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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:
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[*]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]
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0 replies
jlacosta
May 1, 2025
0 replies
UC Berkeley Integration Bee 2025 Bracket Rounds
Silver08   56
N 2 hours ago by Aiden-1089
Regular Round

Quarterfinals

Semifinals

3rd Place Match

Finals
56 replies
Silver08
May 9, 2025
Aiden-1089
2 hours ago
Polynomial with integer coefficients
smartvong   1
N 3 hours ago by alexheinis
Source: UM Mathematical Olympiad 2024
Prove that there is no polynomial $f(x)$ with integer coefficients, such that $f(p) = \dfrac{p + q}{2}$ and $f(q) = \dfrac{p - q}{2}$ for some distinct primes $p$ and $q$.
1 reply
smartvong
3 hours ago
alexheinis
3 hours ago
Existence of scalars
smartvong   0
3 hours ago
Source: UM Mathematical Olympiad 2024
Let $U$ be a finite subset of $\mathbb{R}$ such that $U = -U$. Let $f,g : \mathbb{R} \to \mathbb{R}$ be functions satisfying
$$g(x) - g(y ) = (x - y)f(x + y)$$for all $x,y \in \mathbb{R} \backslash U$.
Show that there exist scalars $\alpha, \beta, \gamma \in \mathbb{R}$ such that
$$f(x) = \alpha x + \beta$$for all $x \in \mathbb{R}$,
$$g(x) = \alpha x^2 + \beta x + \gamma$$for all $x \in \mathbb{R} \backslash U$.
0 replies
smartvong
3 hours ago
0 replies
Invertible matrices in F_2
smartvong   1
N 4 hours ago by alexheinis
Source: UM Mathematical Olympiad 2024
Let $n \ge 2$ be an integer and let $\mathcal{S}_n$ be the set of all $n \times n$ invertible matrices in which their entries are $0$ or $1$. Let $m_A$ be the number of $1$'s in the matrix $A$. Determine the minimum and maximum values of $m_A$ in terms of $n$, as $A$ varies over $S_n$.
1 reply
smartvong
Today at 12:41 AM
alexheinis
4 hours ago
ISI UGB 2025 P3
SomeonecoolLovesMaths   13
N 4 hours ago by iced_tea
Source: ISI UGB 2025 P3
Suppose $f : [0,1] \longrightarrow \mathbb{R}$ is differentiable with $f(0) = 0$. If $|f'(x) | \leq f(x)$ for all $x \in [0,1]$, then show that $f(x) = 0$ for all $x$.
13 replies
SomeonecoolLovesMaths
May 11, 2025
iced_tea
4 hours ago
Lots of Cyclic Quads
Vfire   104
N Today at 5:53 AM by Ilikeminecraft
Source: 2018 USAMO #5
In convex cyclic quadrilateral $ABCD$, we know that lines $AC$ and $BD$ intersect at $E$, lines $AB$ and $CD$ intersect at $F$, and lines $BC$ and $DA$ intersect at $G$. Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$.

Proposed by Kada Williams
104 replies
Vfire
Apr 19, 2018
Ilikeminecraft
Today at 5:53 AM
Goals for 2025-2026
Airbus320-214   107
N Today at 5:02 AM by Jaxman8
Please write down your goal/goals for competitions here for 2025-2026.
107 replies
+1 w
Airbus320-214
May 11, 2025
Jaxman8
Today at 5:02 AM
Evan's mean blackboard game
hwl0304   72
N Today at 3:26 AM by HamstPan38825
Source: 2019 USAMO Problem 5, 2019 USAJMO Problem 6
Two rational numbers \(\tfrac{m}{n}\) and \(\tfrac{n}{m}\) are written on a blackboard, where \(m\) and \(n\) are relatively prime positive integers. At any point, Evan may pick two of the numbers \(x\) and \(y\) written on the board and write either their arithmetic mean \(\tfrac{x+y}{2}\) or their harmonic mean \(\tfrac{2xy}{x+y}\) on the board as well. Find all pairs \((m,n)\) such that Evan can write 1 on the board in finitely many steps.

