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k a August Highlights and 2025 AoPS Online Class Information
jwelsh   0
Yesterday at 2:14 PM
CONGRATULATIONS to all the competitors at this year’s International Mathematical Olympiad (IMO)! The US Team took second place with 5 gold medals and 1 silver - we are proud to say that each member of the 2025 IMO team has participated in an AoPS WOOT (Worldwide Online Olympiad Training) class!

"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all."
— Doreen Dai, parent of IMO US Team Member Tiger Zhang

Interested to learn more about our WOOT programs? Check out the course page here or join a Free Scheduled Info Session. Early bird pricing ends August 19th!:
CodeWOOT Code Jam - Monday, August 11th
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There is still time to enroll in our last wave of summer camps that start in August at the Virtual Campus, our video-based platform, for math and language arts! From Math Beasts Camp 6 (Prealgebra Prep) to AMC 10/12 Prep, you can find an informative 2-week camp before school starts. Plus, our math camps don’t have homework and cover cool enrichment topics like graph theory. Our language arts courses will build the foundation for next year’s challenges, such as Language Arts Triathlon for levels 5-6 and Academic Essay Writing for high school students.

Lastly, Fall is right around the corner! You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US. We’ve opened new Academy locations in San Mateo, CA, Pasadena, CA, Saratoga, CA, Johns Creek, GA, Northbrook, IL, and Upper West Side (NYC), New York.

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0 replies
jwelsh
Yesterday at 2:14 PM
0 replies
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9 AMC 10A vs. AMC 10B
a.zvezda   6
N 2 minutes ago by iwastedmyusername
Usually, I think AMC 10B is harder but, it depends. Also, what are your 2024 AMC 10 scores?
6 replies
a.zvezda
an hour ago
iwastedmyusername
2 minutes ago
J
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one problem from Mathworks 2023 contest
Carmen8102   4
N 28 minutes ago by SpeedCuber7
The letters A and B represent different digits. The 5-digit integer AB5AB = X^3-3X^2 for some two-digit integer X. Find X.

Is there a faster way of solving this problem, not trying so many numbers?
4 replies
Carmen8102
Jul 30, 2025
SpeedCuber7
28 minutes ago
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Circle problem
littleduckysteve   2
N an hour ago by vanstraelen
3 circles are drawn such that they are externally tangent to each other. The circles have radius, 1,2, and 3 respectively. A bigger circle is drawn such that all three circles are internally tangent to it. If we call the center of the bigger circle, $A$, and the centroid of the triangle formed by the centers of the three smaller circles, B. What is the length $AB^2$.
2 replies
littleduckysteve
Yesterday at 2:33 PM
vanstraelen
an hour ago
J
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Find the largest value of p
Darealzolt   8
N an hour ago by P0tat0b0y
It is known that
\[ \sqrt{x-3}+\sqrt{6-x} \leq p \]In which \(x \in \mathbb{R}\), hence find the largest value of \(p\).
8 replies
Darealzolt
Jun 6, 2025
P0tat0b0y
an hour ago
J
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Circle geometry proof
littleduckysteve   1
N 2 hours ago by littleduckysteve
Suppose 3 circles are drawn in the 2-dimensional grid such that no two circles are of the same radius. Now we draw the 2 lines which are both tangent to the smallest circle and the median circle, and call their intersection, A. Now we do the same thing for the biggest circle and the smallest, and finally the biggest and the median circles. Now assume that we call these two points, B and C. Prove that A, B, and C are all colinear regardless of where the circles are.
1 reply
littleduckysteve
Today at 7:29 AM
littleduckysteve
2 hours ago
J
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15 dropped AIME problems from 1983-88 #1 37 | 37abc, 37bca, 37cab
parmenides51   6
N 2 hours ago by littleduckysteve
Determine the number of five digit integers $37abc$ (in base $10$), such that $37abc, 37bca, 37cab$ is divisible by $37$.
6 replies
parmenides51
Jan 22, 2024
littleduckysteve
2 hours ago
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|x_1-x_2|+|x_2-x_3|+....+|x_1-x_{10}|/x_1+...+x_10
nhathhuyyp5c   1
N 2 hours ago by littleduckysteve
Given positive real numbers \( x_1, x_2, \dots, x_{10} \) satisfying \( 1 < x_i < 10 \). Find the maximum value of the expression:

\[ P = \frac{|x_1 - x_{2}| + |x_2 - x_3| + |x_3 - x_4| + \dots + |x_9 - x_{10}| + |x_{10} - x_1|}{x_1 + x_2 + \dots + x_{10}}. \]
1 reply
nhathhuyyp5c
5 hours ago
littleduckysteve
2 hours ago
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2017 preRMO p4, roots (x^2+ax+20)(x^2+17x+b) = 0 are negative integers
parmenides51   9
N 2 hours ago by littleduckysteve
Let $a, b$ be integers such that all the roots of the equation $(x^2+ax+20)(x^2+17x+b) = 0$ are negative integers. What is the smallest possible value of $a + b$ ?
9 replies
parmenides51
Aug 9, 2019
littleduckysteve
2 hours ago
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Divisibility
Ecrin_eren   8
N 3 hours ago by P0tat0b0y


For which n is:

(1ⁿ + 2ⁿ + 3ⁿ + ... + nⁿ) divisible by n!?



