoAOPS
1541 posts
#37 • 2 Y
Y by Leo.Euler, OronSH
I've come with the chaotic neutral solution.

Let $P'$ be the reflection of $P$ over $XY$ and let $M$ be the midpoint of $XY$.

Claim: $(P'XKY)$ is cyclic and tangent to $AX, AY$.
Proof. \[ \measuredangle XKY = \measuredangle XBP + \measuredangle XCY = \measuredangle YXP + \measuredangle PYX = \measuredangle YPX = \measuredangle XP'Y \]gives that it is cyclic.
Then, \[ \measuredangle AXY = \measuredangle BXY = \measuredangle BXP + \measuredangle PXY = \measuredangle BXP + \measuredangle PBX = \measuredangle BPX = \measuredangle YPX \]as desired, symmetry suffices. $\blacksquare$

Claim: The center $G$ of that $(P'XKY)$ lies on $(AXY)$ and the angle bisector of $\angle BAC$.
Proof. $\measuredangle AXG = \measuredangle AYG = 90^\circ$, then SAS gives the angle condition. $\blacksquare$

Claim: $A$ lies on $P'K$.
Proof. Since $AB$ and Quadrilateral $P'XKY$ is harmonic.
As such, since $AX$ and $AY$ are tangents, the result follows. $\blacksquare$
Thus, $XY$, and the tangents from $P'$ and $K$ concur at some point $E$.

Claim: $(EP'MGK)$ is cyclic.
Proof. This follows as \[ \measuredangle EQG = \measuredangle EMG = \measuredangle EP'G = \measuredangle EK'G = 90^\circ \]$\blacksquare$

Claim: $Q$, $D$, $G$ are collinear.
Proof. Length bash. $\blacksquare$

Claim: $E$ lies on $AQ$.
Proof. Radical axis on $(P'MGK)$, $(AQXFGY)$, and $(PX'YK)$. $\blacksquare$

As such, $\measuredangle EQG = \measuredangle AQG$ holds as well, and thus $Q$ lies on $(EP'MGK)$. As such, it follows that \[ \measuredangle QXA = \measuredangle QGA = \measuredangle QGM = \measuredangle QKM = \measuredangle QKP \]as desired.
Attachments:
This post has been edited 2 times. Last edited by YaoAOPS, Sep 7, 2023, 10:36 PM
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math_comb01
662 posts
#38
Y by
Cute Spiral
[asy]
size(8cm);
defaultpen(fontsize(9pt));

pen pri=green+linewidth(0.5);
pen sec=blue+linewidth(0.5);
pen tri=rgb(41, 207, 255)+linewidth(0.5);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;

pair X, Y, O1, O2, P, K, B, C, A, Q, L;
X=(-1, 0);
Y=(1, 0);
O1=(-1, -0.9);
O2=(1, -1.3);
P=intersectionpoints(circle(O1, length(X-O1)), circle(O2, length(Y-O2)))[0];
K=intersectionpoints(circle(O1, length(X-O1)), circle(O2, length(Y-O2)))[1];
B=intersectionpoint(circle(O1, length(X-O1)), (P+(P-Y)*0.01) -- (P+(P-Y)*100));
C=intersectionpoint(circle(O2, length(Y-O2)), (P+(P-X)*0.01) -- (P+(P-X)*100));
A=extension(B, X, C, Y);
Q=intersectionpoints(circumcircle(A, B, C), circumcircle(A, X, Y))[1];
L=intersectionpoint(K -- (K+(K-P)*100), circumcircle(B, C, P));

X=(X-B)/(C-B);
Y=(Y-B)/(C-B);
O1=(O1-B)/(C-B);
O2=(O2-B)/(C-B);
P=(P-B)/(C-B);
K=(K-B)/(C-B);
A=(A-B)/(C-B);
Q=(Q-B)/(C-B);
L=(L-B)/(C-B);
B=(0,0);
C=(1,0);

filldraw(circle(O1, length(X-O1)), fil, pri);
filldraw(circle(O2, length(Y-O2)), fil, pri);
draw(B -- Y -- X -- C, pri);
draw(B -- A -- C -- B, sec);
filldraw(circumcircle(A, B, C), sfil, sec);
filldraw(circumcircle(A, X, Y), sfil, sec);
draw(P -- L, tri);
draw(B -- L -- C, tri);
filldraw(circumcircle(B, P, C), tfil, tri);

dot("$X$", X, dir(100));
dot("$Y$", Y, dir(80));
dot("$P$", P, dir(100));
dot("$K$", K, dir(-30));
dot("$B$", B, dir(210));
dot("$C$", C, dir(-30));
dot("$A$", A, N);
dot("$Q$", Q, dir(120));
dot("$L$", L, S);
[/asy]
Let $K$ go to $L$ under the spiral at $Q$
Claim: $BPCL$ is cyclic
Proof
Claim 2: $P-K-L$
Proof
Now $\measuredangle QKP = \measuredangle QKL = \measuredangle QXA$. Done. $\blacksquare$
This post has been edited 2 times. Last edited by math_comb01, Jan 25, 2024, 2:43 PM
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