Stay ahead of learning milestones! Enroll in a class over the summer!

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

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

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

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
KJMO 2001 P1
RL_parkgong_0106   1
N 7 minutes ago by JH_K2IMO
Source: KJMO 2001
A right triangle of the following condition is given: the three side lengths are all positive integers and the length of the shortest segment is $141$. For the triangle that has the minimum area while satisfying the condition, find the lengths of the other two sides.
1 reply
RL_parkgong_0106
Jun 29, 2024
JH_K2IMO
7 minutes ago
Cotangential circels
CountingSimplex   5
N 10 minutes ago by rong2020
Let $ABC$ be a triangle with circumcenter $O$ and let the angle bisector of $\angle{BAC}$ intersect $BC$
at $D$. The point $M$ is such that $\angle{MCB}=90^o$ and $\angle{MAD}=90^o$. Lines $BM$ and $OA$ intersect at
the point $P$. Show that the circle centered at $P$ and passing through $A$ is tangent to segment
$BC$.
5 replies
CountingSimplex
Jun 23, 2020
rong2020
10 minutes ago
Inspired by old results
sqing   3
N 17 minutes ago by sqing
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
3 replies
sqing
Today at 7:36 AM
sqing
17 minutes ago
Inspired by Philippine 2025
sqing   1
N 22 minutes ago by sqing
Source: Own
Let $ a,b,c,d $ be real numbers . Prove that
$$\frac{(a-1)(b-3)(c-3)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{7+4\sqrt 3}{144}$$$$\frac{(a-1)(b-2)(c-2)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{11+4\sqrt 7}{432}$$


1 reply
sqing
37 minutes ago
sqing
22 minutes ago
Combinatorics
P162008   0
an hour ago
$4$ girls and $4$ boys are to be seated in a line. Find the total number of ways such that boys and girls are alternate and a particular boy and girl are never adjacent to each other in any arrangement.
0 replies
P162008
an hour ago
0 replies
Combinatorics
P162008   0
an hour ago
A cricket team comprising of $11$ players named $A,B,C,\cdots,J,K$ is to be sent for batting. If $A$ wants to bat before $J$ and $J$ wants to bat after $G.$ Then find the total number of batting orders if other players could go in any order.
0 replies
P162008
an hour ago
0 replies
[PMO25 Qualifying II.8] A Square Can't Be A Floor
kae_3   2
N 2 hours ago by tapilyoca
Determine the largest perfect square less than $1000$ that cannot be expressed as $\lfloor x\rfloor + \lfloor 2x\rfloor + \lfloor 3x\rfloor + \lfloor 6x\rfloor$ for some positive real number $x$.

