Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!

Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
G
Topic
First Poster
Last Poster
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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
Close to JMO, but not close enough
isache   6
N 28 minutes ago by mathprodigy2011
Im currently a freshman in hs, and i rlly wanna make jmo in sophmore yr. Ive been cooking at in-person competitions recently (ucsd hmc, scmc, smt, mathcounts) but I keep fumbling jmo. this yr i had a 133.5 on 10b and a 9 on aime. How do i get that up by 20 points to a 240?
6 replies
isache
May 28, 2025
mathprodigy2011
28 minutes ago
Frustration with Olympiad Geo
gulab_jamun   13
N 29 minutes ago by mathprodigy2011
Ok, so right now, I am doing the EGMO book by Evan Chen, but when it comes to problems, there are some that just genuinely frustrate me and I don't know how to deal with them. For example, I've spent 1.5 hrs on the second to last question in chapter 2, and used all the hints, and I still am stuck. It just frustrates me incredibly. Any tips on managing this? (or.... am I js crashing out too much?)
13 replies
gulab_jamun
Thursday at 2:13 PM
mathprodigy2011
29 minutes ago
EGMO (geo) Radical Center Question
gulab_jamun   8
N 34 minutes ago by mathprodigy2011
For this theorem, Evan says that the power of point $P$ with respect to $\omega_1$ is greater than 0 if $P$ lies between $A$ and $B$. (I've underlined it). But, I'm a little confused as I thought the power was $OP^2 - r^2$ and since $P$ is inside the circle, wouldn't the power be negative since $OP < r$?
8 replies
gulab_jamun
May 25, 2025
mathprodigy2011
34 minutes ago
AIME qual outside US?
daijobu   6
N 2 hours ago by Yiyj
Can students outside the US take the AIME if they earn a qualifying score?
6 replies
daijobu
5 hours ago
Yiyj
2 hours ago
Cyclic Sum
P162008   1
N Yesterday at 3:13 PM by aaravdodhia
If $\sum_{cyc} \alpha = 0$ and $\sum_{cyc} \frac{\alpha^4}{2\alpha^2 + \beta\gamma} = 1$ then find the greatest possible value of $\sum_{cyc}\alpha^4.$
1 reply
P162008
May 26, 2025
aaravdodhia
Yesterday at 3:13 PM
Original Problem
wonderboy807   1
N Yesterday at 2:49 PM by jasperE3
f(0)=f(1)=1. \frac{f(n)f(n-m+1)}{f(n-m)} + \frac{f(n+1)f(n-m)}{f(m-n)} = \frac{f(n+2)f(n-m)f(m-n)}{f(n-m+1)f(m-n+1)}. Find f(10).

Answer: Click to reveal hidden text

Solution: Click to reveal hidden text
1 reply
wonderboy807
Yesterday at 11:36 AM
jasperE3
Yesterday at 2:49 PM
Find x^2 + y^2
Darealzolt   3
N Yesterday at 2:42 PM by jasperE3
Let \(x,y\) be positive real numbers that fulfill
\[
\frac{x^2}{y^2}+\frac{4x^2-3xy-4y^2}{2xy-5y^2}=2
\]Hence find the value of \(x^2+y^2\)
3 replies
Darealzolt
Yesterday at 11:45 AM
jasperE3
Yesterday at 2:42 PM
Own Problem
BinariouslyRandom   1
N Yesterday at 2:19 PM by BinariouslyRandom
LuAn is a renowned treasure hunter from Muntinlupa City. One day while patrolling the jungles of Mindanao, he found a safe with the following problem written on it:
[quote]Convert \(50392420515\) (base-10) into base-\(n\), where \(n\) is the smallest integer that has 8 factors. The answer will be the code to the safe.[/quote]
What is the secret password?

