It's February and we'd love to help you find the right course plan!

G
Topic
First Poster
Last Poster
k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1
Monday, Feb 3 - May 19
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10

Prealgebra 1 Self-Paced

Prealgebra 2
Sunday, Feb 16 - Jun 8
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10

Prealgebra 2 Self-Paced

Introduction to Algebra A
Sunday, Feb 16 - Jun 8 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28

Introduction to Algebra A Self-Paced

Introduction to Counting & Probability
Sunday, Feb 9 - Apr 27 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2

Introduction to Counting & Probability Self-Paced

Introduction to Number Theory
Sunday, Feb 16 - May 4
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3

Introduction to Algebra B
Thursday, Feb 13 - May 29
Sunday, Mar 2 - Jun 22
Monday, Mar 17 - Jul 7
Wednesday, Apr 16 - Jul 30

Introduction to Algebra B Self-Paced

Introduction to Geometry
Friday, Feb 14 - Aug 1
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1

Intermediate: Grades 8-12

Intermediate Algebra
Wednesday, Feb 12 - Jul 23
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13

Intermediate Counting & Probability
Monday, Feb 10 - Jun 16
Sunday, Mar 23 - Aug 3

Intermediate Number Theory
Thursday, Feb 20 - May 8
Friday, Apr 11 - Jun 27

Precalculus
Tuesday, Feb 25 - Jul 22
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21

Calculus
Friday, Feb 28 - Aug 22
Sunday, Mar 30 - Oct 5

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Tuesday, Feb 4 - Apr 22
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2

MATHCOUNTS/AMC 8 Advanced
Sunday, Feb 16 - May 4
Friday, Apr 11 - Jun 27

AMC 10 Problem Series
Sunday, Feb 9 - Apr 27
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23

AMC 10 Final Fives
Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)

AMC 12 Problem Series
Sunday, Feb 23 - May 11

AMC 12 Final Fives
Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)

Special AIME Problem Seminar B
Sat & Sun, Feb 1 - Feb 2 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

F=ma Problem Series
Wednesday, Feb 19 - May 7

Programming

Introduction to Programming with Python
Sunday, Feb 16 - May 4
Monday, Mar 24 - Jun 16

Intermediate Programming with Python
Tuesday, Feb 25 - May 13

USACO Bronze Problem Series
Thursday, Feb 6 - Apr 24

Physics

Introduction to Physics
Friday, Feb 7 - Apr 25
Sunday, Mar 30 - Jun 22

Physics 1: Mechanics
Sunday, Feb 9 - Aug 3
Tuesday, Mar 25 - Sep 2

Relativity
Sat & Sun, Apr 26 - Apr 27 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
0 replies
jlacosta
Feb 2, 2025
0 replies
2014-2015 Fall OMO #26
v_Enhance   14
N Feb 9, 2025 by abeot
Let $ABC$ be a triangle with $AB=26$, $AC=28$, $BC=30$. Let $X$, $Y$, $Z$ be the midpoints of arcs $BC$, $CA$, $AB$ (not containing the opposite vertices) respectively on the circumcircle of $ABC$. Let $P$ be the midpoint of arc $BC$ containing point $A$. Suppose lines $BP$ and $XZ$ meet at $M$ , while lines $CP$ and $XY$ meet at $N$. Find the square of the distance from $X$ to $MN$.

Proposed by Michael Kural
14 replies
v_Enhance
Oct 28, 2014
abeot
Feb 9, 2025
2018-2019 Fall OMO Problem 30
trumpeter   17
N Jan 10, 2025 by qwerty123456asdfgzxcvb
Let $ABC$ be an acute triangle with $\cos B =\frac{1}{3}, \cos C =\frac{1}{4}$, and circumradius $72$. Let $ABC$ have circumcenter $O$, symmedian point $K$, and nine-point center $N$. Consider all non-degenerate hyperbolas $\mathcal H$ with perpendicular asymptotes passing through $A,B,C$. Of these $\mathcal H$, exactly one has the property that there exists a point $P\in \mathcal H$ such that $NP$ is tangent to $\mathcal H$ and $P\in OK$. Let $N'$ be the reflection of $N$ over $BC$. If $AK$ meets $PN'$ at $Q$, then the length of $PQ$ can be expressed in the form $a+b\sqrt{c}$, where $a,b,c$ are positive integers such that $c$ is not divisible by the square of any prime. Compute $100a+b+c$.

