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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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[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
jlacosta
Mar 2, 2025
0 replies
geometry
srnjbr   1
N a few seconds ago by removablesingularity
the points f,n,o, t a lie in the plane such that the triangles tfo ton are similar, preserving direction and order, and fano is a parallelogram. show that of×on=oa×ot.
1 reply
srnjbr
an hour ago
removablesingularity
a few seconds ago
D1010 : How it is possible ?
Dattier   8
N 24 minutes ago by Dattier
Source: les dattes à Dattier
Is it true that$$\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0 $$?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
8 replies
Dattier
Mar 10, 2025
Dattier
24 minutes ago
GOTEEM #5: Circumcircle passes through fixed point
tworigami   21
N an hour ago by Ilikeminecraft
Source: GOTEEM: Mock Geometry Contest
Let $ABC$ be a triangle and let $B_1$ and $C_1$ be variable points on sides $\overline{BA}$ and $\overline{CA}$, respectively, such that $BB_1 = CC_1$. Let $B_2 \neq B_1$ denote the point on $\odot(ACB_1)$ such that $BC_1$ is parallel to $B_1B_2$, and let $C_2 \neq C_1$ denote the point on $\odot(ABC_1)$ such that $CB_1$ is parallel to $C_1C_2$. Prove that as $B_1, C_1$ vary, the circumcircle of $\triangle AB_2C_2$ passes through a fixed point, other than $A$.

Proposed by tworigami
21 replies
tworigami
Jan 2, 2020
Ilikeminecraft
an hour ago
Strike the inequality
giangtruong13   1
N an hour ago by arqady
Source: Idk
Let $a,b,c \geq 0$ satisfy that $a+b+c=3$. Prove that $$\sum a\sqrt{b^3+1} \leq 5$$
1 reply
giangtruong13
5 hours ago
arqady
an hour ago
Linear algebra
Dynic   2
N Today at 3:32 PM by loup blanc
Let A and B be two square matrices with the same size. Prove that if AB is an invertible matrix, then A and B are also invertible matrices
2 replies
Dynic
Today at 2:48 PM
loup blanc
Today at 3:32 PM
3-dimensional matrix system
loup blanc   0
Today at 1:04 PM
Let $A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$.
i) Find the matrices $B\in M_3(\mathbb{R})$ s.t. $A^TA=B^TB,AA^T=BB^T$.
ii) Show that each solution of i) is in $M_3(K)$, where $K=\mathbb{Q}[\sin^2(\dfrac{2}{3}\arctan(3\sqrt{3}))]$.
iii) Solve i) when $B\in M_3(\mathbb{C})$.
EDIT. In iii) we again consider the transpose of $B$ and not its conjugate transpose.
0 replies
loup blanc
Today at 1:04 PM
0 replies
Very hard group theory problem
mathscrazy   3
N Today at 9:50 AM by quasar_lord
Source: STEMS 2025 Category C6
Let $G$ be a finite abelian group. There is a magic box $T$. At any point, an element of $G$ may be added to the box and all elements belonging to the subgroup (of $G$) generated by the elements currently inside $T$ are moved from outside $T$ to inside (unless they are already inside). Initially $
T$ contains only the group identity, $1_G$. Alice and Bob take turns moving an element from outside $T$ to inside it. Alice moves first. Whoever cannot make a move loses. Find all $G$ for which Bob has a winning strategy.
3 replies
mathscrazy
Dec 29, 2024
quasar_lord
Today at 9:50 AM
limit of u(pi/45)
EthanWYX2009   0
Today at 7:08 AM
Source: 2025 Pi Day Challenge T5
Let \(\omega\) be a positive real number. Divide the positive real axis into intervals \([0, \omega)\), \([\omega, 2\omega)\), \([2\omega, 3\omega)\), \([3\omega, 4\omega)\), \(\ldots\), and color them alternately black and white. Consider the function \(u(x)\) satisfying the following differential equations:
\[
u''(x) + 9^2u(x) = 0, \quad \text{for } x \text{ in black intervals},
\]\[
u''(x) + 63^2u(x) = 0, \quad \text{for } x \text{ in white intervals},
\]with the initial conditions:
\[
u(0) = 1, \quad u'(0) = 1,
\]and the continuity conditions:
\[
u(x) \text{ and } u'(x) \text{ are continuous functions}.
\]Show that
\[
\lim_{\omega \to 0} u\left(\frac{\pi}{45}\right) = 0.
\]
0 replies
EthanWYX2009
Today at 7:08 AM
0 replies
Integration Bee Kaizo
Calcul8er   42
N Today at 12:57 AM by awzhang10
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
42 replies
Calcul8er
Mar 2, 2025
awzhang10
Today at 12:57 AM
maximum value
Tip_pay   3
N Yesterday at 9:37 PM by alexheinis
Find the value $x\in [0,4]$ at which the function $f(x)=\int_{0}^{\sqrt{x}}\ln \frac{e}{1+t^2}dt$ takes its maximum value
3 replies
Tip_pay
Yesterday at 8:48 PM
alexheinis
Yesterday at 9:37 PM
Spheres and a point source of light
mofidy   3
N Yesterday at 9:22 PM by kiyoras_2001
How many spheres are needed to shield a point source of light?
Unfortunately, I didn't find a suitable solution for it on the page below:
https://artofproblemsolving.com/community/c4h1469498p8521602
Here too, two different solutions are given:
https://math.stackexchange.com/questions/2791186
3 replies
mofidy
Mar 11, 2025
kiyoras_2001
Yesterday at 9:22 PM
Real Analysis
rljmano   4
N Yesterday at 4:38 PM by alexheinis
In [0 1], {f(t)}*{f’(t)-1} =0 and f is continuously differentiable. How do we conclude that either f is identically zero or f’(t) is identically 1 in [0 1]?
4 replies
rljmano
Yesterday at 6:32 AM
alexheinis
Yesterday at 4:38 PM
find the isomorphism
nguyenalex   14
N Yesterday at 1:49 PM by Royrik123456
I have the following exercise:

Let $E$ be an algebraic extension of $K$, and let $F$ be an algebraic closure of $K$ containing $E$. Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$, then $\sigma$ extends to an automorphism of $F$.

