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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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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 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
Update on Basic Forum Rules
What belongs on this forum?
How do I write a thorough solution?
How do I get a problem on the contest page?
How do I study for mathcounts?
Mathcounts FAQ and resources
Mathcounts and how to learn

As always, if you have any questions, you can PM me or any of the other Middle School Moderators. Once again, if you see spam, it would help a lot if you filed a report instead of responding :)

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
Inequality with a,b,c,d
GeoMorocco   5
N a few seconds ago by GeoMorocco
Source: Moroccan Training 2025
Let $ a,b,c,d$ positive real numbers such that $ a+b+c+d=3+\frac{1}{abcd}$ . Prove that :
$$ a^2+b^2+c^2+d^2+5abcd \geq 9 $$
5 replies
GeoMorocco
Apr 9, 2025
GeoMorocco
a few seconds ago
number theory
Levieee   4
N 2 minutes ago by Safal
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).
4 replies
Levieee
2 hours ago
Safal
2 minutes ago
Sequence and prime factors
USJL   7
N 28 minutes ago by MathLuis
Source: 2025 Taiwan TST Round 2 Independent Study 1-N
Let $a_0,a_1,\ldots$ be a sequence of positive integers with $a_0=1$, $a_1=2$ and
\[a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1\]for all $n\geq 2$. Show that if $p$ is a prime less than $2^k$ for some positive integer $k$, then there exists $n\leq k+1$ such that $p\mid a_n$.
7 replies
USJL
Mar 26, 2025
MathLuis
28 minutes ago
powers sums and triangular numbers
gaussious   4
N 42 minutes ago by kiyoras_2001
prove 1^k+2^k+3^k + \cdots + n^k \text{is divisible by } \frac{n(n+1)}{2} \text{when} k \text{is odd}
4 replies
gaussious
Yesterday at 1:00 PM
kiyoras_2001
42 minutes ago
0!??????
wizwilzo   50
N 3 hours ago by wipid98
why is 0! "1" ??!
50 replies
wizwilzo
Jul 6, 2016
wipid98
3 hours ago
Bogus Proof Marathon
pifinity   7603
N 3 hours ago by MathWinner121
Hi!
I'd like to introduce the Bogus Proof Marathon.

In this marathon, simply post a bogus proof that is middle-school level and the next person will find the error. You don't have to post the real solution :P

Use classic Marathon format:
[hide=P#]a1b2c3[/hide]
[hide=S#]a1b2c3[/hide]


Example posts:

P(x)
-----
S(x)
P(x+1)
-----
Let's go!! Just don't make it too hard!
7603 replies
pifinity
Mar 12, 2018
MathWinner121
3 hours ago
Weird Similarity
mithu542   0
3 hours ago
Is it just me or are the 2023 national sprint #21 and 2025 state target #4 strangely similar?
[quote=2023 Natioinal Sprint #21] A right triangle with integer side lengths has perimeter $N$ feet and area $N$ ft^2. What is the arithmetic mean of all possible values of $N$?[/quote]
[quote=2025 State Target #4]Suppose a right triangle has an area of 20 cm^2 and a perimeter of 40 cm. What is
the length of the hypotenuse, in centimeters?[/quote]
0 replies
mithu542
3 hours ago
0 replies
An algebra math problem
AVY2024   6
N 3 hours ago by Roger.Moore
Solve for a,b
ax-2b=5bx-3a
6 replies
AVY2024
Apr 8, 2025
Roger.Moore
3 hours ago
easy olympiad problem
kjhgyuio   5
N 3 hours ago by Roger.Moore
Find all positive integer values of \( x \) such that
\[
\sqrt{x - 2011} + \sqrt{2011 - x} + 10
\]is an integer.
5 replies
kjhgyuio
Yesterday at 2:00 PM
Roger.Moore
3 hours ago
Mathcounts Nationals Roommate Search
iwillregretthisnamelater   37
N 3 hours ago by MathWinner121
Does anybody want to be my roommate at nats? Every other qualifier in my state is female. :sob:
Respond quick pls i gotta submit it in like a couple of hours.
37 replies
iwillregretthisnamelater
Mar 31, 2025
MathWinner121
3 hours ago
EaZ_Shadow
Apr 6, 2025
derekwang2048
4 hours ago
Math and AI 4 Girls
mkwhe   20
N 5 hours ago by fishchips
Hey everyone!