Proposed by Vincent Huang
17 replies
trumpeter
Nov 7, 2018
qwerty123456asdfgzxcvb
Jan 10, 2025
2013-2014 Fall OMO #26
v_Enhance   8
N Jan 8, 2025 by OronSH
Let $ABC$ be a triangle with $AB=13$, $AC=25$, and $\tan  A = \frac{3}{4}$. Denote the reflections of $B,C$ across $\overline{AC},\overline{AB}$ by $D,E$, respectively, and let $O$ be the circumcenter of triangle $ABC$. Let $P$ be a point such that $\triangle DPO\sim\triangle PEO$, and let $X$ and $Y$ be the midpoints of the major and minor arcs $\widehat{BC}$ of the circumcircle of triangle $ABC$. Find $PX \cdot PY$.

Proposed by Michael Kural
8 replies
v_Enhance
Oct 30, 2013
OronSH
Jan 8, 2025
2012-2013 Winter OMO #22
v_Enhance   2
N Dec 28, 2024 by NicoN9
In triangle $ABC$, $AB = 28$, $AC = 36$, and $BC = 32$. Let $D$ be the point on segment $BC$ satisfying $\angle BAD = \angle DAC$, and let $E$ be the unique point such that $DE \parallel AB$ and line $AE$ is tangent to the circumcircle of $ABC$. Find the length of segment $AE$.

Ray Li
2 replies
v_Enhance
Jan 16, 2013
NicoN9
Dec 28, 2024
2012-2013 Winter OMO #11
v_Enhance   2
N Dec 26, 2024 by NicoN9
Let $A$, $B$, and $C$ be distinct points on a line with $AB=AC=1$. Square $ABDE$ and equilateral triangle $ACF$ are drawn on the same side of line $BC$. What is the degree measure of the acute angle formed by lines $EC$ and $BF$?

Ray Li
2 replies
v_Enhance
Jan 16, 2013
NicoN9
Dec 26, 2024
2011-2012 Winter OMO #22
Zhero   4
N Dec 24, 2024 by NicoN9
Find the largest prime number $p$ such that when $2012!$ is written in base $p$, it has at least $p$ trailing zeroes.

Author: Alex Zhu
4 replies
Zhero
Jan 24, 2012
NicoN9
Dec 24, 2024
2013-2014 Fall OMO #29
v_Enhance   22
N Dec 11, 2024 by eg4334
Kevin has $255$ cookies, each labeled with a unique nonempty subset of $\{1,2,3,4,5,6,7,8\}$. Each day, he chooses one cookie uniformly at random out of the cookies not yet eaten. Then, he eats that cookie, and all remaining cookies that are labeled with a subset of that cookie (for example, if he chooses the cookie labeled with $\{1,2\}$, he eats that cookie as well as the cookies with $\{1\}$ and $\{2\}$). The expected value of the number of days that Kevin eats a cookie before all cookies are gone can be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Proposed by Ray Li
22 replies
v_Enhance
Oct 30, 2013
eg4334
Dec 11, 2024
2012-2013 Winter OMO #38
v_Enhance   11
N Oct 29, 2024 by PEKKA
Triangle $ABC$ has sides $AB = 25$, $BC = 30$, and $CA=20$. Let $P,Q$ be the points on segments $AB,AC$, respectively, such that $AP=5$ and $AQ=4$. Suppose lines $BQ$ and $CP$ intersect at $R$ and the circumcircles of $\triangle{BPR}$ and $\triangle{CQR}$ intersect at a second point $S\ne R$. If the length of segment $SA$ can be expressed in the form $\frac{m}{\sqrt{n}}$ for positive integers $m,n$, where $n$ is not divisible by the square of any prime, find $m+n$.

