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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jun 2, 2025
Congratulations to all the mathletes who competed at National MATHCOUNTS! If you missed the exciting Countdown Round, you can watch the video at this link. Are you interested in training for MATHCOUNTS or AMC 10 contests? How would you like to train for these math competitions in half the time? We have accelerated sections which meet twice per week instead of once starting on July 8th (7:30pm ET). These sections fill quickly so enroll today!

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[*]AMC 10 Problem Series[/list]
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0 replies
jlacosta
Jun 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
Bugs Bunny at it again
Rijul saini   8
N 14 minutes ago by quantam13
Source: LMAO 2025 Day 2 Problem 1
Bugs Bunny wants to choose a number $k$ such that every collection of $k$ consecutive positive integers contains an integer whose sum of digits is divisible by $2025$.

Find the smallest positive integer $k$ for which he can do this, or prove that none exist.

Proposed by Saikat Debnath and MV Adhitya
8 replies
Rijul saini
Wednesday at 7:01 PM
quantam13
14 minutes ago
The Bank of Bath
TelMarin   101
N 29 minutes ago by monval
Source: IMO 2019, problem 5
The Bank of Bath issues coins with an $H$ on one side and a $T$ on the other. Harry has $n$ of these coins arranged in a line from left to right. He repeatedly performs the following operation: if there are exactly $k>0$ coins showing $H$, then he turns over the $k$th coin from the left; otherwise, all coins show $T$ and he stops. For example, if $n=3$ the process starting with the configuration $THT$ would be $THT \to HHT  \to HTT \to TTT$, which stops after three operations.

(a) Show that, for each initial configuration, Harry stops after a finite number of operations.

(b) For each initial configuration $C$, let $L(C)$ be the number of operations before Harry stops. For example, $L(THT) = 3$ and $L(TTT) = 0$. Determine the average value of $L(C)$ over all $2^n$ possible initial configurations $C$.

Proposed by David Altizio, USA
101 replies
TelMarin
Jul 17, 2019
monval
29 minutes ago
Intersections and concyclic points
Lukaluce   2
N 29 minutes ago by AylyGayypow009
Source: 2025 Junior Macedonian Mathematical Olympiad P2
Let $B_1$ be the foot of the altitude from the vertex $B$ in the acute-angled $\triangle ABC$. Let $D$ be the midpoint of side $AB$, and $O$ be the circumcentre of $\triangle ABC$. Line $B_1D$ meets line $CO$ at $E$. Prove that the points $B, C, B_1$, and $E$ lie on a circle.
2 replies
Lukaluce
May 18, 2025
AylyGayypow009
29 minutes ago
Calvin needs to cover all squares
Rijul saini   5
N 31 minutes ago by quantam13
Source: India IMOTC 2025 Day 2 Problem 1
Consider a $2025\times 2025$ board where we identify the squares with pairs $(i,j)$ where $i$ and $j$ denote the row and column number of that square, respectively.

Calvin picks two positive integers $a,b<2025$ and places a pawn at the bottom left corner (i.e. on $(1,1)$) and makes the following moves. In his $k$-th move, he moves the pawn from $(i,j)$ to either $(i+a,j)$ or $(i,j+a)$ if $k$ is odd and to either $(i+b,j)$ and $(i,j+b)$ if $k$ is even. Here all the numbers are taken modulo $2025$. Find the number of pairs $(a,b)$ that Calvin could have picked such that he can make moves so that the pawn covers all the squares on the board without being on any square twice.

