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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
JBMO Shortlist 2022 N1
Lukaluce   8
N an hour ago by godchunguus
Source: JBMO Shortlist 2022
Determine all pairs $(k, n)$ of positive integers that satisfy
$$1! + 2! + ... + k! = 1 + 2 + ... + n.$$
8 replies
Lukaluce
Jun 26, 2023
godchunguus
an hour ago
Discuss the Stanford Math Tournament Here
Aaronjudgeisgoat   290
N an hour ago by techb
I believe discussion is allowed after yesterday at midnight, correct?
If so, I will put tentative answers on this thread.
By the way, does anyone know the answer to Geometry Problem 5? I was wondering if I got that one right
Also, if you put answers, please put it in a hide tag

Answers for the Algebra Subject Test
Estimated Algebra Cutoffs
Answers for the Geometry Subject Test
Estimated Geo Cutoffs
Answers for the Discrete Subject Test
Estimated Cutoffs for Discrete
Answers for the Team Round
Guts Answers
290 replies
Aaronjudgeisgoat
Apr 14, 2025
techb
an hour ago
MOP Emails
hellohannah   97
N an hour ago by Pengu14
So mop emails are probably coming tomorrow, feel free to discuss here. I'll probably post when I hear that they're out unless I'm asleep
97 replies
+1 w
hellohannah
Yesterday at 4:59 AM
Pengu14
an hour ago
P(x) | P(x^2-2)
GreenTea2593   4
N an hour ago by GreenTea2593
Source: Valentio Iverson
Let $P(x)$ be a monic polynomial with complex coefficients such that there exist a polynomial $Q(x)$ with complex coefficients for which \[P(x^2-2)=P(x)Q(x).\]Determine all complex numbers that could be the root of $P(x)$.

Proposed by Valentio Iverson, Indonesia
4 replies
GreenTea2593
4 hours ago
GreenTea2593
an hour ago
USEMO P6 (Idk what to say here)
franzliszt   16
N an hour ago by MathLuis
Source: USEMO 2020/6
Prove that for every odd integer $n > 1$, there exist integers $a, b > 0$ such that, if we let $Q(x) = (x + a)^
2 + b$, then the following conditions hold:
$\bullet$ we have $\gcd(a, n) = gcd(b, n) = 1$;
$\bullet$ the number $Q(0)$ is divisible by $n$; and
$\bullet$ the numbers $Q(1), Q(2), Q(3), \dots$ each have a prime factor not dividing $n$.
16 replies
franzliszt
Oct 25, 2020
MathLuis
an hour ago
Predicted AMC 8 Scores
megahertz13   166
N an hour ago by sepehr2010
$\begin{tabular}{c|c|c|c}Username & Grade & AMC8 Score \\ \hline
megahertz13 & 5 & 23 \\
\end{tabular}$
166 replies
megahertz13
Jan 25, 2024
sepehr2010
an hour ago
Prove that the fraction (21n + 4)/(14n + 3) is irreducible
DPopov   110
N 2 hours ago by Shenhax
Source: IMO 1959 #1
Prove that the fraction $ \dfrac{21n + 4}{14n + 3}$ is irreducible for every natural number $ n$.
110 replies
DPopov
Oct 5, 2005
Shenhax
2 hours ago
Let \( a, b, c \) be positive real numbers satisfying \[ a^2 + c^2 = b(a + c). \
Jackson0423   3
N 2 hours ago by Mathzeus1024
Let \( a, b, c \) be positive real numbers satisfying
\[
a^2 + c^2 = b(a + c).
\]Let
\[
m = \min \left( \frac{a^2 + ab + b^2}{ab + bc + ca} \right).
\]Find the value of \( 2024m \).
3 replies
Jackson0423
Apr 16, 2025
Mathzeus1024
2 hours ago
real+ FE
pomodor_ap   3
N 2 hours ago by MathLuis
Source: Own, PDC001-P7
Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that
$$f(x)f(x^2 + y f(y)) = f(x)f(y^2) + x^3$$for all $x, y \in \mathbb{R}^+$. Determine all such functions $f$.
3 replies
pomodor_ap
Yesterday at 11:24 AM
MathLuis
2 hours ago
Inspired by hlminh
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that $$ |a-kb|+|b-kc|+|c-ka|\leq \sqrt{3k^2+2k+3}$$Where $ k\geq 0 . $
1 reply
sqing
2 hours ago
sqing
2 hours ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   77
N 3 hours ago by Ruegerbyrd
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 12th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AMC 12 level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!