Victor Wang
11 replies
v_Enhance
Jan 16, 2013
PEKKA
Oct 29, 2024
2015-2016 Fall OMO #12
pi37   14
N Aug 7, 2024 by eg4334
Let $a$, $b$, $c$ be the distinct roots of the polynomial $P(x) = x^3 - 10x^2 + x - 2015$.
The cubic polynomial $Q(x)$ is monic and has distinct roots $bc-a^2$, $ca-b^2$, $ab-c^2$.
What is the sum of the coefficients of $Q$?

Proposed by Evan Chen
14 replies
pi37
Nov 18, 2015
eg4334
Aug 7, 2024
2017-2018 Fall OMO Problem 18
trumpeter   6
N Aug 5, 2024 by ryanbear
Let $a,b,c$ be real nonzero numbers such that $a+b+c=12$ and \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=1.\]Compute the largest possible value of $abc-\left(a+2b-3c\right)$.

Proposed by Tristan Shin
6 replies
trumpeter
Nov 7, 2017
ryanbear
Aug 5, 2024
2013-2014 Fall OMO #26
v_Enhance   8
N Jan 8, 2025 by OronSH
Let $ABC$ be a triangle with $AB=13$, $AC=25$, and $\tan  A = \frac{3}{4}$. Denote the reflections of $B,C$ across $\overline{AC},\overline{AB}$ by $D,E$, respectively, and let $O$ be the circumcenter of triangle $ABC$. Let $P$ be a point such that $\triangle DPO\sim\triangle PEO$, and let $X$ and $Y$ be the midpoints of the major and minor arcs $\widehat{BC}$ of the circumcircle of triangle $ABC$. Find $PX \cdot PY$.

Proposed by Michael Kural
8 replies
v_Enhance
Oct 30, 2013
OronSH
Jan 8, 2025
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
v_Enhance
6853 posts
#1 • 2 Y
Y by HamstPan38825, Adventure10
Let $ABC$ be a triangle with $AB=13$, $AC=25$, and $\tan  A = \frac{3}{4}$. Denote the reflections of $B,C$ across $\overline{AC},\overline{AB}$ by $D,E$, respectively, and let $O$ be the circumcenter of triangle $ABC$. Let $P$ be a point such that $\triangle DPO\sim\triangle PEO$, and let $X$ and $Y$ be the midpoints of the major and minor arcs $\widehat{BC}$ of the circumcircle of triangle $ABC$. Find $PX \cdot PY$.

Proposed by Michael Kural
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aZpElr68Cb51U51qy9OM
1600 posts
#2 • 1 Y
Y by Adventure10
Sketch
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
quantumman
245 posts
#3 • 2 Y
Y by Adventure10, Mango247
I did two solutions one using terrible law of cosines one using complex does anyone know of a nicer solution considering the answer is the square of one of the sides of the triangle?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
superpi83
1416 posts
#4 • 2 Y
Y by Adventure10, Mango247
Complex bash attached.
Attachments:
OMO 26 Complex Bash.pdf (77kb)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
quantumman
245 posts
#5 • 2 Y
Y by Adventure10, Mango247
Anyone have a non bash solution?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathocean97
606 posts
#6 • 2 Y
Y by Adventure10, Mango247
Sorry, my solution is still a bash solution, and it is also complex numbers, but I just want to mention that most people don't bash with complex numbers like superpi83 does above, and you can simplify calculations a bit (and reduce conjugates, which are generally annoying)

Instead, we set $ABC$ as the unit circle, and lowercase letters denote the complex numbers. Firstly, note that $x^2 = bc$ and $y = -x$ (diametrically opposite) Then $PX \cdot PY = |(p-x)(p-y)| = |p^2-bc|$, since $y = -x$. It's well known that the reflection of $B$ over $AC$, a unit circle chord, is $d = a+c-ac\bar{b} = a+c-\frac{ac}{b}$. Similarly, $e = a+b-\frac{ab}{c}$. By the similarity condition, $\frac{p-0}{d-0} = \frac{e-0}{p-0} \implies p^2 = de$. Therefore,

