Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

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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:
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[*]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.
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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
JBMO TST Bosnia and Herzegovina 2024 P3
FishkoBiH   2
N a few seconds ago by Ianis
Source: JBMO TST Bosnia and Herzegovina 2024 P3
Let $ABC$ be a right-angled triangle where $ACB$=90°.Let $CD$ be an altitude of that triangle and points $M$ and $N$ be the midpoints of $CD$ and $BC$, respectively.If $S$ is the circumcenter of the triangle $AMN$, prove that $AS$ and $BC$ are paralel.
2 replies
FishkoBiH
3 hours ago
Ianis
a few seconds ago
A line parallel to the asymptote of a cubic
kosmonauten3114   0
13 minutes ago
Source: My own, but uninspiring...
Given a scalene triangle $\triangle{ABC}$, let $P$ be a point ($\neq \text{X(4)}$). Let $P'$ be the anticomplement of $P$, and let $Q$ be the $\text{X(1)}$-anticomplementary conjugate of $P$. Prove that the line $P'Q$ is parallel to the real asymptote of the circular pivotal isocubic with pivot $P$.
0 replies
kosmonauten3114
13 minutes ago
0 replies
My Unsolved Problem
ZeltaQN2008   1
N 21 minutes ago by Tung-CHL
Source: IDK
Given a positive integer \( m \) and \( a > 1 \). Prove that there always exists a positive integer \( n \) such that \( m \mid (a^n + n) \).

P/s: I can prove the problem if $m$ is a power of a prime number, but for arbitrary $m$ then well.....
1 reply
ZeltaQN2008
Yesterday at 10:18 AM
Tung-CHL
21 minutes ago
Please be my guest!
Madunglecha   0
an hour ago
Find all the (a,b,c) are integers which suggests
a^4-6a^2b^2+b^4=-c^2
0 replies
Madunglecha
an hour ago
0 replies
Problem of the day
sultanine   11
N 2 hours ago by CJB19
[center]Every day I will post 3 new problems
one easy, one medium, and one hard.
Please hide your answers so others won't be affected
:D :) :D :) :D
11 replies
sultanine
May 23, 2025
CJB19
2 hours ago
9 How many squares do you have memorized
LXC007   103
N 2 hours ago by sultanine
How many squares have you memorized. I have 1-20

Edit: to clarify i mean positive squares from 1 so if you say ten you mean you memorized the squares 1,2,3,4,5,6,7,8,9 and 10
103 replies
LXC007
May 17, 2025
sultanine
2 hours ago
Rational equation
soruz   1
N Today at 10:37 AM by Mathzeus1024
Solve the equation:
$ \frac{x+2}{1011} +\frac{x}{1010} +\frac{x-2}{1009} +\frac{x-4}{1008}...+\frac{x-2016}{2}=2020$.
1 reply
1 viewing
soruz
Feb 23, 2022
Mathzeus1024
Today at 10:37 AM
Penrick Hates 2, 5, and 7
LilKirb   0
Today at 7:59 AM
Penrick writes all positive integers starting from $1$, but skips any number containing digits $2$, $5,$ or $7$. If the last number he writes is $3141$, how many numbers has Penrick written in total?

Answer
Solution
0 replies
LilKirb
Today at 7:59 AM
0 replies
[PMO26 Qualis] I.3 Change every number
aops-g5-gethsemanea2   4
N Today at 3:48 AM by Shinfu
The arithmetic mean of $11$ integers is $10$. After adding $20$ to each of the first four and subtracting $24$ from each of the last seven, what is the new mean?

$\text{(a) }2\qquad\text{(b) }3\qquad\text{(c) }4\qquad\text{(d) }6$

Answer confirmation
4 replies
aops-g5-gethsemanea2
Feb 9, 2025
Shinfu
Today at 3:48 AM
khan academy
Spacepandamath13   17
N Today at 3:41 AM by KF329
I haven't done khan academy in so long but today I had to learn law of sines. I wish khan academy taught competition math because their format, and self paced learning seems a bit better than aops' plus sal's videos for each topic are so good
17 replies
Spacepandamath13
May 18, 2025
KF329
Today at 3:41 AM
9 Favorite topic
A7456321   13
N Today at 2:52 AM by A7456321
What is your favorite math topic/subject?

If you don't know why you are here, go binge watch something!

