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
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
3-var inequality
sqing   0
a minute ago
Source: Own
Let $ a,b>0 $ and $\frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \leq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{1}{a^3+3}+ \frac{1}{b^3+ 3}\leq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
0 replies
1 viewing
sqing
a minute ago
0 replies
Iranians playing with cards module a prime number.
Ryan-asadi   2
N 17 minutes ago by AshAuktober
Source: Iranian Team Selection Test - P2
.........
2 replies
Ryan-asadi
2 hours ago
AshAuktober
17 minutes ago
Coloring plane in black
Ryan-asadi   1
N 18 minutes ago by AshAuktober
Source: Iran Team Selection Test - P3
..........
1 reply
Ryan-asadi
2 hours ago
AshAuktober
18 minutes ago
An analytic sequence
Ryan-asadi   1
N 19 minutes ago by AshAuktober
Source: Iran Team Selection Test - P1
..........
1 reply
Ryan-asadi
3 hours ago
AshAuktober
19 minutes ago
AD=BE implies ABC right
v_Enhance   115
N 31 minutes ago by Adywastaken
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
115 replies
v_Enhance
Apr 10, 2013
Adywastaken
31 minutes ago
Geometry
gggzul   6
N 35 minutes ago by Captainscrubz
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
6 replies
gggzul
Yesterday at 8:22 AM
Captainscrubz
35 minutes ago
Need help on this simple looking problem
TheGreatEuler   0
40 minutes ago
Show that 1+2+3+4....n divides 1^k+2^k+3^k....n^k when k is odd. Is this possible to prove without using congruence modulo or binomial coefficients?
0 replies
TheGreatEuler
40 minutes ago
0 replies
Geometry
Lukariman   5
N an hour ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
5 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
an hour ago
inq , not two of them =0
win14   0
an hour ago
Let a,b,c be non negative real numbers such that no two of them are simultaneously equal to 0
$$\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} \ge \frac{5}{2\sqrt{ab + bc + ca}}.$$
0 replies
win14
an hour ago
0 replies
IMO Genre Predictions
ohiorizzler1434   62
N an hour ago by ehuseyinyigit
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
62 replies
ohiorizzler1434
May 3, 2025
ehuseyinyigit
an hour ago
Number theory
MathsII-enjoy   5
N 2 hours ago by MathsII-enjoy
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
5 replies
MathsII-enjoy
Monday at 3:22 PM
MathsII-enjoy
2 hours ago
Number theory
Foxellar   0
2 hours ago
It is known that for all positive integers $k$,
\[
1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k + 1)(2k + 1)}{6}
\]Find the smallest positive integer $k$ such that $1^2 + 2^2 + 3^2 + \ldots + k^2$ is divisible by 200.
0 replies
Foxellar
2 hours ago
0 replies
Combinatorics
P162008   4
N 3 hours ago by cazanova19921
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
4 replies
P162008
Today at 5:38 AM
cazanova19921
3 hours ago
Aime type Geo
ehuseyinyigit   4
N 3 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
4 replies
ehuseyinyigit
Monday at 9:04 PM
ehuseyinyigit
3 hours ago
IMO 2008, Question 1
orl   155
N Apr 25, 2025 by Ilikeminecraft
Source: IMO Shortlist 2008, G1
Let $ H$ be the orthocenter of an acute-angled triangle $ ABC$. The circle $ \Gamma_{A}$ centered at the midpoint of $ BC$ and passing through $ H$ intersects the sideline $ BC$ at points $ A_{1}$ and $ A_{2}$. Similarly, define the points $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$.

Prove that the six points $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia
155 replies
orl
Jul 16, 2008
Ilikeminecraft
Apr 25, 2025
IMO 2008, Question 1
G H J
Source: IMO Shortlist 2008, G1
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
orl
3647 posts
#1 • 29 Y
Y by Davi-8191, Smita, microsoft_office_word, OlympusHero, stayhomedomath, Jc426, centslordm, Adventure10, jhu08, mathlearner2357, megarnie, HWenslawski, bjump, lian_the_noob12, Mango247, ItsBesi, Sedro, Tastymooncake2, Rounak_iitr, ehuseyinyigit, DEKT, AlexCenteno2007, and 7 other users
Let $ H$ be the orthocenter of an acute-angled triangle $ ABC$. The circle $ \Gamma_{A}$ centered at the midpoint of $ BC$ and passing through $ H$ intersects the sideline $ BC$ at points $ A_{1}$ and $ A_{2}$. Similarly, define the points $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$.