$|p^2-x^2| = |de-bc| = \left|\left(a+c-\frac{ac}{b}\right)\left(a+b-\frac{ab}{c}\right)-bc\right| = \\
|(bc-(ac-ab))(bc+(ac-ab))-b^2c^2| = |(ac-ab)^2| = |b-c|^2 = BC^2$, by difference of squares. Now $BC^2$ is easy to compute with LOC.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
john0512
4166 posts
#7
Y by
We claim that $PX\cdot PY=BC^2$. Let $A=a^2$ etc with $(ABC)$ as the unit circle. We have $$d=a^2+c^2-\frac{a^2c^2}{b^2}$$and similarly for $e$. Now, by triangle similarity, we have $p^2=de$.

Note that $$(PX\cdot PY)^2=(p-bc)(\overline{p-bc})(p+bc)(\overline{p+bc})$$$$=(p^2-b^2c^2)(\overline{p^2-b^2c^2}).$$We then have $$p^2=de=\frac{(a^2b^2+b^2c^2-a^2c^2)(a^2c^2+b^2c^2-a^2b^2)}{b^2c^2},$$so $$p^2-b^2c^2=\frac{(a^2b^2+b^2c^2-a^2c^2)(a^2c^2+b^2c^2-a^2b^2)}{b^2c^2}-b^2c^2$$$$=\frac{2a^4b^2c^2-a^4b^4-a^4c^4}{b^2c^2}=\frac{-a^4(b^2-c^2)^2}{b^2c^2}.$$Conjugating this gives $$\overline{p^2-b^2c^2}=\frac{-(b^2-c^2)^2}{a^4b^2c^2}.$$Thus, $$(PX\cdot PY)^2=\frac{-a^4(b^2-c^2)^2}{b^2c^2}\cdot \frac{-(b^2-c^2)^2}{a^4b^2c^2}=\frac{(b^2-c^2)^4}{b^4c^4}.$$
Note that $$BC^2=(b^2-c^2)(\frac{1}{b^2}-\frac{1}{c^2})=\frac{-(b^2-c^2)^2}{b^2c^2}.$$Since $(PX\cdot PY)^2=BC^4$, we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8793 posts
#8
Y by
Very quick complex bash. Let $a = x^2$, $b = y^2$, $c = z^2$, so the arc midpoints are $yz$ and $-yz$. The similarity condition implies $p = de$, where $d = a+c-\frac{ac}b$ and $e = a+b-\frac{ab}c$. So we want to compute
\begin{align*}
|p-yz||p+yz| &= |p^2-y^2z^2| \\
&= \left|\left(a+c-\frac{ac}b\right)\left(a+b-\frac{ab}c\right)-bc\right| \\
&= \left|\frac{a^2(b-c)^2}{bc}\right| = |b-c|^2 = BC^2,
\end{align*}which is easy.
This post has been edited 1 time. Last edited by HamstPan38825, Mar 11, 2024, 9:19 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OronSH
1688 posts
#9 • 5 Y
Y by GrantStar, peace09, Zhaom, CyclicISLscelesTrapezoid, bjump
We claim the answer is $BC^2$.

Let $\Omega$ be the circumcircle. Let $Z$ be the antipode of $A$, delete $X,Y$ and let $X,Y$ instead be points on $\Omega$ with $BX\perp AC,CY\perp AB$. Let $S$ be on $\Omega$ with $AS\parallel BC$.

Fix $\Omega$ and $A$ and vary $B,C$ fixing $\measuredangle BAC$. Then $X,Y$ are fixed. Since $D$ is the orthocenter of $\triangle AXC$ it follows that $D$ lies on the reflection $\Omega_D$ of $\Omega$ over $AX$. Define $\Omega_E$ similarly, and let them intersect at $T$.