If you forgot why you are here, go to a hospital! :)

If you know why you are here and have voted, maybe say why you picked the option that you picked in a response) :thumbup:

Timeline

Oh yeah and you see that little thumb in the top right corner? The one that upvotes when you press it? Yeah. Press it. Thaaaaaaaanks! :D
13 replies
A7456321
Friday at 11:53 PM
A7456321
Today at 2:52 AM
Summer math contest prep
Abby0618   14
N Today at 1:59 AM by CJB19
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
14 replies
Abby0618
May 22, 2025
CJB19
Today at 1:59 AM
Solve this
DhruvJha   3
N Today at 1:18 AM by tintin21
Len is playing a Arkansas-styled basketball game with his friend, Dawson. The game ends whenever a player has a 2 point lead over the other player. In Arkansas styled basketball, points can only be scored in increments of two. Whenever Len has possession of the ball, he scores at a rate of 60 percent. However, Dawson is slightly worse and when he has possession of the ball, he scores at a rate of only 40 percent. Given that Len starts with possession first each game, what is the expected amount of games he wins if they play 38 total?
3 replies
DhruvJha
Yesterday at 9:20 PM
tintin21
Today at 1:18 AM
polynomial division vs simplify
Miranda2829   1
N Today at 12:58 AM by CJB19
the question is

4x² - 4x +1 divide by 2x + 1 write in fraction format

the answer is 2x-3 + 4/2x+1

so my question is can we simiplfy this -4x with 2x become -2x

then answer 4x² -2x+1?

Im confused , when do usual fraction division we can simplify, but in above question doesn't seem to work.

many thanks
1 reply
Miranda2829
Today at 12:31 AM
CJB19
Today at 12:58 AM
Computer too strong
Eyed   62
N May 19, 2025 by AR17296174
Source: 2020 ISL G6
Let $ABC$ be a triangle with $AB < AC$, incenter $I$, and $A$ excenter $I_{A}$. The incircle meets $BC$ at $D$. Define $E = AD\cap BI_{A}$, $F = AD\cap CI_{A}$. Show that the circumcircle of $\triangle AID$ and $\triangle I_{A}EF$ are tangent to each other
62 replies
Eyed
Jul 20, 2021
AR17296174
May 19, 2025
Source: 2020 ISL G6
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AdventuringHobbit
164 posts
#54 • 1 Y
Y by taki09
WLOG $AB<AC$. Let $N$ be the midpoint of major arc $BAC$. Let $H$ be the orthocenter. Let $S=ND\cap (ABC)$. Then $\angle ASD = \angle ASN = \frac{\angle B - \angle C}{2}$. We can also compute $$180-\angle AID = \angle IAH = \frac{\angle A}{2}-(90-\angle B)=\frac{\angle B - \angle C}{2},$$Thus $AIDS$ are concyclic. Now let $X=NI\cap (AIDS) \neq I$. Now we have $\angle NBC = \angle NCB = \angle NSD$, so $NB$ is tangent to $(BSD)$, and $NB^2=NS\cdot ND = NI \cdot NX$. Since it is well-known that $NB$ is tangent to $BIC$, we have by Power of a Point that $X \in (BIC)$. Since $NB$ and $NC$ are tangent to $(BICX)$, and $N,I,X$ are collinear, we have $(B,C;I,X)=-1$.Now let $R$ be the midpoint of $NI_A$. Consider the circle $\omega$ centered at $R$ with diameter $NI_A$. $NA \perp AI_A \rightarrow A \in \omega$. Also, we have $90 = \angle IXI_A = \angle NXI_A$, so $X \in \omega$. Now let $G$ be the midpoint of $EF$. Here is a brief interlude showing $G \in \omega$. Here are coordinates of various points. They can be computed in this order.
$D=(0:s-c:s-b)$
$I_A=(-a:b:c)$
$E=-a(s-c):b(s-c):b(s-b)$
$F=-a(s-b):c(s-c):c(s-b)$
$G=-2a^2:b^2-c^2+ab+ac:c^2-b^2+ab+ac$
Now using the formula for perpendicular bisectors, $a^2(z-y)+x(c^2-b^2)=0$ we can easily check $G$ is on the bisector of $BC$. We can similarly straightforwardly check that $GI_A || BC$ since in normalized barycentric coordinates they both have $A$ coordinate $\frac{-a}{-a+b+c}$. This implies $NG \perp GI_A$, so $G \in \omega$. Now projecting through $I_A$ we get $-1=(B,C;I,X)=(E,F;A,Y)$, where $Y=I_AX \cap AD$. We now have by Power of a Point that $XY \cdot YI_A = YG \cdot YA = YE \cdot YF$. This implies that $EFI_AX$ is cyclic. Now angle chasing gives $$\angle RAI = \angle RI_AA = \angle NI_AA = \angle NXA = \angle IXA,$$so $RA$ is tangent to $AIDX$. Since $RA=RX$, $RX$ is the other tangent.
Now we do one last angle chase to show $RI_A$ is tangent to $EXFI_A$. Let $f(\ell)$ be the unit complex number corresponding to the direction of $\ell$. Equalities are modulo rotation by $\pi$. Now it suffices to check $$f(RI_A)f(EF)=f(I_AF)f(I_AE)$$$$\iff f(I_AB)f(I_AC)=f(RI_A)f(AG)$$$$\iff f(IB)f(IC)=f(GI_A)f(AN)$$$$\iff \frac{f(IB)}{f(AN)}=\frac{f(BC)}{f(IC)}.$$Both of these are equal to $\frac{\angle C}{2}$, so we are done. Now $RI_A$ is tangent to $EI_AFX$, which means $RX$ is as well. Since $RX$ is tangent to both circles, we are done.
This post has been edited 1 time. Last edited by AdventuringHobbit, Oct 16, 2023, 5:31 AM
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popop614
271 posts
#55 • 1 Y
Y by taki09
The whole point of the problem is classifying $T$, the point of tangency.