Answer Confirmation
2 replies
kae_3
Feb 21, 2025
tapilyoca
2 hours ago
solve in R
zolfmark   1
N 2 hours ago by Mathzeus1024
x+1/y=9/2 and y+1/z=11/4 and z+1/x=12/5
1 reply
zolfmark
Feb 16, 2016
Mathzeus1024
2 hours ago
not so hard problem (own)
BinariouslyRandom   1
N 3 hours ago by BinariouslyRandom
An open rectangular box is made from exactly $900 m^2$ of cardboard, with 5 rectangular faces instead of 6 and there is no leftover cardboard. What is the largest volume this box can hold given that $100 m^2$ of cardboard is needed to cover the box?
1 reply
BinariouslyRandom
3 hours ago
BinariouslyRandom
3 hours ago
Minimize z.
Entrepreneur   3
N 3 hours ago by no_room_for_error
Minimize $z = 6x + 3y.$ Subject to the constraints:
$$\begin{cases}
4x+y\ge80,\\
x+5y \ge115,\\
3x+2y\le150,\\
x,y\ge0.
\end{cases}$$
3 replies
Entrepreneur
May 21, 2025
no_room_for_error
3 hours ago
Inequalities
sqing   18
N Today at 7:48 AM by sqing
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
18 replies
sqing
May 21, 2025
sqing
Today at 7:48 AM
Inequalities
sqing   18
N Today at 7:20 AM by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
18 replies
sqing
May 13, 2025
sqing
Today at 7:20 AM
helpppppppp me
stupid_boiii   1
N Today at 6:56 AM by vanstraelen
Given triangle ABC. The tangent at ? to the circumcircle(ABC) intersects line BC at point T. Points D,E satisfy AD=BD, AE=CE, and ∠CBD=∠BCE<90 ∘ . Prove that D,E,T are collinear.
1 reply
stupid_boiii
Yesterday at 4:22 AM
vanstraelen
Today at 6:56 AM
Algebra Polynomials
Foxellar   2
N Today at 5:43 AM by Foxellar
The real root of the polynomial \( p(x) = 8x^3 - 3x^2 - 3x - 1 \) can be written in the form
\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]where \( a, b, \) and \( c \) are positive integers. Find the value of \( a + b + c \).
2 replies
Foxellar
Today at 4:52 AM
Foxellar
Today at 5:43 AM
k line tangent and parallel at the same time.
Mr.C   8
N Apr 19, 2021 by Rg230403
Source: Iranian RMM TST 2021 Day1 P1
Suppose that two circles $\alpha, \beta$ with centers $P,Q$, respectively , intersect orthogonally at $A$,$B$. Let $CD$ be a diameter of $\beta$ that is exterior to $\alpha$. Let $E,F$ be points on $\alpha$ such that $CE,DF$ are tangent to $\alpha$ , with $C,E$ on one side of $PQ$ and $D,F$ on the other side of $PQ$. Let $S$ be the intersection of $CF,AQ$ and $T$ be the intersection of $DE,QB$. Prove that $ST$ is parallel to $CD$ and is tangent to $\alpha$
8 replies
Mr.C
Apr 16, 2021
Rg230403
Apr 19, 2021
line tangent and parallel at the same time.
G H J
G H BBookmark kLocked kLocked NReply
Source: Iranian RMM TST 2021 Day1 P1
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mr.C
539 posts
#1
Y by
Suppose that two circles $\alpha, \beta$ with centers $P,Q$, respectively , intersect orthogonally at $A$,$B$. Let $CD$ be a diameter of $\beta$ that is exterior to $\alpha$. Let $E,F$ be points on $\alpha$ such that $CE,DF$ are tangent to $\alpha$ , with $C,E$ on one side of $PQ$ and $D,F$ on the other side of $PQ$. Let $S$ be the intersection of $CF,AQ$ and $T$ be the intersection of $DE,QB$. Prove that $ST$ is parallel to $CD$ and is tangent to $\alpha$
Z Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
iman007
270 posts
#2 • 1 Y
Y by Mgh
[asy]
import graph; size(11.3cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -2.29, xmax = 5.24, ymin = -1.65, ymax = 2.88;  /* image dimensions */
pen uququq = rgb(0.25,0.25,0.25); pen xdxdff = rgb(0.49,0.49,1); 
pair P = (0,0), A = (0.76,0.66), B = (0.88,-0.48), Q = (1.21,0.13), C = (1.26,0.82), D = (1.16,-0.56), F = (0.43,-0.9), T = (0.98,-0.29), G = (-1,0.08), H = (0.83,-0.06), I = (1,-0.08), J = (-0.81,2.46); 
 /* draw figures */
draw(circle(P, 1)); 
draw(circle(Q, 0.69)); 
draw(D--F); 
draw(C--(0.15,0.99)); 
draw((1.03,0.34)--T); 
draw(D--(0.15,0.99)); 
draw(C--F); 
draw(A--Q); 
draw(B--Q); 
draw(A--P); 
draw(P--B); 
draw(C--D); 
draw(D--G); 
draw(G--C); 
draw(A--B); 
draw(D--A); 
draw(C--B); 
draw(G--I); 
draw(J--A); 
draw(J--G); 
draw((0.15,0.99)--J); 
 /* dots and labels */
dot(P,uququq); 
label("$P$", (0.02,0.03), NE * labelscalefactor,uququq); 
dot(A,xdxdff); 
label("$A$", (0.77,0.68), NE * labelscalefactor,xdxdff); 
dot(B,xdxdff); 
label("$B$", (0.9,-0.44), NE * labelscalefactor,xdxdff); 
dot(Q,uququq); 
label("$Q$", (1.23,0.16), NE * labelscalefactor,uququq); 
dot(C,xdxdff); 
label("$C$", (1.28,0.85), NE * labelscalefactor,xdxdff); 
dot(D,xdxdff); 
label("$D$", (1.17,-0.53), NE * labelscalefactor,xdxdff); 
dot((0.15,0.99),uququq); 
label("$E$", (0.17,1.02), NE * labelscalefactor,uququq); 
dot(F,uququq); 
label("$F$", (0.45,-0.87), NE * labelscalefactor,uququq); 
dot((1.03,0.34),uququq); 
label("$S$", (1.05,0.37), NE * labelscalefactor,uququq); 
dot(T,uququq); 
label("$T$", (0.96,-0.42), NE * labelscalefactor,uququq); 
dot(G,uququq); 
label("$G$", (-0.98,0.11), NE * labelscalefactor,uququq); 
dot(H,uququq); 
label("$H$", (0.72,-0.13), NE * labelscalefactor,uququq); 
dot(I,uququq); 
label("$I$", (1.06,-0.08), NE * labelscalefactor,uququq); 
dot(J,uququq); 
label("$J$", (-0.79,2.49), NE * labelscalefactor,uququq); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
[/asy]
ez

Note that $DA \perp GC$ and $BC \perp GD$ so we get that $I$ is the orthocenter of $\triangle GCD$ so it also lies on the circle $\alpha$.
now let the tangent at $I$ to $\alpha$ intersect the line $BQ$ at $T'$. now by applying pascal and duality lemma we get that $B-T'-H-E-J$ are collinear so $T'=T$. doing the same thing for $S$ we get that $S$ also lies on the tangent from $I$ to $\alpha$ so $ST$ is tangent to $\alpha$.
now we are done since $GI$ is perpendicular to both $\overline{ST}$ and $\overline{CD}$.$\blacksquare$



