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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 3:18 PM
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.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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 events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Yesterday at 3:18 PM
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?
How do I write a thorough solution?
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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
Something nice
KhuongTrang   25
N 4 minutes ago by KhuongTrang
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
25 replies
1 viewing
KhuongTrang
Nov 1, 2023
KhuongTrang
4 minutes ago
Inequality from China
sqing   1
N 33 minutes ago by SunnyEvan
Source: lemondian(https://kuing.cjhb.site/thread-13667-1-1.html)
Let $x\in (0,\frac{\pi}{2}) . $ Prove that $$tanx\ge x^k$$Where $ k=1,2,3,4.$
1 reply
sqing
3 hours ago
SunnyEvan
33 minutes ago
Burak0609
Burak0609   0
43 minutes ago
So $b=x+y+z$ $x^3+a=-3y-3z$ $P(t)=t^3-3t+a-3b\implies x+y+z=0$ from vieta $x^3+a=-3y-3z,y^3+a=-3x-3z\implies x^2+xy+y^2=3\implies y^2+xy+(x^2-3)=0\implies \Delta\ge 0 \implies 12\ge 3x^2 \implies x\in [-2,2]$. Notice in similar fashion we get $y\in [-2,2] , z\in [-2,2]$, also its easy to observe that none of $\{x,y,z\} \in \{-2,2,0\}$. So we have$\boxed{a \in (-2,2)-\{0\}}$
0 replies
Burak0609
43 minutes ago
0 replies
Problem 1 IMO 2005 (Day 1)
Valentin Vornicu   92
N an hour ago by Baimukh
Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.

Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent.

Bogdan Enescu, Romania
92 replies
Valentin Vornicu
Jul 13, 2005
Baimukh
an hour ago
9 MATHCOUNTS STATE difficulty
Eddie_tiger   61
N 4 hours ago by DhruvJha
I personally thought the problems were much easier than last year, but I didn't really improve as much as I would of liked to improve.
61 replies
Eddie_tiger
Apr 1, 2025
DhruvJha
4 hours ago
Easy Counting?
orangefronted   2
N 4 hours ago by Mathematicalprodigy37
Lucien randomly picks two distinct numbers at the same time from the following group of numbers: $\frac{1}{8}, \frac{1}{4},\frac{1}{2},1,2,4,8$. If the product of the 2 numbers is also a member of this group, how many selections can be done?
2 replies
orangefronted
Today at 6:45 AM
Mathematicalprodigy37
4 hours ago
Factoring Marathon
pican   1454
N 4 hours ago by AVY2024
Hello guys,
I think we should start a factoring marathon. Post your solutions like this SWhatever, and your problems like this PWhatever. Please make your own problems, and I'll start off simple: P1
1454 replies
pican
Aug 4, 2015
AVY2024
4 hours ago
Square Track
orangefronted   1
N 4 hours ago by Lankou
Ziron and Zane are running around a square track at constant speeds. Ziron runs at a speed of 2 meters per second, while Zane runs twice as fast as Ziron. They begin at the same point and start running in opposite directions. If they meet after exactly 8 seconds, what is the area enclosed by the square track?
1 reply
orangefronted
Today at 6:46 AM
Lankou
4 hours ago
Rationalizing
mithu542   2
N Today at 6:18 AM by KevinKV01
Rationalize the following fraction:

$$\dfrac{5\sqrt{2}-2+\sqrt{6}}{24-10\sqrt{2}}.$$
2 replies
mithu542
Yesterday at 2:02 AM
KevinKV01
Today at 6:18 AM
Tricky summation
arfekete   12
N Today at 6:17 AM by KevinKV01
If $\dots = 7$, what is the value of $1 + 2 + 3 + \dots + 100$?
12 replies
arfekete
Yesterday at 2:15 AM
KevinKV01
Today at 6:17 AM
Probability NOT a perfect square
orangefronted   4
N Today at 5:12 AM by ilikemath247365
Mike decides to play a game with himself. He begins with a score of 0 and proceeds to flip a fair coin. If he lands on heads, he adds 2 to his score. If he lands on tails, he subtracts 1 from his score. After 5 flips, what is the probability that Mike’s score is not a perfect square?
4 replies
orangefronted
Tuesday at 5:37 PM
ilikemath247365
Today at 5:12 AM
prime factorization formula questions
Soupboy0   8
N Today at 3:29 AM by martianrunner
If i have a number, say $N$ with prime factorization $p_1^{e_1}p_2^{e_2}...p_n^{e_n}$, and I want to find $3, 4, 5, ..., k$ numbers that multiply to $N$, does anybody know a formula for this?
8 replies
Soupboy0
Today at 2:54 AM
martianrunner
Today at 3:29 AM
2025 MATHCOUNTS State Hub
SirAppel   217
N Today at 2:26 AM by morestuf
Previous Years' "Hubs": (2022) (2023) (2024)Please Read