We set $T = \overline{NI} \cap (BIC)$ where $N$ is the major arc midpoint. Let $M$ be the other arc midpoint. Since $NB$ and $NC$ are tangent to $(BIC)$, we observe that $(B, C; I, T) = -1$.

$\textbf{Claim.}$ $AIDT$ is cyclic.
$\textit{Proof.}$ Let $\overline{I_aT} \cap \overline{BC} = S$. Perspectivity at $I_a$ reveals that $S$ is the intersection point of the external bisector of $BAC$ with $BC$. Thus, $SA \perp AI$, and since $ST \perp TI$ we have $SATI$ cyclic. But then since $ID \perp DS$ we also have $D$ on this circle too. $\square$

$\textbf{Claim.} TEFI_a$ is cyclic.
$\textit{Proof.}$ Let $E' = (BTD) \cap \overline{BI_a}$. We will show $E = E'$, and by symmetry we get a similar result for $F$.
Indeed this isn't hard to show:
\[ \measuredangle TDE' = \measuredangle TBE' = \measuredangle TIM = \measuredangle TIA = \measuredangle TDA, \]so $E'$ lies on $AD$. Thus $E' = E$. It now follows that $T$ is the center of spiral similarity taking $BC$ to $EF$, whence $TEFI_a$ is cyclic. $\square$

To finish, since we have $S-T-I_a$ just simply note that
\[ \measuredangle SDT = \measuredangle BDT = \measuredangle BET = \measuredangle I_aET \]which shows that the arcs have the same measure, hence the tangency is proven.
This post has been edited 2 times. Last edited by popop614, Dec 30, 2023, 8:33 PM
Reason: asdfasdf
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wu2481632
4239 posts
#56 • 1 Y
Y by taki09
Let $L$ and $M_A$ be the midpoints of arcs $BC$ containing and not containing $A$, respectively. Suppose that $LI$ meets $(BICI_A)$ again at $X$. We claim $X$ is the desired tangency point.

First, we show that $AIDX$ is cyclic. Note that $\angle LAI_A = \angle LXI_A = 90^{\circ}$, so $LAXI_A$ is cyclic. Then by the Radical Center theorem, we know that $LA, BC, XI_A$ concur at some point $Y$. Thus $A, D, X$ all lie on the circle with diameter $IY$, so $AIDX$ is cyclic.

Next, we show that $I_AFXE$ is cyclic. Observe that $\angle BXD = \angle IXD + \angle IXB = \angle DAI + \frac{1}{2}\angle C$. But $\angle BEA = 90^{\circ} - \angle(BI, AD) = 90^{\circ} - (\angle(BI, AI) -\angle DAI)$, and the right-hand-side reduces to $90^{\circ} - (90^{\circ} -\frac{1}{2}\angle C -\angle DAI) = \angle DAI + \frac{1}{2}\angle{C}$. Thus $BDEX$ is cyclic, and so is $CDFX$, by symmetry, and $X$ must be the spiral center mapping $FE$ to $CB$, which implies $XEI_AF$ is cyclic.