Another idea: there could be another way using $\textit{moving points}$ and $\textit{steiner conic}$ theorem but i wasn't able to complete that idea, unfortunately.
Z Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TheBarioBario
132 posts
#3
Y by
here is a proof using mostly pole-polar:
lemma 1: let $ABCD$ be a cyclic quadrilateral and let $AD\cap BC=E, AC\cap BD=F, AA\cap BB=X, CC\cap DD=Y$ where by $AA$ we mean the tangent to the circle at point $A$. then $EFXY$ are collinear.

proof: using pascal's theorem in $AADBBC$ and $DDACCB$ we get $E, X, F$ and $E, Y, F$ are collinear so $EXYF$ are collinear.

back to the problem:
define $\rho (X)$ the polar line of $X$ with respect to $\alpha$.
we know that $D \in \rho (C)$ because the midpoint of $DC$ is on the radical axis of $C,\alpha$ (because $QB^2=QC^2$). similarly $C\in \rho (D)$. because we know that $CE,DF$ are tangent to $\alpha$, we get that $\rho (C)=DE, \rho (D)=CF$
let $Y=CB\cap AD$. with simple angle chasing we conclude that $Y$ lies on $/\alpha$. on the other hand, $PY$ is perpendicular to $CD$ as a direct consequence of lemma 1 for points $ACDB$. now:
$$T=\rho (B)\cap\rho (C) \Rightarrow BC=\rho (T)$$$$S=\rho (A)\cap\rho (D) \Rightarrow AD=\rho (S)$$$$\Rightarrow Y=AD\cap BC=\rho (T)\cap \rho (S) \Rightarrow ST=\rho (Y)$$so because $Y\in\alpha$, $\rho (Y)$ is tangent to $\alpha$ and perpendicular to $PY$ (thus parallel to $CD$) so $ST$ is tangent to $\alpha$ and parallel to $CD$.
Z Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rickyminer
343 posts
#4
Y by
Again, this problem is from American Mathematical Monthly, 12245. See 2021 Issue 4 here.

worth noting that it's currently ongoing.
Z Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
UKR3IMO
26 posts
#5
Y by
Taking the problems from some source without saying them is wrong ?
Z Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NTstrucker
164 posts
#6 • 1 Y
Y by Mango247
UKR3IMO wrote:
Taking the problems from some source without saying them is wrong ?

You’d better change “?” with “.”
Z Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mr.C
539 posts
#7
Y by
Rickyminer wrote:
Again, this problem is from American Mathematical Monthly, 12245. See 2021 Issue 4 here.

worth noting that it's currently ongoing.

The problem is from Iran Tst 2018. :) and the magazine you are refering is from 2021.
So how about just enjoying the problem? All i see is un needed comments about where the problem is from and i dont know how this helps this tread.
If you want a source its from Iran Tst 2018 Day 1 . :)
This post has been edited 1 time. Last edited by Mr.C, Apr 17, 2021, 2:07 PM
Z Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rickyminer
343 posts
#8 • 1 Y
Y by amar_04
You are right.

I was writing these things just for your information. As you can see, 2 out of 9 problems are spotted to be from AMM. And this one has the exactly same wording with that AMM problem, making it suspicious.

Then it turns out that AMM problem is copied from Iranian TST problem, and Iranians used it again. This is stranger.

If you do find the discussion here useless, I shall request deleting them and lock the whole thread, since itself is a duplicate one.
Z Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rg230403
222 posts
#9
Y by
Solved with p_square.

Let $PC\cap \beta=X, PD\cap \beta=Y, BC\cap AD=Z$.

Claim 1: $Z\in \alpha$
Proof: Observe that we want $\angle BCD+\angle ADC=\angle AEB$ but we have $2\angle BCD+2\angle ADC=180-\angle BQA$ and $2\angle AEB=\angle ABP$ but $ABPQ$ is cyclic so we are done.

Claim 2: $X,E,D$ collinear.
Observe that $\triangle PEC$ is right angled and $PX\cdot PC=PE^2$. Thus, $\angle EXC=90$. Let $EX\cap \beta=D'$ then $\angle CXD'=90$ and thus $D\equiv D'$.

Claim 3: $Y,F,C$ collinear.
Proof: Same as claim 2.

Claim 4: $BTPZX$ cyclic.
Proof: $\angle TBP=\angle TXP=90$, thus $TBXP$ cyclic. Also, $PB^2=PZ^2=PX\cdot PC$ and $B,Z,P$ collinear. Thus, $P,B,Z,X$ cyclic. Thus, the claim follows.

Now, we get that $\angle TZP=90$ and similarly $\angle SZP=90$. So we get the desired tangency. Now, we just want $TZ\perp CD$ for the parallel condition.

Let $AC\cap BD=Z'$. Now, as before $Z'\in \alpha$. Also, $\angle ZBZ'=90$. Thus, $ZZ'$ is diameter of $\alpha$ and thus $P\in ZZ'$.

Thus, in cyclic quadrilateral $ABCD$, let $AB\cap CD=R$. Now, we have $QR\perp ZZ'$($Q$ is center of $(ABCD)$) but $QR=CD$ and $ZZ'=PZ$ and we are done.

Very nice problem!!
Z Y
G
H
=
a