Thank you to our lead sponsor, Jane Street!

IMAGE
77 replies
TennesseeMathTournament
Mar 9, 2025
Ruegerbyrd
3 hours ago
Is this FE solvable?
ItzsleepyXD   3
N 3 hours ago by jasperE3
Source: Original
Let $c_1,c_2 \in \mathbb{R^+}$. Find all $f : \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that for all $x,y \in \mathbb{R^+}$ $$f(x+c_1f(y))=f(x)+c_2f(y)$$
3 replies
ItzsleepyXD
Yesterday at 3:02 AM
jasperE3
3 hours ago
PQ bisects AC if <BCD=90^o, A, B,C,D concyclic
parmenides51   2
N 3 hours ago by venhancefan777
Source: Mathematics Regional Olympiad of Mexico Northeast 2020 P2
Let $A$, $B$, $C$ and $D$ be points on the same circumference with $\angle BCD=90^\circ$. Let $P$ and $Q$ be the projections of $A$ onto $BD$ and $CD$, respectively. Prove that $PQ$ cuts the segment $AC$ into equal parts.
2 replies
parmenides51
Sep 7, 2022
venhancefan777
3 hours ago
Inequality with three conditions
oVlad   3
N 3 hours ago by sqing
Source: Romania EGMO TST 2019 Day 1 P3
Let $a,b,c$ be non-negative real numbers such that \[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]Prove that $a^2+b^2+c^2\leqslant 2abc+1.$
3 replies
oVlad
Yesterday at 1:48 PM
sqing
3 hours ago
Moving P(o)in(t)s
bobthegod78   69
N Apr 2, 2025 by akliu
Source: USAJMO 2021/4
Carina has three pins, labeled $A, B$, and $C$, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance $1$ away. What is the least number of moves that Carina can make in order for triangle $ABC$ to have area 2021?

(A lattice point is a point $(x, y)$ in the coordinate plane where $x$ and $y$ are both integers, not necessarily positive.)
69 replies
bobthegod78
Apr 15, 2021
akliu
Apr 2, 2025
Moving P(o)in(t)s
G H J
G H BBookmark kLocked kLocked NReply
Source: USAJMO 2021/4
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IAmTheHazard
5001 posts
#59 • 1 Y
Y by channing421
exactly 2021 not at least 2021
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channing421
1353 posts
#60
Y by
sketch
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ihatemath123
3443 posts
#61
Y by
basic idea
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huashiliao2020
1292 posts
#62
Y by
kind of a boring problem, construction is just weird

The answer is 128. First note that the area of a triangle is at most half of the area of the box around it (with sides parallel to the axes), because if you place three of the vertices on the corners of the box, moving any of them decreases the area. Now note that in n moves, by AM-GM this box has area at most $(\tfrac{n}{2})^2$. Indeed, this implies that for a triangle with 2021 area, you need a box of area at least 4042, so n is at least $2\sqrt{4042}$, or 128 moves. This intuition can be checked also because 63x64/2 is not enough, while 64x64/2 is more than 2021.