Prove that the six points $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia
This post has been edited 4 times. Last edited by orl, Jul 20, 2008, 8:14 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Jan
606 posts
#2 • 23 Y
Y by kgo, microsoft_office_word, Williamgolly, OlympusHero, myh2910, centslordm, Adventure10, Adventure10, jhu08, EpicBird08, Mango247, Tastymooncake2, Sedro, Adi1005247, MS_asdfgzxcvb, and 8 other users
Call $ M_1,M_2,M_3$ the midpoints of $ BC,AC$ and $ AB$.

We know that $ BP \perp AC$, so $ BP$ and $ M_1M_3$ are also perpendicular, and $ B$ lies on the radical axis of $ \Gamma_A$ and $ \Gamma_C$. It follows that $ BC_1\cdot BC_2 = BA_1 \cdot BA_2$, so $ A_1,A_2,C_1$ and $ C_2$ all lie on a circle, whose center is clearly $ O$, the circumcenter of $ \Delta ABC$.

Now we can show that $ B_1$ and $ B_2$ also lie on the circle centered at $ O$, passing through $ A_1$ and $ A_2$, from which the conclusion follows.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TomciO
552 posts
#3 • 27 Y
Y by mathmaster2012, AdBondEvent, Arthur., pad, myh2910, Lifefunction, OlympusHero, MiraclesINmaths, Adventure10, SPHS1234, jhu08, Numbertheorydog, rayfish, EpicBird08, Mango247, Tastymooncake2, Sedro, and 10 other users
The radical axis of $ \Gamma_{A}$ and $ \Gamma_{B}$ is perpendicular to the line connecting the centers of these circles, i.e. it is perpendicular to $ AB$. Since both circles pass trough $ H$ their radical axis is a height from $ C$, so $ C$ lies on a radical axis. It means that $ CA_1CA_2 = CB_1CB_2$ so $ A_1, A_2, B_2, B_1$ lie on one circle. From the same reasoning $ B_1, B_2, C_2, C_1$ and $ C_1, C_2, A_2, A_1$ lie on a one circle as well. Suppose that this three circles doesn't coincide. Then we obtain a contradiction since radical axis of these circles - the sides of the triangle $ ABC$ - don't intersect in one point.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lepuslapis
78 posts
#4 • 3 Y
Y by jhu08, Adventure10, Mango247
Another possibility to prove that, let's say, $ A_1A_2B_1B_2$ is cyclic consists in transforming the relation $ CA_1\cdot CA_2 = CB_1\cdot CB_2$ into $ CD^2 - DH^2 = CE^2 - EH^2$ (where $ D$ and $ E$ are the respective midpoints of $ [BC]$ and $ [AC]$). Adding $ CH^2$ on both sides and applying the generalized version of Pythagoras' theorem (the one with cosines, I ignore the correct English name), we get a condition which is obviously true (simplifying cosines in right-angled triangles)...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
plane geometry
467 posts
#5 • 4 Y
Y by Mathmick51, jhu08, Adventure10, Mango247
A,B,C lie on the radix axis of each two circles
so conclusion apparently follows
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TTsphn
1313 posts
#6 • 10 Y
Y by Mobashereh, jhu08, Adventure10, Mango247, Tastymooncake2, AlexCenteno2007, and 4 other users
I have a same solution with you .
Call $ M,N,P$ is the midpoint of $ BC,CA,AB$
From condition we have :
$ BA_1.BA_2=BM^2-HM^2$
$ BC_1.BC_2=BN^2-HN^2$
But from $ MN||AC,BH\perp AC$ therefore $ BH\perp MN$
It gives $ BM^2-BN^2=HM^2-HN^2$
Therefore $ A_1,B_1,A_2,B_2$ are cyclic on circle $ O_c$
Similar for $ A_1,A_2,C_1,C_2$ lie on $ O_b$and $ B_1,B_2,C_1,C_2$ lie on circle $ O_a$
Easy to check that three circle are coincide .
So problem claim.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pohoatza
1145 posts
#7 • 11 Y
Y by iarnab_kundu, trumpeter, Jc426, Adventure10, jhu08, Tastymooncake2, and 5 other users
Indeed, the proof involving the three radical axis is the best. In any case, the circle belongs to Droz-Farny as Darij well mentioned (see http://www.pandd.demon.nl/drozf.htm; scroll down till figure 3).