My attempt:

Theorem (*): Suppose that $E$ is an algebraic extension of the field $K$, $F$ is an algebraically closed field, and $\sigma: K \to F$ is an embedding. Then, there exists an embedding $\tau: E \to F$ that extends $\sigma$. Moreover, if $E$ is an algebraic closure of $K$ and $F$ is an algebraic extension of $\sigma(K)$, then $\tau$ is an isomorphism.

Back to our main problem:

Since $K \subset E$ and $F$ is an algebraic extension of $K$, it follows that $F$ is an algebraic extension of $E$. Assume that there exists an embedding $\sigma : E \to F$ such that $\sigma(c) = c$ for all $c \in K$. By Theorem (*), there exists an embedding $\tau : F \to F$ that extends $\sigma$. Since $F$ is algebraically closed, $\tau(F)$ is also an algebraically closed field.

Furthermore, because $\sigma(c) = c$ for all $c \in K$ and $\tau$ is an extension of $\sigma$, we have
$$K = \sigma(K) \subset K \subset \sigma(E) \subset \tau(F) \subset F.$$
This implies that $F$ is an algebraic extension of $\tau(F)$. We conclude that $F = \tau(F)$, meaning that $\tau$ is an automorphism. (Finished!!)

Let choose $F = A$ be the field of algebraic numbers, $K=\mathbb{Q}$. Consider the embedding $\sigma: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}) \subset A$ defined by
$$
a + b\sqrt{2} \mapsto a - b\sqrt{2}.
$$Then, according to the exercise above, $\sigma$ extends to an isomorphism
$$
\bar{\sigma}: A \to A.
$$How should we interpret $\bar{\sigma}$?
14 replies
nguyenalex
Mar 12, 2025
Royrik123456
Yesterday at 1:49 PM
find the convex sets satisfied
nguyenalex   2
N Yesterday at 9:54 AM by ILOVEMYFAMILY
In $\mathbb{R}^2$, let $B = \{(x, y) \mid x \geq 0\}$. Find all convex sets $C$ such that

\[\mathcal{E}(B \cup C) = B \cup C.\]
2 replies
nguyenalex
Yesterday at 8:57 AM
ILOVEMYFAMILY
Yesterday at 9:54 AM
An I for an I
Eyed   65
N Today at 12:51 AM by joshualiu315
Source: 2020 ISL G8
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$. Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$. Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$, and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$. Rays $PM$ and $QN$ meet at $X$. Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$.

Show that $A,X,Y$ are collinear.
65 replies
Eyed
Jul 20, 2021
joshualiu315
Today at 12:51 AM
An I for an I
G H J
Source: 2020 ISL G8
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dolphinday
1310 posts
#57
Y by
Let $K$ be the intersection point of the tangent at $I$ and $\Gamma$. Then $\angle BIC = \angle BIK + \angle CIK = \angle IMB + \angle CNI = \angle YCI + \angle YBI$. Note that $\angle BIC = 90^{\circ} + \frac{\angle A}{2}$ so $\angle BYC = 360^{\circ} - (180^{\circ} - \angle A) = 180^{\circ} - \angle A \implies ABYC$ is cyclic.
Notice that $MNC_1B_1$ is an isosceles trapezoid due to the fact that the perpendicular bisectors of $MB_1$ and $NC_1$ coincide. Note that $PBCQ$ is cyclic.
We can then prove that $\triangle XMN \sim \triangle XQP$. $\angle XNM = 180^{\circ} - (\angle MNC_1 + \angle C_1NQ) = 180^{\circ} - (\angle MBB_1 + 180^{\circ} - \angle C_1CQ)$ $= \angle BPM + \angle C_1CQ - 180^{\circ} = \angle BPM - \angle BPQ = \angle XPQ$.
By using similar triangle ratios, we find that $XM \cdot XP = XN \cdot XQ$, so $AX$ is the radical axis of $\omega_B$ and $\omega_C$. Similarly, we have $XN \cdot XQ = XM \cdot XP \implies AX$ is the radical axis of $(AMP)$ and $(ANP)$. All that is left is to show that $AY$ is the radical axis of $(AMP)$ and $(ANP)$ which is equivalent to $AY$ tangent to both circle. Notice that $PAYB$ is cyclic. Then $\angle PMB = 180^{\circ} - PBY = \angle PAY$, which proves the tangency, so we are done.
This post has been edited 1 time. Last edited by dolphinday, Feb 6, 2024, 7:04 PM
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jp62
53 posts
#58
Y by
Angle chase :sleeping:

We use directed angles mod $180^\circ$ throughout this solution.

ABCY is cyclic
I is the incenter of BCNM
(AMP) and (ANQ) are tangent at A
PMNQ is cyclic
AXY collinear

:wallbash: mfw g8 is just a massive chase
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awesomeming327.
1664 posts
#59
Y by
https://cdn.aops.com/images/a/6/b/a6b5883bce870c8c62780cb1b6b03ad25f549101.png
Take an inversion about I.