As for the construction, WLOG A=(0,0); noticing that 2048 maximal area is not far from 2021, we still want our vertices close to 64,0 and 0,64 to see if it's possible, which can easily be guessed and checked by Shoelace to find B and C have coordinates 10,63 and 64,-1.
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Jndd
1416 posts
#63
Y by
First, to get from the origin to some point, say $(x,y)$, the minimum number of moves to get to there is $|x|+|y|$. One possible construction is $A=(-1, 54)$, $B=(63,0)$, $C=(0,-10)$, and we can verify the area is $2021$ using the Shoelace Theorem, and the number of moves is \[\lvert-1\rvert+|54|+|63|+|0|+|0|+\lvert-10\rvert=128.\]We claim that $128$ is in fact the minimum possible number of moves, and we'll prove it. First, create a rectangle $R$ with sides parallel to the axes such that it's the smallest rectangle that contains all of the points $A, B, C$, and clearly, one of $A$, $B$, and $C$ must be a vertex of $R$. WLOG, let $C$ be a vertex of $R$, $A$ be on a line parallel to the x-axis, and $B$ be on a line parallel to the y-axis. Then, draw a line $\ell_1$ parallel to the y-axis from $A$, draw a line $\ell_2$ parallel to the x-axis from $B$, and call their intersection $P$.

Now, we show it's optimal for us to start at $P$. Let $O$ be our starting point, so that it's a distance $d_1$ from $\ell_1$ and a distance $d_2$ from $\ell_2$. Notice that shifting $O$ onto line $\ell_1$ will decrease the number of moves to get from $O$ to the points $A$ and $B$ by $d_1$, while increasing the number of moves to get from $O$ to $C$ by $d_1$, if $O$ is to the right of $\ell_1$. Otherwise, if $O$ is to the left of $\ell_1$, it will decrease the number of moves to get from $O$ to the points $A$ and $C$ by $d_1$, while increasing the number of moves to get from $O$ to $B$ by $d_1$. Either way, the total decrease of the number of moves is $d_1+d_1-d_1=d_1$, which is good. Similarly, shifting $O$ onto $\ell_2$ decreases the number of total moves needed by $d_2$. So, once we shift $O$ onto both $\ell_1$ and $\ell_2$, we have $O=P$.

Then, we see that the total number of moves starting from $P$ is $w+\ell$ where $w$ is the width of $R$, and $\ell$ is the length (or height) of $R$. Now, we show that the maximum area of $ABC$ is half the area of $R$. Let $\ell_1$ intersect the bottom edge of $R$ at $P_1$ and let $\ell_2$ intersect the left edge of $R$ at $P_2$. Then, it's not hard to show that $[AP_1CB]=[ABP_2C]=[R]/2$, so since $[AP_1CB] = [ABC]+[AP_1C]$ and $[ABP_2C]=[ABC]+[BP_2C]$, meaning $[ABC]\leq [R]/2$, as desired.

Now, in the case of $A=(-1, 54)$, $B=(63,0)$, $C=(0,-10)$, we have $w=\ell=64$, which gives $[ABC]\leq 64^2/2 = 2048$, which has a possible construction since $2021<2048$. However, if we had $w+\ell < 128$, then the maximum possible value of $[R]/2$ would be $w\ell/2 = 63\cdot 64/2 = 2016 < 2021$, which implies there is no possible construction if $w+\ell < 128$, so we're done.
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peace09
5417 posts
#64 • 1 Y
Y by OronSH
Silly Bound
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asdf334
7586 posts
#65
Y by
oh no i have ptsd1
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joshualiu315
2513 posts
#66
Y by
Define the $\textit{bounding box}$ of $ABC$ to be the smallest axis-parallel rectangle that contains $ABC$.


Lemma: The area of $ABC$ is at most half the area of the bounding box.

Proof: Suppose the bounding box is $AXYZ$. If we fix $C$ and only vary $B$ around the perimeter of the rectangle, the maximal area must be achieved when $B$ coincides with one of the corners of the rectangles.

Manually checking each case, we find that the area of $ABC$ is at most half the area of $AXYZ$. $\square$


We proceed by making a claim.


Claim: After $n$ moves, the bounding box has area at most $\frac{n^2}{4}$.

