Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

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
Inspired by 2025 Beijing
sqing   10
N 22 minutes ago by ytChen
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
10 replies
1 viewing
sqing
Yesterday at 4:56 PM
ytChen
22 minutes ago
Three operations make any number
awesomeming327.   0
2 hours ago
Source: own
The number $3$ is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
0 replies
awesomeming327.
2 hours ago
0 replies
IMO 2017 Problem 4
Amir Hossein   116
N 2 hours ago by cj13609517288
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
116 replies
Amir Hossein
Jul 19, 2017
cj13609517288
2 hours ago
A sharp one with 3 var
mihaig   10
N 2 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
10 replies
mihaig
May 13, 2025
mihaig
2 hours ago
Another right angled triangle
ariopro1387   1
N 2 hours ago by lolsamo
Source: Iran Team selection test 2025 - P7
Let $ABC$ be a right angled triangle with $\angle A=90$. Point $M$ is the midpoint of side $BC$ And $P$ be an arbitrary point on $AM$. The reflection of $BP$ over $AB$ intersects lines $AC$ and $AM$ at $T$ and $Q$, respectively. The circumcircles of $BPQ$ and $ABC$ intersect again at $F$. Prove that the center of the circumcircle of $CFT$ lies on $BQ$.
1 reply
ariopro1387
Today at 4:13 PM
lolsamo
2 hours ago
four points lie on a circle
pohoatza   78
N 2 hours ago by ezpotd
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
78 replies
pohoatza
Jun 28, 2007
ezpotd
2 hours ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   2
N 3 hours ago by Stear14
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
2 replies
FishkoBiH
Today at 1:38 PM
Stear14
3 hours ago
Does there exist 2011 numbers?
cyshine   8
N 3 hours ago by TheBaiano
Source: Brazil MO, Problem 4
Do there exist $2011$ positive integers $a_1 < a_2 < \ldots < a_{2011}$ such that $\gcd(a_i,a_j) = a_j - a_i$ for any $i$, $j$ such that $1 \le i < j \le 2011$?
8 replies
cyshine
Oct 20, 2011
TheBaiano
3 hours ago
D1036 : Composition of polynomials
Dattier   1
N 3 hours ago by Dattier
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
1 reply
Dattier
Yesterday at 1:52 PM
Dattier
3 hours ago
number sequence contains every large number
mathematics2003   3
N 3 hours ago by sttsmet
Source: 2021ChinaTST test3 day1 P2
Given distinct positive integer $ a_1,a_2,…,a_{2020} $. For $ n \ge 2021 $, $a_n$ is the smallest number different from $a_1,a_2,…,a_{n-1}$ which doesn't divide $a_{n-2020}...a_{n-2}a_{n-1}$. Proof that every number large enough appears in the sequence.
3 replies
mathematics2003
Apr 13, 2021
sttsmet
3 hours ago
IMO ShortList 2002, geometry problem 2
orl   28
N 3 hours ago by ezpotd
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
28 replies
orl
Sep 28, 2004
ezpotd
3 hours ago
Arc Midpoints Form Cyclic Quadrilateral
ike.chen   56
N 3 hours ago by ezpotd
Source: ISL 2022/G2
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
56 replies
ike.chen
Jul 9, 2023
ezpotd
3 hours ago
Non-linear Recursive Sequence
amogususususus   4
N 3 hours ago by GreekIdiot
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
4 replies
amogususususus
Jan 24, 2025
GreekIdiot
3 hours ago
Russian Diophantine Equation
LeYohan   2
N 3 hours ago by RagvaloD
Source: Moscow, 1963
Find all integer solutions to

$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$.
2 replies
LeYohan
Yesterday at 2:59 PM
RagvaloD
3 hours ago
an acute angled triangle with $\angle{BAC}=60^0$
N.T.TUAN   22
N Mar 12, 2024 by AshAuktober
Source: APMO 2007
Let $ABC$ be an acute angled triangle with $\angle{BAC}=60^\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$ . Prove that $2\angle{AHI}= 3\angle{ABC}$.
22 replies
N.T.TUAN
Mar 31, 2007
AshAuktober
Mar 12, 2024
an acute angled triangle with $\angle{BAC}=60^0$
G H J
G H BBookmark kLocked kLocked NReply
Source: APMO 2007
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
N.T.TUAN
3595 posts
#1 • 5 Y
Y by Adventure10, Mango247, Rounak_iitr, DEKT, and 1 other user
Let $ABC$ be an acute angled triangle with $\angle{BAC}=60^\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$ . Prove that $2\angle{AHI}= 3\angle{ABC}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
e.lopes
349 posts
#2 • 3 Y
Y by vsathiam, Adventure10, Mango247
Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anuj kumar
61 posts
#3 • 1 Y
Y by Adventure10
in fact $B,C,H,O,I$ are concyclic and that leads to an easy solution.
by the way i was not a contestent at apmo .
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
saywhattt
14 posts
#4 • 3 Y
Y by aats411, Adventure10, Mango247
Here is my solution:

Easy angle chasing yields $\angle BIC=120^\circ$, while more angle chasing yields $\angle BHC=120^\circ$. Hence $BIHC$ is cyclic, and if we denote the foot of the perpendicular from $C$ to $AB$ as $D$, then $\angle IHD=\angle IBC$. Now, let $\angle ABC=2\beta$. It is easy to find $\angle DCB=90-2\beta$, hence $\angle AHD=2\beta$. Therefore, $3\angle ABC=6\beta=2\angle AHI$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aditya21
717 posts
#5 • 1 Y
Y by Adventure10
we first prove a lemma
lemma = let $ABC$ be a triangle with $D$ as midpoint of minor arc $BC$ of circumcircle of $ABC$
also let $I$ be incentre of triangle $ABC$ than we prove that $D$ is centre of circumcircle of $BIC$.

proof = it is evident that $A$ excentre of triangle $ABC$ lie on circumcircle of $BIC$
also $BD=DC$ and thus $\angle CBD=A/2 = \angle CAD=\angle BAI$
thus $A,I,D$ are collinear.
now $\angle IBD = 90-C/2=\angle JCB=\angle BID$ where $J$ is $A$ excentre.
and thus $BD=DI$
thus $D$ is circumcentre of $BIC$.

main proof = by this lemma and $\angle A = 60$ we get that $\angle BOC=\angle BIC=\angle BHC = 120$
and hence $B,H,O,I,C$ are concyclic.
by concyclicity of $HIJB$ we have $\angle HBJ = \angle AIH = 120-C-B/2=60+B/2$
in triangle $AHI$ we thus get $\angle AHI=3B/2$ since $\angle HAI=60-B$
and thus $2\angle AHI = 3\angle B$.

we are done :D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HMed
2 posts
#6 • 3 Y
Y by igli.2001, Adventure10, Mango247
A nice problem .
We will show that points B,I, H and C are concyclic wich is too easy since angle<BIC = 90°+A/2 = 120°=180°-A=<BHC.
Let (w) the cercle ( BIHC). Let Y be the intersection point of line AH with (w);then angle <YBC = <ABC .
Prof
By simple angle chasing <BAY = <HCB =<HYB = <AYB and (AY) perpendicular to (BC) so triangle BAY is isoceles then the equality follows.
Now denote by X the center of (w) ,then :
2<AHI=360°-2<IHY = <IXY = 2<IBY =<ABC+ 2<CBY = <ABC + 2<ABC = 3<ABC , QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AMN300
563 posts
#7 • 2 Y
Y by Mai-san, Adventure10
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Abdollahpour
63 posts
#9 • 2 Y
Y by Adventure10, Mango247
This problem is easy.
First we should know $B,I,H,C$ are concyclic.
$<BHC=<CIB=120$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dikhendzab
108 posts
#11 • 3 Y
Y by Mango247, Mango247, Mango247
Let altitudes intersect sides $a,b,c$ in points $F,D,E$. Now, $\angle IBH=\angle ABD-\angle ABI=30^\circ-\frac{\beta}{2}$.
$\angle ICH=\angle ACI-\angle ACH=30^\circ-\frac{\beta}{2} \implies \angle IBH=\angle ICH$. Using that quadrilateral $BCHI$ is cyclic we obtain:
$\angle BCI=\angle BHI=\frac{\gamma}{2}$ and $\angle FHB=90^\circ-\angle FBH=90^\circ-\beta+\angle ABH=120^\circ-\beta$
Finally, $\angle AHI=180^\circ-\angle BHI-\angle FHB=60^\circ-\frac{\gamma}{2}+\beta=60^\circ-\frac{120^\circ-\beta}{2}+\beta=\frac{3\beta}{2} \implies 2\angle AHI=3\beta$ :)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dchenmathcounts
2443 posts
#13
Y by
Note $\angle BHC=\angle BIC=\angle BOC=120^{\circ},$ so $H,O$ lie on $(BIC).$