The 2025 MA4G competition is now open!

Apply Here: https://xmathandai4girls.submittable.com/submit


Visit https://www.mathandai4girls.org/ to get started!

Feel free to PM or email mathandai4girls@yahoo.com if you have any questions!
20 replies
mkwhe
Apr 5, 2025
fishchips
5 hours ago
k real math problems
Soupboy0   60
N Today at 2:12 PM by Soupboy0
Ill be posting questions once in a while. Here's the first question:

What fraction of numbers from $1$ to $1000$ have the digit $7$ and are divisible by $3$?
60 replies
Soupboy0
Mar 25, 2025
Soupboy0
Today at 2:12 PM
Website to learn math
hawa   29
N Today at 5:50 AM by Andrew2019
Hi, I'm kinda curious what website do yall use to learn math, like i dont find any website thats fun to learn math
29 replies
hawa
Apr 9, 2025
Andrew2019
Today at 5:50 AM
Parallel
Fang-jh   10
N Aug 23, 2023 by starchan
Source: All Russian Mathematical Olympiad 2008. 10.6
In a scalene triangle $ ABC$ the altitudes $ AA_{1}$ and $ CC_{1}$ intersect at $ H, O$ is the circumcenter, and $ B_{0}$ the midpoint of side $ AC$. The line $ BO$ intersects side $ AC$ at $ P$, while the lines $ BH$ and $ A_{1}C_{1}$ meet at $ Q$. Prove that the lines $ HB_{0}$ and $ PQ$ are parallel.
10 replies
Fang-jh
Jun 13, 2008
starchan
Aug 23, 2023
Source: All Russian Mathematical Olympiad 2008. 10.6
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Fang-jh
237 posts
#1 • 2 Y
Y by Adventure10, Mango247
In a scalene triangle $ ABC$ the altitudes $ AA_{1}$ and $ CC_{1}$ intersect at $ H, O$ is the circumcenter, and $ B_{0}$ the midpoint of side $ AC$. The line $ BO$ intersects side $ AC$ at $ P$, while the lines $ BH$ and $ A_{1}C_{1}$ meet at $ Q$. Prove that the lines $ HB_{0}$ and $ PQ$ are parallel.
Z K Y
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pohoatza
1145 posts
#2 • 2 Y
Y by Adventure10, Mango247
Since the reflection $ D$ of $ H$ in $ B_{0}$ is the second intersection of $ BO$ with the circumcircle, the lines $ HB_{0}$ and $ PQ$ are parallel if and only if the pencil $ (PQ, PH, PB_{0}, PD)$ is harmonic. The conclusion follows from the fact that $ BH \cap (PQ, PH, PB_{0}, PD) = (B, Q, H, B_{1}) = - 1$, where $ B_{1}$ is the foot of the $ B$-altitude in triangle $ ABC$.
This post has been edited 1 time. Last edited by pohoatza, Jun 13, 2008, 7:07 PM
Z K Y
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28121941
152 posts
#3 • 2 Y
Y by Adventure10, Mango247
Just a question to clarify the brilliant solution of Pohoatza: is point D the second intersection of BO with the circumcircle? Thanks
Z K Y
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pohoatza
1145 posts
#4 • 1 Y
Y by Adventure10
Indeed. I've edited now. Thanks.
Z K Y
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28121941
152 posts
#5 • 2 Y
Y by Adventure10, Mango247
My solution is metric. If B_1 is the foot of the altitude from B, we can consider triangles B_1B_0H and B_1PQ. Then it suffices to prove the equality B_1Q/B_1H = B_1P/B_1B_0 . Both quotients are equal to 2sinAsinC/cos(A-C).
Z K Y
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Sailor
256 posts
#6 • 2 Y
Y by Adventure10, Mango247
If $ \{T\}=HB_0\cap BO$ then $ \frac{BH}{BT}=\frac{2R\cos{B}}{2R}=\cos{B}$.
In addition, $ \triangle{BC_1Q}\sim \triangle{BCP}$, thus $ \frac{BQ}{BP}=\frac{BC_1}{BC}=\cos{B}$.
Thus, $ \frac{BH}{BT}=\frac{BQ}{BP} \Longleftrightarrow PQ||HB_0$.
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windrock
46 posts
#7 • 1 Y
Y by Adventure10
Construct the diameter $ BOT$. In fact that the quadrilateral $ AHCT$ is a parrallelogram. We have :
$ \frac{BP}{PH}=\frac{[BMN]}{[HMN]}=\frac{BM.BN}{HM.HN}$
$ \frac{BQ}{BT}=\frac{[BAC]}{[TAC]}=\frac{BA.BC}{HA.HC}$
Easy to proof that $ \frac{BM.BN}{HM.HN}=\frac{BA.BC}{HA.HC}$
Hence, by Thales theorem we get that $ PQ // HT  \Rightarrow PQ // HB_{0}$
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sunken rock
4384 posts
#8 • 3 Y
Y by Pluto1708, Adventure10, Mango247
Let BT be a diameter of (ABC); it’s easy to see that $ \triangle BAT \sim  \triangle BA_{1}H$ and P and Q are homologues points, hence the conclusion follows.

