,
,
,
,
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.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
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 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
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 
given by the mapping 
is undefined for 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
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 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In ![\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*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
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 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
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 ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
and 
Prove that
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be non-negative real numbers such that 
Prove that
such that![\[
\sqrt{x - 2011} + \sqrt{2011 - x} + 10
\]](http://latex.artofproblemsolving.com/4/8/b/48b82d13b733947420a0d763c6af74636770835a.png)
is an integer.
-
?
students want to share 
pencils. If every student gets at least one pencil, how many ways are there to distribute the pencils?
feet and area 
Y by megarnie, mathematicsy, jhu08, centslordm, Adventure10, Mango247
The inscribed circle of triangle 
touches the sides 
, 
and 
at 
, 
and 
respectively. Let 
denote the other point of intersection of 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
touches the sides 
, 
and 
at 
, 
and 
respectively. Let 
denote the other point of intersection of 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
,
,
.
.
,![\[ t = \frac{AQ}{AD} = \frac{a(s-a)^2}{c^2(s-c)+b^2(s-b)-a(s-b)(s-c)}. \]](http://latex.artofproblemsolving.com/e/0/b/e0b2ec4aae9f7eed9bc7b28954c4ead44de41209.png)
![\[ Q = \left(1-t, t \cdot \frac{s-c}{a}, t \cdot \frac{s-b}{a} \right). \]](http://latex.artofproblemsolving.com/6/8/b/68bad710fc1e008302367c6827d12c32b67aa8c2.png)
as the midpoint of 
,
,![\[ \det \left[ \begin{array}{ccc} 1-t & t \cdot \frac{s-c}{a} & t \cdot \frac{s-b}{a} \\ s-c & 0 & s-a \\ c+(s-b) & s-a & 0 \end{array} \right] = 0 \]](http://latex.artofproblemsolving.com/4/d/2/4d2159cff8b9bf1848c3f13d8880cc4a4ab8f66f.png)
![\[ \left[ c(s-a)(s-c) + 2(s-a)(s-b)(s-c) \right] = a(s-a)^2 \left( \frac{1}{t} - 1 \right) \iff (s-a)(s-c)(c+2(s-b)) = a(s-a)^2 \left( \frac{1}{t} - 1 \right)\]](http://latex.artofproblemsolving.com/9/0/7/907dc41896d240bbc687ba6470625d3c53ad589e.png)
,![\[ c^2(s-c) + b^2(s-b) - a(s-b)(s-c) - a(s-a)^2 = (s-a)(s-c)(c+2(s-b)) \]](http://latex.artofproblemsolving.com/2/f/1/2f1f1719219e12a25e1f342d67cdeb659c4456a2.png)
,
,
.
,
and 
.
gives
.
.
.
and 
.
is harmonic, so 
.
is the midpoint, then 
,
,
,
,
;
,
.
,
,
,
.
be the incricle. Define 
such that 
.
.
since 
.
,
.
,
But 
,
and 
.
.
is the point at infinity along line 
.
,
which implies the problem.
Notice ![\[-1=(Q,D;E,F)\stackrel{E}=(P,\overline{DE}\cap\overline{AB};A,F)\]](http://latex.artofproblemsolving.com/1/a/d/1ad45a237a0b0f6b5dcc322c6a5122702fcddd80.png)
so 
if and only if 
But 
so 
is equivalent to 
as desired. 
,
is harmonic. Let 
.
is parallel to 
so 
,
but since 
.
is the polar of 
.
Thus, since 
,
is the point at infinity along line 
iff 
.
and 
be the point on the incircle such that 
.
is the midpoint iff 
.