Note: If n > 10, A = 11, B = 12, and so on.
1 reply
BinariouslyRandom
Thursday at 11:21 AM
BinariouslyRandom
Yesterday at 2:19 PM
Sipnayan 2009 HS Eliminations (Difficult, #5)
BinariouslyRandom   1
N Yesterday at 2:14 PM by BinariouslyRandom
A round number is an integer whose base 2 representation has at least as many zeros as ones (the number of zeros is greater than or equal to the number of ones). How many round numbers are there less than or equal to 100?
1 reply
BinariouslyRandom
Thursday at 11:19 AM
BinariouslyRandom
Yesterday at 2:14 PM
[PMO27 Areas] I.11 Polynomial interpolation
aops-g5-gethsemanea2   3
N Yesterday at 2:05 PM by BinariouslyRandom
A polynomial $f(x)$ with nonnegative integer coefficients satisfies $f(1)=24$ and $f(9)=2024$. Find $f(5)$.

Answer confirmation
3 replies
aops-g5-gethsemanea2
Jan 25, 2025
BinariouslyRandom
Yesterday at 2:05 PM
Is problem true?!?!?!?
giangtruong13   0
Yesterday at 2:00 PM
Let $ABC$ be a triangle, $I$ is incenter of triangle $ABC$. Draw $IM$ perpendicular to $AB$ at $M$ and $IN$ perpendicular to $AC$ at $N$, $IM=IN=m$. Prove that: Area of triangle $ANM$ $\geq 2m^2$
0 replies
giangtruong13
Yesterday at 2:00 PM
0 replies
Logarithms
P162008   0
Yesterday at 1:48 PM
Let $a = \frac{(\log_{2} 3 - \log_{5} 7)(\log_{2} 3 - \log_{7} 9)}{(\log_{3} 5 - \log_{5} 7)(\log_{3} 5 - \log_{7} 9)}, b = \frac{(\log_{2} 3 - \log_{3} 5)(\log_{2} 3 - \log_{7} 9)}{(\log_{5} 7 - \log_{3} 5)(\log_{5} 7 - \log_{7} 9)}$ and $c = \frac{(\log_{2} 3 - \log_{3} 5)(\log_{2} 3 - \log_{5} 7)}{(\log_{7} 9 - \log_{3} 5)(\log_{7} 9 - \log_{5} 7)}.$
Find the value of $\lfloor a + b + c \rfloor$ where $\lfloor.\rfloor$ denotes greatest integer function.
0 replies
P162008
Yesterday at 1:48 PM
0 replies
Logarithms
P162008   0
Yesterday at 1:40 PM
Let $a = \log_{3} 5, b = \log_{3} 4$ and $c = -\log_{3} 20.$
Evaluate $\sum_{cyc} \frac{a^2 + b^2}{a^2 + b^2 + ab}.$
0 replies
P162008
Yesterday at 1:40 PM
0 replies
Binomial Sum
P162008   0
Yesterday at 1:36 PM
Let $\alpha = \sum_{k=0}^{1006} \frac{2012 - 2k}{(k+1) \binom{2013}{k+1}}.$
Evaluate $\alpha + \frac{1}{\binom{2013}{1006}}.$
0 replies
P162008
Yesterday at 1:36 PM
0 replies
sussy baka stop intersecting in my lattice points
Spectator   24
N Apr 29, 2025 by ilikemath247365
Source: 2022 AMC 10A #25
Let $R$, $S$, and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of $R$ and the right edge of $S$ are on the $y$-axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$. The top two vertices of $T$ are in $R \cup S$, and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S$. See the figure (not drawn to scale).

IMAGE

The fraction of lattice points in $S$ that are in $S \cap T$ is 27 times the fraction of lattice points in $R$ that are in $R \cap T$. What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$?

$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
24 replies
Spectator
Nov 11, 2022
ilikemath247365
Apr 29, 2025
sussy baka stop intersecting in my lattice points
G H J
Source: 2022 AMC 10A #25
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Spectator
657 posts
#1 • 1 Y
Y by megarnie
Let $R$, $S$, and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of $R$ and the right edge of $S$ are on the $y$-axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$. The top two vertices of $T$ are in $R \cup S$, and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S$. See the figure (not drawn to scale).

