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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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0 replies
jlacosta
May 1, 2025
0 replies
Close to JMO, but not close enough
isache   12
N 2 hours ago by LearnMath_105
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?
12 replies
isache
Yesterday at 11:37 PM
LearnMath_105
2 hours ago
Bring Back Downvotes
heheman   99
N 2 hours ago by heheman
i would like to start a petition to bring back downvote, it you agree then write "bbd $    $" in threads
99 replies
heheman
Yesterday at 7:21 PM
heheman
2 hours ago
[TEST RELEASED] OMMC Year 5
DottedCaculator   180
N 3 hours ago by fuzimiao2013
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
180 replies
DottedCaculator
Apr 26, 2025
fuzimiao2013
3 hours ago
[$10K+ IN PRIZES] Poolesville Math Tournament (PVMT) 2025
qwerty123456asdfgzxcvb   19
N 3 hours ago by Ruegerbyrd
Hi everyone!

After the resounding success of the first three years of PVMT, the Poolesville High School Math Team is excited to announce the fourth annual Poolesville High School Math Tournament (PVMT)! The PVMT team includes a MOPper and multiple USA(J)MO and AIME qualifiers!

PVMT is open to all 6th-9th graders in the country (including rising 10th graders). Students will compete in teams of up to 4 people, and each participant will take three subject tests as well as the team round. The contest is completely free, and will be held virtually on June 7, 2025, from 10:00 AM to 4:00 PM (EST).

Additionally, thanks to our sponsors, we will be awarding approximately $10K+ worth of prizes (including gift cards, Citadel merch, AoPS coupons, Wolfram licenses) to top teams and individuals. More details regarding the actual prizes will be released as we get closer to the competition date.

Further, newly for this year we might run some interesting mini-events, which we will announce closer to the competition date, such as potentially a puzzle hunt and integration bee!

If you would like to register for the competition, the registration form can be found at https://pvmt.org/register.html or https://tinyurl.com/PVMT25.

Additionally, more information about PVMT can be found at https://pvmt.org

If you have any questions not answered in the below FAQ, feel free to ask in this thread or email us at falconsdomath@gmail.com!

We look forward to your participation!

FAQ
19 replies
qwerty123456asdfgzxcvb
Apr 5, 2025
Ruegerbyrd
3 hours ago
Inequalities
sqing   21
N Today at 12:37 AM by sqing
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
21 replies
sqing
May 21, 2025
sqing
Today at 12:37 AM
Polar Coordinates
pingpongmerrily   4
N Today at 12:11 AM by K124659
Convert the equation $r=\tan(\theta)$ into rectangular form.
4 replies
pingpongmerrily
Today at 12:02 AM
K124659
Today at 12:11 AM
Calculus, sets
wl8418   2
N Yesterday at 11:18 PM by wl8418
Is empty set a proper set of an non empty set? Why or why not? Any clarification or insight is appreciated. Thanks in advance!
2 replies
wl8418
Yesterday at 6:11 AM
wl8418
Yesterday at 11:18 PM
Geometry
AlexCenteno2007   1
N Yesterday at 11:13 PM by ohiorizzler1434
Given triangle ABC, it is true that BD = CF where D and F are points in the same half-plane with respect to line BC and it is also known that BD is parallel to AC and CF is parallel to AB. Show that BF, CD and the interior bisector of A are concurrent.
1 reply
AlexCenteno2007
Yesterday at 10:02 PM
ohiorizzler1434
Yesterday at 11:13 PM
How do I prove this? - Sets and symmetric difference
smadadi1000   1
N Yesterday at 8:21 PM by KSH31415
For sets A, B and C, prove that (A Δ B) Δ C = (A Δ C) Δ (A \ B).

The textbook - Proofs by Jay Cummings - had this definition for symmetric difference:
A Δ B = ( A U B)/(A ∩ B)

this is exercise 3.37 (e).
1 reply
smadadi1000
Yesterday at 3:23 PM
KSH31415
Yesterday at 8:21 PM
help me ..
exoticc   2
N Yesterday at 5:00 PM by sodiumaka
Find all pairs of functions (f,g) : R->R that satisfy:
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
2 replies
exoticc
Yesterday at 9:26 AM
sodiumaka
Yesterday at 5:00 PM
21st PMO National Orals #9
yes45   0
Yesterday at 3:22 PM
In square $ABCD$, $P$ and $Q$ are points on sides $CD$ and $BC$, respectively, such that $\angle{APQ} = 90^\circ$. If $AP = 4$ and $PQ = 3$, find the area of $ABCD$.