Proof: The sum of the width and height of the bounding box increases by at most $1$ each move. Hence, letting the length equal $l$ and the width $w$, we obtain $l+w=n$, at which point the result follows by AM-GM. $\square$


The claim implies a bounding box of area $A$ requires at least $\lceil 2 \sqrt{A} \rceil$ moves, implying that $n \ge 128$.

Now, we must find a construction for $n=128$. The following construction works:

\begin{align*}
    A &= (-3,-18) \\
    B&= (61,0) \\
    C &= (0,46)
\end{align*}
so the answer is $n=128$. $\square$
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shendrew7
794 posts
#67
Y by
The following two results can be easily proven with some unfortunately bashy case analysis:
  • The area of a triangle is at most half of the area its "bounding rectangle", or the smallest rectangle with axes-parallel sides containing the triangle.
  • The sum of the taxicab distances from a point to the three vertices of a triangle is at least the sum of the dimensions of its "bounding rectangle". We can set this optimal point to be our origin.

Note our the area of our bounding rectangle is at least $2 \cdot 2021 = 4042$, from which AM-GM gives us a bound of $\lceil 2\sqrt{2024} \rceil = \boxed{128}$. A construction of this case uses the points $(0,55)$, $(58,0)$, and $(-6,-9)$ with bounding rectangle of dimensions $64 \times 64$. $\blacksquare$
This post has been edited 1 time. Last edited by shendrew7, Dec 20, 2023, 11:28 PM
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EpicBird08
1747 posts
#68
Y by
When mocking this JMO I thought the maximum area was $133$ :noo: This time around I had to look at hints from ARCH :skull:

The answer is $128.$

First, define the surrounding rectangle of a triangle as the smallest rectangle with sides parallel to the axes that contains a triangle. It is pretty easy to see (e.g. by an area bash) that the area of the surrounding rectangle is at least twice as large as the area of the triangle. Every move, the sidelength of exactly one side increases by $1$. Thus the area of the surrounding rectangle after every move is at most equal to $\frac{n^2}{2},$ so the area of the triangle is at most equal to $\frac{n^2}{4}.$ This must be greater than or equal to $2021,$ so $n \ge 128.$

A construction for $n = 128$ is $(-2,-27), (0, 62), (37, 0).$
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Markas
105 posts
#70
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Denote the surrounding rectangle of a triangle as the smallest rectangle with sides parallel to the axes that contains the triangle. The area of the triangle is at most half of the area of the surrounding rectangle, because if you place three of the vertices on the corners of the rectangle, moving any of them decreases the area. The sidelength of exactly one side increases by 1 $\Rightarrow$ the area of the surrounding rectangle after all n moves is at most $(\frac{n}{2})^2 = \frac{n^2}{4}$ $\Rightarrow$ the area of the triangle is at most $\frac{n^2}{8}$. But $\frac{n^2}{8} \geq 2021$, so $n \ge 128$.

A construction for n = 128 is A = (0,0); B = (1,64); C = (64,54).
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Siddharthmaybe
106 posts
#71
Y by
coolmath2017 wrote:
It's not 133 :(

lol I was so excited to see the answer and this the first thing that pops up :(
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happypi31415
742 posts
#72 • 1 Y
Y by peace09
We claim that the answer is $\boxed{128}$. This can be achieved by taking the points $(-9,-6), (0.58),$ and $(55,0)$. (This construction can be found by just assuming that two points are always on the axes and playing around until the construction is found.)

For the bound, it is easy to see that if we let two points, starting from the origin, move $n$ units in total, then the rectangle whose diagonal is formed by these two points will be at least twice the triangles area, which implies that the triangle has area at most $2n^2$ by say, AM-GM. The minimum positive integer $n$ that satisfies this is $n=128$ so we're done.
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gladIasked
648 posts
#73
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The answer is $\boxed{128}$ moves, achieved with $(0, 58)$, $(-55, 0)$, and $(9, -6)$.