By the way, it is obvious that the triangle doesn't need to be acute-angled. I guess the coordinators didn't want messy diagrams :roll:.

In addition, here is a natural generalization:

Theorem. Let $ P$, $ Q$ be two isogonal points with respect to a given triangle $ ABC$. Let $ P_{A}$, $ P_{B}$, $ P_{C}$ be the orthogonal projections of $ P$ on the sidelines $ BC$, $ CA$ and $ AB$, respectively. Denote by $ \Gamma_{A}$ the circle centered at $ P_{A}$ which passes through $ Q$ and let $ A_{1}$, $ A_{2}$ be the intersections points of $ \Gamma_{A}$ with the sideline $ BC$. Similarly, define $ B_{1}$, $ B_{2}$, $ C_{1}$, $ C_{2}$. Then, the points $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$, $ C_{2}$ are on a same circle centered at $ P$.

I guess it would have been a better IMO problem, at least to avoid the "well-known"-type commentaries.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ahiles
374 posts
#8 • 10 Y
Y by thunderz28, microsoft_office_word, Adventure10, Mango247, Tastymooncake2, and 5 other users
It is enough to prove that points $ A_1,A_2,B_1,B_2$ are concyclic. Let $ O$ be circumcenter of the triangle $ ABC$. We'll prove that $ O$ is circumcenter of quadrilateral $ A_1A_2B_1B_2$. Then $ O$ is situated on the perpendicular bisectors of the segments $ A_1A_2$ and $ B_1B_2$ $ \Longrightarrow OA_1 = OA_2, OB_1 = OB_2$. So it is enough to prove that $ OA_1 = OB_1$.
Call $ r_1,r_2$ circumcenters of the $ \Gamma_a$ and $ \Gamma_b$ respectively and $ A',B'$ midpoints of the segments $ BC,CA$. Then $ \left\|\begin{array}{cc} OA_1 = r_1^2 + OA'^2 \\
OB_1 = r_2^2 + OB'^2 \end{array} \right\|$.
By Stewart we have
$ \boxed{\begin{array}{cc} 4r_1^2 = 2(CH^2 + BH^2) - a^2 = \\
= 8R^2(\cos^2{B} + \cos^2{C}) - 4R^2\sin^2{A} = \\
= 4R^2(2\cos^2{B} + 2\cos^2{C} - \sin^2{A})\end{array}}$.

By analogy $ 4r_2^2 = 4R^2(2\cos^2{A} + 2\cos^2{C} - \sin^2{B})$.

But $ \left\| \begin{array}{cc} OA'^2 = \frac {1}{4}a^2\cos^2{A} = R^2\cot^2{A} \\
OB'^2 = \frac {1}{4}b^2\cos^2{B} = R^2\cot^2{B} \end{array} \right\|$.

Then $ OA_1^2 = 4R^2(2\cos^2{B} + 2\cos^2{C} - \sin^2{A} + \cot^2{A})$ and $ OB_1^2 = 4R^2(2\cos^2{A} + 2\cos^2{C} - \sin^2{B} + \cot^2{B})$.

