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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)!

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[*]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
Inequalities
sqing   1
N 13 minutes ago by Royal_mhyasd
Let $ a,b>0 $ and $\frac{a}{a^2+3}+ \frac{b}{b^2+ 3} \geq \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{a}{a^3+3}+ \frac{b}{b^3+ 3}\geq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
1 reply
sqing
3 hours ago
Royal_mhyasd
13 minutes ago
Hard Inequality
William_Mai   10
N 29 minutes ago by sqing
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
10 replies
William_Mai
May 3, 2025
sqing
29 minutes ago
GCD of consecutive terms
nsato   38
N an hour ago by heocoi
The numbers in the sequence 101, 104, 109, 116, $\dots$ are of the form $a_n = 100 + n^2$, where $n = 1$, 2, 3, $\dots$. For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n + 1}$. Find the maximum value of $d_n$ as $n$ ranges through the positive integers.
38 replies
nsato
Mar 14, 2006
heocoi
an hour ago
Geometry Proof
strongstephen   11
N an hour ago by jasperE3
Proof that choosing four distinct points at random has an equal probability of getting a convex quadrilateral vs a concave one.
not cohesive proof alert!

NOTE: By choosing four distinct points, that means no three points lie on the same line on the Gaussian Plane.
NOTE: The probability of each point getting chosen don’t need to be uniform (as long as it is symmetric about the origin), you just need a way to choose points in the infinite plane (such as a normal distribution)

Start by picking three of the four points. Next, graph the regions where the fourth point would make the quadrilateral convex or concave. In diagram 1 below, you can see the regions where the fourth point would be convex or concave. Of course, there is the centre region (the shaded triangle), but in an infinite plane, the probability the fourth point ends up in the finite region approaches 0.

Next, I want to prove to you the area of convex/concave, or rather, the probability a point ends up in each area, is the same. Referring to the second diagram, you can flip each concave region over the line perpendicular to the angle bisector of which the region is defined. (Just look at it and you'll get what it means.) Now, each concave region has an almost perfect 1:1 probability correspondence to another convex region. The only difference is the finite region (the triangle, shaded). Again, however, the actual significance (probability) of this approaches 0.

If I call each of the convex region's probability P(a), P(c), and P(e) and the concave ones P(b), P(d), P(f), assuming areas a and b are on opposite sides (same with c and d, e and f) you can get:
P(a) = P(b)
P(c) = P(d)
P(e) = P(f)

and P(a) + P(c) + P(e) = P(convex)
and P(b) + P(d) + P(f) = P(concave)

therefore:
P(convex) = P(concave)
11 replies
strongstephen
Yesterday at 4:54 AM
jasperE3
an hour ago
Number Theory
Foxellar   4
N 2 hours ago by Shan3t
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.
4 replies
Foxellar
Today at 9:00 AM
Shan3t
2 hours ago
inequality
danilorj   1
N 4 hours ago by sqing
A triangle with perimeter 1 has side lengths \( a \), \( b \), and \( c \). Show that
\[
a^2 + b^2 + c^2 + 4abc < \frac{1}{2}.
\]
1 reply
danilorj
5 hours ago
sqing
4 hours ago
Inequality
nhhlqd   27
N 4 hours ago by sqing
Given that $a,b$ are positive real number such that $a\geq 1$. Prove that
$$\dfrac{b^2}{a+b}+\dfrac{a}{b^2+b}+\dfrac{1}{a+1}\geq \dfrac{3}{2}$$
27 replies
nhhlqd
Feb 20, 2020
sqing
4 hours ago
A rather difficult question
BeautifulMath0926   4
N 4 hours ago by BeautifulMath0926
I got a difficult equation for users to solve:
Find all functions f: R to R, so that to all real numbers x and y,
1+f(x)f(y)=f(x+y)+f(xy)+xy(x+y-2) holds.
4 replies
BeautifulMath0926
Apr 13, 2025
BeautifulMath0926
4 hours ago
Polynomial Minimization
ReticulatedPython   2
N Today at 7:46 AM by lgx57
Consider the polynomial $$p(x)=x^{n+1}-x^{n}$$, where $x, n \in \mathbb{R+}.$

(a) Prove that the minimum value of $p(x)$ always occurs at $x=\frac{n}{n+1}.$
2 replies
ReticulatedPython
Yesterday at 5:07 PM
lgx57
Today at 7:46 AM
Compilation of functions problems
Saucepan_man02   0
Today at 6:15 AM
Could anyone post some handout/compilation of problems related to functions (difficulty similar to AIME/ARML/HMMT etc)?