In the inverted image, we have $I$ is the orthocenter of $\triangle ABC$, $B,P,M$ collinear and $C,Q,N$ collinear. Note that \[\angle APM=\angle ACB=180^\circ-\angle AIB=\angle AMP\]so $AP=AM$ and similarly $AQ=AN$. This implies that $PMNQ$ is an isosceles trapezoid and $(MAP)$ is tangent to $(NAQ)$. In our uninverted image, note that since $PMNQ$ is cyclic, $X$ is on the radical axis of $\omega_B$ and $\omega_C$. Furthermore, it is on the radical axis of $(MAP)$ and $(NAQ)$. It is also on the common tangents from $I$ to $\omega_B$ and $\omega_C$ and from $A$ to $(MAP)$ and $(NAQ)$.
Note that
\[\angle BIC+\angle MIN=180^\circ-\angle IBC-\angle ICB+\angle MIX+\angle XIN=180^\circ-\angle IBC-\angle ICB+\angle IBM+\angle ICN=180^\circ\]so $\angle BIM+\angle CIN = 180^\circ\implies \angle BPM+\angle CQN=180^\circ\implies \angle ABY+\angle ACY=180^\circ$, so $Y$ is on $(ABC)$.
Furthermore, \[\angle AQN=\angle AQC-\angle NQC=180^\circ-\angle AYC-\angle ACY=\angle YAC\]so $YA$ is tangent to $(NAQ)$. Similarly, it is tangent to $(MAP)$ and so $Y$ is also on the common tangent from $A$ to $(MAP)$ and $(NAQ)$. We are done.
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awesomehuman
496 posts
#60 • 2 Y
Y by OronSH, dolphinday
Invert about $I$ and rename $I$ to $H$ to get the following equivalent statement:

Let $\triangle ABC$ have orthocenter $H$.
Let $\ell_B$, $\ell_C$ be paralell lines through $B$ and $C$, respectively.
Let $P$ and $Q$ be the intersections of
$\ell_B$ and $\ell_C$ with $(ABC)$, respectively.
Let $M$ and $N$ be the intersections of
$\ell_B$ and $\ell_C$ with $(AHB)$ and $(AHC)$, respectively.
Let $X$ be the intersection of $(HPM)$ and $(HQN)$.
Let $Y$ be the second intersection of the circle through $H$ tangent to $\ell_B$ at $B$ and the circle through $H$ tangent to $\ell_C$ at $C$.
Then, $AHYX$ is cyclic.

We will prove this statement instead of the original.
Let $H'$ be the reflection of $H$ over the angle bisector of $BC$ and $PQ$.
Then, $\triangle BHC \cong PH'Q$.

Claim:
$X$ is the intersection of $HH'$ with $(H'PQ)$.

Proof:
We have
\[\angle PXQ = \angle PXH + \angle HXQ = \angle BMH + \angle HNC\]\[= \angle BAH + \angle HAC = \angle BAC = \angle CHB = \angle PH'Q.\]Thus, $X$ lies on $(PH'Q)$.
We have
\[\angle PXH = \angle PMH = \angle BMH = \angle BAH = \angle HCB = \angle PQH' = \angle PXH'.\]Thus, $X$ is on $HH'$.

Claim:
$A$ is the orthocenter of $\triangle PXQ$.

Proof:
Let $D$ be the intersection of $PQ$ and $HH'$.
We have
\[\angle APQ = \angle ACQ = \angle BCQ + \angle ACB\]\[\angle PQX = \angle DXQ + \angle QDX = \angle HCB + \angle QCB.\]Thus,
\[\angle APQ + \angle PQX = \angle BCQ + \angle QCB + \angle ACB + \angle HCB = 90.\]Repeating the same logic on the other side, $A$ is the orthocenter of $\triangle PXQ$.

Claim:
$AHYX$ is cyclic

Proof:
We have
\[\angle AXH = \angle AXP + \angle PXH' = \angle PQA + \angle PQH'\]\[= \angle PBA + \angle HCB = \angle PBC + \angle CBA + \angle HCB = \angle PBC + 90\]\[= \angle PBC + \angle CBH + \angle ACB = \angle PBH + \angle ACB = \angle BYH + \angle AYB\]\[= \angle AYH.\]
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kiemsibongtoi
25 posts
#62
Y by
Eyed wrote:
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$. Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$. Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$, and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$. Rays $PM$ and $QN$ meet at $X$. Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$.

Show that $A,X,Y$ are collinear.

