Stay ahead of learning milestones! Enroll in a class over the summer!

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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:
[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]
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0 replies
jlacosta
May 1, 2025
0 replies
Weird locus problem
Sedro   2
N 19 minutes ago by ReticulatedPython
Points $A$ and $B$ are in the coordinate plane such that $AB=2$. Let $\mathcal{H}$ denote the locus of all points $P$ in the coordinate plane satisfying $PA\cdot PB=2$, and let $M$ be the midpoint of $AB$. Points $X$ and $Y$ are on $\mathcal{H}$ such that $\angle XMY = 45^\circ$ and $MX\cdot MY=\sqrt{2}$. The value of $MX^4 + MY^4$ can be expressed in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
2 replies
Sedro
Yesterday at 3:12 AM
ReticulatedPython
19 minutes ago
dirichlet
spiralman   0
4 hours ago
Let n be a positive integer. Consider 2n+1 distinct positive integers whose total sum is less than (n+1)(3n+1). Prove that among these 2n+1 numbers, there exist two numbers whose sum is 2n+1.
0 replies
spiralman
4 hours ago
0 replies
Inequalities
sqing   0
5 hours ago
Let $ a, b, c >0, a^2 + \frac{b}{a}  = 8 $ and $ 3a + b + c \geq  9\sqrt{3} .$ Prove that $$   ab + c^2\geq 18$$
0 replies
sqing
5 hours ago
0 replies
Plz help
Bet667   2
N Today at 7:42 AM by jasperE3
f:R-->R for any integer x,y
f(yf(x)+f(xy))=(x+f(x))f(y)
find all function f
(im not good at english)
2 replies
Bet667
Jan 28, 2024
jasperE3
Today at 7:42 AM
No more topics!
polynomial with inequality
nhathhuyyp5c   1
N Apr 18, 2025 by matt_ve
Given the polynomial \( P(x) = x^3 + ax^2 + bx + c \), where \( a, b, c \) are real numbers. Suppose that \( P(x) \) has three distinct real roots and the polynomial \( Q(x) = P(x^2 + 12x - 32) \) has no real roots. Prove that
\[
P(1) > 69^3.
\]
1 reply
nhathhuyyp5c
Apr 18, 2025
matt_ve
Apr 18, 2025
polynomial with inequality
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nhathhuyyp5c
62 posts
#1
Y by
Given the polynomial \( P(x) = x^3 + ax^2 + bx + c \), where \( a, b, c \) are real numbers. Suppose that \( P(x) \) has three distinct real roots and the polynomial \( Q(x) = P(x^2 + 12x - 32) \) has no real roots. Prove that
\[
P(1) > 69^3.
\]
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The post below has been deleted. Click to close.
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matt_ve
3 posts
#2
Y by
Since $Q(x)$ has no real roots, and $x^2+12x-32$ takes all real values $\geq -68$, the real roots of $P$ must all be $<-68$.
By Viete's then, we have $a>3\cdot 68$, $b>3\cdot 68^2$, $c>68^3$ --- and so
$$P(1)>1+3\cdot 68+3\cdot 68^2+68^3=69^3.$$
This post has been edited 1 time. Last edited by matt_ve, Apr 18, 2025, 6:17 PM
Reason: typo
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