Proposed by Yannick Yao
72 replies
hwl0304
Apr 18, 2019
HamstPan38825
Today at 3:26 AM
9 JMO<200?
DreamineYT   4
N Today at 3:16 AM by megarnie
Just wanted to ask
4 replies
DreamineYT
May 10, 2025
megarnie
Today at 3:16 AM
Group Theory
Stephen123980   3
N Yesterday at 9:01 PM by BadAtMath23
Let G be a group of order $45.$ If G has a normal subgroup of order $9,$ then prove that $G$ is abelian without using Sylow Theorems.
3 replies
Stephen123980
May 9, 2025
BadAtMath23
Yesterday at 9:01 PM
calculus
youochange   2
N Yesterday at 7:46 PM by tom-nowy
$\int_{\alpha}^{\theta} \frac{d\theta}{\sqrt{cos\theta-cos\alpha}}$
2 replies
youochange
Yesterday at 2:26 PM
tom-nowy
Yesterday at 7:46 PM
ISI UGB 2025 P1
SomeonecoolLovesMaths   6
N Yesterday at 5:10 PM by SomeonecoolLovesMaths
Source: ISI UGB 2025 P1
Suppose $f \colon \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable and $| f'(x)| < \frac{1}{2}$ for all $x \in \mathbb{R}$. Show that for some $x_0 \in \mathbb{R}$, $f \left( x_0 \right) = x_0$.
6 replies
SomeonecoolLovesMaths
May 11, 2025
SomeonecoolLovesMaths
Yesterday at 5:10 PM
Cute matrix equation
RobertRogo   3
N Yesterday at 2:23 PM by loup blanc
Source: "Traian Lalescu" student contest 2025, Section A, Problem 2
Find all matrices $A \in \mathcal{M}_n(\mathbb{Z})$ such that $$2025A^{2025}=A^{2024}+A^{2023}+\ldots+A$$Edit: Proposed by Marian Vasile (congrats!).
3 replies
RobertRogo
May 9, 2025
loup blanc
Yesterday at 2:23 PM
Integration Bee Kaizo
Calcul8er   63
N Yesterday at 1:50 PM by MS_asdfgzxcvb
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
63 replies
Calcul8er
Mar 2, 2025
MS_asdfgzxcvb
Yesterday at 1:50 PM
Problem 22
evt917   16
N Feb 20, 2025 by Radio2
Source: AMC 12B 2024 Problem 22
Let $\triangle{ABC}$ be a triangle with integer side lengths and the property that $\angle{B} = 2\angle{A}$. What is the least possible perimeter of such a triangle?

$
\textbf{(A) }13 \qquad
\textbf{(B) }14 \qquad
\textbf{(C) }15 \qquad
\textbf{(D) }16 \qquad
\textbf{(E) }17 \qquad
$
16 replies
evt917
Nov 13, 2024
Radio2
Feb 20, 2025
Problem 22
G H J
Source: AMC 12B 2024 Problem 22
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evt917
2418 posts
#1
Y by
Let $\triangle{ABC}$ be a triangle with integer side lengths and the property that $\angle{B} = 2\angle{A}$. What is the least possible perimeter of such a triangle?

$
\textbf{(A) }13 \qquad
\textbf{(B) }14 \qquad
\textbf{(C) }15 \qquad
\textbf{(D) }16 \qquad
\textbf{(E) }17 \qquad
$
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rama_bulusu
1 post
#2 • 2 Y
Y by OronSH, alexanderhamilton124
:)

Claim: $AC^2 = BC(AB + BC)$.

Proof: Let $D$ be the unique point such that $ABCD$ is an isosceles trapezoid with $AB \parallel CD$. Then, triangle $CDA$ is isosceles, so we must have $CD = AD = BC$, and the conclusion follows from Ptolemy's Theorem.