Answer Confirmation
Solution
0 replies
yes45
Yesterday at 3:22 PM
0 replies
Original Problem (qrxz17)
qrxz17   0
Yesterday at 1:32 PM
Problem. Suppose that
\begin{align*}
    (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 &= 138 \\
    (a^2+b^2+c^2)^2 &=100.
\end{align*}Find \(a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1)\).
Answer: Click to reveal hidden text
Solution. Subtracting the two given equations, we get
\begin{align*}
        a^4+b^4+c^4=38.
    \end{align*}Taking the square root of the second equation, we get

\begin{align*}
        a^2+b^2+c^2 = 10.
    \end{align*}
Then,
\begin{align*}
        a^4+b^4+c^4-(a^2+b^2+c^2) &= a^4-a^2+b^4-b^2+c^4-c^2 \\
        &= a^2(a^2-1)+b^2(b^2-1)+c^2(c^2-1) = \boxed{28}.
    \end{align*}
0 replies
qrxz17
Yesterday at 1:32 PM
0 replies
17th PMO Area Stage #5
yes45   0
Yesterday at 1:31 PM
Triangle \(ABC\) has a right angle at \(B\), with \(AB = 3\) and \(BC = 4\). If \(D\) and \(E\) are points
on \(AC\) and \(BC\), respectively, such that \(CD = DE = \frac{5}{3}\), find the perimeter of quadrilateral
\(ABED\).
Answer Confirmation
Solution
0 replies
yes45
Yesterday at 1:31 PM
0 replies
Sipnayan 2019 SHS Orals Semifinal Round B Difficult \#3
qrxz17   0
Yesterday at 1:30 PM
Problem. Suppose that \(a\), \(b\), \(c\) are positive integers such that \((a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 63\). Find all possible values of \(a+b+c\).
Answer. Click to reveal hidden text
Solution. Expanding and cancelling "like" terms, we get

\begin{align*}
        (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) &= (a^4 +b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2) - 2(a^4+b^4+c^4) \\
        &= 2a^2b^2+2a^2c^2+2b^2c^2 - a^4 -b^4-c^4
    \end{align*}
We multiply the entire equation by \(-1\) to rearrange the terms and place the higher-degree terms at the front of the expression in order to have a more recognizable form.

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 = -63.
    \end{align*}
Simplifying the left-hand expression, we get

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 &=a^4-2a^2(b^2+c^2)+b^4-2b^2c^2+c^4 \\
        &= a^4-2a^2(b^2+c^2)+(b^2+c^2)^2 - 4b^2c^2 \\
        &= [a^2 - (b^2+c^2)]^2-(2bc)^2 \\
        &= (a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)\\
        &=[a^2-(b-c)^2][a^2-(b+c)^2] \\
        &= (a+b-c)(a-b+c)(a+b+c)(a-b-c) = -63.
    \end{align*}
If \(a=b=c\), we get
\begin{align*}
     -3a^4 &=-63 \\
     a^4 &= 21
    \end{align*}

in which there are no real solutions for \(a\).

If we either have \(a=b\), \(a=c\), or \(b=c\), we get something of the form

\[
    a^2(a+2b)(a-2b)=-63.
    \]
Since \(1\) and \(9\) are the only square factors of \(63\), we have \(a = \{1, 3\}\).

In this case, the solutions for \((a, b, c)\) are \((1, 4, 4)\) and \((3, 2, 2)\), including all their permutations.

If \(a, b, c\) are all distinct positive integers, there are no solutions.

Therefore, all possible values of \(a+b+c\) are \(\boxed{7 \text{ and }9}\).
0 replies
qrxz17
Yesterday at 1:30 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
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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
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peelybonehead
6290 posts
#3
Y by
You guys had time to solve P25?!!?
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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
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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.
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mathboy100
675 posts
#6
Y by
Can we use Pick's theorem on this?
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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
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ihatemath123
3449 posts
#8
Y by
Uhhh what would you need picks for here
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Spectator
657 posts
#9
Y by
can someone include asy code for this
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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.
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HamstPan38825
8868 posts
#11
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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
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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
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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.
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saturnrocket
1306 posts
#14
Y by
I didn't even get to 15
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Dino76
707 posts
#15 • 1 Y
Y by Mango247
That's a lot better than me. I didn't even bother to take it...
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qwerty123456asdfgzxcvb
1088 posts
#16
Y by
me when not 340
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mannshah1211
652 posts
#17
Y by
if There is one more Baka in AMC Titles, i will Riot.
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Bluesoul
899 posts
#19
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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
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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:
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Amkan2022
2027 posts
#21
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.........
This post has been edited 1 time. Last edited by Amkan2022, Apr 29, 2025, 4:07 AM
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IbrahimNadeem
888 posts
#22
Y by
aops being wild today
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IbrahimNadeem
888 posts
#23
Y by
why am I getting an irrational side for T?
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ESAOPS
263 posts
#26
Y by
oops
sol
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xHypotenuse
787 posts
#27
Y by
This feels harder than average aime #12s and I lwk have 0 clue on how to approach it
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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
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ilikemath247365
262 posts
#29
Y by
I was actually able to solve this problem......except it took me like 30 minutes. :rotfl:
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