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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
Nonlinear System
worthawholebean   39
N an hour ago by GeoKing
Source: AIME 2010I Problem 9
Let $ (a,b,c)$ be the real solution of the system of equations $ x^3 - xyz = 2$, $ y^3 - xyz = 6$, $ z^3 - xyz = 20$. The greatest possible value of $ a^3 + b^3 + c^3$ can be written in the form $ \frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m + n$.
39 replies
worthawholebean
Mar 17, 2010
GeoKing
an hour ago
Beams inside a cube
AOPS12142015   32
N 3 hours ago by cursed_tangent1434
Source: USOMO 2020 Problem 2, USOJMO 2020 Problem 3
An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:
[list=]
[*]The two $1 \times 1$ faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are $3 \cdot {2020}^2$ possible positions for a beam.)
[*]No two beams have intersecting interiors.
[*]The interiors of each of the four $1 \times 2020$ faces of each beam touch either a face of the cube or the interior of the face of another beam.
[/list]
What is the smallest positive number of beams that can be placed to satisfy these conditions?

Proposed by Alex Zhai
32 replies
AOPS12142015
Jun 21, 2020
cursed_tangent1434
3 hours ago
Goals for 2025-2026
Airbus320-214   60
N 3 hours ago by Math4Life2020
Please write down your goal/goals for competitions here for 2025-2026.
60 replies
Airbus320-214
Yesterday at 8:00 AM
Math4Life2020
3 hours ago
Summer internships/research opportunists in STEM
o99999   8
N 4 hours ago by Craftybutterfly
Hi, I am a current high school student and was looking for internships and research opportunities in STEM. Do you guys know any summer programs that do research such as RSI, but for high school freshmen that are open?
Thanks.
8 replies
o99999
Apr 22, 2020
Craftybutterfly
4 hours ago
No more topics!
Linecense
fruitmonster97   13
N Apr 25, 2025 by Ilikeminecraft
Source: 2024 AMC 8 Problem 23
Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point $(0,4)$ to point $(2,0)$ and colors the $4$ cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point $(2000,3000)$ to point $(5000,8000)$. How many cells will he color this time?

IMAGE

$\textbf{(A) }6000\qquad\textbf{(B) }6500\qquad\textbf{(C) }7000\qquad\textbf{(D) }7500\qquad\textbf{(E) }8000$
13 replies
fruitmonster97
Jan 25, 2024
Ilikeminecraft
Apr 25, 2025
Linecense
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G H BBookmark kLocked kLocked NReply
Source: 2024 AMC 8 Problem 23
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fruitmonster97
2492 posts
#1
Y by
Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point $(0,4)$ to point $(2,0)$ and colors the $4$ cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point $(2000,3000)$ to point $(5000,8000)$. How many cells will he color this time?

[asy]
filldraw((0,4)--(1,4)--(1,3)--(0,3)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((0,3)--(1,3)--(1,2)--(0,2)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((1,2)--(2,2)--(2,1)--(1,1)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((1,1)--(2,1)--(2,0)--(1,0)--cycle, gray(.75), gray(.5)+linewidth(1));

draw((-1,5)--(-1,-1),gray(.9));
draw((0,5)--(0,-1),gray(.9));
draw((1,5)--(1,-1),gray(.9));
draw((2,5)--(2,-1),gray(.9));
draw((3,5)--(3,-1),gray(.9));
draw((4,5)--(4,-1),gray(.9));
draw((5,5)--(5,-1),gray(.9));

draw((-1,5)--(5, 5),gray(.9));
draw((-1,4)--(5,4),gray(.9));
draw((-1,3)--(5,3),gray(.9));
draw((-1,2)--(5,2),gray(.9));
draw((-1,1)--(5,1),gray(.9));
draw((-1,0)--(5,0),gray(.9));
draw((-1,-1)--(5,-1),gray(.9));


dot((0,4));
label("$(0,4)$",(0,4),NW);

dot((2,0));
label("$(2,0)$",(2,0),SE);

draw((0,4)--(2,0));

draw((-1,0) -- (5,0), arrow=Arrow);
draw((0,-1) -- (0,5), arrow=Arrow);

[/asy]