So we have to prove that:
$ \boxed{\begin{array}{cc} 2\cos^2{B} + 2\cos^2{C} - \sin^2{A} + \cot^2{A} = 2\cos^2{A} + 2\cos^2{C} - \sin^2{B} + \cot^2{B} \\
\cos^2{B} + (\cos^2{B} + \sin^2{B}) + \cot^2{A} = \cos^2{A} + (\cos^2{A} + \sin^2{A}) + \cot^2{B} \\
\cos^2{B} + \cot^2{A} = \cos^2{A} + \cot^2{B} \\
\cos^2{B} + \frac {\cos^2{A}}{\sin^2{A}} = \cos^2{A} + \frac {\cos^2{B}}{\sin^2{B}} \\
\cos^2{B}\sin^2{A} + \cos^2{A} = \cos^2{A}\sin^2{B} + \cos^2{B} \\
\cos^2{B}(\sin^2{A} - 1) = \cos^2{A}(\sin^2{B} - 1) \\
- \cos^2{B}\cos^2{A} = - \cos^2{B}\cos^2{A} \end{array} }$
and 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.
cyshine
236 posts
#9 • 2 Y
Y by Adventure10, Mango247
Fix $ O$ as center of a coordinate vector system, so that $ H = A + B + C$. If $ M = \frac{A + B}2$ we have to prove that $ MC_1^2 + OM^2$ is symmetric in $ A$, $ B$ and $ C$ (it would be the square of the radius of the circle).

But

$ MC_1^2 + OM^2 = HM^2 + OM^2 = \left(A + B + C - \frac{A+B}2\right)\cdot \left(A + B + C - \frac{A+B}2\right) + \left(\frac{A+B}2\right)\cdot \left(\frac{A+B}2\right) = \frac12A\cdot A + \frac12B\cdot B + C\cdot C + A\cdot B + B\cdot C + C\cdot A = 2R^2 + A\cdot B + B\cdot C + C\cdot A$

and 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.
msecco
154 posts
#10 • 2 Y
Y by Adventure10, Mango247
No no no!!
More one trivial Geometry.
Denote AB=2c, AC=2b and BC=2a.
Denote BA_1=x and AC_1=y.Denote too M, N and P the midpoints of BC, AB and AC respectively.
Then, MA_1=a-x=MA_2. We have to prove that BA_1.BA_2=BC_2.BC_1.
Applying the Cosines'Law in triangles BNH and BMH, we have:

a²-2ax+x²=a²+BH²-2a.BH.sen<C and c²-2cy+y²=c²+BH²-2c.BH.sen<A <=>
2ax-x²=2a.BH.sen<C-BH² and 2cy-y²=2c.BH.sen<A-BH².
But 2ax-x²=BA_1.BA_2 and 2cy-y²=BC_2.BC_1.
Then, we have to prove that: 2c.BH.sen<A-BH²=2a.BH.sen<C-BH² <=>2a.sen<C=2c.sen<A (obviously by Sines'law).
Hence, A_1,A_2,C_1,C_2 are concyclics and by the same form, the other 2 quadruples of points are concyclics.
Conclusion: The six points are concyclics!

Have fun!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
not_trig
2049 posts
#11 • 5 Y
Y by Adventure10, Mango247, DEKT, and 2 other users
What is a sideline!? What is sideline $ BC$?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kaloi01
20 posts
#12 • 6 Y
Y by Adventure10, Tastymooncake2, and 4 other users
See point $ A$ first. let $ X$ is the midpoint of $ AB$ and $ Y$ is the midpoint $ AC$. It follows that $ XY \perp AH$. Thus, $ AH$ is the radical axis of circle $ B_1B_2H$ and $ C_1C_2H$. It follows that $ AC_1*AC_2 = AB_1*AB_2 \Leftrightarrow B_1B_2C_1C_2$ cylcic, by similar argument on $ B$ and $ C$ we´re done. :lol: :blush:
This post has been edited 1 time. Last edited by kaloi01, Jul 16, 2008, 6:33 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
campos
411 posts
#13 • 6 Y
Y by Adventure10, Tastymooncake2, and 4 other users
it's enough to prove that $ O$ is at the same distance of any of the six points... obviously $ OA_1 = OA_2$, so it's enough to prove that the expression $ OA_1$ is independent from $ A$. In fact, $ OA_1^2 = OM^2 + MA_1^2 = R^2\cos^2A + MH^2$...