Finally, let $LX$ meet $(FI_AEX)$ again at $Z$. Showing that the circumcircles of $AID$ and $I_AEF$ are tangent is equivalent to showing $IY \parallel ZI_A$. But $\angle YIX = \angle YDX= \angle BEX = 180^{\circ} - \angle XEI_A = \angle XZI_A$, so we are done.
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BlizzardWizard
108 posts
#57 • 2 Y
Y by ihatemath123, taki09
Here's another complex bash (first one was post #13) using a different unit circle and a different characterization of the tangency point, and much less synthetic work.

Setup and computing the three lines that bound I_AEF
Computing (I_AEF)
Computing the other intersection of (I_AEF) and (BICI_A)
Showing it lies on (AID)
Showing the circles are tangent
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AngeloChu
471 posts
#58 • 1 Y
Y by taki09
first, by miquel point, circles $BI_AC$, $BDE$, and $EI_AF$ intersect at one point $G$, and it is obvious $I$ lies on $BI_AC$
we can also easily prove that $II_A$ is a diameter of $BI_AC$, so $IG$ is perpendicular to $GI_A$
construct a point $J$ such that $JE$ is perpendicular to $I_AE$, and that $JF$ is perpendicular to $I_AF$, so $J$ lies on $EI_AF$
then, $JG$ is perpendicular to $GI_A$ and $JGI_A$ are collinear
extend $GI_A$ to intersect $BC$ at $H$, so we get that $IGH=90$, and since $ID$ is perpendicular to $DH$, $IDGH$ are concyclic
then, $HIG=HDG=BEG=GJI_A$, so $IH$ and $JI_A$ are parallel and homothety yields that the circumcircles of $\triangle AID$ and $\triangle I_AEF$ are tangent to each other
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Thapakazi
68 posts
#59 • 2 Y
Y by taki09, ABYSSGYAT
Let $T = (BDE) \cap (I_AEF)$. Then, by Miquel's theorem, we see that $(CDF)$ intersects $(BDE)$ at $T$ as well. Furthermore as

\[\angle BTC = \angle BTD + \angle DTC = \angle BED + \angle DFC = \angle FEI_A + \angle EFI_A = \angle BI_AC\]
$T$ lies on $(BIC)$ too. Moreover, as

\[\angle ITD = \angle BTD - \angle BTI = \angle BEA - \angle BI_AI = \angle IAD\]
$T$ lies on $(AID)$ too. Let $\ell$ be the tangent at $T$ to $(ATD).$ First we notice,

\[\angle TAI = 180 - \angle IDT = 180 - (90 + \angle BDT) = 90 - \angle BDT = 90 - \angle TFC.\]
To finish,

$$\angle(\ell, TI_A) = \angle ITI_A - \angle IAT = 90 - \angle IAT = \angle TFC$$
implying $\ell$ is the tangent to $(I_AEF)$ as needed.
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bin_sherlo
733 posts
#60 • 1 Y
Y by taki09
A similar one.
Let $N$ be the midpoint of the arc $BAC$ on $(ABC)$. $NA\cap BC=K,KI_A\cap  (BI_AC)=P,PI\cap (BI_AC)=S$

Since $KI\perp SI_A,I_AI\perp KA,IP\perp KI_A$ we get that $(K,N,I_A,I)$ is an orthogonal system. Hence $N,I,P$ are collinear.
$K,P,D,I,A$ lye on the circle with diameter $KI$.
\[\angle BPD=\angle BPI+\angle IPD=\frac{\angle C}{2}+\angle IAD=\angle EI_AI+\angle I_AAE=\angle BEA=\angle BED\]Thus, $P\in (BED)$. Since $P$ is on $(BCI_A)$ and $(BDE),$ it is the miquel point of $DEI_AC$. So $P\in (EFI_A)$.
\[\angle PI_AE+\angle DKP=\angle DBE=\angle DPE\]If $l$ is tangent to $PEI_A$ at $P,$ then
\[\angle DPE-\angle (PD,l)=\angle (l,PE)=\angle PI_AE=\angle DPE-\angle DKP\]Hence $l$ is also tangent to $(DKP)$. These show that $(KPDIA)$ and $(PEI_AF)$ are tangent to each other at $P$ as desired.$\blacksquare$
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ihatemath123
3449 posts
#61 • 1 Y
Y by taki09
Let $K$ be where the external bisector of $\angle A$ meets line $BC$,and let $P$ be the foot from $I$ to line $KI_A$. We claim the tangency point is $P$.