We claim that $AHO$ is isosceles with $AH=AO.$ We use complex numbers. Put $ABC$ on the unit circle and note that $H=A+B+C=A+D,$ so $AH=OD=AO.$ Now note that $\angle CAH=\angle BAO=90^{\circ}-\angle C,$ so $\angle HAO=\angle A-(180^{\circ}-\angle 2C)=\angle C-\angle B=2\angle C-120^{\circ}.$ Thus $\angle AHO=150^{\circ}-\angle C=\angle B+30^{\circ}.$

Also note $\angle HIO=180^{\circ}-\angle CHO=180^{\circ}-\angle A+\angle B=120^{\circ}+\angle B,$ so $\angle IHO = 30^{\circ}-\frac{1}{2}\angle B.$ Thus $\angle AHI=\frac{3}{2}\angle B,$ as desired.
This post has been edited 1 time. Last edited by dchenmathcounts, Jun 15, 2020, 5:43 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
star32
165 posts
#14 • 1 Y
Y by TwilightZone
solution sketch
This post has been edited 1 time. Last edited by star32, Jul 17, 2020, 5:29 PM
Reason: typo:(
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
snakeaid
125 posts
#15 • 1 Y
Y by TwilightZone
Let $M$ be the midpoint of minor arc $BC$ of $(ABC)$ and $O$ be its center. We have that $\triangle COM$ is equilateral and hence $OM=AH$. Also evidently $AH || OM$, hence $AOMH$ is a parallelogram. But since $OM=OA$ it's a rhombus, whence $OM=MH$. Now from the excenter/incenter lemma we have that $CHIOB$ is cyclic. We have $\angle AHI=\angle AOI=\angle AOC-\angle COH=\angle AOC-\angle IBC=\angle AOC-\frac{1}{2}\angle ABC=\frac{3}{2}\angle ABC$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kevinmathz
4680 posts
#16
Y by
Note that we have $$\angle BIC = 180^{\circ}-\angle IBC - \angle ICB = 180^{\circ} - \frac{\angle ABC + \angle ACB}{2} = 180^{\circ} - \frac{180^{\circ}-\angle A}{2}$$and since $\angle A = 60^{\circ}$, then $\angle BIC = 120^{\circ}$. Now, it is well-known that $$\angle BHC = 180^{\circ}-\angle A = 180^{\circ}-60^{\circ}=120^{\circ}$$so as a result we have $\angle BIC = \angle BHC$ so $BIHC$ is cyclic. Now, reflecting $A$ over $BC$ and letting that point be $A'$ we know $A'$ is on the circumcircle of $BIHC$, so $$\angle AHI = 180^{\circ}- \angle A'HI = \angle A'BI = \angle B + \angle CBI = \frac32\angle B$$so $2\angle AHI = 3 \angle B$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
PCChess
548 posts
#17
Y by
First note that $BIC$ are concyclic. I claim that $H$ lies on this circle. Let $\Gamma_1$ denote the circumcircle of $\triangle ABC$ and $\Gamma_2$ denote $(BIC)$. Then, arc $\overarc {BAC}$ has measure $240^{\circ}$. Then, since $\angle BIC=90^{\circ}+\frac{\angle A}{2}=120^{\circ}$, looking at $\Gamma_2$, $\overarc{BC}$ not containing $I$ has measure $240^{\circ}$ as well. Hence, $\Gamma_1$ and $\Gamma_2$ are reflections of each other across $BC$. Since the reflection of $H$ across $BC$ lies on $\Gamma_1$, $H$ must lie on $\Gamma_2$.

Now,
\[3\angle ABC=6\angle IBC=6(180-\angle IHC)=1080-6(180-\angle AHI+\angle ABC)=6\angle AHI-6\angle ABC,\]which implies the desired result
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maths_1729
390 posts
#18 • 2 Y
Y by shalomrav, Mango247
Points $I, H, O, B, C$ Will lie on same circle..
And then by some angle chase we will get the desired results
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
franzliszt
23531 posts
#19 • 1 Y
Y by JustinLee2017
We start with a lemma.

Lemma: In a triangle $ABC$ with orthocenter $H$, $AH=2R\cos A$.