We'll prove this problem follows these claims :
$\textbf{1/}$ Let $B'$ be the intersection of line $BI$ with $\omega_c$ ($B' \neq I$); $C'$ be the intersection of line $CI$ with $\omega_b$ ($C' \neq I$)
$\hspace{1cm}$$M'$ be the intersection of lines $C'M$ and $AC$; $N'$ be the intersection of lines $B'N$ and $AB$
Then $N'MM'N$ is a isosceles trapezoid and $I$ lie on circumcircle of $N'MM'N$
$\textit{Prove :}$
$\, \,$Cuz $M$, $I$, $B$, $C$ are concyclic and $I$ is incenter of $\triangle ABC$, we see that :
$$ \angle AM'M = \angle MC'C + \angle ACC' = \angle IBC + \angle ICB = \angle C'IB = \angle C'MB $$$\, \,$So $\triangle AMM'$ is a isosceles triangle at $A$. Similar, $\triangle ANN'$ is a isosceles triangle at $A$
$\, \,$Therefore, $N'MM'N$ is a isosceles trapezoid
$\, \,$Next, let $\ell$ be the line which tangent both $\omega_b$, $\omega_c$ at $I$, we see that :
$$ \angle MIN = \angle MI\ell + \angle NI\ell = \angle MBI + \angle NCI = \angle IBC + \angle ICB = \angle AM'M $$$\, \,$Therefore, $M'$ lies on circle $(IMN)$. Similar, $N'$ lies on circle $(IMN)$.
$\textbf{2/}$ $M$, $N$, $P$, $Q$ are concyclic
$\textit{Prove :}$
$\, \,$Let $\omega_b$, $\omega_c$ meet line $BC$ at $T$, $L$ respectively ($T \neq B$, $L \neq C$)
$\hspace{1cm}$$O_b$, $O_c$ be center of $\omega_b$, $\omega_c$ respectively
$\, \,$Cuz $CI$ is the internal bisector of $\angle NCL$ so $O_cI \perp NL$. Similar, $O_bI \perp MT$
$\, \,$But $O_b$, $I$, $O_c$ are colibnear, so $NL$ $\|$ $MT$
$\, \,$Combine with claim $\textbf{1/}$, we see that :
$$\angle N'IN = 2\angle NIA = 2(\angle AIC - \angle NIC) = 180^\circ + \angle BAC - 2\angle NIC = \angle BAC + \angle NQC - \angle NLC = \angle NQC - \angle BMT$$$\, \,$So
$$ \angle PMN = \angle PMA + \angle AMN = \angle PC'B + \angle N'IN = \angle PBT + \angle NQC = 180^\circ - (\angle PQC - \angle NQC) = 180^\circ - \angle PQN$$$\, \,$Therefore, $M$, $N$, $P$, $Q$ are concyclic
$\textbf{3/}$ Let $V$ be the midpoint of arc $BAC$ of $\Gamma$
Then $V$ is the intersection of lines $B'Q$, $C'P$
$\textit{Prove :}$
$\, \,$Cuz $Q$ is the second intersection of $\omega_c$ and $\Gamma$ so $\angle B'QC = \angle BIC = 90^\circ + \frac12 \angle BAC = \angle VQC$
$\, \,$Lead to $V$, $B'$, $Q$ are colinear. Similar, $P$, $C'$, $P$ are colinear
$\textbf{4/}$$\triangle AXI$ is a isosceles triangle at $X$
$\textit{Prove :}$
$\, \,$Let $V_b$, $V_c$ be the intersections of lines $BI$, $CI$ with $\Gamma$ respectively ($V_b \neq B$, $V_c \neq C$)
$\hspace{0.7cm}$$U_b$ be the intersection of lines $V_bQ$ and $B'N$
$\, \,$We ez to see that $V_bV_c$ is the perpendicular bisector of $AI$, so we just need to prove that $V_b$, $X$, $V_c$ are colinear
$\, \,$Cuz $M$, $N$, $P$, $Q$ are concyclic from $\textbf{2/}$, so $X$ lie on radical axis of $\omega_b$ and $\omega_c$
$\, \,$Which means $XI$ tangent with both $\omega_b$, $\omega_c$ at $I$
$\, \,$Lead to $\triangle XNI \sim \triangle XIQ$. As a result : $\dfrac{XN}{XQ} = \dfrac{XN}{XI} .\dfrac{XI}{XQ} = \left( \dfrac{IN}{IQ} \right)^2 $
$\, \,$In the order hand :
$\hspace{0.5cm}$From claim $\textbf{3/}$ and $NB' \perp AI$ at claim $\textbf{1/}$, we see that :
$$\angle V_bQB' = \angle V_bQV = 90^\circ - \frac12(\angle BAC + \angle ABC) = 90^\circ - \angle AIB' = \angle NB'I = \angle V_bB'U_b$$$\hspace{0.5cm}$Lead to $\triangle V_bU_bB' \sim \triangle V_bB'Q$. As a result : $\dfrac{V_bU_b}{V_bQ} = \dfrac{V_bU_b}{V_bB'} .\dfrac{V_bB'}{V_bQ} = \left( \dfrac{B'U_b}{B'Q} \right)^2 $
$\, \,$Therefore
$$ \dfrac{XN}{XQ} = \left( \dfrac{IN}{IQ} \right)^2 = \left( \dfrac{\sin \angle NQI}{\sin \angle INQ} \right)^2 = \left( \dfrac{\sin \angle V_bB'U_b}{\sin \angle V_bB'Q} \right)^2 = \left( \dfrac{V_bU_b}{B'V_b} .\dfrac{V_bB'}{V_bQ} \right)^2 =  \dfrac{V_bU_b}{V_bQ} $$$\, \,$Hence, follow Thales theorem, we see that $XV_b$ $\|$ $NB'$. Similar, $XV_c$ $\|$ $MC'$
$\, \,$But from claim $\textbf{1/}$, we already have that $B'N$ $\|$ $C'M$, so $X$, $V_b$, $V_c$ are colinear
$\textbf{Finally, from these claims, we have : }$
$\, \,$ Cuz $\triangle XMI \sim \triangle XIP$ and $AX = XI$, so $\triangle XMA \sim \triangle XAP$
$\, \,$Lead to
$$\angle MAX = \angle APM = \angle APB - \angle MPB = 180^\circ - \angle ACB - \angle MTC = \angle LNC = \angle BCY = \angle BAY $$$\, \,$Similar, $\angle XAN = \angle CAY$. Therefore $A$, $X$, $Y$ are colinear, done :)
Attachments:
imosl g8 2020.pdf (74kb)
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bin_sherlo
661 posts
#63 • 2 Y
Y by ehuseyinyigit, swynca
Claim: $Y$ lies on $(ABC)$.
Proof:
\[\measuredangle BYC= 180-\measuredangle CBY-\measuredangle YCB=180-(\measuredangle IMB-\frac{\measuredangle B}{2})-(\measuredangle CNI-\frac{\measuredangle C}{2})=270-\frac{\measuredangle A}{2}-\measuredangle CIB=180-\measuredangle A\]Which gives that $Y\in (ABC)$.$\square$

Claim: $I$ is the $A-$excenter on $\triangle AMN$.
Proof: Let $O$ be the circumcenter of $(IMN)$. Since $\measuredangle MIN=\measuredangle MBI+\measuredangle ICN=90-\frac{\measuredangle A}{2},$ we have $\measuredangle MON=180-\measuredangle MAN$ hence $O$ lies on $(AMN)$. If $AI$ intersects $(IMN)$ at $J$ for second time, then
\[\measuredangle IMB=\measuredangle MIJ+\frac{\measuredangle A}{2}=\measuredangle MNJ+\measuredangle ONM=\measuredangle ONJ=\measuredangle NJO=\measuredangle NMI\]Similarily we get $\measuredangle CNI=\measuredangle INM$ so $I$ is the $A-$excenter on $\triangle AMN$.$\square$

