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
Additive combinatorics (re Cauchy-Davenport)
mavropnevma   3
N 17 minutes ago by Orzify
Source: Romania TST 3 2010, Problem 4
Let $X$ and $Y$ be two finite subsets of the half-open interval $[0, 1)$ such that $0 \in X \cap Y$ and $x + y = 1$ for no $x \in X$ and no $y \in Y$. Prove that the set $\{x + y - \lfloor x + y \rfloor : x \in X \textrm{ and } y \in Y\}$ has at least $|X| + |Y| - 1$ elements.

***
3 replies
mavropnevma
Aug 25, 2012
Orzify
17 minutes ago
Ducks can play games now apparently
MortemEtInteritum   34
N an hour ago by HamstPan38825
Source: USA TST(ST) 2020 #1
Let $a$, $b$, $c$ be fixed positive integers. There are $a+b+c$ ducks sitting in a
circle, one behind the other. Each duck picks either rock, paper, or scissors, with $a$ ducks
picking rock, $b$ ducks picking paper, and $c$ ducks picking scissors.
A move consists of an operation of one of the following three forms:

[list]
[*] If a duck picking rock sits behind a duck picking scissors, they switch places.
[*] If a duck picking paper sits behind a duck picking rock, they switch places.
[*] If a duck picking scissors sits behind a duck picking paper, they switch places.
[/list]
Determine, in terms of $a$, $b$, and $c$, the maximum number of moves which could take
place, over all possible initial configurations.
34 replies
MortemEtInteritum
Nov 16, 2020
HamstPan38825
an hour ago
Floor sequence
va2010   87
N an hour ago by Mathgloggers
Source: 2015 ISL N1
Determine all positive integers $M$ such that the sequence $a_0, a_1, a_2, \cdots$ defined by \[ a_0 = M + \frac{1}{2}   \qquad  \textrm{and} \qquad    a_{k+1} = a_k\lfloor a_k \rfloor   \quad \textrm{for} \, k = 0, 1, 2, \cdots \]contains at least one integer term.
87 replies
va2010
Jul 7, 2016
Mathgloggers
an hour ago
INMO 2019 P3
div5252   45
N an hour ago by anudeep
Let $m,n$ be distinct positive integers. Prove that
$$gcd(m,n) + gcd(m+1,n+1) + gcd(m+2,n+2) \le 2|m-n| + 1. $$Further, determine when equality holds.
45 replies
div5252
Jan 20, 2019
anudeep
an hour ago
Inequalities
sqing   1
N 2 hours ago by DAVROS
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
1 reply
sqing
5 hours ago
DAVROS
2 hours ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   3
N 5 hours ago by SomeonecoolLovesMaths
This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3
3 replies
SomeonecoolLovesMaths
Today at 8:16 AM
SomeonecoolLovesMaths
5 hours ago
an algebra problem
Asyrafr09   2
N 6 hours ago by pooh123
Determine all real number($x,y,z$) that satisfy
$$x=1+\sqrt{y-z^2}$$$$y=1+\sqrt{z-x^2}$$$$z=1+\sqrt{x-y^2}$$
2 replies
Asyrafr09
Today at 10:05 AM
pooh123
6 hours ago
Inequalities
sqing   1
N 6 hours ago by sqing
Let $ a,b,c\geq 0 ,   2a +ab + 12a bc \geq 8. $ Prove that
$$  a+  (b+c)(a+1)+\frac{4}{5}  bc \geq 4$$$$  a+  (b+c)(a+0.9996)+ 0.77  bc \geq 4$$
1 reply
sqing
Today at 5:23 AM
sqing
6 hours ago
When to look at solutions - pre calc
omerrob13   1
N Today at 10:51 AM by abartoha
Hey all.
I am doing the precalc book, and unfortunately, im getting into the habit of looking in the solutions quite fast on a problem I did not able to make any progress on.
My goal is mainly to develop problem solving and reasonning skills.

I divide the problems in AOPS to 2:

- Challenge problems at the end of the of each chapter.
- The problems that teach you the material itself, and the problems at the end of each section (1.1,1.2, etc...)