Proof. Consult problem Exercise 4.4.7 in the AoPS Precalculus Book. $\square$

Now note that $AH=2R\cos A=2R\cdot \frac12=R=AO$. Also, $AI=AI$ and $\measuredangle HAI=\measuredangle IAO$ since $H$ and $O$ are isogonal conjugates. Thus, $\triangle AHI\cong \triangle AOI$.

Now observe the following angle equalities: \begin{align*} \angle BHC&=180^\circ-\angle A=120^\circ\\ \angle BIC&=90^\circ+\frac{\angle A}2=120^\circ\\ \angle BOC&=2\angle A=120^\circ \end{align*}
so hexagon $CHIOBI_A$ is cyclic.

The rest is just angle chasing: $$\measuredangle AHI=\measuredangle IOA=\measuredangle IOC+\measuredangle COA=\measuredangle IBC+2\measuredangle CBA=4\measuredangle CBI- \measuredangle CBI=3\measuredangle CBI$$and multiplying both sides by $2$ gives the desired $2\measuredangle AHI=6\measuredangle CBI=3\measuredangle CBA$. $\blacksquare$


Remarks
This post has been edited 4 times. Last edited by franzliszt, Mar 8, 2021, 2:06 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
geometry6
304 posts
#20
Y by
APMO 2007 P2 wrote:
Let $ABC$ be an acute angled triangle with $\angle{BAC}=60^\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$ . Prove that $2\angle{AHI}= 3\angle{ABC}$.
Solution. Since $\angle A=60^{\circ}$, we have that $BIHC$ is cyclic. Now just by angle chasing, we get that:
$$\angle AHI=180^{\circ}-\angle IHD=180^{\circ}-(\angle IHB+\angle BHD)=180^{\circ}-(\frac{\angle C}{2}+\angle C)=180^{\circ}-\frac{3\angle C}{2}$$$$2\angle AHI=360^{\circ}-3\angle C=3(120^\circ-\angle C)=3\angle B.\blacksquare$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mogmog8
1080 posts
#21 • 1 Y
Y by centslordm
Claim: $BIHC$ is cyclic.

Proof. Notice that \begin{align*}\angle HCI&=\tfrac{1}{2}\angle C-(90-60)\\&=\tfrac{1}{2}\angle C-30+\tfrac{1}{2}(\angle B+\angle C-120)\\&=\tfrac{1}{2}\angle B-(90-\angle C)\\&=\angle HBI.\end{align*}$\blacksquare$

Hence, $$\angle AHI=(180-\angle IHC)+(180-AHC)=\tfrac{1}{2}\angle B+180-\angle A-\angle C=\tfrac{3}{2}\angle B.$$$\square$
This post has been edited 3 times. Last edited by Mogmog8, Dec 17, 2021, 3:51 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mahdi_Mashayekhi
696 posts
#22
Y by
Let AD,BE and CF be altitudes of triangle. ∠EHF = 180 - 60 = 120 = ∠BHC and ∠BIC = ∠A + ∠B/2 + ∠C/2 = 120 so BIHC is cyclic.
∠AHI = ∠IHF + ∠FHA = ∠IBC + ∠DHC = ∠IBC + ∠ABC ---> ∠AHI = 3/2 ∠ABC.
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.
dwip_neel
40 posts
#23 • 1 Y
Y by Mango247
$BHIC$ is cyclic since $\angle BHC = 180^{\circ} - \angle A = 120^{\circ} = 90^{\circ} + \frac{\angle A}{2}$. $D$ be the foot from $A$ to $BC$. Consider $DHIC$ where $\angle HDC = 90^{\circ}$, $\angle IBC = \frac{\angle B}{2}$, $\angle HIC = 180^{\circ} - 90^{\circ} + \angle B = 90^{\circ} + \angle B$. Now, trivial calculation gives, $\angle AHI = \frac{3}{2}\angle B$ and check that, this is the result we wanted.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lifeismathematics
1188 posts
#24
Y by
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(42cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -10.982013289036544, xmax = 16.091971207087482, ymin = -7.555732004429678, ymax = 8.00945071982281;  /* image dimensions */

