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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
k i A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
k i Marathon Threads
LauraZed   0
Jul 2, 2019
Due to excessive spam and inappropriate posts, we have locked the Prealgebra and Beginning Algebra threads.

We will either unlock these threads once we've cleaned them up or start new ones, but for now, do not start new marathon threads for these subjects. Any new marathon threads started while this announcement is up will be immediately deleted.
0 replies
LauraZed
Jul 2, 2019
0 replies
k i Basic Forum Rules and Info (Read before posting)
jellymoop   368
N May 16, 2018 by harry1234
f (Reminder: Do not post Alcumus or class homework questions on this forum. Instructions below.) f
Welcome to the Middle School Math Forum! Please take a moment to familiarize yourself with the rules.

Overview:
[list]
[*] When you're posting a new topic with a math problem, give the topic a detailed title that includes the subject of the problem (not just "easy problem" or "nice problem")
[*] Stay on topic and be courteous.
[*] Hide solutions!
[*] If you see an inappropriate post in this forum, simply report the post and a moderator will deal with it. Don't make your own post telling people they're not following the rules - that usually just makes the issue worse.
[*] When you post a question that you need help solving, post what you've attempted so far and not just the question. We are here to learn from each other, not to do your homework. :P
[*] Avoid making posts just to thank someone - you can use the upvote function instead
[*] Don't make a new reply just to repeat yourself or comment on the quality of others' posts; instead, post when you have a new insight or question. You can also edit your post if it's the most recent and you want to add more information.
[*] Avoid bumping old posts.
[*] Use GameBot to post alcumus questions.
[*] If you need general MATHCOUNTS/math competition advice, check out the threads below.
[*] Don't post other users' real names.
[*] Advertisements are not allowed. You can advertise your forum on your profile with a link, on your blog, and on user-created forums that permit forum advertisements.
[/list]

Here are links to more detailed versions of the rules. These are from the older forums, so you can overlook "Classroom math/Competition math only" instructions.
Posting Guidelines
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What belongs on this forum?
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Mathcounts and how to learn

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Marathons!
Relays might be a better way to describe it, but these threads definitely go the distance! One person starts off by posting a problem, and the next person comes up with a solution and a new problem for another user to solve. Here's some of the frequently active marathons running in this forum:
[list][*]Algebra
[*]Prealgebra
[*]Proofs
[*]Factoring
[*]Geometry
[*]Counting & Probability
[*]Number Theory[/list]
Some of these haven't received attention in a while, but these are the main ones for their respective subjects. Rather than starting a new marathon, please give the existing ones a shot first.

You can also view marathons via the Marathon tag.

Think this list is incomplete or needs changes? Let the mods know and we'll take a look.
368 replies
jellymoop
May 8, 2015
harry1234
May 16, 2018
Interesting inequalities
sqing   0
4 minutes ago
Source: Own
Let $ a,b> 0 ,   a+b+a^2+b^2=2.$ Prove that
$$ab+ \frac{k}{a+b+ab} \geq \frac{3-k+(k-1)\sqrt{5}}{2}$$Where $ k\geq 2. $
$$ab+ \frac{2}{a+b+ab} \geq \frac{1+\sqrt{5}}{2}$$$$ab+ \frac{3}{a+b+ab} \geq  \sqrt{5} $$
0 replies
sqing
4 minutes ago
0 replies
Stability of Additive Cauchy Equation
doanquangdang   1
N 2 hours ago by jasperE3
Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies
$$
|f(x+y)-f(x)-f(y)-x y| \leq \varepsilon\left(|x|^p+|y|^p\right)
$$for some $\varepsilon>0,$ $p \in[0,1)$ and for all $x, y \in \mathbb{R}$, then there exists a unique solution $a: \mathbb{R} \rightarrow \mathbb{R}$ of the functional equation $a(x+y)=$ $a(x)+a(y)$ for all $x, y \in \mathbb{R}$ such that
$$
\left|f(x)-a(x)-\frac{1}{2} x^2\right| \leq \frac{2}{2-2^p} \varepsilon|x|^p
$$for all $x \in \mathbb{R}$.
1 reply
doanquangdang
Aug 16, 2024
jasperE3
2 hours ago
Summer math contest prep
Abby0618   6
N 2 hours ago by A7456321
School is almost out, so I have a lot of time in the summer. I want to be able to make DHR on AMC 8 in 7th grade

(current 6th grader) and hopefully get an average score in AMC 10. What should I do during the summer to achieve

these goals? For context, I have many books from AOPS, have already taken the Intro to Algebra A course, and took

AMC 8 for the first time as a 6th grader. If there are any challenging math problems you think would benefit learning,

please post them here. Thank you! :-D
6 replies
Abby0618
Yesterday at 8:52 PM
A7456321
2 hours ago
Polynomials with common roots and coefficients
VicKmath7   10
N 2 hours ago by math-olympiad-clown
Source: Balkan MO SL 2020 A3
Let $P(x), Q(x)$ be distinct polynomials of degree $2020$ with non-zero coefficients. Suppose that they have $r$ common real roots counting multiplicity and $s$ common coefficients. Determine the maximum possible value of $r + s$.