For non challenging problems, It takes around 20 mins of me not be able to solve a problem, and look at the solutions for it

Is it too little?
My goal is mainly to develop problem solving and reasoning skills.
I'm not sure if it's too little time to bring to a regular problem, or its ok to give 20 mins to a problem and continue if making no progress.
1 reply
omerrob13
Today at 9:36 AM
abartoha
Today at 10:51 AM
f_n(x)=\sum sin(nx)/n
Urumqi   4
N Today at 9:11 AM by Urumqi
$F_n(x)=\sum_{k=1}^{n}\frac{\sin (kx)}{k}$, prove that for all $x \in (0,\pi), F_n(x)>0$.

Thanks.
4 replies
Urumqi
Today at 2:13 AM
Urumqi
Today at 9:11 AM
How many pairs
Ecrin_eren   3
N Today at 8:21 AM by Ecrin_eren


Let n be a natural number and p be a prime number. How many different pairs (n, p) satisfy the equation:

p + 2^p + 3 = n^2 ?



3 replies
Ecrin_eren
May 2, 2025
Ecrin_eren
Today at 8:21 AM
Hard Inequality
William_Mai   8
N Today at 8:17 AM by DAVROS
Given $a, b, c \in \mathbb{R}$ such that $a^2 + b^2 + c^2 = 1$.
Find the minimum value of $P = ab + 2bc + 3ca$.

Source: Pham Le Van
8 replies
William_Mai
Yesterday at 2:13 PM
DAVROS
Today at 8:17 AM
How many triangles
Ecrin_eren   2
N Today at 8:15 AM by Ecrin_eren


"Inside a triangle, 2025 points are placed, and each point is connected to the vertices of the smallest triangle that contains it. In the final state, how many small triangles are formed?"


2 replies
Ecrin_eren
May 2, 2025
Ecrin_eren
Today at 8:15 AM
Summation
Saucepan_man02   2
N Today at 3:54 AM by P162008
$\sum_{r=1}^{\infty}\frac{12r^2+1}{64r^6-48r^4+12r^2-1}$
2 replies
Saucepan_man02
Mar 6, 2025
P162008
Today at 3:54 AM
Two circles, a tangent line and a parallel
Valentin Vornicu   104
N Friday at 9:58 PM by cubres
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
104 replies
Valentin Vornicu
Oct 24, 2005
cubres
Friday at 9:58 PM
Two circles, a tangent line and a parallel
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Math_.only.
22 posts
#96 • 1 Y
Y by cubres
arqady wrote:
let MN and AB intesect at K. Then $KA^2=KM*KN=KB^2.$ Hence KA=KB.
CD||AB. Hence PM=QM.
$\angle AEB=\angle ECD=\angle MAB,$ $\angle EBA=\angle EDC=\angle MBA.$
hence$\triangle AMB$ and$\triangle AEB$ congruity.
Hence $EM\perp AB.$ Hence $EM\perp PQ.$
Hence $EP=EQ.$

Why angles AEB and ECD are equal ?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MagicalToaster53
159 posts
#97 • 1 Y
Y by cubres
Notice that $MN$ bisects $\overline{AB},$ as the point where they intersect has equivalent powers to the respective circles. From this we discover that $M$ is the midpoint of $\overline{PQ}$ due to the homothety at $N$ mapping $\overline{PQ} \mapsto \overline{AB}.$ It then suffices to show $EM \perp PQ$.

Claim: $\triangle EAB \cong \triangle MAB.$
Proof: Observe that $AB$ bisects $\angle EAM:$ \[\angle EAB = \angle ECM = \angle ACM = \angle BAM. \]Similarly by symmetry, we find that $AB$ bisects $\angle EBM.$ Then by SAS, we conclude that $\triangle EAB \cong \triangle MAB,$ as was to be shown. $\square$

