,
,
,
,
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 
,
.
.
.
satisfying 
,
.
,
be the circumcentre, 
be the orthocentre, and 
be the centroid.
satisfy the following condition:
is constant.
Y by Adventure10, PikaPika999
Suppose that 
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
,
-
and 
are collinear; therefore, the perpendicularity of 
and 
is equivalent with the fact that 
.
(the power of 
and so 
.
.
and 
(since the circumcircle of 
is the nine-point center of 
cuts the circumcircle 
of 
and 
is a circle with center 
cuts 
is cyclic and the angle 
is right. Since 
is a diameter of 
and 
goes through 
)
is perpendicular to 
,we are done.
.
and 
have 
as the radical axis and as 
are concurrent, then 
is cyclic. So 
is 
to 
as 
is the diameter of 
.
;
w.r.t. 
we get (already proven 
are collinear): 
,
is cyclic, i.e. 
.
,
,
are collinear, and we are done.
is the midpoint of the arc 
,
is the 
intersects 
is the incenter of 
and 
,
and ![\[\angle B_{1}BI\\=\angle IBC+\angle B_{1}BC
\\=\angle IBC+\angle B_{1}AC
\\=\angle IBC+(\angle R-\frac{1}{2}\angle AB_{1}C)
\\=\angle R+(\angle IBC-\frac{1}{2}\angle ABC)
\\=\angle R\cdot \cdot \cdot (6)\]](http://latex.artofproblemsolving.com/6/7/7/677420e109fa1a2b2bed326bd115276ddb6cd335.png)
and 
,
and 
are collinea
,
be a point such that 
and 
,
are concyclic
and 
and 
are concyclic
, 
and 
are concyclic
and 
,
and 
are collinear
(
is a radical ceneter)
,
,
and 
,
.
.
be the line through 
perpendicular to 
.
.
are collinear on the external angle bisector of 
are collinear by the definition. Now consider the circles 
.
while 
since 
and 
since 
is a diameter, where 
is the 
.
is 
,
is 
,
is 
,
are collinear, showing 
.
,
,
has coordinates 
.
be the 
excenter, By Brokard's Theorem on Quadrilateral 
and 
inversion, 
swap, so we're done because then we must have
which means that 
.
.
.
is cyclic.
,
.
is cyclic too. As 
Its consequence leads to the fact that 
.
)
the intersection of line 
with the circumcircle of 
.![\[B_2B \cdot B_2B_1 = B_2C \cdot B_2A = B_2P \cdot B_2I_B.\]](http://latex.artofproblemsolving.com/9/6/4/96421bc720fe61b02e3919130ca77c79e47d729b.png)
Thus 
are concyclic. Now by the incenter/excenter lemma, we have that 
.![\[\angle I_BPI = 90 = \angle I_BBB_1 = \angle I_BPB_1.\]](http://latex.artofproblemsolving.com/a/2/7/a27b81ccfacba67e9030325fdeee756a43c97e95.png)
Therefore 
are collinear, and 
,
.![\[BI\cdot BI_B = BB_1\cdot BB_2.\]](http://latex.artofproblemsolving.com/e/1/f/e1f805bf6fd4471a5ebaeb9de8300a2c7d69bce5.png)
Let 
and 
be the intersections of line 
with 
.
,
.
and 
,
.![\[BB_1\cdot BB_2 = BD\cdot BM = BI\cdot BI_B\]](http://latex.artofproblemsolving.com/3/5/7/3579b2a835b31fdcea59a6196ca4462e6417e23f.png)
as desired.
is simply 
.![\[\measuredangle IAI_b=\measuredangle ICI_b=90^\circ\]](http://latex.artofproblemsolving.com/d/4/b/d4b74190fff0dbce1ca7c2ecc0aefacb690144b9.png)
and thus, 
.
,
also lies on 
.
.![\[B_2I\cdot B_2B_3' = B_2B \cdot B_2B_1 = B_2A\cdot B_2C\]](http://latex.artofproblemsolving.com/d/4/0/d400d91248e047569cc5dd80d05fe0750e552886.png)
Thus, 
must also lie on 
and indeed, 
as required.
![[asy]
import olympiad; size(400);
draw(unitcircle);
pair B = dir(130), A = dir(200), C = dir(-20), M = midpoint(A--C), O = origin, B1 = intersectionpoint(O--(O+3*(O-M)),unitcircle);
pair I = incenter(A,B,C), L = intersectionpoint(B--(I+3*(I-B)),unitcircle), Ib = L + L-I;
pair exB = rotate(90,B) * I, B2 = extension(exB, B, A, C);
dot("$A$",A); dot("$B$",B); dot("$C$",C); dot("$I$",I); dot("$O$",O); dot("$M$",M); dot("$L$",L); dot("$I_B$",Ib); dot("$B_1$",B1); dot("$B_2$",B2);
draw(C--B2--B--C^^A--B--Ib^^Ib--B1--L^^B2--extension(B2,I,B1,Ib));
draw(B--B1);
// draw(incircle(A,B,C);
[/asy]](http://latex.artofproblemsolving.com/7/e/4/7e4b00a850d18aa54a1429fef14ca203eb05877b.png)
(since 
is diameter), all 
,
(from right triangle 
)
(from cyclic quad 
)
so from similar triangles 
.
,
so 
.
wrt 
and 
be 
.
are concyclic by incenter/excenter lemma we have that 
and because 
,
are concyclic.Then it succifies to show that 
are collinear which is trivial by the radical axis concurrence lemma on 
,
are concurrent at 
and 
.
lie on the same line, and 