Claim: $M,N,P,Q$ are concyclic.
Proof: Let $MN$ and incircle of $ABC$ be tangent to each other at $S$. Take the inversion centered at $I$ with radius $IS$. Let $D,E,F$ be the tangency points of the incircle with $BC,CA,AB$ respectively. $A^*,B^*,C^*,M^*,N^*$ are the midpoints of $EF,FD,DE,SF,SE$. $P^*,Q^*$ are the intersections of $B^*M^*,C^*N^*$ with $(A^*B^*C^*)$. Note that $M^*B^*\parallel SD\parallel N^*C^*$ and $M^*N^*\parallel EF\parallel B^*C ^*$.
\[\measuredangle M^*P^*Q^*=180-\measuredangle Q^*P^*B^*=\measuredangle B^*C^*N^*=\measuredangle N^*M^*P^*\]Thus, $N^*M^*P^*Q^*$ is an isosceles trapezoid which yields $M,N,P,Q$ lie on a circle.$\square$

Claim: $A,X,Y$ are collinear.
Proof: Let $PM$ and $QN$ intersect $(ABC)$ at $K,L$ for second time.
\[\measuredangle YCA=\measuredangle YCN=\measuredangle CQN=\measuredangle CQL \]\[\measuredangle ABY=\measuredangle MBY=\measuredangle MPB=\measuredangle KPB\]These give that $AY=CL$ and $AY=BK$. Hence $YK\parallel AB$ and $YL\parallel AC$. Also
\[\measuredangle KLN=\measuredangle KLQ=\measuredangle KPQ=\measuredangle MPQ=\measuredangle MNL\]Thus, $MN\parallel KL$. By Desargues on $\triangle AMN$ and $\triangle YKL$, since $AM\cap YK=AB_{\infty},MN\cap KL=MN_{\infty},NA\cap LY=AC_{\infty}$ are collinear, they are perspective. $AY,MK,NL$ are concurrent. $MK$ and $NL$ intersect at $X$ so $A,X,Y$ are collinear.$\blacksquare$
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GeorgeRP
130 posts
#64
Y by
Why did I solve a G8 synthetically???

Let's denote with $t$ the common tangent of $w_b$ and $w_c$.
Also let $U,V$ be the intersections of $BC$ with $w_b$ and $w_c$

Claim 1: $Y$ lies on $(ABC)$
Proof. $\angle BIU=\angle BCY=\phi$ and $\angle CIV=\angle CBY=\psi, \angle UIt=\angle IBC=\frac{\beta}{2}, \angle VIt=\angle ICB=\frac{\gamma}{2}$
Now:
$$90+\frac{\alpha}{2}=\angle BIC=\angle BIU+\angle UIV+\angle CIV= \phi+\psi+90-\frac{\alpha}{2}$$So $\angle BYC=180-\phi-\psi=180- \alpha=\angle BAC \text{. }\square$

Claim 2: $MU$ || $NV$ || $t$
Proof. $\angle MUI=\angle MBI= \angle UBI = \angle UIt \Rightarrow $ $MU$ || $t$ and similarly $NV$ || $t$. $\square$

Claim 3: $MUVN$ is an isosceles trapezoid
Proof. As $I$ is the midpoint of arcs $MU$ and $NV$ in $w_b,w_c$, this means that $MI=UI, IN=IV \Rightarrow \triangle MIN$ and $\triangle UIV $ are congruent $\Rightarrow MN=UV$. $\square$

Let $Q'$ and $P'$ be the intersections of $QX$ and $PX$ with $(ABC)$.

Claim 4: $BQ'$ || $QC$ || $t$
Proof. Reims theorem on $(ABC), (NQCV)$ and lines $QX,CV$ gives that $BQ'$ || $NV$ and similarly $CP'$ || $UM$. $\square$

Let $K;G$ be the intersections of $AB;BC$ and $NV;MU$

Claim 5: $PMAG$ and $QNAK$ are cyclic
Proof. $\angle KNQ=\angle VCQ=180-\angle BQ'Q=180-\angle BAQ=\angle KAQ$ and similarly $\angle PMG=\angle PAG$. $\square$

Claim 6:
$(PMAG)$ and $(NAKQ)$ are tangent at $A$
Proof. As $GMNK$ is a trapezoid we have that the circumcenters of $GMA$ and $KAN$ lie on a line with $A$. $\square$

Claim 7: $PMNQ$ is cyclic
Proof. $\angle MPQ= \angle BPQ-\angle BPM=(180-\angle BCQ)-\angle MUV=$
$(180-\angle KNQ)-\angle UMN=\angle XNK-\angle MNK=\angle MNX$. $\square$