Now $EAMB$ is a kite, so that $EM \perp AB \implies EM \perp PQ,$ as $AB \parallel PQ$. Hence $\triangle EPQ$ is isosceles, as desired. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lnzhonglp
120 posts
#98 • 1 Y
Y by cubres
Observe that $$\measuredangle EBA = \measuredangle EDC = \measuredangle ABM$$and $$\measuredangle EAB = \measuredangle ECM = \measuredangle MAB,$$so $\triangle EAB \cong \triangle MAB$ and $EM \perp AB$. From radical axis we see that $MN$ bisects $AB$, so by homothety at $N$ we find that $M$ is the midpoint of $PQ$. Then since $EM \perp PQ$ it follows that $EP = EQ$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Likeminded2017
391 posts
#99 • 1 Y
Y by cubres
Let $MN$ and $AB$ intersect at $K.$ As $K$ lies on the radical axis we have $KA=KB.$ Taking the homothety centered at $N$ that maps $M$ to $K$ we find that $MP=MQ.$ Then $\angle BAM=\angle MCA=\angle BAE$ and similarly $\angle EBA=\angle ABM$ so by ASA congruence $\triangle EAB \cong \triangle MAB.$ Thus $EM$ is perpendicular to $AB$ and thus $PQ$ so $EP=EQ.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ezpotd
1262 posts
#100 • 1 Y
Y by cubres
Letting $AB$ meet $MN$ at $K$, we see $KA^2 = KM \cdot KN = KB^2$, so $K$ is the midpoint of $AB$, taking the homothety from $AB$ to $PQ$ this gives $MP = MQ$, so it suffices to prove $EM \perp PQ$, or $EM \perp AB$. However, angle chasing gives $\angle EAB = \angle ECM = \angle AMC = \angle BAM, \angle EBA = \angle EDM = \angle BMD = \angle ABM$, so $\triangle EAB \cong \angle MAB$, so $E,M$ are reflections about $AB$, done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
optimusprime154
21 posts
#101 • 2 Y
Y by SorPEEK, cubres
let \(Z \)= \(MN  \cap AB\) from PoP we know \(ZB^2 = ZM * ZN\) and \(ZA^2 = ZM * ZN\) which means \(ZA = ZB\) and since \(PQBA\) is a trapezoid, \(PM = MQ\)
now we prove \(EM \perp CD\) and we're done. label \(\angle ACM = x, \angle BDM = y\) we know from \(AB \parallel CD\) that \(\angle EAB = x\) and \(\angle EDC = y\)
we also know from \(AB\) being tangent that \(\angle MBA\ = y\) and \(\angle MAB = x\) the above facts indicate the similarity of \(\triangle MAB , \triangle EAB \) so \(EM \perp AB\) => \(EM \perp CD\) and we're done
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#102 • 1 Y
Y by cubres
Extend $MN$ to intersect $AB$ at $K,$ then clearly $AK=BK.$ Therefore $PM=MQ$ by the parallel lines.

Now, notice that $\angle EAB = \angle ACM = \angle BAM = \angle AMC.$ Therefore $AB$ bisects $\angle EAM$ and $AC=AM.$ Similarly $AB$ bisects $EBM$ and $BD=BM.$ Therefore, $\triangle EAB \cong \triangle MAB$ so $A, B$ are the circumcenters of $\triangle EMC, \triangle EMD$ respectively so by Thales' Theorem it follows that $\angle EMP = \angle EMQ = 90^\circ.$ Hence $EM$ is the perpendicular bisector of $PQ,$ and we are done. QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ehuseyinyigit
815 posts
#104 • 1 Y
Y by cubres
WELL-KNOWN. Since $AB\parallel CD$, we have $$\angle ACM=\angle AMC=\angle MAB=\angle BAE$$Thus, $AC=AM$ and $AB$ bisects $\angle EAM$. Similarly $BD=BM$ and $AB$ bisects $\angle EBM$ hence $ABME$ is a kite $\Rightarrow AE=AM=AC$ implying $EM\perp CD$. On the other hand, for point $F$ being intersection of lines $AB=MN$, $AF=BF$ implies $PM=MQ$. We obtain $EP=EQ$.

WELL-KNOWN. We will show $EN$ bisects $\angle CND$. This is true because $$\angle ENB=\angle EAB=\angle ENC$$and $\angle ANE=\angle ABE=\angle BND$. Thus, $EN$ bisects $\angle CND$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathwiz_1207
96 posts
#105 • 1 Y
Y by cubres
We have
\[\measuredangle EAB = \measuredangle ACM = \measuredangle ANM = \measuredangle BAM = \measuredangle CMA\]\[\measuredangle EBA = \measuredangle BDM = \measuredangle BNM = \measuredangle ABM = \measuredangle DMB\], implying that $\triangle AMB \cong \triangle AEB$, and that $\triangle AMC, \triangle BDM$ are isosceles. Therefore,
\[AE = AM = AC\]so $A$ is the center of $(EMC)$, and similarly $B$ is the cetner of $(EMD)$. Thus, $EM \perp CD$. Now, let $L = MN \cap AB$, by Radical Axis,
\[LA = LB\]and since $\triangle NPQ \sim \triangle NAB$, we have $MP = MQ \implies EQ = EP$, so we are done.
This post has been edited 1 time. Last edited by mathwiz_1207, Feb 23, 2025, 11:53 PM
Reason: forgot to add last line
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
study1126
554 posts
#106 • 1 Y
Y by cubres
All angles will be directed in this solution.