Proposed by Tejaswi Navilarekallu
5 replies
1 viewing
Rijul saini
Wednesday at 6:35 PM
quantam13
31 minutes ago
9 How orz are you? (not troll)
pingpongmerrily   64
N 2 hours ago by Yihangzh
Before voting, please look at the AoPS competition ratings. Your skill level is defined as the hardest level of question you can consistently solve correctly. I have enabled revoting, so you can change your vote after you improve. This is to approximately gauge the orzness of the AoPS community.
I voted 4 lol.
64 replies
pingpongmerrily
Yesterday at 2:13 PM
Yihangzh
2 hours ago
hi guys.
unicornix   2
N 2 hours ago by ohiorizzler1434
if x=(69^2+69)/69 & y=69(x+1)/69-2 then wut is y
2 replies
unicornix
Yesterday at 11:36 PM
ohiorizzler1434
2 hours ago
Problem of the week
evt917   62
N 3 hours ago by nxchman
Whenever possible, I will be posting problems randomly! They will be roughly of AMC 8 difficulty. Have fun solving! Also, these problems are all written by myself!

Also occasionally I may also post theorems that I recently learned, or already knew about.

First problem:

$20^{16}$ has how many digits?
62 replies
evt917
Mar 5, 2025
nxchman
3 hours ago
9 Favorite topic
A7456321   49
N 3 hours ago by K124659
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:

if ur here for any reason whatsoever, CLICK ME YOU KNOW YOU WANT TO
Timeline
49 replies
A7456321
May 23, 2025
K124659
3 hours ago
Trouble focusing
GallopingUnicorn45   18
N 5 hours ago by LXC007
Hi all,

So I'm currently hard-grinding for AIME in AMC 10 this year (I'm taking both A and B) and I'm having a hard time focusing and my productivity is slipping; I can't finish all of the stuff I plan daily and weekly. Before, during the school year, I was also grinding and listening to K-pop while working, and now I have songs stuck in my head as I work, which also makes me unable to focus.

Any tips on how to concentrate for longer periods of time? Thanks!
18 replies
GallopingUnicorn45
Jun 2, 2025
LXC007
5 hours ago
The daily problem!
Leeoz   220
N Yesterday at 11:18 PM by Yihangzh
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!
220 replies
Leeoz
Mar 21, 2025
Yihangzh
Yesterday at 11:18 PM
Another Nice Geo Problem
Rice_Farmer   1
N Yesterday at 10:50 PM by Aaronjudgeisgoat
In parallelogram $ABCD, AB=13,BC=14,$ and the distance between $\overline{AD}$ and $\overline{BC}$ is $12.$ Point $E$ lies on $\overline{BD}$ such that $\angle{ECB}=90^\circ.$ What is $CE?$
1 reply
Rice_Farmer
Yesterday at 10:01 PM
Aaronjudgeisgoat
Yesterday at 10:50 PM
Nice Geo Problem
Rice_Farmer   0
Yesterday at 10:00 PM
$ABCD$ is a quadrilateral with $BA=BC,DA=DA,AC=10,$ and $BD=18.O$ is the circumcenter of $\triangle{ABD}.$ If $\angle{ACO}=90^\circ,$ what is $CO?$
0 replies
Rice_Farmer
Yesterday at 10:00 PM
0 replies
Area of Polygon
AIME15   58
N Yesterday at 8:27 PM by Yihangzh
The area of polygon $ ABCDEF$, in square units, is

IMAGE

\[ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 46 \qquad
\textbf{(D)}\ 66 \qquad
\textbf{(E)}\ 74
\]
58 replies
AIME15
Jan 12, 2009
Yihangzh
Yesterday at 8:27 PM
An annoying math problem
Wolfpierce   10
N Yesterday at 8:20 PM by Capybara7017
Okay so what is 2/((√3+1)((3 to the 1/4)+1)((3 to the 1/8)+1)((3 to the 1/16)+1)) to the power of 32?
10 replies
Wolfpierce
Yesterday at 12:17 AM
Capybara7017
Yesterday at 8:20 PM
Prove excircle is tangent to circumcircle
sarjinius   8
N Apr 24, 2025 by Lyzstudent
Source: Philippine Mathematical Olympiad 2025 P4
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
8 replies
sarjinius
Mar 9, 2025
Lyzstudent
Apr 24, 2025
Prove excircle is tangent to circumcircle
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Source: Philippine Mathematical Olympiad 2025 P4
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sarjinius
249 posts
#1 • 4 Y
Y by MathLuis, mpcnotnpc, JollyEggsBanana, Rounak_iitr
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
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ilovemath0402
192 posts
#2
Y by
bump bump this problem is so nice
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sarjinius
249 posts
#3
Y by
ilovemath0402 wrote:
bump bump this problem is so nice