 /* draw figures */
draw((-0.2750542635658916,5.839758582502767)--(-9.51982945736434,-4.371944629014396), linewidth(2)); 
draw((-0.2750542635658916,5.839758582502767)--(2.932316722037652,-0.5513997785160583), linewidth(2)); 
draw((-9.51982945736434,-4.371944629014396)--(2.932316722037652,-0.5513997785160583), linewidth(2)); 
draw(circle((-0.9208974895600552,1.0958517060424493), 2.7050286774217995), linewidth(2)); 
draw((xmin, -3.25925925925926*xmin + 4.943285427176898)--(xmax, -3.25925925925926*xmax + 4.943285427176898), linewidth(2)); /* line */
draw((xmin, 0.5018450184501846*xmin + 0.405534360659222)--(xmax, 0.5018450184501846*xmax + 0.405534360659222), linewidth(2)); /* line */
draw((xmin, -0.9053117782909931*xmin + 2.103261087624264)--(xmax, -0.9053117782909931*xmax + 2.103261087624264), linewidth(2)); /* line */
draw((-0.2750542635658916,5.839758582502767)--(-0.9208974895600541,1.0958517060424489), linewidth(2)); 
draw((-0.9208974895600541,1.0958517060424489)--(1.2064943515156186,1.0110075407556212), linewidth(2)); 
draw((-0.9208974895600541,1.0958517060424489)--(-9.51982945736434,-4.371944629014396), linewidth(2)); 
draw((-0.9208974895600541,1.0958517060424489)--(2.932316722037652,-0.5513997785160583), linewidth(2)); 
 /* dots and labels */
dot((-0.2750542635658916,5.839758582502767),dotstyle); 
label("$A$", (-0.1807198228128475,6.07559468438538), NE * labelscalefactor); 
dot((-9.51982945736434,-4.371944629014396),dotstyle); 
label("$B$", (-9.425495016611295,-4.136108527131783), NE * labelscalefactor); 
dot((2.932316722037652,-0.5513997785160583),dotstyle); 
label("$C$", (3.0266511627906953,-0.3155636766334448), NE * labelscalefactor); 
dot((-0.9208974895600541,1.0958517060424489),dotstyle); 
label("$I$", (-0.8174772978959038,1.3352890365448493), NE * labelscalefactor); 
dot((-0.12745211158699393,-1.490192488832711),dotstyle); 
label("$D$", (-0.03921816168327944,-1.2589080841638987), NE * labelscalefactor); 
dot((-2.9262243575179765,2.9112977389281434),dotstyle); 
label("$E$", (-2.822084163898118,3.151227021040973), NE * labelscalefactor); 
dot((1.4967668262111673,2.3091444991970103),dotstyle); 
label("$F$", (1.5880509413067534,2.538053156146178), NE * labelscalefactor); 
dot((1.2064943515156186,1.0110075407556212),dotstyle); 
label("$H$", (1.3050476190476172,1.240954595791804), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]


Claim:-points $B,I,H,C$ are concylic

Proof:- we notice that $\angle{BHC}=180^{\circ}-\angle{A}=120^{\circ}$ and $\angle{BIC}=90^{\circ}+\frac{A}{2}=120^{\circ}$ so we get $\angle{BHC}=\angle{BIC}$ hence points $B,I,H,C$ are concylic $\blacksquare$

now we have $\angle{AHI}+\angle{AHC}+\angle{CHI}=360^{\circ} \implies \angle{AHI}=360^{\circ}-\left(180^{\circ}-\frac{\angle{B}}{2}\right)-(180^{\circ}-\angle{B}) \implies \angle{AHI}=\frac{3\angle{B}}{2} \implies 2\angle{AHI}=3\angle{ABC}$ $\square$
This post has been edited 1 time. Last edited by lifeismathematics, Apr 6, 2023, 9:42 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
john0512
4190 posts
#25
Y by
Let $\angle ABC=2\theta$ and $\angle ACB=120-2\theta$. We wish to show that $\angle AHI=3\theta$.

Claim: $HIBC$ is cyclic. This is because $$\angle BHC=180-\angle A=120$$and $$\angle BIC=90+(\angle A/2)=120.$$
Then, $$\angle IHB=\angle ICB=60-\theta.$$Let $E$ be the foot from $B$ to $AC$. Then, $$\angle AHE=\angle C=120-2\theta,$$so $$\angle AHI=180-\angle AHE-\angle IHB=180-(120-2\theta)-(60-\theta)=3\theta,$$hence done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1009 posts
#26
Y by
Observe that $B, C, H, I$ are concyclic, as $\angle BHC = 120^\circ = \angle BIC$ from some simple angle chasing.
Now we have $\angle AHC = 180^\circ - B, \angle CHI = 180^\circ - \frac{B}{2} \implies \angle AHI = \frac{3B}{2}$, and we are done. $\square$
Z K Y
N Quick Reply
G
H
=
a