First, $\angle EAB=\angle ECB=\angle BAM$, since $AB||CD$ and $AB$ is tangent to the two circles. Similarly, $\angle EBA=\angle ABM$. Combined with $AB=AB$ gives $\triangle EAB\cong \triangle MAB$.

This implies $EAMB$ is a kite, so $EM\perp AB$, which means $EM\perp CD$. Since $M$ is on the radical axis, extending $NM$ to $X$ on $AB$ gives $X$ is the midpoint of $AB$. Homothety shows $M$ is the midpoint of $PQ$. Thus, $MP=MQ$, so by congruent triangle $\triangle EPM\cong \triangle EQM$, $EP=EQ$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joshualiu315
2534 posts
#107 • 1 Y
Y by cubres
It is well-known that $\overline{MN}$ bisects $\overline{AB}$, and since $\overline{AB} \parallel \overline{CD}$, we also know that $\overline{MN}$ bisects $\overline{PQ}$. This implies $M$ is the midpoint of $\overline{PQ}$.

We have $\angle EAB = \angle ACM = \angle MAB$ and similarly, $\angle EBA = \angle BDM = \angle MBA$. This means $\triangle EAB \cong \triangle MAB$, which implies that $\overline{EM} \perp \overline{AB}$. Because $\overline{AB} \parallel \overline{CD}$, we have $\overline{EM} \perp \overline{PQ}$, so $\overline{EM}$ is the perpendicular bisector of $\overline{PQ}$. Thus, $EP=EQ$, as desired. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
LeYohan
41 posts
#108 • 1 Y
Y by cubres
First IMO #1 solve and it's geo!! :D

Claim: $PM = MQ$.
Let $NM$ intersect $AB$ at $X$. It's well known that the radical axis intersects the common tangent, hence $AX = XB$. By the Thales Theorem, we inmediately get the result. $\square$

Now doing some angle chase using the fact that $CD \parallel AB$ and $AB$ is tangent to both circles, we get:
$\angle EBA = \angle BDM = \angle BNM = \angle ABM = \angle BMD$, and $\angle EAB = \angle ACM = \angle ANM = \angle BAM = \angle AMC$. As a result, $\triangle ACM$ and $\triangle BMC$ are isosceles. Furthermore, $\triangle MAB \cong \triangle EAB \implies EB = BD, EA = AC$. This means that $\angle EMC = \angle EMD = 90$, and notice that we originally proved that $PM = MQ$, so $\triangle EPQ$ is isosceles and we're done. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Avron
37 posts
#109 • 1 Y
Y by cubres
$MN$ is the radical axis of $G_1,G_2$ so it bisects $AB$ so we also know that $M$ is the midpoint of $PQ$, thus it is enough to prove $ME\perp CD$.
Now notice that $\angle MAB=\angle ACM = \angle EAB$ and similarly $\angle MBA = \angle ABE$ so $EAB \cong MAB$ and we're done
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zuat.e
55 posts
#110 • 1 Y
Y by cubres
Easy angle chasing yields the cyclicity of $(BEAN)$. We claim the following:

Claim: $EA=AM=AC$ and similarly $EB=BM=BD$
Proof: Let $O_1$ be the center of $G_1$. It is clear that $O_1A$ is the side bisector of $MC$, hence $AM=AC$.
Furthermore, note that $\measuredangle BAE=\measuredangle DCE= \measuredangle BMC=\measuredangle MAB$ and similarly $\measuredangle EBA =\measuredangle ABM$, hence $\triangle EAB\cong MAB$ (because they also share $AB$).

It now follows that $AC=AM=AE$ and analogously $BD=BM=BE$
Finally, as $\measuredangle MCA= \measuredangle MNA=\measuredangle ANC$, we have $AP\cdot AN=AC^2=AE^2$, hence $\measuredangle CEP =\measuredangle ENA=\measuredangle EBA=\measuredangle EDC$ and similarly $\measuredangle QED=\measuredangle PCE$, yielding $\measuredangle CPE =\measuredangle EQD$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cubres
118 posts
#111
Y by
ACGN
MOHS Hardness Scale
Daily IMO Problem #1 (May 2) - IMO 2001 p1
Z K Y
N Quick Reply
G
H
=
a