Now that it's April and we're allowed to discuss, and no one else has made this yet ...
[list=disc]
[*] CA: 43 (45 44 43 43 43 42 42 41 41 41)
[*] NJ: 43 (45 44 44 43 39 42 40 40 39 38) *
[*] NY: 42 (43 42 42 42 41 40)
[*] TX: 42 (43 43 43 42 42 40 40 38 38 38)
[*] MA: 41 (45 43 42 41)
[*] WA: 41 (41 45 42 41 41 41 41 41 41 40) *
[*] FL: 39 (42 41 40 39 38 37 37)
[*] IN: 39 (41 40 40 39 ?? 35)
[*] NC: 39 (42 42 41 39)
[*] IL: 38 (41 40 39 38 38 38)
[*] OR: 38 (44 41? 38 38)
[*] PA: 38 (41 40 40 38 38 37 36 36 34 34) *
[*] MD: 37 (43 39 39 37 37 37)
[*] CT: 36 (44 39? 38 36 34 34 34 34)
[*] MI: 36 (39 41 41 36 37 37 36 36 36 36) *
[*] MN: 36 (40 36 36 36 35 35 35 34)
[*] CO: 35 (41 37 37 35 35 35 ?? 31 31 30) *
[*] GA: 35 (38 37 36 35 34 34 34 34 34 33)
[*] OH: 35 (41 37 36 35)
[*] AR: 34 (46 45 35 34 33 31 31 31 29 29)
[*] WI: 34 (40 37 37 34 35 30 28 29 29 29) *
[*] NH: 31 (42 35 33 31 30)
[*] DE: 30 (34 33 32 30 30 29 28 27 26? 24)
[*] SC: 30 (33 33 31 30)
[*] IA: 29 (33 30 31 29 29 29 29 29 29 29 29 29) *
[*] NE: 28 (34 30 28 28 27 27 26 26 25 25)
[*] SD: 22 (30 29 24 22 22 22 21 21 20 20)
[/list]
Cutoffs Unknown

* means that CDR is official in that state.

Notes

For those asking about the removal of the tiers, I'd like to quote Jason himself:
[quote=peace09]
learn from my mistakes
[/quote]

Help contribute by sharing your state's cutoffs!
As per last year's guidelines, refrain from problem discussion until their official release on the MATHCOUNTS website.
217 replies
SirAppel
Apr 1, 2025
morestuf
Today at 2:26 AM
I cant find one problem
tanujkundu   9
N Today at 1:47 AM by vsarg
Does anybody know which problem is about when a number is a meteor and when a number is a shooting star?
9 replies
tanujkundu
Sep 17, 2024
vsarg
Today at 1:47 AM
Easy geometry
Bluesoul   13
N Mar 30, 2025 by AshAuktober
Source: CJMO 2022 P1
Let $\triangle{ABC}$ has circumcircle $\Gamma$, drop the perpendicular line from $A$ to $BC$ and meet $\Gamma$ at point $D$, similarly, altitude from $B$ to $AC$ meets $\Gamma$ at $E$. Prove that if $AB=DE, \angle{ACB}=60^{\circ}$
(sorry it is from my memory I can't remember the exact problem, but it means the same)
13 replies
Bluesoul
Mar 12, 2022
AshAuktober
Mar 30, 2025
Easy geometry
G H J
G H BBookmark kLocked kLocked NReply
Source: CJMO 2022 P1
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Bluesoul
890 posts
#1
Y by
Let $\triangle{ABC}$ has circumcircle $\Gamma$, drop the perpendicular line from $A$ to $BC$ and meet $\Gamma$ at point $D$, similarly, altitude from $B$ to $AC$ meets $\Gamma$ at $E$. Prove that if $AB=DE, \angle{ACB}=60^{\circ}$
(sorry it is from my memory I can't remember the exact problem, but it means the same)
Z K Y
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awesomeming327.
1685 posts
#2
Y by
:( lucky imagine getting a doable geo