Next define $V,W$ as the antipodes of $X,Y$ on $\Omega_D,\Omega_E$ respectively. Then $\Omega'=(VTW)$ is the reflection of $\Omega$ over the perpendicular bisector of $AT$. Finally, denote by $Q$ the image of $S$ under the translation taking $\Omega$ to $\Omega'$.

First, since $\Omega_D,\Omega_E,\Omega$ are congruent, $A,T,X,Y$ is an orthocentric system. Since $T$ is the orthocenter of $AXY$ it follows that $T,Z$ are reflections over $XY$. Thus we get \[\measuredangle ZYT=2\measuredangle ZYX=\measuredangle YAX=\measuredangle YZX=\measuredangle YOA=\measuredangle YOT,\]implying $ZY$ is tangent to $(YOT)$ and thus we have \[AO\cdot QS=AO\cdot ZT=ZO\cdot ZT=ZY^2=BC^2.\]
The main claim is that \[\measuredangle OAE=\measuredangle ODQ.\]To show this, move $B$ on $\Omega$ with degree $2$. Then $C,D,E$ move with degree $2$ as well (e.g. since $\overrightarrow{BE}$ has length $BC$ and is perpendicular to $AY$ and thus is fixed). Next $S$ moves with degree $4$, so $Q$ does as well. Now lines $OA,AE,OD,DQ$ have degrees at most $0,1,2,6$ respectively. Thus $e^{2i\measuredangle OAE}$ and $e^{2i\measuredangle ODQ}$ have degrees at most $1,8$ so to show they are equal we need to check $1+8+1=10$ cases.

First, if $A=S$ we have $\measuredangle EBA=\measuredangle ABC=\measuredangle BCS=\measuredangle BYS$ implies $EB$ is tangent to $\Omega$. Similarly $DC$ is tangent as well, so \begin{align*}OD\cdot OE&=\sqrt{(BO^2+BE^2)(CO^2+CD)^2}=AO^2+BC^2\\&=AO^2+AO\cdot ZT=OA\cdot(AO+SQ)=OA\cdot OQ.\end{align*}By symmetry $\measuredangle EOA=\measuredangle QOD$, so $\triangle OAE\sim\triangle ODQ$ implying $\measuredangle OAE=\measuredangle ODQ$. This happens for two distinct positions of $BC$, so we have two cases.

Next, consider the two cases when $Z=S$. Checking, we see that the two cases are when $B,C$ are opposite arc midpoints of $AY,AX$ respectively on $\Omega$. Then from $BO\perp AY$ and $BE\perp AY$ and $OA=OY$, we get line $OBE$ perpendicularly bisects $AY$, so $E$ is an arc midpoint of $AY$ on $\Omega_E$. Similarly $D,E$ are opposite arc midpoints of $AX,AY$ on $\Omega_D,\Omega_E$. Now let $D'$ be the antipode of $D$ on $\Omega_D$ so that $D',E$ are reflections over $AO$ by symmetry. Thus we get \[\measuredangle OAE=\measuredangle TAE=\measuredangle D'AT=\measuredangle D'DT=\measuredangle ODT=\measuredangle ODQ\]for these cases.

When $B=A$ since $BYCZ$ is a trapezoid we get $C$ is the reflection of $Y$ over the line through $O$ perpendicular to $AO$, and $S=C$ so $Q=W$. However, $AY\perp BE$ implies $E=W$ as well. Next since $C$ is the antipode of $X$ we must also have $D=A$, so $\measuredangle OAE=\measuredangle ODQ$. Similarly when $C=A$ we get $E=A$ and $Q=D=V$, still giving $\measuredangle OAE=\measuredangle ODQ$. This gives two more cases.