Define the bounding box of $\triangle ABC$ as the smallest rectangle with sides parallel to the axes containing all three vertices. I claim that a triangle's area is at most twice the area of its bounding box. We have two cases; the case where one of the sides of the triangle coincides with a side of the bounding box is trivial to see. The other case is not much harder:
[asy]
     /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(5cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -5.320243582333011, xmax = 13.300111666175955, ymin = -2.57973075125075, ymax = 6.877837547794034;  /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0); 

draw((0,0)--(0,3)--(4,3)--(4,0)--cycle, linewidth(1) + zzttqq); 
draw((4,0)--(0.9561426524876215,3.02725616889723)--(0,1.2831331838785818)--cycle, linewidth(1) + zzttqq); 
 /* draw figures */
draw((0,0)--(0,3), linewidth(1) + zzttqq); 
draw((0,3)--(4,3), linewidth(1) + zzttqq); 
draw((4,3)--(4,0), linewidth(1) + zzttqq); 
draw((4,0)--(0,0), linewidth(1) + zzttqq); 
draw((4,0)--(0.9561426524876215,3.02725616889723), linewidth(1) + zzttqq); 
draw((0.9561426524876215,3.02725616889723)--(0,1.2831331838785818), linewidth(1) + zzttqq); 
draw((0,1.2831331838785818)--(4,0), linewidth(1) + zzttqq); 
 /* dots and labels */
dot((0,0),linewidth(4pt) + dotstyle); 
label("$F$", (0.047233491280721025,0.09786650743985113), NE * labelscalefactor); 
dot((0,3),dotstyle); 
label("$E$", (0.047233491280721025,3.1193753406411715), NE * labelscalefactor); 
dot((4,3),dotstyle); 
label("$D$", (4.0513468230841,3.1193753406411715), NE * labelscalefactor); 
dot((4,0),dotstyle); 
label("$A$", (4.0513468230841,0.12243161990490252), NE * labelscalefactor); 
dot((0.9561426524876215,3.02725616889723),dotstyle); 
label("$B$", (1.0052728774177258,3.1562230093387487), NE * labelscalefactor); 
dot((0,1.2831331838785818),dotstyle); 
label("$C$", (-0.2,1.15), W * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
Let $DE=b$, $AD=a$, $BE=x$, and $CE=y$. We have $[ABC]=ab-\left(\frac{xy+a(b-x)+b(a-y)}{2}\right)=\frac{ax+by-xy}{2}=\frac{ab}2-\frac{(a-y)(b-x)}{2}<\frac{ab}2$, as desired.

Note that a move increases one of the dimensions of the bounding box by exactly $1$ unit. Therefore, we have a lower bound of $\boxed{128}$ moves, which can produce a bouncing box with dimensions $64\times 64$ (less moves will produce a smaller bounding box).
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akliu
1795 posts
#74
Y by
Consider the rectangular "bounding box" with perimeter $P$ of triangle $ABC$ bounded by lines $y = y_{min}$, $y = y_{max}$, $x = x_{min}$, and $x = x_{max}$, where $x_{min}, x_{max}$ are the smallest and largest $x$ coordinates out of pins $A$, $B$, and $C$, and $y_{min}$ and $y_{max}$ are defined similarly with respect to $y$ coordinates.

It can be proven by Shoelace that $[ABC] \leq \frac{1}{2} R$, where $R$ is the area of the rectangular bounding box. Since the rectangular bounding box's area is at most $\frac{P^2}{16}$, we have a bound of $[ABC] \leq \frac{P^2}{32}$. Since $[ABC] = 2021$, this gives us $P \geq 256$, since $P$ must be even.

Every move that Carina makes can increase the value of $P$ by at most $2$, meaning that we have a minimum of $128$ moves. A configuration that works would be $A = (5, 1), B = (-52, 0), C = (0, -70)$.
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