Thanks..
0 replies
Saucepan_man02
Today at 6:15 AM
0 replies
Nice Number Theory Question (Inspired by AIME)
MathRook7817   5
N Today at 4:13 AM by jasperE3
Let $n$ be a positive integer. Find the smallest value of $n$ such that:

$2^n + 3^n - n$ is divisible by $216$.

5 replies
MathRook7817
Today at 1:57 AM
jasperE3
Today at 4:13 AM
Inequalities
sqing   12
N Today at 4:08 AM by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
12 replies
sqing
Jul 12, 2024
sqing
Today at 4:08 AM
A pentagon inscribed in a circle of radius √2
tom-nowy   4
N Today at 4:03 AM by jasperE3
Can a pentagon with all rational side lengths be inscribed in a circle of radius $\sqrt{2}$ ?
4 replies
tom-nowy
Yesterday at 2:37 AM
jasperE3
Today at 4:03 AM
A Collection of Good Problems from my end
SomeonecoolLovesMaths   24
N Today at 12:46 AM by ReticulatedPython
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

Problem 4

Problem 5
24 replies
SomeonecoolLovesMaths
May 4, 2025
ReticulatedPython
Today at 12:46 AM
Basic geometry
AlexCenteno2007   7
N Apr 30, 2025 by KAME06
Given an isosceles triangle ABC with AB=BC, the inner bisector of Angle BAC And cut next to it BC in D. A point E is such that AE=DC. The inner bisector of the AED angle cuts to the AB side at the point F. Prove that the angle AFE= angle DFE
7 replies
AlexCenteno2007
Feb 9, 2025
KAME06
Apr 30, 2025
Basic geometry
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AlexCenteno2007
150 posts
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Given an isosceles triangle ABC with AB=BC, the inner bisector of Angle BAC And cut next to it BC in D. A point E is such that AE=DC. The inner bisector of the AED angle cuts to the AB side at the point F. Prove that the angle AFE= angle DFE
This post has been edited 1 time. Last edited by AlexCenteno2007, Feb 9, 2025, 4:16 AM
Reason: Error
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AlexCenteno2007
150 posts
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A pretty nice problem
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sunken rock
4392 posts
#4
Y by
AlexCenteno2007 wrote:
A pretty nice problem

Where is $E$?
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AlexCenteno2007
150 posts
#5
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sunken rock wrote:
AlexCenteno2007 wrote:
A pretty nice problem

Where is $E$?

$E$ is about $AC$
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Jackson0423
81 posts
#6
Y by
someone draw pls
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mathafou
417 posts
#7 • 1 Y
Y by AlexCenteno2007
it is equivalent to prove that AEDF is a rhombus
that is without the F stuff, with just the first bissector, that CDE is isosceles
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vanstraelen
9011 posts
#8 • 1 Y
Y by AlexCenteno2007
$\triangle AEF$ is isosceles, so $ACDF$ is a trapezoid.
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KAME06
158 posts
#9 • 1 Y
Y by AlexCenteno2007
Let $F'$ the intersection of $\angle ACB$'s angle bisector and $AB$. We'll prove $F'=F$.
By the symmetry of an isosceles triangle $ABC$, we know that $ACDF'$ must be an isosceles trapezoid (the two opposite angle bisectors are symmetric) where $DC=F'A$ and $AC \parallel F'D$
An isosceles trapezoid is cyclic, so $ACDF'$ is cyclic, then $\angle ACF'=\angle ADF'=\angle F'AD$ , so $F'D=F'A$.
$AD$ is the perpendicular bisector of $F'E$, so $\triangle DF'E$ must be isosceles. By $LLL$, $\triangle DF'E \cong \triangle AF'E$.
They are congruent, then $\angle DEF' = \angle AEF'$, so $EF'$ is the inner angle bisector of $\angle AED$. We conclude that $F=F'$.
Finally, $\triangle DFE \cong \triangle AFE \Rightarrow \angle DFE = \angle AFE$.


https://imgur.com/4I5wPKd.jpeg
This post has been edited 1 time. Last edited by KAME06, Apr 30, 2025, 4:40 PM
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