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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.

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[*]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
A Collection of Good Problems from my end
SomeonecoolLovesMaths   8
N 2 hours ago by Math-lover1
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
8 replies
SomeonecoolLovesMaths
Yesterday at 8:16 AM
Math-lover1
2 hours ago
find number of elements in H
Darealzolt   1
N 3 hours ago by alexheinis
If \( H \) is the set of positive real solutions to the system
\[
x^3 + y^3 + z^3 = x + y + z
\]\[
x^2 + y^2 + z^2 = xyz
\]then find the number of elements in \( H \).
1 reply
Darealzolt
Today at 1:50 AM
alexheinis
3 hours ago
primes and perfect squares
Bummer12345   0
4 hours ago
If $p$ and $q$ are primes, then can $2^p + 5^q + pq$ be a perfect square?
0 replies
Bummer12345
4 hours ago
0 replies
simple trapezoid
gggzul   0
5 hours ago
Let $ABCD$ be a trapezoid. By $x$ we denote the angle bisector of angle $X$ . Let $P=a\cap c$ and $Q=b\cap d$. Prove that $ABPQ$ is cyclic.
0 replies
gggzul
5 hours ago
0 replies
geometry
JetFire008   0
5 hours ago
Given four concyclic points. For each subset of three points take the incenter. Show that the four incentres form a rectangle.
0 replies
JetFire008
5 hours ago
0 replies
36x⁴ + 12x² - 36x + 13 > 0
fxandi   0
5 hours ago
Prove that for any real $x \geq 0$ holds inequality $36x^4 + 12x^2 - 36x + 13 > 0.$
0 replies
fxandi
5 hours ago
0 replies
Inequalities
sqing   11
N Today at 3:02 PM by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
11 replies
sqing
Jul 12, 2024
sqing
Today at 3:02 PM
A rather difficult question
BeautifulMath0926   3
N Today at 2:23 PM by evankuang
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.
3 replies
BeautifulMath0926
Apr 13, 2025
evankuang
Today at 2:23 PM
The return of an inequality
giangtruong13   4
N Today at 1:26 PM by sqing
Let $a,b,c$ be real positive number satisfy that: $a+b+c=1$. Prove that: $$\sum_{cyc} \frac{a}{b^2+c^2} \geq \frac{3}{2}$$
4 replies
giangtruong13
Mar 18, 2025
sqing
Today at 1:26 PM
Looking for users and developers
derekli   10
N Today at 1:22 PM by Jackson0423
Guys I've been working on a web app that lets you grind high school lvl math. There's AMCs, AIME, BMT, HMMT, SMT etc. Also, it's infinite practice so you can keep grinding without worrying about finding new problems. Please consider helping me out by testing and also consider joining our developer team! :P :blush:

Link: https://stellarlearning.app/competitive
10 replies
derekli
Yesterday at 12:57 AM
Jackson0423
Today at 1:22 PM
Polynomial
kellyelliee   1
N Today at 1:19 PM by Jackson0423
Let the polynomial $f(x)=x^2+ax+b$, where $a,b$ integers and $k$ is a positive integer. Suppose that the integers
$m,n,p$ satisfy: $f(m), f(n), f(p)$ are divisible by k. Prove that:
$(m-n)(n-p)(p-m)$ is divisible by k
1 reply
kellyelliee
Today at 3:57 AM
Jackson0423
Today at 1:19 PM
Sum of digits is 18
Ecrin_eren   14
N Today at 1:11 PM by jestrada
How many 5 digit numbers are there such that sum of its digits is 18
14 replies
Ecrin_eren
May 3, 2025
jestrada
Today at 1:11 PM
IOQM 2022-23 P-7
lifeismathematics   2
N Today at 12:09 PM by Adywastaken
Find the number of ordered pairs $(a,b)$ such that $a,b \in \{10,11,\cdots,29,30\}$ and
$\hspace{1cm}$ $GCD(a,b)+LCM(a,b)=a+b$.
2 replies
lifeismathematics
Oct 30, 2022
Adywastaken
Today at 12:09 PM
Inequalities
sqing   7
N Today at 10:31 AM by sqing
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
7 replies
sqing
Yesterday at 12:46 PM
sqing
Today at 10:31 AM
Simiplifying a Complicated Expression
phiReKaLk6781   6
N Apr 23, 2025 by lbh_qys
Simplify: $ \frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}$
6 replies
phiReKaLk6781
Mar 15, 2010
lbh_qys
Apr 23, 2025
Simiplifying a Complicated Expression
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phiReKaLk6781
2074 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Simplify: $ \frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}$
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varunrocks
1134 posts
#2 • 2 Y
Y by Adventure10, Mango247
There are too many steps but I am getting a+b+c.
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Truffles
1459 posts
#3 • 2 Y
Y by Adventure10, Mango247
You could use the fact that the fractions in the expression are symmetric?
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Vo Duc Dien
341 posts
#4 • 2 Y
Y by Adventure10 and 1 other user
After expanding the expression, just divide the numerator by the denominator and we will get a + b + c.
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dgreenb801
1896 posts
#5 • 1 Y
Y by Adventure10
If we combine it into one fraction we get $ \frac{a^3b-ab^3+b^3c-bc^3+c^3a-ca^3}{(a-b)(a-c)(b-c)}$. But in the numerator if $ a=b$ then the numerator would would $ 0$, thus $ a-b$ must be a factor of the numerator, similarly $ a-c$ and $ b-c$ must be also, dividing them out we get $ a+b+c$.
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P162008
181 posts
#6
Y by
Let $$\frac{a^3}{(a - b)(a - c)} + \frac{b^3}{(b - c)(b - a)} + \frac{c^3}{(c - a)(c - b)} = t$$
$$t = -\frac{a^3(b - c) + b^3(c - a) + c^3(a - b)}{(a - b)(b - c)(c - a)}$$
Now, $$ N_r = a^3(b - c) + b^3(c - a) + c^3(a - b)$$
$$= a^3 b - ab^3 + b^3c - ca^3 + c^3a - bc^3$$
$$= ab(a^2 - b^2) - c(a^3 - b^3) + c^3(a - b)$$
$$= (a - b)[ab(a + b) - c(a^2 + ab + b^2) + c^3]$$
$$= (a - b)[a²b + ab² - ca² - abc - b²c + c³]$$
$$= (a - b)[a²(b - c) + ab(b - c) - c(b² - c²)]$$
$$= (a - b)(b - c)[a² + ab - c(b + c)]$$
$$= (a - b)(b - c)[a² + ab - bc - c²)]$$
$$= (a - b)(b - c)[-b(c - a) - (c² - a²)]$$
$$= -(a - b)(b - c)(c - a)(a + b + c)$$
Therefore, $$t = \frac{(a - b)(b - c)(c - a)(a + b + c)}{(a - b)(b - c)(c - a)} = \boxed{a + b + c}$$
This post has been edited 1 time. Last edited by P162008, Apr 22, 2025, 12:49 AM
Reason: Typo
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lbh_qys
558 posts
#7
Y by
This is the coefficient of the \(x^2\) term in \(x^3 - (x-a)(x-b)(x-c)\) by Lagrange interpolation.
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