Claim 8: $A,X,Y$ are collinear
Proof. By radical axises on $(PMNQ), (PMAG), (NAKQ)$ we get that $AX$ is the tangent of $(PMAG)$ and $(NAKQ)$.
$\angle YAC=\angle YBC= \angle BMU=\angle Q'BM=\angle Q'QA=\angle XAC$. $\square$
Attachments:
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L13832
245 posts
#65 • 2 Y
Y by radian_51, S_14159
Amazing problem! also my first G8 :showoff:
We show $AY$ is tangent to $\odot(AMP), \odot(ANQ)$
$$\angle YAM+\angle MAP = 180^{\circ} - \angle YBP=180^{\circ}-\angle PBI-\angle BPI = \angle BIP =  180^{\circ} - \angle PMA$$Note that $Y\in{\odot(ABC)}$as
\begin{align*}
& \angle BIC=\angle IMB+\angle INC=\angle IBY+\angle ICY=90^{\circ}+\frac{\angle A}{2}\\ 
& \angle BYC=360-\angle YBI-\angle BIC-\angle ICY=360^{\circ}-2\left(90^{\circ}+\frac{\angle A}{2}\right)=180^{\circ}-\angle A
\end{align*}By another angle chase we show $\odot(I)$ is the incircle of $MNCB$. $$\angle MIN = \angle MBI + \angle NCI = \angle IBC + \angle ICB = 180^{\circ}-\angle BIC=90-\frac{\angle A}{2}$$If we show $\odot(MPNQ)$ we get $X$ as the radical center of $(MNQP)$, $(AMP)$, and $(ANQ)$, which will clearly lie on radax of $(AMP)$, and $(ANQ)$ which is clearly $\overline{AY}$ implying $\overline{A-X-Y}$ is collinear.
So finally we go on to another angle chase to prove $\odot(MPNQ)$ which took me a lotta time to figure out :stretcher:
\begin{align*}
\angle PQN=\angle CQN-\angle CQP&= 180^{\circ}-\angle CIN-\angle CQP\\&=180^{\circ}-(180^{\circ}-\angle INC-\angle ICN)-\angle CQP\\&=180^{\circ}-(180^{\circ}-\angle ICY-\angle ICN)-\angle CQP\\&=\angle ICY+\angle ICN-\angle CQP\\&=\angle ACY-\angle CQP\\&=(180^{\circ}-\angle ABY)-(180^{\circ}-\angle CBP)\\&=\angle CBP-\angle ABY\\&=\angle CBA+\angle ABP-\angle CBA-\angle YBC\\&=\angle ABP-\angle YBC\\
\angle PMN=\angle PMB+\angle BMN&=\angle PIB+2\angle BMI\\&=180^{\circ}-(\angle IPB+\angle PBI)+2\angle BMI\\&=180^{\circ}-(\angle IBY+\angle PBI)+2\angle BMI\\&=180^{\circ}-\angle PBY+2\angle BMI\\&=180^{\circ}-\angle PBY+2\angle CBY+\angle CBA\\&=180^{\circ}-\angle PBA+\angle CBY
\end{align*}and we are finally done! :stretcher: :wacko:
This post has been edited 2 times. Last edited by L13832, Nov 21, 2024, 5:31 AM
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Saucepan_man02
1294 posts
#66
Y by
Here's my solution (angle-chase+invert+radical-axis+trig) :

Claim-1: $Y \in (ABC)$

Notice that: $$\angle ABY+\angle ACY = 360^\circ - (\angle MIB + \angle NIC) =  \angle MIC + \angle BIC = \angle MIX+\angle XIN + 90^\circ + \angle A/2$$$$ = \angle MBI + \angle NCI + 90^ + \angle A/2 = 180^\circ.$$
Claim-2: $MNQP$ is cyclic + $(AMP), (ANQ)$ are tangent.

Invert around incircle. Let $X'$ denote the inverse of point $X$ in this inversion. Notice that:

-$I$ is the orthocenter of $\triangle A'B'C'$.
-$M'A'B'I$ and $N'A'C'I$ are cyclic ($A, M, B$ and $A, N, C$ are collinear).
-$M', P', B'$ and $N', Q', C'$ are collinear ($MPIB$ and $NIAC$ are cyclic).
-$M'P'B' \parallel N'Q'C'$.
-$A',P',B',C',Q'$ lie on same circle.

We will prove $M'N'Q'P'$ is cyclic. It suffice to show $M'N'=P'Q'$. Note that: $\angle M'IB'+\angle N'IC' = 360^\circ - (\angle B'M'I + \angle C'N'I) - (\angle M'B'I + \angle N'C'I) = 180^\circ.$
Now, we have: $$M'B' = \frac{B'I \sin \angle M'IB'}{\sin \angle M'B'I}$$Note that: $$\frac{B'I}{\sin \angle M'B'I} = 2R  = \frac{C'I}{\sin \angle N'C'I}.$$Thus, we have: $\frac{B'I \sin \angle M'IB'}{\sin \angle M'B'I} = \frac{C'I \sin \angle N'IC'}{\sin \angle N'C'I} = N'C'$ which implies $M'B'C'N'$ is a parallelogram.

Note that, from the following sub-claim, we will have: $P'Q' \parallel M'N' \parallel BC$ and $(A'M'P'), (A'N'Q')$ to be tangent at $A'$ which finishes the main-claim.

Sub-Claim: $\triangle A'M'P'$ and $\triangle A'N'Q'$ are isosceles.
Note that: $\angle A'M'B' =  180^\circ - \angle A'IB' = \angle A'C'B'$ and $\angle A'P'M' = \angle A'C'B'$ which shows that $A'M' = A'P'$. Similarly, $A'N'=A'P'$ and we are done.
Hence, Claim-2 is true.

Thus, from Claim-2, $AX$ is the radical axis of $(APM), (ANQ)$ along with $AX$ being tangent to both circles $(APM), (ANQ)$.

Note that, the following claim finishes the problem:

Claim-3: $AY$ is tangent to both circles $(APM), (ANQ)$.

Let $R$ be a point of line $AP$ such that $A, R$ are on different sides with-respect to $P$.
We will show it with an one-liner angle-chase:
$$\angle BAY = 180^\circ  - \angle AYB - \angle ABY = 180^\circ - \angle RPB - \angle BPM  = \angle APM.$$Thus, $AY$ is tangent to $(APM)$. Similarly, $AY$ is tangent to $(ANQ)$ and we are done.