When $B=Y$ we get $C=Z$ so $CD\perp AX\perp XZ$ gives $D=X$ and $BE\perp AY$ so $E$ is the antipode of $A$ on $\Omega_E$. Next $S$ is the antipode of $Y$ on $\Omega$ so $Q=V$. Since $AOY$ forms a parallelogram with the center of $\Omega_E$ we have \[\measuredangle OAE=\measuredangle AOY=\measuredangle XOA=\measuredangle OXV=\measuredangle ODQ.\]
When $B=Z$ and $C=X$ we similarly get $E=Y,Q=W$ and $D$ is the antipode of $A$ on $\Omega_D$. Since $AXDV$ is a rectangle and $OA=OX$ we get $OV=OD$ and clearly by symmetry $OV=OW$, so $O$ is the center of $(DVW)$ and thus \[\measuredangle OAE=\measuredangle XAO=\measuredangle DVW+90^\circ=\measuredangle ODW=\measuredangle ODQ.\]
When $C=Y$ then from $BX\perp AC$ and $BE\perp AY$ we get $B,E,X$ collinear. Next since $\Omega,\Omega_D,\Omega_E$ are congruent, $A,T,X,Y$ are orthocentric, so $T$ is the reflection of $Z$ over $XY$. Thus $TX\parallel YZ$ so $T$ lies on line $BEX$ as well. Additionally, since this line is perpendicular to $AY$ we see $T,B$ and $E,X$ are reflections over $AY$. From $Y=C$ we get $D=T$ since $A,T,X,Y$ and $A,D,X,C$ are orthocentric. Thus we get \begin{align*}\measuredangle OAE&=\measuredangle TAE=\measuredangle TYE=\measuredangle XYB=\measuredangle XYA+\measuredangle AYB\\&=\measuredangle AXY+\measuredangle YBS=\measuredangle AZS=\measuredangle OTQ=\measuredangle ODQ.\end{align*}
When $B$ is the antipode of $X$ then from $AC\perp BX$ implies $ABC$ is isosceles. Since $D$ is the reflection of $B$ over $AC$ we get $B,O,X,D$ collinear. Next $YA=AX=XC$ so $AXCY$ is a trapezoid and $AC=XY$. Next $VD\perp XD\perp AC$. Now letting $Q'$ be the point on $DV$ with $WV=WQ'$ we have $WQ'=WV=XY=AC=BS$ and \begin{align*}\measuredangle WQ'V&=\measuredangle DVW=\measuredangle DOA=\measuredangle CBA\\&=2\measuredangle BCA=\measuredangle BSA+\measuredangle SAC=\measuredangle(BS,AC)\end{align*}so $WQ'\parallel BS$ and thus $WQ'SB$ is a parallelogram. However, $\overrightarrow{BW}=\overrightarrow{ZT}$ so $Q'=Q$. Now we have $AE\perp BY\parallel AO$. Thus \[\measuredangle OAE=90^\circ=\measuredangle XDV=\measuredangle ODQ.\]
This is all ten cases, so the claim is proven.

We similarly get $\measuredangle OAD=\measuredangle OEQ$, so $O$ must be the Miquel point of $ADQE$.

Now since $\triangle ODP\sim\triangle OPE$ we have $OP$ bisects $\angle DOE$ and $OP^2=OD\cdot OE$, so $OP$ bisects $\angle AOQ$ and $OP^2=OA\cdot OQ$ as well, so $\triangle OAP\sim\triangle OPQ$.

Now we delete all points from the diagram, except $O,A,S,P,Q$. Redefine $X,Y$ to be the intersections of the perpendicular bisector of $AS$ with $\Omega$. We claim $QS\cdot AO=PX\cdot PY$, which would imply $PX\cdot PY=BC^2$ as desired.

To show this, first let $P'$ be the reflection of $P$ over $O$. From the similarity we know $O$ is the $P$-Dumpty point of $\triangle APQ$ so it follows that $APQP'$ is cyclic and harmonic. Next as $A$ moves on $\Omega$, fixing $X,Y,P$, both $S,Q$ move on circles with the opposite angular velocity of $A$, so $SQ$ has fixed length. Thus we can assume $A=X$, and we want to show $AP\cdot AP'=AO\cdot AQ$, but this is because force-overlay inversion at $A$ swapping $P,P'$ swaps $O$ and $Q$, so we are done.
Z K Y
N Quick Reply
G
H
=
a