Claim: $P$ lies on $(AID)$.
Proof: The antipode of $I$ in $(AID)$ is $K$ – since $\angle IPK = 90^{\circ}$, $P$ also belongs to $(AIDK)$.

Claim: $P$ lies on $(EI_AF)$.
Proof: It would suffice to show that $PBDE$ is cyclic, since then it follows that $P$ is the Miquel point of concave quadrilateral $BCFE$. This cyclic quadrilateral follows from
\[\angle PBE = \angle PII_A = 180^{\circ} - \angle PIA = 180^{\circ} - \angle PDA = \angle PDE.\]
Claim: $(AID)$ is tangent to $(EI_AF)$.
Proof: We have
\[ \angle KDP = \angle BEP = \angle PFI_A.\]remark
This post has been edited 3 times. Last edited by ihatemath123, Dec 22, 2024, 7:31 PM
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kiemsibongtoi
25 posts
#62 • 1 Y
Y by taki09
Eyed wrote:
Let $ABC$ be a triangle with $AB < AC$, incenter $I$, and $A$ excenter $I_{A}$. The incircle meets $BC$ at $D$. Define $E = AD\cap BI_{A}$, $F = AD\cap CI_{A}$. Show that the circumcircle of $\triangle AID$ and $\triangle I_{A}EF$ are tangent to each other

Let $\Gamma$ be the circumcircle of $\triangle ABC$ and $N$ be the midpoint of $\overarc{BAC}$ of $\Gamma$
$\hspace{0.4cm}$$W$ be the intersection of line $NI_a$ and $\Gamma$ ($W \neq N$)
$\textbf{First}$, we'll prove that $\angle WAC = \angle DAB$ :
$\, \,$Let $M$, $L$ be the intersections of line $I_aA$, $I_aC$ with $\Gamma$ respectively ($M \neq A$, $L \neq C$)
$\, \,$Cuz $\triangle I_aWA \sim \triangle IaMN$, $\triangle I_aWC \sim \triangle I_aLN$, $\triangle I_aAC \sim \triangle I_aLM$ we have that :
$$ \dfrac{CW}{AW} = \dfrac{NL. \dfrac{I_aW}{I_aL}}{MN . \dfrac{I_aW}{I_aM}} = \dfrac{NL}{MN}.\dfrac{I_aM}{I_aL} = \sin \angle NML .\dfrac{I_aC}{I_aA}$$$\, \,$In the order hand, we ez to have that : $\angle NML = \angle MAC + \angle ICA = 90^\circ - \angle IBD$ and $\triangle AIB \sim \triangle ACI_a$
$\, \,$Therefore, $$\dfrac{CW}{AW} = \cos \angle IBD. \dfrac{AI}{BI} = \dfrac{BD}{AB}$$$\, \,$Combine with $\angle ABC = \angle AWC$, we see that : $\triangle ABD \sim \triangle AWC$. Lead to $\angle BAD = \angle WAC$ $(*)$
$\textbf{Next}$, let $V$ be the intersection of lines $AN$, $BC$ and $T$ be the interserction of lines $IN$, $VI_a$
$\hspace{1.4cm}$$\omega$ be the circle go through $T$ and tangent with $NI_a$ at $NI_a$, we'll prove that $\omega$ tangent with circle $(AID)$ :
$\, \,$Since $\angle VAI = \angle VDI = 90^\circ$, we have that $A$, $D$ are lie on the circle with diameter $IV$
$\, \,$Combine with $\angle BAD = \angle WAC$ at $(*)$, we have that $\angle IVD = \angle IAD = \angle MAW = \angle MNW$
$\, \,$But $MN \perp VD$, so $NW \perp IV$. Lead to $I$ is the orthocenter of $\triangle NI_aV$, $NI \perp VI_a$
$\, \,$So $T$ lies on circle $(AID)$. Combine with $NI_a$ tangent $\omega$ at $I_a$ and $IV \perp NI_a$, it's clearly that $(AID)$ tangle $\omega$ at $T$
$\textbf{In the final}$, we just need to prove that $E$, $F$ all line on $\omega$ :
$\, \,$Let $Y$ be the intersection of line $BI$ and $(AID)$ ($Y \neq I$); $E'$ be the intersection of lines $AD$, $YT$
$\, \,$Use Pascal theorem for $\binom{A\ V\ Y}{T\ I\ D}$, we have that $E'$, $B$, $I_a$ are colinear, which means $E$, $T$, $Y$ are colinear
$\, \,$Therefore, $\angle ETI_a = \angle VTY = \angle VIB = \angle BI_aN$ (Cuz $IB \perp BI_a$, $IV \perp NI_a$
$\, \,$Combine with $NI_a$ tangent with $\omega$ at $I_a$, we see that $E$ lies on $\omega$. Similar, $F$ lie on $\omega$, done
Attachments:
imosl g6 2020.pdf (73kb)
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L13832
268 posts
#63 • 1 Y
Y by taki09
Easy for a g6
Solution
This post has been edited 1 time. Last edited by L13832, Dec 16, 2024, 8:16 AM
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lelouchvigeo
183 posts
#64 • 1 Y
Y by taki09
How is this a G6 :roll:
SKetch
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HamstPan38825
8868 posts
#65 • 1 Y
Y by taki09
oops I solved this problem again -- but here's a solution by pure angle chasing (without adding any points (!!))