$\textbf{(A) }6000\qquad\textbf{(B) }6500\qquad\textbf{(C) }7000\qquad\textbf{(D) }7500\qquad\textbf{(E) }8000$
This post has been edited 1 time. Last edited by fruitmonster97, Oct 11, 2024, 4:43 PM
Z K Y
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Countmath1
180 posts
#2
Y by
Using the below lemmas (or lemmata idk), the answer is Click to reveal hidden text
Click to reveal hidden text
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fruitmonster97
2492 posts
#3
Y by
Or just note that this is the same as up $3$, over $4$ 1000 times and calc the number of squares that goes through, which also gives C.
Z K Y
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Countmath1
180 posts
#4
Y by
i like your solution more
This post has been edited 1 time. Last edited by Countmath1, Jan 25, 2024, 4:45 PM
Reason: clarity
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MC413551
2228 posts
#5 • 1 Y
Y by axsolers_24
5000/3000=5/3 and the number of squares that are crossed by the diagonal of a 5 by 3 rectangle is 7 so multiply by 1000 to get 7000
Z K Y
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blueprimes
354 posts
#6
Y by
It is well-known that a line passing from $(0, 0)$ to the point $(m, n)$ when $m, n$ are relatively prime positive integers intersects $m + n - 1$ unit squares. Translating, the number of unit squares intersected from $(2000, 3000)$ to $(5000, 8000)$ is the same as the number of unit squares intersected from $(0, 0)$ to $(3000, 5000)$. This is simply $1000$ times the number of squares intersected from $(0, 0)$ to $(3, 5)$, so our final answer is $\boxed{1000(3 + 5 - 1) = \textbf{(C)}~7000}$.
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Squidget
433 posts
#7
Y by
I solved it the same way that @blueprimes did, but I also tried the Pythagorean theroem(yeah… no).
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dbnl
3377 posts
#8
Y by
i can confirm c but i misbubbled sadly
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yambe2002
1726 posts
#9
Y by
i got c too, i think i guessed that there were 7 when i drew (2,3) and (5,8) it looked like that, then multiplied by 1000
Z K Y
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DuoDuoling0
3864 posts
#10
Y by
The answer is C) 7000.

Solution
This post has been edited 2 times. Last edited by DuoDuoling0, Jan 26, 2024, 1:28 AM
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Genevieve2
1924 posts
#11
Y by
Ich habe auch C, 700.
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pi_is_3.14
1437 posts
#12 • 1 Y
Y by WannabeUSAMOkid
Video Solution:
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pog
4917 posts
#13 • 1 Y
Y by fruitmonster97
Quote:
Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point $(0,4)$ to point $(2,0)$ and colors the $4$ cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point $(2000,3000)$ to point $(5000,8000)$. How many cells will he color this time?

[asy]
filldraw((0,4)--(1,4)--(1,3)--(0,3)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((0,3)--(1,3)--(1,2)--(0,2)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((1,2)--(2,2)--(2,1)--(1,1)--cycle, gray(.75), gray(.5)+linewidth(1));
filldraw((1,1)--(2,1)--(2,0)--(1,0)--cycle, gray(.75), gray(.5)+linewidth(1));

draw((-1,5)--(-1,-1),gray(.9));
draw((0,5)--(0,-1),gray(.9));
draw((1,5)--(1,-1),gray(.9));
draw((2,5)--(2,-1),gray(.9));
draw((3,5)--(3,-1),gray(.9));
draw((4,5)--(4,-1),gray(.9));
draw((5,5)--(5,-1),gray(.9));

draw((-1,5)--(5, 5),gray(.9));
draw((-1,4)--(5,4),gray(.9));
draw((-1,3)--(5,3),gray(.9));
draw((-1,2)--(5,2),gray(.9));
draw((-1,1)--(5,1),gray(.9));
draw((-1,0)--(5,0),gray(.9));
draw((-1,-1)--(5,-1),gray(.9));


dot((0,4));
label("$(0,4)$",(0,4),NW);

dot((2,0));
label("$(2,0)$",(2,0),SE);

draw((0,4)--(2,0));

draw((-1,0) -- (5,0), arrow=Arrow);
draw((0,-1) -- (0,5), arrow=Arrow);

[/asy]

$\textbf{(A) }6000\qquad\textbf{(B) }6500\qquad\textbf{(C) }7000\qquad\textbf{(D) }7500\qquad\textbf{(E) }8000$
Z K Y
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Ilikeminecraft
631 posts
#14
Y by
Solved with ST2009 and Awesomeness_in_a_bun

it is $1000(5 + 8 - 1) = 12000$ then multiply by 7/12 by intuition which is C
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