The minimum perimeter is therefore achieved with $AB = 5$, $BC = 4$, and $AC = 6$. The answer is $\boxed{15}$, or C.
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Bluesoul
898 posts
#3 • 1 Y
Y by OronSH
Draw the circumcircle of $\triangle{ABC}$ and let the angle bisector of $\angle{B}$ meet the circle at $D$

By fact 5 we have $CD=AD, \angle{CBD}=\angle{ABD}=\angle{CAB}=\angle{DAC}, \angle{CBA}=\angle{DAB}$, thus $AC=BD, CD=CB$

By Ptolemy, we have $c^2=(a+b)CD, CD=\frac{c^2}{a+b}=a, a^2+ab=c^2$. Try some numbers and the answer is $(4,5,6)\implies \boxed{15}$
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sixoneeight
1138 posts
#4 • 1 Y
Y by Airbus320-214
Memorize 4-5-6 triangle
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plang2008
337 posts
#5
Y by
1968 IMO/1 thonk
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vsamc
3789 posts
#6 • 1 Y
Y by centslordm
its like 2mn, n^2, 4m^2-n^2 by law of sines (let cos A = m/n, sin 2A/sin A = 2cosA = 2m/n, sin 3A/sin A = 4cos^2 A - 1 = 4m^2/n^2-1), then just try small n until u get n=4, m=3 works
This post has been edited 1 time. Last edited by vsamc, Nov 13, 2024, 5:42 PM
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pqr.
174 posts
#7
Y by
Law of sines give 15
This post has been edited 1 time. Last edited by pqr., Nov 13, 2024, 5:41 PM
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centslordm
4786 posts
#8 • 11 Y
Y by pqr., clarkculus, mygoodfriendusesaops, idksomething, Riemann123, geodash2, nikenikenike, pog, Yolandayu, abeot, bachkieu
By the sixteenth diagram, the answer is $\boxed{15}.$
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megarnie
5608 posts
#9 • 1 Y
Y by akliu
Let $a,b,c$ be the side lengths. Note that the cosines of all the angles are rational (LOC), so taking cosines of both sides, we have $\cos(\angle B) = 2 \cos^2( \angle A) - 1$, so \[ \cos(\angle B) + 1 =   \frac{a^2 + c^2 - b^2}{2ac} + 1= \frac{(a+c)^2 - b^2) }{2ac}= \frac{(a+b+c)(a + c - b)}{2ac} \]must be two times a square. Multiplying this by $2$ gives that \[ f(a,b,c) = \frac{(a+b+c)(a + c - b)}{ac} \]is the square of a rational number.


Claim: If $p$ is an odd prime dividing $a + b + c$ so that $\nu_p(a+b+c)$ is odd, then $p$ divides $abc$.
Proof: Assume not. Then $\nu_p(f(a,b,c)) = 1$, which is a contradiction since $f(a,b,c)$ is the square of a rational. $\square$

Now, looking at the answer choices, we get that $13$ doesn't work because it requires one of $a,b,c$ to be a multiple of $13$ and $14$ doesn't work because it requires one of $a,b,c$ to be a multiple of $7$ (thus triangle inequality won't hold), so $a + b + c \ge 15$. Now, we see that $3$ must divide one of $a,b,c$ and $5$ must divide one of $a,b,c$, so we could have $3, 5, 7$ or $4, 5, 6$.


Case 1: $\{a,b,c\} = \{3,5,7\}$
Taking the $\nu_7$ of $f(a,b,c)$ gives that $7$ can't divide $ac$, so $b = 7$ and $\cos(\angle B) + 1$ is $\frac 12$, meaning $\angle B = 120^{+}$. This is impossible as the sum of the angles would exceed $180^{\circ}$.

Case 2: $\{a,b,c\} = \{4,5,6\}$.
Then since $\angle B$ isn't the smallest in the triangle, we cannot have $b = 4$. If $b = 5$, then $f(a,b,c) = \frac{15 \cdot 5}{24} = \frac{15}{8}$, which is not a square. Therefore, $b = 6$ and $f(a,b,c) = \frac{15 \cdot 3}{20} = \frac{9}{4}$, which is a square. Then, $2 \cos(\angle A)^2=  \cos \angle B  + 1 = \frac{9}{8}$, so $\cos(\angle A) = \frac{3}{4} $ (must be positive as $\angle A < \angle B$), which can be verified to be true as $\cos\angle A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{45}{60} = \frac 34$.