Remarks
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Likeminded2017
391 posts
#67
Y by
We prove that $Y$ lies on $(ABC)$ through angle chasing - $\angle ABY+\angle ACY=360-\angle MIB-\angle NIC=180^\circ.$ Next, $YA$ is tangent to both $(ANQ)$ and $(AMP)$ and is thus the radical axis, as $\angle APM=\angle BPA-\angle BPM=(180-\angle BYA)-\angle YBA=\angle YAB=\angle YAM$ and the same works for $(ANQ).$ Finally, we show $X$ lies on the radical axis by showing $MNQP$ are concyclic to finish. Let $\omega_b$ and $\omega_c$ intersect $BC$ at $R,S$ respectively. Observe by Fact 5 $\triangle IRS \cong \triangle IMN$ so $\angle INM=\angle ISR=\angle INC.$
\begin{align*}
    \angle QPM &= \angle MPB-\angle QPB \\
    &= \angle NIC+\angle QCB-180 \\
    &= \angle NIC+\angle BCI-\angle QNI \\
    &=\angle NIC+\angle NCI-\angle QNI \\
    &=180-\angle INC-\angle QNI \\
    &=180-\angle INM-\angle QNI \\
    &=180-\angle QNM
\end{align*}so we are done.
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cj13609517288
1857 posts
#68
Y by
Part 1 - The spiral similarity
First, note that
\[\angle BIC=\angle BMI+\angle CNI=\angle YBI+\angle YCI.\]Therefore,
\[\angle BYC=360^{\circ}-\angle BIC-\angle BIC=180^{\circ}-\angle A.\]Therefore, $Y\in (ABC)$.

Now note that
\[\angle PBY=180^{\circ}-\angle PMB=\angle PMA\]\[\angle PYB=\angle PAB=\angle PAM\]so $\triangle PBY\sim\triangle PMA$.

So there exists a spiral similarity centered at $P$ sending $AM$ to $YB$.

Part 2 - Redefining $X$
Redefine $X$ to be the intersection of the perpendicular bisector of $AI$ with $AY$. Then $\angle IXY=2\angle IAX$, so $IX$ is parallel to the isogonal of $AY$ wrt angle $BAC$. In particular,
\[\angle(BA,IX)=\angle YAC=\angle YBC=\angle(YB,CB)\Longrightarrow
\angle(BI,IX)=\angle(YB,IB)\Longrightarrow \angle BIX=180^{\circ}-\angle YBI=180^{\circ}-\angle BPI.\]Therefore, $XI$ is tangent to $(BPI)$.

Part 3 - The finish
By our spiral similarity from earlier,
\[\angle YAM=\angle YAB=\angle YPB=\angle APM.\]Therefore, $AY$ is tangent to $(APM)$. Then the powers from $X$ to $(APM)$ and $(BPM)$ are $AX^2$ and $IX^2$, respectively, which are equal, so $X$ lies on their radical axis $PM$. Similarly, $X$ lies on $QN$, as desired. $\blacksquare$
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swynca
10 posts
#69 • 1 Y
Y by bin_sherlo
A different approach,
After a simple angle chasing, it appears that $K=AC \cap YP$ and $L=AB \cap YQ$ is on $(MNPQ)$. Using pascal with points $P,K,N,Q,L,M$ gives the desired result.
This post has been edited 1 time. Last edited by swynca, Feb 13, 2025, 8:23 PM
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Iveela
115 posts
#70
Y by
The ultimate angle chasing.

Let $O_B$ and $O_C$ be the circumcenters of $\omega_B$ and $\omega_C$ respectively. Recall that $\angle BIC = 90^{\circ} + \frac{\angle A}{2}$ which implies $\angle O_BIB + \angle O_CIC = 90^{\circ} - \frac{\angle A}{2}$ and hence
\[\angle IBY + \angle ICY = \angle IPB + \angle IQC = 180 - \angle O_BIB - \angle O_CIC = 90^{\circ} + \frac{\angle A}{2}.\]Consequently, $\angle ABY + \angle ACY = 180^{\circ}$ or $ABCY$ cyclic.

Let $T = BP \cap CQ$. Since $TP \cdot TB = TQ \cdot TC$, it must lie on the radical axis of $\omega_B$ and $\omega_C$. In other words, $TI$ is tangent to both circles. We will now show that the quadrilateral $TPQX$ is cyclic. Indeed, we have
\[\angle XQC = \angle NQC = \angle ACY = 180^{\circ} - \angle ABY = 180^{\circ} - \angle XPB = \angle TPX\]as desired. Now, observe that
\[\angle XTQ = \angle QPX = \angle QPB - \angle ABY = \angle BCY - \angle ACQ = \angle ICY - \angle ICQ = \angle ACQ \]which implies that $T$, $X$, and $I$ are collinear. In particular, $XI$ is tangent to $\omega_B$ and $\omega_C$. This implies $XM \cdot XP = XI^2 = XN \cdot XQ$ or $MNPQ$ cyclic.

Now, we are well equipped to prove that $XA = XI$. We once again, angle chase. Recall that $XI$ is tangent to $\omega_B$ and $\omega_C$. Indeed, we have
\[\angle XIA = \angle AIB - \angle XIB = (90^{\circ} + \frac{\angle C}{2}) + \angle IBY - 180^{\circ} = \angle YAC - \frac{\angle A}{2} = \angle YAI.\]Recall that $XN \cdot XQ = XI^2 = XA^2$. Therefore, $(ANQ)$ is tangent to $XA$. Angle chasing one final time, we receive
\[\angle XAC = \angle AQN = \angle AQC - \angle NQC = \angle BAC + \angle ACB - \angle ACY = \angle CAY\]which implies the result.
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EeEeRUT
48 posts
#71
Y by
Denote $\ell$ a line tangent to $\omega_B$ and $\omega_C$ at $I$.