it can be proven that $ MH^2 = R^2(4\cos^2B\cos^2C + \sin^2(B - C))$

so, $ OA_1^2 = R^2(\cos^2A + 4\cos^2B\cos^2C + \sin^2(B - C))$...

finally, after some work (you can try $ \cos A = \sin B\sin C - \cos B\cos C$) it follows that it equals $ R^2(1-4\cos A\cos B\cos C)$, and we're done. :D
This post has been edited 1 time. Last edited by campos, Jul 17, 2008, 5:21 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Virgil Nicula
7054 posts
#14 • 7 Y
Y by Adventure10, Mango247, Tastymooncake2, Rounak_iitr, and 3 other users
orl wrote:
Let $ H$ be the orthocenter of an acute triangle $ ABC$ . The circle centered at the midpoint of $ BC$ and passing through $ H$ intersects

the line $ BC$ at $ A_{1}$ , $ A_{2}$ . Similarly define the pairs $ B_{1}$, $ B_{2}$ and $ C_{1}$ , $ C_{2}$ . Prove that $ A_{1}$ , $ A_{2}$, $ B_{1}$, $ B_{2}$ , $ C_{1}$ , $ C_{2}$ are concyclically.


Proof. Denote the midpoints $ D$ , $ E$ , $ F$ of the sides $ [BC]$ , $ [CA]$ , $ [AB]$ respectively. Prove easily or is well-known that

$ \boxed {\ HA^2 + a^2 = HB^2 + b^2 = HC^2 + c^2 = 4R^2\ }\ (*)$ . Thus, $ 4\cdot\overline {BA_1}\cdot\overline {BA_2} = 4\cdot\left(BD^2 - HD^2\right) =$ $ a^2 - 4\cdot HD^2 =$

$ a^2 - 2\cdot\left(HB^2 + HC^2\right) + a^2$ $ \stackrel {(*)}{\ \ \implies\ \ }$ $ \overline {BA_1}\cdot\overline {BA_2} = \frac 12\cdot \left(a^2 + b^2 + c^2\right) - 4R^2$ $ \implies$ $ \boxed {\ \overline {BA_1}\cdot\overline {BA_2} = 4R^2\prod\cos A\ }$

because prove easily or is well-known that $ \boxed {\ \left(a^2 + b^2 + c^2\right) - 8R^2 = 8 R^2\prod\cos A\ } > 0$ (remark that $ \{A_1,A_2\}\subset (BC)$ a.s.o.).

Thus, $ \overline {BA_1}\cdot \overline {BA_2} = - \overline {A_1B}\cdot\overline {A_1C} = R^2 - OA_1^2\ \implies\ 4R^2\prod\cos A = R^2 - OA_1^2$ $ \implies$ $ OA_1^2 = R^2\left(1 - 4\prod\cos A\right)$

(symmetrically in $ a$ , $ b$ , $ c$ ). In conclusion, $ \boxed {\ \rho = OA_1 = OA_2 = OB_1 = OB_2 = OC_1 = OC_2 = R\sqrt {1 - 4\cos A\cos B\cos C}\ }$ .

Remark. Since $ 1 - 8\prod\cos A\ge 0$ obtain easily that $ \rho\ge 2R\sqrt {\prod\cos A}$ .

Pohoatza wrote:
I guess it would have been a better IMO problem, at least to avoid the "well-known"-type commentaries.
All right !
This post has been edited 23 times. Last edited by Virgil Nicula, Jul 17, 2008, 1:47 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Akashnil
736 posts
#15 • 12 Y
Y by iarnab_kundu, AnonymousBunny, Adventure10, Adventure10, Mango247, Mango247, Mango247, Tastymooncake2, and 4 other users
Let $ O$ circumcenter, $ D$ midpoint of $ BC$,$ N$ nine point center.
$ OA_1^2=OD^2+DA_1^2=OD^2+DH^2=2(DN^2+ON^2)=const.$
Z K Y
G
H
=
a