Demetres Christofides, Cyprus
10 replies
+1 w
VicKmath7
Sep 9, 2021
math-olympiad-clown
2 hours ago
Serbian selection contest for the IMO 2025 - P3
OgnjenTesic   1
N 2 hours ago by korncrazy
Source: Serbian selection contest for the IMO 2025
Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that:
- $f$ is strictly increasing,
- there exists $M \in \mathbb{N}$ such that $f(x+1) - f(x) < M$ for all $x \in \mathbb{N}$,
- for every $x \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$ such that
\[
            f(y) = \frac{f(x) + f(x + 2024)}{2}.
        \]Proposed by Pavle Martinović
1 reply
OgnjenTesic
Yesterday at 4:06 PM
korncrazy
2 hours ago
The daily problem!
Leeoz   192
N 3 hours ago by valenbb
Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! :)

Please hide solutions and answers, hints are fine though! :)

Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!

Past Problems!
192 replies
Leeoz
Mar 21, 2025
valenbb
3 hours ago
max number of candies
orangefronted   30
N 5 hours ago by Creeperboat
A store sells a strawberry flavoured candy for 1 dollar each. The store offers a promo where every 4 candy wrappers can be exchanged for one candy. If there is no limit to how many times you can exchange candy wrappers for candies, what is the maximum number of candies I can obtain with 100 dollars?
30 replies
orangefronted
Apr 3, 2025
Creeperboat
5 hours ago
Help to make it clear on basic Concept
Miranda2829   2
N 6 hours ago by UberPiggy
Where I use extraneous solution in equation?
Why square of both side we will have plus and minus answer?
2 replies
Miranda2829
Yesterday at 8:32 PM
UberPiggy
6 hours ago
If $x = (4096)^{7+4\sqrt{3}},$ then which of the following equals $64:$ (A) $\fr
Vulch   4
N 6 hours ago by pieMax2713
If $x = (4096)^{7+4\sqrt{3}},$ then which of the following equals $64:$
(A) $\frac{x^{\frac{7}{2}}}{x^{\frac{4}{\sqrt{3}}}}$
(B) $\frac{x^{7}}{x^{4\sqrt{3}}}$
(C) $\frac{x^{\frac72}}{x^{2\sqrt{3}}}$
(D) $\frac{x^{7}}{ x^{2\sqrt{3}}}$
4 replies
Vulch
Yesterday at 9:27 PM
pieMax2713
6 hours ago
Worst math problems
LXC007   6
N 6 hours ago by A7456321
What is the most egregiously bad problem or solution you have encountered in school?
6 replies
LXC007
Wednesday at 11:42 PM
A7456321
6 hours ago
Quick Question
b2025tyx   13
N Yesterday at 8:44 PM by b2025tyx
During my math final today at school, the question said stated "When every integer is raised to the power of zero, it is equal to 1". The answers were multiple choice and were : Always, sometimes, never, and I don't know.

I ended up putting the first one, and was informed that it was incorrect. My teacher told me that $0^0$ is not equal to one. I looked it up, and it said $0^0 = 1$. Can someone confirm and prove this. Thanks!
13 replies
b2025tyx
May 20, 2025
b2025tyx
Yesterday at 8:44 PM
AP calc?
Thayaden   22
N Yesterday at 7:53 PM by Inaaya
How are we all feeling on AP calc guys?
22 replies
Thayaden
May 20, 2025
Inaaya
Yesterday at 7:53 PM
What's the chance that two AoPS accounts generate with the same icon?
Math-lover1   19
N Yesterday at 7:30 PM by ZMB038
So I've been wondering how many possible "icons" can be generated when you first create an account. By "icon" I mean the stack of cubes as the first profile picture before changing it.