[asy]
//kaaaaaaaaaante314
size(8cm);
import olympiad;
label(scale(.8)*"$y$", (0,60), N);
label(scale(.8)*"$x$", (60,0), E);
filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white);
label(scale(1.3)*"$R$", (55/2,55/2));
filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white);
label(scale(1.3)*"$S$",(-14,14));
filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white);
label(scale(1.3)*"$T$",(3.5,25/2));
draw((0,-10)--(0,60),EndArrow(TeXHead));
draw((-34,0)--(60,0),EndArrow(TeXHead));[/asy]

The fraction of lattice points in $S$ that are in $S \cap T$ is 27 times the fraction of lattice points in $R$ that are in $R \cap T$. What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$?

$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
This post has been edited 3 times. Last edited by Spectator, Dec 24, 2022, 12:54 PM
Reason: Asy
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
peelybonehead
6290 posts
#3
Y by
You guys had time to solve P25?!!?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
iamhungry
149 posts
#4 • 22 Y
Y by Paul10, balllightning37, bobthebuilder1234, andrewwang2623, bestzack66, michaelwenquan, Ritwin, ivyshine13, eibc, aidan0626, Lamboreghini, plang2008, mahaler, mathboy100, megarnie, mathmax12, spiritshine1234, the_mathmagician, EpicBird08, aidensharp, akliu, Sedro
I looked at this problem, saw colored squares and the word "lattice points" and immediately went back to the previous problems
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Snowyowl2005
57 posts
#5 • 8 Y
Y by Spectator, wamofan, iamhungry, Mogmog8, Lamboreghini, cosmicgenius, Aopamy, centslordm
Solution at the end of https://randommath.app.box.com/s/fnajq15ew2gvx0bmt8ucri8karksulyo.

Suppose that the square $R$ has $r^2$ lattice points (i.e. has coordinates ranging from $(0, 0)$ to $(r - 1, r - 1)$). Similarly, suppose that the square $S$ has $s^2$ lattice point (i.e. has coordinates ranging from $(0, 0)$ to $(1 - s, s - 1)$).
Then $\frac{r^2}{s^2} = \frac{9}{4}$, and so $r = 3x$ and $s = 2x$ for some positive integer $x$.
Furthermore, the union of $R$ and $S$ contains $r^2 + s^2 - s$ lattice points since they share $s$ lattice points of the form $(0, 0), (0, 1), (0, 2), \ldots, (0, s - 1)$. Therefore:
\[t^2 = \frac{r^2 + s^2 - s}{4} = \frac{13x^2 - 2x}{4}\]and so:
\[13x^2 - 2x = 4t^2\]Since the right hand side is even, $13x^2$ is even, and so we can write $x = 2y$ for some integer $y$. Thus:
\[52y^2 - 4y = 4t^2\]\[13y^2 - y = y(13y - 1) = t^2\]But by the Euclidean algorithm, $y$ and $13y - 1$ are relatively prime, and so both $y$ and $13y - 1$ must be perfect squares. Write:
\[y = m^2\]\[13y - 1 = 13m^2 - 1 = n^2\]Suppose that the top left corner of $T$ is given by $(1 - k, 0)$. Then the top right corner must be given by $(t - k, 0)$. We have:
\[\frac{|S \cap T|}{|S|} = 27 \cdot \frac{|R \cap T|}{|R|}\]where $|A|$ denotes the number of lattice points in a region $A$. Plugging in the relevant values, we get:
\[\frac{k \cdot t}{s^2} = 27 \cdot \frac{(t - k + 1) \cdot t}{r^2}\]Cross-multiplying, we get:
\[k \cdot t \cdot r^2 = 27 \cdot (t - k + 1) \cdot t \cdot s^2\]Simplifying and using $r^2 = \frac{9}{4}s^2$, we get:
\[9 \cdot k = (t - k + 1) \cdot 4 \cdot 27\]\[12 \cdot (t - k + 1) = k\]\[12 \cdot (t + 1) = 13k\]Therefore, for any $t \equiv 12 \pmod{13}$, we have a solution for $k$. But:
\[y(13y - 1) \equiv -y \equiv t^2 \equiv 1 \pmod{13}\]and so $y \equiv 12 \pmod{13}$. Since $y = m^2$, we have:
\[m^2 + 1 \equiv 0 \pmod{13}\]and that $13m^2 - 1$ is a perfect square. The smallest $m$ satisfying both conditions is $m = 5$. Plugging this in, we get:
\[y = m^2 = 25\]\[x = 2y = 50\]and so $r = 3x = 150$, $s = 2x = 100$, and $t^2 = y(13y - 1) = 25 \cdot 324$, so $t = 5 \cdot 18 = 90$, and:
\[(r - 1) + (s - 1) + (t - 1) = 150 + 100 + 90 - 3 = 340 - 3 = 337\]so the answer is B.
This post has been edited 2 times. Last edited by Snowyowl2005, Nov 11, 2022, 6:09 PM
Reason: The problem is asking for the sum of the edge lengths, which are $r - 1, s - 1, t - 1$, instead of $r + s + t$. It has been fixed above.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathboy100
675 posts
#6
Y by
Can we use Pick's theorem on this?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bryanguo
1032 posts
#7
Y by
Answer E is incorrect, they did sum of coordinates which happened to give an answer $3$ larger than what it's supposed to be.