,
passing through the intersection of the tangents through 
.
is cyclic, by Reim's 
should be as well, hence by power of a point 
,
is harmonic. Let ![\[ (EF;QD) \overset{E}= (A,F;M,\overline{ED} \cap \overline{AB}) = -1. \]](http://latex.artofproblemsolving.com/8/4/d/84df8505c76cb88d3de9bccbea3ffc172b8b64c3.png)
This in turn in only holds if 
which holds when ![\[ \frac{AC}{BC} = \frac{EC}{DC} = 1. \]](http://latex.artofproblemsolving.com/9/8/6/986a3472e684914b7b769a69b91a63bd444701fb.png)
is harmonic. Projecting through
.
is a harmonic pencil. On the other hand 
.
and we are done. 
As its given 
by POP we get
hence 
.
and 
.
are tangents and 
we have that,![\[(QD;EF)=-1\]](http://latex.artofproblemsolving.com/0/b/5/0b57400ce853755d55c9c2ac3348359a72f619cd.png)
Now, if 
.![\[(AF;MP_\infty) \overset{E}{=}(EF;QD)=-1\]](http://latex.artofproblemsolving.com/4/1/e/41eb1b90b1042af2ceb696e730916872202a9e6d.png)
This means that ![\[-1=(EF;QD) \overset{E}{=} (AF;MP)\]](http://latex.artofproblemsolving.com/3/0/3/3036bd81204aeada5eb1b940b285e6990272b771.png)
where 
.
which implies that 
and thus, 
is isoceles with 
![[asy]
import olympiad; import cse5;
defaultpen(fontsize(10pt));
usepackage("amsmath"); usepackage("amssymb"); usepackage("gensymb"); usepackage("textcomp");
size(8cm);
pair MRA (pair B, pair A, pair C, real r, pen q=anglepen) {
r=r/2;
pair Bp=unit(B-A)*r+A;
pair Cp=unit(C-A)*r+A;
pair P=Bp+Cp-A;
D(Bp--P--Cp,q);
return A;
}
pointpen=black+linewidth(2);
pen polyline=linewidth(pathpen)+rgb(0.6,0.2,0);
pen polyfill=polyline+opacity(0.1);
pen angleline=linewidth(pathpen)+rgb(0,0.4,0);
pen anglefill=angleline+opacity(0.4);
markscalefactor=0.01;
size(12cm);
pair A=dir(180+50),B=dir(-50),C=dir(90);
// filldraw(A--B--C--cycle,polyfill,polyline);
D(A--B--C--cycle,polyline);
pair I=incenter(A,B,C);
pair D=foot(I,B,C),E=foot(I,C,A),F=foot(I,A,B);
pair Q=2*foot(I,A,D)-D;
pair M=(A+F)/2;
D(CP(I,D));
D(A--D);
D(E--M,pathpen+dashed);
D("A",D(A),A);
D("B",D(B),B);
D("C",D(C),C);
D("I",D(I),dir(90));
D("D",D(D),unit(D-I));
D("E",D(E),unit(E-I));
D("F",D(F),unit(F-I));
D("Q",D(Q),dir(174));
D("M",D(M),S);
[/asy]](http://latex.artofproblemsolving.com/6/d/4/6d401bcc37e4157a91878d96feae70826e172c7b.png)
![\[-1 = (DQ; EF) \stackrel{E}{=} (DE \cap AB, EQ \cap AF; AF)\text{.}\]](http://latex.artofproblemsolving.com/4/d/0/4d0f84eaa6e3add938b0b232d8df4489ab479516.png)
Now if 
so 
must necessarily be 
then 
and so 
.
is a harmonic quadrilateral, we have 
so 
.
.
,
.
and from 
,
Thus, 
is the midpoint of 
.
we know by similar triangles 
be 
be 
be 
as ![\[-1 = (A, F; \overline{QE} \cap \overline{AF}, P_\infty) \stackrel E= (A, \overline{EF} \cap \overline{AD}; Q, \overline{EP_\infty} \cap \overline{AD}).\]](http://latex.artofproblemsolving.com/f/b/f/fbffa4bac924aa9fcb1f58e8cad4755598f0eb78.png)
But 