Claim: The Miquel points of convex quadrilateral $CI_AED$ and self-intersecting $EFI_AB$ coincide at a point $P$ on $(II_A)$.

Proof: Let $P$ be the Miquel point of $EFI_AB$. Then $\measuredangle PI_AC = \measuredangle PED = \measuredangle PBC$, so $P$ lies on $(CI_ABI)$. Then $\measuredangle DFP = \measuredangle BI_AP = \measuredangle DCP$ so $P$ lies on $(CDF)$, which implies the result. $\blacksquare$

Denote $\theta = \angle BAD$.

Claim: $P$ lies on $(AID)$.

Proof: More angle chasing. \[\angle DPI = \angle CPI - \angle CPD = \frac B2 - \angle CFD = \frac A2 - \theta = \angle IAD. \ \blacksquare\]
So now we show that $(AID)$ and $(I_AEF)$ are tangent at $P$. This ends up being yet another angle chase:
\begin{align*}
\angle DIP + \angle EI_AP &= 180^\circ - \angle DPI-\angle IDP \\
&= 90^\circ - \angle DAI - \angle BDP + \angle DCP \\
&= 90^\circ - \left(\frac A2 - \theta\right) - \angle CFD \\
&= 90^\circ - \left(\frac A2 - \theta\right) - \left(\theta+B+\frac C2 - 90^\circ\right) \\
&= 90^\circ - \frac B2 = \angle EBD = \angle EPD.
\end{align*}Thus done.
This post has been edited 3 times. Last edited by HamstPan38825, Dec 31, 2024, 4:03 AM
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EthanWYX2009
871 posts
#66 • 3 Y
Y by S_14159, ohiorizzler1434, taki09
The fastest way to calculate should be using Casey Theorem on $\odot (I_AEF),$ we only need
$$I_AF\cdot\sqrt{I_AA\cdot I_AI}+I_AE\cdot\sqrt{FA\cdot FD}=EF\sqrt{I_AA\cdot I_AI}.$$Clearly all segments are easy to calculate.$\Box$
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Haris1
77 posts
#67 • 2 Y
Y by taki09, TopGbulliedU
This is what happens when u skip basic ideas: Let $R$ be $ID\cap (IBC)=R$. We have pairs $(B,C)(I,ID\cap(IBC))(AD\cap(IBC),AD\cap(IBC)_{2})$ this is true because $D$ swaps them all. Now Taking vertex $I_{a}$ and putting the pairs to line $AD$ we get $(E,F)(A,I_{a}R\cap AD), (AD\cap(IBC),AD\cap(IBC)_{2})$. Proving $X,A,Ia,I_{a}R\cap AD$ cyclic is trivial since $I_{a}R||BC$ and then just angle chase. The point which swaps these is $AD\cap I_{a}$ which also swaps $(AD\cap(IBC),AD\cap(IBC)_{2})$ so by ddit it also swaps $(E,F)$ so we have $XFI_{a}E$ cyclic. And the rest follows like everyone.
This post has been edited 1 time. Last edited by Haris1, Mar 31, 2025, 11:19 AM
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AR17296174
9 posts
#68
Y by
Nice problem!
Here is my solution
Attachments:
2020G6.pdf (281kb)
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