Rough sketch is to label side length of $S$ as $x,$ then $(x+1)^2$ lattice points in $S$ and $\tfrac{9}{4}(x+1)^2$ lattice points in $R,$ so side length of $R$ is $\tfrac{3x+1}{2},$ from there just use the other two given conditions are you get modular congruences to give side lengths $(89,99,149) \implies 337.$
This post has been edited 1 time. Last edited by bryanguo, Nov 11, 2022, 5:35 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihatemath123
3449 posts
#8
Y by
Uhhh what would you need picks for here
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Spectator
657 posts
#9
Y by
can someone include asy code for this
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Snowyowl2005
57 posts
#10 • 1 Y
Y by bryanguo
bryanguo wrote:
Answer E is incorrect, they did sum of coordinates which happened to give an answer $3$ larger than what it's supposed to be.

Rough sketch is to label side length of $S$ as $x,$ then $(x+1)^2$ lattice points in $S$ and $\tfrac{9}{4}(x+1)^2$ lattice points in $R,$ so side length of $R$ is $\tfrac{3x+1}{2},$ from there just use the other two given conditions are you get modular congruences to give side lengths $(89,99,149) \implies 337.$

The only error seems to be the answer extraction - it has been fixed above.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8868 posts
#11
Y by
This problem is essentially how quickly you can do number theory.

Let the side length of $R$ be $x$, so the number of lattice points in $R$ is $(x+1)^2$, the number of lattice points in $S$ is $\frac 49(x+1)^2$. It follows that the side length of $S$ is $\frac 23 x - \frac 13$.

Now, let $a$ and $b$ be the lengths of the portions of the sides of $T$ in $S$ and $R$, respectively. We have $$\frac{(a+1)(k+1)}{\frac 49(x+1)^2} = 27 \cdot \frac{(b+1)(k+1)}{(x+1)^2}$$by the third condition, so $a = 12b+11$, and the side length $k$ of $T$ is of the form $13b+11$ for some $b$.