Thanks, I proposed this problem :)
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SimplisticFormulas
132 posts
#4
Y by
what’s the solution? I am completely stuck
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MathLuis
1562 posts
#5 • 4 Y
Y by drago.7437, sarjinius, Mysteriouxxx, radian_51
Well this geo is really amazing I have to say...solved in around 30 mins but I think this could even be around 30-35 MOHS because the way to find things on this problem requires deep intuition.
Let $BI \cap (ABC)=M_B$ and $CI \cap (ABC)=M_C$, also let $N_A$ be midpoint of arc $BAC$ on $(ABC)$, now let reflections of $D$ over $EX, FY, BI, CI, Y \infty_{\perp CI}, X \infty_{\perp BI}$ be $D_B, D_C, L', K', K, L$ respectively now let reflection of $D_B$ over $AX$ be $L_1$ and reflection of $D_C$ over $AY$ be $K_1$.
Using the paralelogram we can easly see from the direction of the reflections that $KK'$ and $LL'$ are diameters on $(Y, YD), (X, XD)$ respectively, now let $I_B, I_C$ be the $B,C$ excenters of $\triangle ABC$ then notice we have $\measuredangle II_CA=\measuredangle CBI=\measuredangle CDY=\measuredangle YK'C$ which implies $I_CAK'Y$ cyclic and similarily $I_BAL'X$ is cyclic however since $\measuredangle CDY=\measuredangle YD_CF$ we also get that $I_CAK'YD_C$ is cyclic and similarily $I_BAL'XD_B$ is cyclic, however it doesn't end here...
Now notice that $YK'=YD_C$ so $Y$ is midpoint of arc $K'D_C$ on $(I_CAK')$ however $D, K'$ are symetric in $CI$ which means both $I_CD, I_CD'$ are reflections of $I_CK'$ over $CI$ and thus $I_C, D, D_C$ are colinear, and similarily $I_B, D, D_B$ are colinear.
Now $\measuredangle CDY=\measuredangle YD_CA=\measuredangle AK_1Y$ which means $CK_1YD$ is cyclic and similarily we have $L_1BXD$ cyclic, but also note that $\measuredangle L_1DL=\measuredangle L_1L'L=\measuredangle AI_BX=\measuredangle ACI=\measuredangle K_1DY$ which means that $L_1, D, K_1$ are colinear.
Now from here notice that $\measuredangle DL_1A=\measuredangle DXI=\measuredangle IYD=\measuredangle AK_1D$ which does in fact show that $\triangle L_1AK_1$ is isosceles and therefore $AK_1=AL_1$, and from reflections this gives $AD_B=AD_C$, but notice from other reflections we have $D_BG=DG=D_CG$ where $EX \cap FY=G$ (clearly then $G$ is A-excenter of $\triangle EAF$), but now also note that we have $\measuredangle AD_BG=\measuredangle GDE=\measuredangle AD_CG$ which means that $AD_BGD_C$ is cyclic but by summing arcs we end up realising $AG$ is diameter and in fact now this means $(D_BDD_C)$ is $\omega$ from the tangencies.
To finish let $J$ be the miquelpoint of $L_1BCK_1$ then $J$ lies on $(ABC)$ but also from Reim's we get $N_A, D, J$ colinear and then Reim's twice gives $M_CX \cap M_BY=J$ and from double Reim's once again we have that $(AL'X) \cap (AK'Y)=J$ and this is excellent news because now we can note that $\measuredangle D_CJD_B=\measuredangle D_CJA+\measuredangle AJD_B=\measuredangle D_CI_CA+\measuredangle AI_BD_B=\measuredangle D_CDD_B$ which shows that $J$ lies on $\omega$ as well, but since $N_A, D, J$ are colinear from the converse of Archiemedes Lemma (or just shooting Lemma/homothety) we have that $\omega, (ABC)$ are tangent at $J$ as desired thus we are done :cool:
This post has been edited 1 time. Last edited by MathLuis, Mar 13, 2025, 8:16 PM
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AndreiVila
212 posts
#6 • 1 Y
Y by Lyzstudent
Notice that $X$ and $Y$ are the incenters of $\triangle ABE$ and $\triangle ACF$. Let $X'$ and $Y'$ be the projections of $X$ and $Y$ onto $BC$. Let $T$ and $S$ be the projections of $X$ onto $AE$ and $AB$ respectively, and let $K$ be the tangency point of $\omega$ with $AE$.

