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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.

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0 replies
jlacosta
May 1, 2025
0 replies
usamOOK geometry
KevinYang2.71   106
N Yesterday at 11:54 PM by jasperE3
Source: USAMO 2025/4, USAJMO 2025/5
Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$.
106 replies
KevinYang2.71
Mar 21, 2025
jasperE3
Yesterday at 11:54 PM
Geo #3 EQuals FReak out
Th3Numb3rThr33   106
N Yesterday at 10:56 PM by BS2012
Source: 2018 USAJMO #3
Let $ABCD$ be a quadrilateral inscribed in circle $\omega$ with $\overline{AC} \perp \overline{BD}$. Let $E$ and $F$ be the reflections of $D$ over lines $BA$ and $BC$, respectively, and let $P$ be the intersection of lines $BD$ and $EF$. Suppose that the circumcircle of $\triangle EPD$ meets $\omega$ at $D$ and $Q$, and the circumcircle of $\triangle FPD$ meets $\omega$ at $D$ and $R$. Show that $EQ = FR$.
106 replies
Th3Numb3rThr33
Apr 18, 2018
BS2012
Yesterday at 10:56 PM
Aime ll 2022 problem 5
Rook567   4
N Yesterday at 9:02 PM by Rook567
I don’t understand the solution. I got 220 as answer. Why does it insist, for example two primes must add to the third, when you can take 2,19,19 or 2,7,11 which for drawing purposes is equivalent to 1,1,2 and 2,7,9?
4 replies
Rook567
Thursday at 9:08 PM
Rook567
Yesterday at 9:02 PM
USAJMO problem 2: Side lengths of an acute triangle
BOGTRO   59
N Yesterday at 4:41 PM by ostriches88
Source: Also USAMO problem 1
Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_1, a_2, \hdots, a_n$ with $\text{max}(a_1,a_2,\hdots,a_n) \leq n \cdot \text{min}(a_1,a_2,\hdots,a_n)$, there exist three that are the side lengths of an acute triangle.
59 replies
BOGTRO
Apr 24, 2012
ostriches88
Yesterday at 4:41 PM
No more topics!
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
6291 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
3446 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
8862 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
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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
1086 posts
#16
Y by
me when not 340
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mannshah1211
651 posts
#17
Y by
if There is one more Baka in AMC Titles, i will Riot.
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Bluesoul
898 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
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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
2015 posts
#21
Y by
.........
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
262 posts
#26
Y by
oops
sol
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xHypotenuse
778 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
182 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
253 posts
#29
Y by
I was actually able to solve this problem......except it took me like 30 minutes. :rotfl:
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