Now, using the second condition, $$\frac{(k+1)^2}{\frac {13}9(x+1)^2 - \frac 23(x+1)} = \frac 14.$$Note Then $$36(13b+12)^2 = (x+1)(13x+7).$$Notice that $\gcd(x+1, 13x+7) = \gcd(x+1, 6) = 6$, so actually
\begin{align*}
x+1 &= 6r_1^2 \\
13x+ 7 &= 6r_2^2
\end{align*}for some positive integers $r_1, r_2$. These $r_1, r_2$ satisfy $$13r_1^2 - r_2^2 = 1.$$At this point, there isn't really a better way than to test out values for $r_1$, as we know by the answer choices that the answers cannot be too big. Indeed, $(r_1, r_2) = (5, 18)$ is a solution. Then, $x = 149$, so $$(k+1)^2 = \frac{150 \cdot 1944}{36} = 324 \cdot 25,$$which yields $k = 89$. Finally, the last side length is $99$, so the answer is $89+99+149 = \boxed{337}$.
This post has been edited 2 times. Last edited by HamstPan38825, Nov 11, 2022, 7:32 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Somersett
71 posts
#12
Y by
peelybonehead wrote:
You guys had time to solve P25?!!?

Bruh this test went by so fast i didn’t even get to look at this problem
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Math_Shisa
158 posts
#13
Y by
All the solutions look so long I don’t think I would have time to even answer this one.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
saturnrocket
1306 posts
#14
Y by
I didn't even get to 15
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Dino76
707 posts
#15 • 1 Y
Y by Mango247
That's a lot better than me. I didn't even bother to take it...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
qwerty123456asdfgzxcvb
1088 posts
#16
Y by
me when not 340
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mannshah1211
652 posts
#17
Y by
if There is one more Baka in AMC Titles, i will Riot.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Bluesoul
899 posts
#19
Y by
Not necessary to solve exact R,S,T, just compute R+S+T mod 13

Denote the side lengths of $R,S,T$ as $r,s,t$ respectively. According to the first condition, we have $\frac{9}{4}(s+1)^2=(r+1)^2, 3s+1=2r$

Second condition yields $4(t+1)^2=(s+1)^2+(r+1)^2-(s+1)$. Plug $r=\frac{3s+1}{2}$, we get $16(t+1)^2=13(s+1)^2-4(s+1)=(s+1)(13s+9)$

Let the length of $T$ in negative x-axis as $x$, the third condition implies $\frac{(x+1)(t+1)}{(s+1)^2}=27\frac{(t+1)(t-x+1)}{(r+1)^2}$ simplify this to $x+1=12(t-x+1), 12t+11=13x$. From here, we can attain $t\equiv -2\pmod{13}$. Consider equation $16(t+1)^2=(s+1)(13s+9)$, we have $(s+1)(13s+9)\equiv 9(s+1)\equiv 3\pmod{13}$

Thus, we have $s\equiv 8\pmod{13}$

Finally, we have $r=\frac{3s+1}{2}$ which yields $r\equiv 6\pmod{13}$. Consequently, $r+s+t\equiv 8+6-2\equiv 12\pmod{13}$ while only $\boxed{B}$ satisfies.
This post has been edited 1 time. Last edited by Bluesoul, Oct 25, 2023, 12:05 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MathyMathMan
250 posts
#20
Y by
mannshah1211 wrote:
if There is one more Baka in AMC Titles, i will Riot.

Bouta say, but what is that title? :huuh:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Amkan2022
2027 posts
#21
Y by
.........
This post has been edited 1 time. Last edited by Amkan2022, Apr 29, 2025, 4:07 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IbrahimNadeem
888 posts
#22
Y by
aops being wild today
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IbrahimNadeem
888 posts
#23
Y by
why am I getting an irrational side for T?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ESAOPS
263 posts
#26
Y by
oops
sol
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
xHypotenuse
788 posts
#27
Y by
This feels harder than average aime #12s and I lwk have 0 clue on how to approach it
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MC_ADe
183 posts
#28
Y by
write out each condition and turn it into different forms, since you know each is an integer you can use divisibility,modulo, euclidean algorithm and other methods to find the values
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ilikemath247365
267 posts
#29
Y by
I was actually able to solve this problem......except it took me like 30 minutes. :rotfl:
Z K Y
N Quick Reply
G
H
=
a