,
,
,
,
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 
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
,
is an homoemorphisme too.
,
and 
are both integers.
be real numbers such that 
Prove that 
Let 
Prove that 
MathJuice bottles and Bob starts with 
MathJuice bottles. An unfair coin is then flipped, with probability 
of landing heads. If the coin lands heads, Alice gives Bob a bottle; otherwise, Bob gives Alice a bottle. This process repeats until someone runs out of bottles.
rest on a plane. A sphere with radius 
is tangent to all of them, but does not intersect nor lie on the plane. A sphere with radius 
lies on the plane, and is tangent to all three spheres with radius 
Compute the shortest distance between the sphere with radius 
Y by mathuz, Davi-8191, jhu08, Adventure10
Let 
to be an acute triangle. Also, let 
and 
to be the two intersections of the perpendicular from 
with respect to side 
with the circle of diameter , with closer to than . Analogously, 
and 
are the two intersections of the perpendicular from 
with respect to side 
with the circle of diamter , with closer to than . Prove that the intersection of 
and 
lies on 
.
to be an acute triangle. Also, let 
and 
to be the two intersections of the perpendicular from 
with respect to side 
with the circle of diameter , with closer to than . Analogously, 
and 
are the two intersections of the perpendicular from 
with respect to side 
with the circle of diamter , with closer to than . Prove that the intersection of 
and 
lies on 
.
be the orthocenter of 
be its projections onto 
respectively. Note that the circle with diameter 
,
.![\[
HY\cdot HX=HD\cdot HA=HK\cdot HL
\]](http://latex.artofproblemsolving.com/5/a/5/5a5c03a9a4915824d271084f2574e0fd76ed56fb.png)
is cyclic. Additionally, since 
are perpendicular bisectors of 
,
is the center of this circle. Now since 
intersect at 
,
are inverses with respect to this circle.
,
is the midpoint of 
,
and 
and 
,
is mapped to 
be the projections from 
,
,
,
is tangent to the circumcircle of 
,
,
is also tangent.
and 
are tangent to the circumcircle of 
gives 
,
,
are collinear, which is the same as 
,
,
intersects 
with 
.
intersects 
with 
.
,
are concurrent.
,
and 
are collinear. As 
,
is cyclic, as 
,
.
and 
,
also lies on the polar of 
lies on 
,
.
;
.
.
are the circle of diameter 
and 
meet at 
.
.
are on the same circle. Also, 
and 
.
,
.
.
,
.
.
.
.
.
is similar to 
.
is similar to 
.
and 
.
.
are collinear. So 
.
is a cyclic quadrilateral. Thus, 
.
so 
is cyclic
as well as 
so it is the center of 
be the orthocenter of 
.
lie on the polar of 
,
. ... 
lie on the polar of 
,
. ... 
.
,
;
fixes each of 
,
and thus quadrilateral 
are tangent to 
,
clearly 
to 
respectively.
But this is easy, we have 
Hence we know that 
Consequently, 
Similarly, we can deduce that 
Therefore, not only do we know that 
are concyclic, but we also know that the polar of 
which is precisely 
On top of that, if 
and 
then by Brocard the polar of 
Therefore, the lines 
which is the intersection of 
lies on 
as desired.
be the feet of the altitude from 
.
is cyclic. By Power of a Point, 
hence shown.
is side 
,
and 
are tangent to 
.
is a harmonic quadrilateral. Let 
,
Finally, let 
Similarly, 
Hence, 
,
and 
are cyclic. By radical center on these two circles, it follows that 
and 
,
.
,
where 
is also cyclic, we have 
as well, hence 
is the polar of 
,
be a line and 
are concurrent at a point 
and 
such that 
and 
.
lies on the perpendicular to 
and 
at 
respectively. Then 
,
as desired.
bisects 
and 
,
![\[HK \cdot HM = HB \cdot HE = HC \cdot HG = HL \cdot HN,\]](http://latex.artofproblemsolving.com/e/7/d/e7d8ccb56406eaa3723a57af90ad569aeada441d.png)
,![\[CH \cdot CG = CF \cdot CB = CK^2.\]](http://latex.artofproblemsolving.com/2/f/4/2f48b385bd321532dc249e3c30ea5ec79dd12650.png)
lies on 
;
,
has polar 
concur at the orthocenter 
.
and 
.
.
tangents to 
so 
is a harmonic quadrilateral. We also have 
and similarly 
.
which implies that 
which finishes by symmetry.
centered problems are... annoying. Great problem though 
,
intersects 
.
are concurrent.
are altitudes. Let 
be the third altitude and 
.
is a cyclic kite, which means that 
.
with perspective at 
.
.
,
and 
.
onto line 
,
onto line 
.
.
concur at a point. 
by some famous harmonic bundle lemma we have 
.
Analogously, we find that 
,![\[CK = CM = \sqrt{CF \cdot CB} = \sqrt{CE \cdot CA} = CL = CN.\]](http://latex.artofproblemsolving.com/c/1/1/c112948b8434d2504a5809670c228766145acf87.png)
,
are collinear.
,
,
giving us that 
bisector of 
so 
which is 
must lie on this polar which is what we needed to show.
concur in 
since the perpendicular bisectors 
of 
meet in 
and 
,
sends 
so 
is cyclic. Furthermore, 
since 
so 
Thus, 
which implies 
os tangent to 
finishes.