I don't know a lot about how AoPS icons generate, so I have a few questions:
- Do the colors on AoPS icons generate through a preset of colors or the RGB (red, green, blue in hexadecimal form) scale? If it generates through the RGB scale, then there may be greater than $256^3 = 16777216$ different icons.
- Do the arrangements of the stacks of blocks in the icon change with each account? If so, I think we can calculate this through considering each stack of blocks independently.
19 replies
Math-lover1
May 2, 2025
ZMB038
Yesterday at 7:30 PM
A Variety of Math Problems to solve
FJH07   41
N Yesterday at 7:22 PM by FJH07
Hi, so people can post different math problems that they think are hard, and I will post some (I think middle school math level) problems so that the community can help solve them. :)
41 replies
FJH07
Yesterday at 1:36 AM
FJH07
Yesterday at 7:22 PM
Points and triangles in the plane
pablock   6
N Feb 23, 2025 by awesomeming327.
Source: Brazilian Mathematical Olympiad 2018 - Q6
Consider $4n$ points in the plane, with no three points collinear. Using these points as vertices, we form $\binom{4n}{3}$ triangles. Show that there exists a point $X$ of the plane that belongs to the interior of at least $2n^3$ of these triangles.
6 replies
pablock
Nov 16, 2018
awesomeming327.
Feb 23, 2025
Points and triangles in the plane
G H J
Source: Brazilian Mathematical Olympiad 2018 - Q6
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pablock
168 posts
#1 • 10 Y
Y by UK2019Project, anantmudgal09, rcorreaa, rkm0959, lminsl, nguyendangkhoa17112003, Aarth, kiyoras_2001, Adventure10, Mango247
Consider $4n$ points in the plane, with no three points collinear. Using these points as vertices, we form $\binom{4n}{3}$ triangles. Show that there exists a point $X$ of the plane that belongs to the interior of at least $2n^3$ of these triangles.
Z K Y
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MellowMelon
5850 posts
#2 • 13 Y
Y by rkm0959, stroller, DanDumitrescu, ZeusDM, nguyendangkhoa17112003, Aarth, kiyoras_2001, lminsl, yayups, edfearay123, wateringanddrowned, CyclicISLscelesTrapezoid, Adventure10
Let $S$ be the set of $4n$ points.

1. Let $T$ be the set of all lines through the origin. For each $\ell \in T$, let $f(\ell)$ be a line parallel to $\ell$ that does not pass through any point in $S$ and has $2n$ points of $S$ on each side. Such an $f(\ell)$ exists for each $\ell$ since we can move the line continuously in the plane and have the number of points on each side of the line vary continuously in the discrete sense.

2. Fix an $\ell \in T$. Let $m$ be a variable line in $T$ that starts as $\ell$ and rotates clockwise continuously until it becomes $\ell$ again. At any time, $f(\ell)$, $f(m)$ split the plane into four sections, and the number of points in each section varies continuously in the discrete sense. Initially the number of points in each section is $0, 2n, 0, 2n$, and at the end it is $2n, 0, 2n, 0$. This means there exists some line $m'$ such that $f(\ell)$ and $f(m')$ split the plane into four sections each with exactly $n$ points.

3. Choose $X$ to be the intersection of $f(\ell)$ and $f(m')$. Consider each of the $n^4$ ways to choose one point from each of the four sections formed by $f(\ell)$ and $f(m')$. These four points form four triangles. Exactly two of the four triangles contain $X$. This method of counting will obtain each triangle exactly $n$ times. This gives a total of $n^4 \cdot 2 / n = 2n^3$ triangles containing $X$.
Z K Y
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Aarth
137 posts
#3 • 2 Y
Y by Adventure10, Mango247
MellowMelon wrote:

2. Fix an $\ell \in T$. Let $m$ be a variable line in $T$ that starts as $\ell$ and rotates clockwise continuously until it becomes $\ell$ again. At any time, $f(\ell)$, $f(m)$ split the plane into four sections, and the number of points in each section varies continuously in the discrete sense. Initially the number of points in each section is $0, 2n, 0, 2n$, and at the end it is $2n, 0, 2n, 0$. This means there exists some line $m'$ such that $f(\ell)$ and $f(m')$ split the plane into four sections each with exactly $n$ points.

Why does the number of points in each section vary continuously?
This post has been edited 1 time. Last edited by Aarth, Nov 26, 2019, 3:51 PM
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Letteer
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#4 • 2 Y
Y by wateringanddrowned, Adventure10
Solution. part
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Eyed
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#5
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I claim there exists two lines, $\ell_{1}, \ell_{2}$, such that the $4$ regions divided by $\ell_1, \ell_2$ each contains $n$ points. Define $S$ as the set of the $4n$ points.