Best regards,
sunken rock
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AlastorMoody
2125 posts
#9 • 2 Y
Y by Adventure10, Mango247
Quote:
In a scalene triangle $ ABC$ the altitudes $ AA_{1}$ and $ CC_{1}$ intersect at $ H, O$ is the circumcenter, and $ B_{0}$ the midpoint of side $ AC$. The line $ BO$ intersects side $ AC$ at $ P$, while the lines $ BH$ and $ A_{1}C_{1}$ meet at $ Q$. Prove that the lines $ HB_{0}$ and $ PQ$ are parallel.
Solution: Let $BP \cap HB_0=X$. Let $\ell || HB_0$ through $P$ intersects $BB_1$ at $Y$
$$-1=(H, ~ X ; ~B_0, ~ \infty_{HX}) \overset{P}{=} (H, ~ B ; ~ B_1, ~Y)$$But, $$-1=(A , ~ C ; ~ A_1C_1 ~ \cap ~ AC, ~B_1)  \overset{B}{=} (C_1, ~ A_1 ; ~A_1C_1 ~ \cap ~ AC , ~ Q)  \overset{C}{=}   (H , ~ B ; ~B_1, ~Q)$$Hence, $Y \equiv Q$
This post has been edited 5 times. Last edited by AlastorMoody, Jun 23, 2019, 8:02 PM
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HamstPan38825
8857 posts
#11
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Just compute everything!

Let $D$ be the foot of the $A$-altitude, and let $A'$ be the $A$-antipode; it suffices to show $\frac{AQ}{AP} = \frac{AH}{AA'}$. Notice that $\angle AQB_1 = 90^\circ - B + C$, so $$\frac{AQ}{AB_1} = \frac{\sin B}{\cos(B-C)} \implies AQ = \frac{2R \cos A \sin B\sin C}{\cos(B-C)}.$$Also, $$AP = \frac{AD}{\cos(B-C)} = \frac{2R\sin B\sin C}{\cos(B-C)}.$$It is evident now that $\frac{AQ}{AP} = \cos A = \frac{AH}{AA'}$, as needed.

Remark. It turns out that $\overline{AQ}$ and $\overline{AP}$ are similar parts in the inverse similarity $\triangle AB_1C_1 \sim \triangle ABC$, which explains a lot.
This post has been edited 1 time. Last edited by HamstPan38825, Jul 18, 2023, 8:02 AM
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starchan
1603 posts
#12
Y by
huh
solution
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