Claim 1. The $A$-excircle of $\triangle AEF$ is tangent to $EF$ in $D$.
Proof: Since $IXDY$ is a parallelogram, by projecting onto $BC$ we get that $BX'+BY'=BT+BD$. This is equivalent to $$BE+AB-AE+2BF+FC+AF-AC=BA+BC-AC+2BD.$$Simplifying yields $AE+ED=AF+FD$, which is equivalent to $D$ being the tangency point of the excircle.

Claim 2. Circle $\omega$ is tangent to $(ABC)$.
Proof: By Casey's Theorem, we need to prove that $$b\cdot BD + c\cdot CD = a\cdot AK.$$But $$AK=AT+TK=AT+X'D=AT-BX'+BD=AS-BS+BD=c-2BS+BD.$$With Thales' Theorem, $\frac{BS}{p-b}=\frac{BX}{BI}=\frac{BD}{a},$ so $BS=\frac{BD(p-b)}{a},$ thus getting $AK=\frac{ac+BD(b-c)}{a},$ and the conclusion follows.
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SimplisticFormulas
132 posts
#7
Y by
I found that $X,Y$ are in centres, $XE$ meets $YF$ in $Z=$$A$- excentre of $AEF$ and that$A$ appears to be Miquel point of $IXYZ$.
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markam
4 posts
#10
Y by
sarjinius, what solution did you have in mind at first, when you proposed this problem?
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Lyzstudent
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#11
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AndreiVila wrote:
Notice that $X$ and $Y$ are the incenters of $\triangle ABE$ and $\triangle ACF$. Let $X'$ and $Y'$ be the projections of $X$ and $Y$ onto $BC$. Let $T$ and $S$ be the projections of $X$ onto $AE$ and $AB$ respectively, and let $K$ be the tangency point of $\omega$ with $AE$.

Claim 1. The $A$-excircle of $\triangle AEF$ is tangent to $EF$ in $D$.
Proof: Since $IXDY$ is a parallelogram, by projecting onto $BC$ we get that $BX'+BY'=BT+BD$. This is equivalent to $$BE+AB-AE+2BF+FC+AF-AC=BA+BC-AC+2BD.$$Simplifying yields $AE+ED=AF+FD$, which is equivalent to $D$ being the tangency point of the excircle.

Claim 2. Circle $\omega$ is tangent to $(ABC)$.
Proof: By Casey's Theorem, we need to prove that $$b\cdot BD + c\cdot CD = a\cdot AK.$$But $$AK=AT+TK=AT+X'D=AT-BX'+BD=AS-BS+BD=c-2BS+BD.$$With Thales' Theorem, $\frac{BS}{p-b}=\frac{BX}{BI}=\frac{BD}{a},$ so $BS=\frac{BD(p-b)}{a},$ thus getting $AK=\frac{ac+BD(b-c)}{a},$ and the conclusion follows.
Excellent!!!Much better than the solution above.
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