Let $\ell_{1}$ be an arbitrary balancing line, so each side of $\ell_{1}$ contains $2n$ points. Then, consider a point $P\in \ell_{1}$, and consider the line $\ell$ going through $P$ that will have $n$ points on its left side, and $n$ points on the right side of $\ell_{1}$. If we put $P$ at the very beginning of line $\ell_{1}$, then the amount of points on the left side of $\ell_{1}$ and the left side of $\ell$ is $2n$. However, moving $P$ to the very end of line $\ell_{1}$ gives $0$ points on the lefts ide of $\ell_{1}, \ell$. Thus, since by moving $P$, at each step we change at most $1$ in the number of points on the left side of $\ell, \ell_1$, this means at some point, there are $n$ points on the left side of $\ell, \ell_{1}$. Letting $\ell_2 = \ell$ gives the desired result.

Now, if $O = \ell_1\cap \ell_2$, we can choose $O$ such that it does not lie on any of the lines joining $2$ points in $S$. Consider the four regions $A,B,C,D$, each with $n$ points, and in clockwise order (so $A,C$ are opposite, same with $B,D$). Now, for each $n^2$ pairs of points in $A,C$, if $E\in A, F\in C$, then either for any point $X\in B$, we have $O\in \triangle EFX$, or for any point in $Y\in D$ we have $O\in \triangle EFY$ (this is because either $O$ lies above or below line $EF$. Therefore, there are $n^3$ triangles that contain $O$ with two vertices in $A,C$. Similarly, there are $n^3$ triangles that contain $O$ with two vertices in $B,D$, so we conclude that there are at least $2n^3$ triangles that $O$ is a part of.
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Leo.Euler
577 posts
#6
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Let $S$ be the set of $4n$ points in the plane.

Claim: There exist lines $\ell_1$ and $\ell_2$ that divide the plane into $4$ regions with $n$ points contained in each region and no points on either line.
Proof. Consider all of the lines through any two points of $S$, and take $\ell$ to be a line parallel to none of these lines. Realize that if we first take it to be arbitrarily far away from the points in $S$, and then translate it in the direction closer to the points, then the number of points on the side that initially has no points starts with $0$ and increases by exactly $1$ whenever it increases. Since it eventually reaches $4n$ points, we can move it to contain exactly $2n$ points on both sides. Call this line $\ell_1$.

Now, color the points on the left side of $\ell_1$ blue and the points on the right side of the line red. Let $P_0$ be a point of minimal distance to $\ell_1$. Then slightly rotate $\ell_1$ so that no points other than $P_0$ lie on the line or switch sides, and $P_0$ lies through the rotated line. Consider the following windmill process. First, rotate $P_0$ clockwise until it passes through another point of different color, $P_1$. Then for the $i$th iteration of the process, when $\ell_1$ is $\overline{P_iP_{i+1}}$, rotate it clockwise until it hits a new point $P_{i+2}$ with color different from $P_{i+1}$, and then rotate $\ell_1$ about $P_{i+1}$ so that it becomes $\overline{P_{i+1}P_{i+2}}$. Eventually, we can get the line to be in a position in which it passes through a red point and a blue point, dividing the plane into quadrants with $(n-1, n, n-1, n)$ points in them. Let this line intersect the original $\ell_1$ at point $X$. Then rotate the line by a sufficiently small angular displacement about $X$ so that it passes through no point in $S$. Call the resulting line $\ell_2$. Then $\ell_1$ and $\ell_2$ divide the plane into quadrants with $n$ points in each quadrant, as claimed
:yoda:

Consider the two lines as in the above claim, and let their intersection be $X$. Color the points in each quarter-plane red, blue, green, and yellow, in the clockwise order. I claim that the size $|T|$ of the set $T$ of triangles with all vertices of distinct colors containing the point $X$ in their interior is exactly $2n^3$. Let $Q$ be the set of quadrilaterals with all vertices having distinct colors. Then for each quadrilateral $q \in Q$, draw its diagonals and consider the $4$ subtriangles it is split into. Since $X$ lies in exactly one of those triangles, there are exactly $2$ triangles whose vertices are a subset of those of $Q$ that contain $X$ in its interior. Thus, $n|T|=2|Q|=2n^4$, so $|T|=2n^3$ as desired.
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awesomeming327.
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#7
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Another way to think about the problem after finding the first balancing line (assumed to be y=0) is to take a variable shear causing the median of the x-values on either side of the line to shift towards each other continuous, until we have n points in each quadrant.

Since isometries and shears are affine transforms, which preserve area ratios, the statement “X lies in ABC” is preserved, so we can make any translation, rotation, or shear to the original configuration that we want.
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