Claim : $\angle MIB + \angle NIC = 180^{\circ}$
Proof : Consider \begin{align*}
\angle MIB + \angle NIC &= 360^{\circ} - \angle MIN - \angle BIC\\ &= 360^{\circ} - (\angle IBM + \angle ICN) - (180^{\circ} - \angle IBC - \angle ICB)\\ &= 360^{\circ} - (\angle IBC + \angle ICB) - (180^{\circ} - \angle IBC - \angle ICB) = 180^{\circ}
\end{align*}Let $BP$ intersect $CQ$ at $T$. By radical axis, $T$ lies on $\ell$. Note that $$\angle TPX + \angle TQX = 360^{\circ} - (\angle MIB + \angle NIC) = 180^{\circ}$$Thus, $TPXQ$ is cyclic.

Claim : $X$ lies on $\ell$.
Proof : Let $\ell$ intersect $BC$ at $S$.
Consider $$\angle BMX = 180^{\circ} - \angle PIB = 180^{\circ} - \angle ISB$$Hence, $PXSB = \Omega_B$ is cyclic. Similarly, $QXSC = \Omega_C$ is cyclic.
That is the radical axis of these $2$ circles is $XS$
By radical center of $\Omega_B, \Omega_C$ and $(BCPQ)$ is $T$. Thus, $T,X,S$ are collinear, which is the line $\ell$.

Claim : $XA = IA$.
Proof : Consider \begin{align*}
\angle MPA + \angle NQA &= \angle APB - \angle BPM + \angle AQC - \angle CQN\\
&= -\angle ABC - \angle ACB + \angle MIB + \angle NIC\\
&= \angle CAB
\end{align*}Thus, $(PMA)$ is tangent to $(QNA)$
By power of point, the segment $XA = XI$ is tangent to $(PMA), (QNA)$.

Claim : $A,B,C,Y$ cyclic.
Proof : $$\angle ABY + \angle YCA = 360^{\circ} - \angle MIB - \angle NIC = 180^{\circ}$$Thus, $A,B,C,Y$ are concyclic.

Let $\angle XAI = \gamma$
Now, we are left to consider the following angle chasing \begin{align*} 
\angle XAC &= \angle CAB - \angle BAX\\ 
&= \angle CAB -(360^{\circ} - \angle AMI - \angle PIX - \angle IXA)\\
&= \angle CAB + 180^{\circ} - \angle IAM - \gamma - \angle IBC\\
&= \angle CAB - \angle IMB - \gamma\\
&= \angle CAB - \angle IBY\\
&= \angle CAB - \angle YAB\\
&= \angle YAC
\end{align*}
Hence, $ A,X,Y$ are collinear as desired.

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 /* end of picture */
[/asy]
This post has been edited 1 time. Last edited by EeEeRUT, Mar 4, 2025, 9:44 AM
Reason: Diagram
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joshualiu315
2513 posts
#72 • 1 Y
Y by OronSH
first g8 yippee! :coolspeak:


Denote the centers of $\omega_B$ and $\omega_C$ as $O_B$ and $O_C$, respectively.


Claim: $Y$ lies on $\Gamma$.

Proof: Notice that

\begin{align*}
\angle ABY + \angle ACY &= \angle BPM + \angle CQN \\
&= (180^\circ - \angle BIM) + (180^\circ - \angle CIN) \\
&= \angle BIC + \angle MIN \\
&= \angle BIC + \angle MBI + \angle NCI \\
&= \angle BIC + \angle IBC + \angle ICB = 180^\circ,
\end{align*}
so $ABYC$ is cyclic. $\square$


Let $D \neq B = \omega_B \cap \overline{BC}$ and $E \neq C = \omega_C \cap \overline{BC}$.


Claim: $MDEN$ is an isosceles trapezoid.

Proof: Since $\overline{BI}$ and $\overline{CI}$ are angle bisectors, $I$ is the midpoint of $\widehat{MD}$ in $\omega_B$ and the midpoint of $\widehat{NE}$ in $\omega_C$. Thus, $\overline{O_BO_C}$ is the perpendicular bisector of both $\overline{MD}$ and $\overline{NE}$. $\square$


Claim: Points $M$, $N$, $P$, and $Q$ are concyclic.

Proof: It suffices to show that $\triangle XMN \sim \triangle XQP$, or that $\angle XMN = \angle XQP$. We proceed with angle chasing:

\begin{align*}
\angle XMN &= 180^\circ - \angle PMD - \angle DMN \\
&= \angle PBD - \angle MDE \\
&= \angle PBD - \angle NEC \\
&= (180^\circ - \angle PQC) - (180^\circ - \angle NQC) \\
&= \angle NQC - \angle PQC = \angle NQP,
\end{align*}
as desired. $\square$


Then, consider the radical axis of $(APM)$ and $(AQN)$, denoted as $\ell$.


Claim: Points $A$, $X$, and $Y$ all lie on $\ell$

Proof: Since $A$ lies on both circles, it obviously lies on the radical axis. Also, we have $XM \cdot XP = XN \cdot XQ$ from Power of a Point with respect to $(MNQP)$. This implies $X$ has equal power with respect to $(APM)$ and $(AQN)$, meaning $X$ lies on $\ell$.

Finally, it suffices to show that $Y$ lies on $\ell$. We prove the stronger statment that $\overline{AY}$ is the common tangent of $(APM)$ and $(AQN)$. Notice that

\begin{align*}
\angle YAP &= 180^\circ - \angle YBP \\
&= 180^\circ - (\angle PBD + \angle YBD) \\
&= 180^\circ - (180^\circ - \angle PMD + \angle BMD) \\
&= \angle PMD - \angle BMD \\
&= \angle PMB = 180^\circ - \angle AMP.
\end{align*}
Thus, $\overline{AY}$ is tangent to $(APM)$, and similarly we can find that $\overline{AY}$ is tangent to $(AQN)$. This means $Y$ lies on $\ell$ as well. $\square$


Since $A$, $X$, and $Y$ all lie on $\ell$, they are obviously collinear. $\blacksquare$
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