our geo was so hard

anyway, AB=DE implies AE || BD which implies $\angle EAD =\angle EBD$ which by orthocenter reflection stuff blah blah implies BHD and AHE both equilateral so we're done
Z K Y
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Bluesoul
890 posts
#3 • 1 Y
Y by Vladimir_Djurica
This is my solution provided during the exam
solution
Z K Y
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parmenides51
30629 posts
#4
Y by
Let $ABC$ be an acute angled triangle with circumcircle $\Gamma$. The perpendicular from $A$ to $BC$ intersects $\Gamma$ at $D$, and the perpendicular from $B$ to $AC$ intersects $\Gamma$ at $E$. Prove that if $|AB| = |DE|$, then $\angle ACB = 60^o$.
This post has been edited 4 times. Last edited by parmenides51, May 6, 2024, 11:14 AM
Z K Y
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samrocksnature
8791 posts
#5 • 4 Y
Y by Vladimir_Djurica, Mango247, Mango247, Mango247
jamboard solve!!!

pretty easy??, nothing used beyond the fact that angles intercepting the same arc are equal + angles in a triangle sum to 180.
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Mogmog8
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#6 • 2 Y
Y by centslordm, Vladimir_Djurica
Notice $\triangle ABE\cong\triangle BED$ by SAS so $H=\overline{AD}\cap\overline{BE}$ is the center of $\Gamma.$ Hence, $$2\angle C=\angle AHB=180-\angle C$$and $\angle C=60.$ $\square$
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brainee-chan
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#7
Y by
Let $AD \cap BE = H$, $AC \cap BE = X$, $BC \cap AD = Y$. Using the fact that angles subtended by arcs of equal lengths are equal, $\angle DBE = \angle ADB$ so $HB = HD$. Since $AXYB$ is cyclic, $\angle HBY = \angle HAX$. By construction $ABDC$ is cyclic so $\angle HAX = \angle DBC$ so $BY$ is the perpendicular bisector of $\triangle HBD$ thus $BD = HB$. Hence $\triangle HBD$ is equilateral and $\angle ACB = \angle ADB = \angle HDB = 60^\circ$.
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yofro
3146 posts
#8 • 2 Y
Y by Mango247, Mango247
Let $X$ be the foot from $A$ to $BC$ and $Y$ be the foot from $B$ to $AC$. By reflecting the orthocenter, $XY=\frac{1}{2}ED=\frac{1}{2}AB$. Since $AYXB$ is cyclic, we have $\triangle CYX\sim \triangle CBA$ so $CX=\frac{1}{2}AC$ and hence $\cos C=\frac{1}{2}\implies C=60^{\circ}$.
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ike.chen
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#9
Y by
Let $AD \cap BC = X$ and $BE \cap CA = Y$. The Orthocenter Reflection Lemma yields $$XY = \frac{DE}{2} = \frac{AB}{2}.$$Now, since $CXY \overset{-}{\sim} CAB$, we know $$| \cos ACB | = \frac{CX}{CA} = \frac{XY}{AB} = \frac{1}{2}$$so $\angle ACB = 60^{\circ}$ or $\angle ACB = 120^{\circ}$ must hold. $\blacksquare$


Remarks: I'm pretty sure $\angle ACB = 120^{\circ}$ is possible too.

Also, what is the CJMO? Personally, I only know of one CJMO, and this problem isn't from there.
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samrocksnature
8791 posts
#10 • 2 Y
Y by oolite, ike.chen
Canada Junior Math Olympiad?
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ZETA_in_olympiad
2211 posts
#11
Y by
ike.chen wrote:
Also, what is the CJMO? Personally, I only know of one CJMO, and this problem isn't from there.

Canadian Junior Math Olympiad
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parmenides51
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#12
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Notation : $(XY)$ stands for small arc $XY$

$(DE)= (EC)+(CD)=2\angle EAC+2<DBC=2(180^o-\angle AEB) + 2(180^o-\angle DBC) = 180^o  +180^o-4(AB)-4(AB)=360^o-8(AB)=360^o- 4\angle ACB=360^o-4 \cdot  60^o=360^o -240^o=120^o =\angle (DCE) \Rightarrow AB= DE$

my solution might be modified to prove that the converse is also true,
and so the problem could have been asked with iff condition
This post has been edited 4 times. Last edited by parmenides51, May 10, 2024, 7:03 PM
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DensSv
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#13
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Sol.
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AshAuktober
958 posts
#14
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Angle chase using $\angle AEB= \angle EBD$ and the orthocentre reflection lemma.
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