Therefore, $(4,5,6)$ works, so it is possible for the perimeter to be $\boxed{\textbf{(C)}\ 15}$.
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brainfertilzer
1831 posts
#10
Y by
this problem sucks smth..

Let $\angle B$-bisector hit $AC$ at $P$. Then $AP = \tfrac{bc}{a+c}$ and $CP = \tfrac{ba}{a+c}$. From the angle condition, $\triangle BPA$ is isosceles so $BP = AP = \tfrac{bc}{a+c}$. Now use Stewart with $BP$ as the cevian:
\[ \frac{ab^3c}{(a+c)^2} + \frac{b^3c^2}{(a+c)^2} = \frac{a^2bc + c^2ba}{a+c}\]\[\iff ab^3c + b^3c^2 = (a+c)^2abc\]\[\iff (a+c)b^2 = (a+c)^2a\]\[\iff b^2 = a(a+c)\]now just check stuff to find that 4,5,6 works -> C. screwed this up by being messing up random division and stuff when simplifying the stewart equation and thus getting ridiculous diophantines. skull
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pi_is_3.14
1437 posts
#11 • 1 Y
Y by pog
Super clean no geometry solution :)
By the law of sines, the sides of the triangle are in the ratio \(\sin(\theta)\), \(\sin(2\theta)\), and \(\sin(3\theta)\), which when expanded give
\[
\sin(\theta) : 2\sin(\theta)\cos(\theta) : 3\sin(\theta) - 4\sin^3(\theta)
\]or
\[
1 : 2\cos(\theta) : 3 - 4\sin^2(\theta)
\]
Note that \(\cos(\theta)\) must be rational; otherwise, it's not possible to have integer side lengths for the triangle (and rational \(\cos(\theta)\) makes \(\sin^2(\theta)\) rational as well). Now, we just test some values of \(\cos(\theta)\).

\(\cos(\theta) = \frac{1}{2}\) fails because it makes the last term 0, \(\frac{1}{3}\) fails, \(\frac{2}{3}\) gives \(1 : \frac{4}{3} : \frac{7}{9}\), which gives \(9 : 12 : 7\), or \(28\), too large.

\(\frac{1}{4}\) fails because it makes the last term negative, but \(\frac{3}{4}\) gives \(1 : \frac{3}{2} : \frac{5}{4} \rightarrow 4 : 5 : 6\) triangle.

We see that for denominators of \(5\) and above, the fraction denominator becomes large, and it's impossible to have a smaller perimeter than \(15\).
This post has been edited 2 times. Last edited by pi_is_3.14, Nov 13, 2024, 6:55 PM
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LikMCAMC
318 posts
#12
Y by
holy 4,5,6 strikes again
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apotosaurus
79 posts
#13
Y by
pi_is_3.14 wrote:
Super clean no geometry solution :)
By the law of sines, the sides of the triangle are in the ratio \(\sin(\theta)\), \(\sin(2\theta)\), and \(\sin(3\theta)\), which when expanded give
\[
\sin(\theta) : 2\sin(\theta)\cos(\theta) : 3\sin(\theta) - 4\sin^3(\theta)
\]or
\[
1 : 2\cos(\theta) : 3 - 4\sin^2(\theta)
\]
Note that \(\cos(\theta)\) must be rational; otherwise, it's not possible to have integer side lengths for the triangle (and rational \(\cos(\theta)\) makes \(\sin^2(\theta)\) rational as well). Now, we just test some values of \(\cos(\theta)\).

\(\cos(\theta) = \frac{1}{2}\) fails because it makes the last term 0, \(\frac{1}{3}\) fails, \(\frac{2}{3}\) gives \(1 : \frac{4}{3} : \frac{7}{9}\), which gives \(9 : 12 : 7\), or \(28\), too large.

\(\frac{1}{4}\) fails because it makes the last term negative, but \(\frac{3}{4}\) gives \(1 : \frac{3}{2} : \frac{5}{4} \rightarrow 4 : 5 : 6\) triangle.

We see that for denominators of \(5\) and above, the fraction denominator becomes large, and it's impossible to have a smaller perimeter than \(15\).
To do the last part without manual computation:
First, rewrite $3-4\sin^2(\theta)=4\cos^2(\theta)-1$.