8 replies
Ecrin_eren
Jul 23, 2025
P0tat0b0y
3 hours ago
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Challenge: Make as many positive integers from 2 zeros
Biglion   51
N Today at 1:55 PM by Shan3t
How many positive integers can you make from at most 2 zeros, any math operation and cocatination?
New Rule: The successor function can only be used at most 3 times per number
Starting from 0, 0=0
51 replies
Biglion
Jul 2, 2025
Shan3t
Today at 1:55 PM
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KVS IOQM P2
akv_6721   9
N Today at 1:34 PM by littleduckysteve
If $ABCD$ is a rectangle and $P$ is a point inside it such that $AP=33, BP=16, DP=63$.
Find $CP$.
9 replies
akv_6721
Jan 30, 2021
littleduckysteve
Today at 1:34 PM
J
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Number theory problem
littleduckysteve   4
N Today at 1:16 PM by littleduckysteve
A number in base 16 has 2025 digits, and no letters in its representation. The sum of the digits in this number is greater than 10126. How many such numbers are there, if the base 10 representation is a palindrome.
4 replies
littleduckysteve
Yesterday at 2:55 PM
littleduckysteve
Today at 1:16 PM
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Inequalities
sqing   15
N Today at 12:36 PM by sqing
Let $ a,b \geq 0, 2a +b^2=1 . $ Prove that
$$ \frac{\sqrt{a+b}}{a(b+1)} \geq \sqrt{2} $$Let $ a,b \geq 0, a^2 +2b^2 =2 . $ Prove that
$$ \frac{\sqrt{a+2b}}{a(b+1)} \geq \frac{9}{8\sqrt{2}} $$Let $ a,b \geq 0, 2a^2 +b^2 =1 . $ Prove that
$$\dfrac{a^2}{a+b}\leq\frac{1}{\sqrt{2}} $$$$ \frac{7}{10}>\frac{a+b}{a^2+b^2+1} \geq\frac{\sqrt{2}}{3} $$
15 replies
sqing
Jul 31, 2025
sqing
Today at 12:36 PM
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Inequalities
sqing   14
N Today at 12:25 PM by sqing
Let $ a,b \geq 0,a +b =1 . $ Prove that
$$ \sqrt{a^4 + \frac{3}{4}ab} + \sqrt{b^4 + \frac{3}{4}ab} \geq 1$$Let $ a,b \geq 0,a +b =4 . $ Prove that
$$ \sqrt{a^4 + 12ab} + \sqrt{b^4 + 12ab} \geq 16$$Let $ a,b,c \geq 0,a +b +c=1 . $ Prove that
$$ a^2 + b^2 + c^2 + \frac{1}{2}\sqrt{3abc} \geq \frac{1}{2}$$Let $ a,b,c \geq 0,a +b +c=3 . $ Prove that
$$ a^2 + b^2 + c^2 +\frac{3}{2} \sqrt{abc} \geq \frac{9}{2}$$
14 replies
sqing
Yesterday at 12:19 PM
sqing
Today at 12:25 PM
J
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J
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Challenging Optimization Problem
Shiyul   5
N Apr 22, 2025 by exoticc
Let $xyz = 1$. Find the minimum and maximum values of $\frac{1}{1 + x + xy}$ + $\frac{1}{1 + y + yz}$ + $\frac{1}{1 + z + zx}$

Can anyone give me a hint? I got that either the minimum or maximum was 1, but I'm sure if I'm correct.
5 replies
Shiyul
Apr 21, 2025
exoticc
Apr 22, 2025
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Challenging Optimization Problem
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Reveal topic tags
symmetry
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Shiyul
22 posts
Shiyul
#1 h  Apr 21, 2025, 8:20 PM
Y by
Let $xyz = 1$. Find the minimum and maximum values of $\frac{1}{1 + x + xy}$ + $\frac{1}{1 + y + yz}$ + $\frac{1}{1 + z + zx}$

Can anyone give me a hint? I got that either the minimum or maximum was 1, but I'm sure if I'm correct.
This post has been edited 2 times. Last edited by Shiyul, Apr 21, 2025, 8:23 PM
Reason: latex problem
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SPQ
5 posts
SPQ
#2 h  Apr 21, 2025, 9:16 PM
Y by
Hi! You're absolutely right—the minimum and maximum are both 1.

Here’s a little hint that might help clarify things:

The expression 1 + x + xy can be rewritten as:
1 + x(1 + y) = 1 + xyz(1 + y)/yz.
Since xyz = 1, this simplifies to:
1 + x + xy = 1 + (1 + y)/yz = (1 + y + yz)/yz.

In a similar way, you can show that:
1 + y + yz = (1 + z + zx)/zx.

Hope that helps.
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aidan0626
2113 posts
aidan0626
#3 h  Apr 22, 2025, 12:18 AM
Y by
try plugging in $z=\frac{1}{xy}$ into the expression, you might notice things simplify quite nicely :)
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lbh_qys
682 posts
lbh_qys
#4 h  Apr 22, 2025, 2:22 AM
Y by
Try let $x =\frac ba$.
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vanstraelen
9190 posts
vanstraelen
#5 h  Apr 22, 2025, 9:43 AM
Y by
$P(x,y,z)=\frac{1}{1 + x + xy}+\frac{1}{1 + y + yz}+\frac{1}{1 + z + zx}=1$, there is no maximum/minimum.
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exoticc
9 posts
exoticc
#6 h  Apr 22, 2025, 12:28 PM
Y by
The value of the expression is 1
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