Now consider two cases. If $\cos(\theta) = \frac{p}{2q}$ for coprime $p$ and $2q$, the primitive triangle would be $q^2:qp:p^2-q^2$, for a perimeter of $p(p+q)$. We must have $q<p<2q$, so the minimal solution is given by $p=2$, $q=3$ for 15. This is the 4-5-6 triangle.

If $\cos(\theta) = \frac{p}{q}$ for coprime $2p$ and $q$, the primitive triangle would be $q^2 : 2pq : 4p^2-q^2$, for a perimeter of $2p(2p+q)$. Now $q<2p<2q$ So the minimal solution is given by $q=3$ and $p=2$, for a perimeter of 28.

Thus 15 is the smallest.
This post has been edited 1 time. Last edited by apotosaurus, Nov 13, 2024, 8:08 PM
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YaoAOPS
1541 posts
#14 • 8 Y
Y by Sedro, megarnie, OronSH, RedFireTruck, akliu, Jack_w, peace09, pog
The night before the AMCs, I was having trouble sleeping. That is, more trouble sleeping than usual on the nights before AMCs, because there was rapping on the door to my room. I told myself "tis some visitor, nothing more." When I opened the door, my visitor was a $4-5-6$ triangle. "I'm here to tell you all about 4-5-6 triangle facts. Did you know I'm the minimal scalene triangle which can be used to form a disphenoid? Did you know my area is $\frac{15\sqrt{7}}{4}$? Did you know that I'm the minimal triangle with one angle twice the other angle? Did you know my perimeter is $15$?" It told me this and two more questions, but I was barely conscious. I asked the queer floating triangle, "What do you want from me? Why are you telling me this?" The triangle told me "Thrice on the MAA tests will you see a $4-5-6$ triangle. If you mess up all three times, I will smite you and rend you to dust."

Then I woke up and forgot everything. Alas, I failed to solve AMC 12A 2024 Problem 24 and this problem. I then remembered this encounter. I'm scared that the AIME will contain a problem about a $4-5-6$ triangle that I won't be able to solve. Help! What do I do?
This post has been edited 2 times. Last edited by YaoAOPS, Nov 13, 2024, 10:29 PM
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djmathman
7938 posts
#15
Y by
LikMCAMC wrote:
holy 4,5,6 strikes again
What if I told you the two 4-5-6 triangle problems were originally on the same exam :|
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nsato
15654 posts
#16
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This is very similar to USAMO 1991, #1. (The main difference is that on the USAMO, the triangle is obtuse.)
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Radio2
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#17
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Simply similar triangles:

Draw the angle bisector of $\angle B$ to meet $\overline{AC}$ at $D$. Let $BC = a$, $AC = b$, $AB = c$ as usual, and let $AD = y$, so $CD = b-y$. Now $BD = AD = y$ as $\angle A = \angle ABD = \dfrac12 \angle B$, and $\angle BDC = 2\angle A$, so $\triangle ABC \sim \triangle BDC$ by AA.

Thus $\dfrac{CD}{BC} = \dfrac{BD}{AB} = \dfrac{BC}{AC}$ gives us $\dfrac{b-y}{a} = \dfrac{y}{c} = \dfrac{a}{b}$.

The right equation gives $y = \dfrac{ac}{b}$, so $\dfrac{b - ac/b}{a} = \dfrac{a}{b}$. Rearranging gives $b^2 - ac = a^2$ so $a(a+c) =b^2$.

We now do casework on $b$: clearly $b$ needs to be composite to avoid $c = 0$ or $a + c = b$. Trying $b = 4$ gives us $a(a+c) = 16$, which gives us a degenerate solution of $a = 2, c= 6$. Now when $b = 6$, we get $a = 4, c = 5$, which works and gives us $\boxed{\textbf{(C) }15}$.

The next composite number is $b = 8$, which would require $a + c \ge 9$, so we have achieved our minimum.
This post has been edited 1 time. Last edited by Radio2, Feb 20, 2025, 7:52 AM
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