,
,
,
,
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 
.
,
,
be the midpoints of the sides 
,
respectively. A circle 
passing through 
and tangent to 
and 
at 
and 
,
and 
are symmetric to 
meets 
and 
at 
and 
meets 
.
of acute triangle 
.
.
be 
.
,
,
,
,
,
similarly. Prove that 
,
,
are concurrent.
sigmas = 
looksmaxxers, and 
looksmaxxers = 
skibdis, how many skibidis are in a sigma? Express your answer as a common fraction
![$P(x) \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/4/5/3/453a624c3b002c1b0e78e0023b24dd22ddd03557.png)
such that:
with 
is an 
th prime and 
is an arbitrary positive integer.
grid with each cell colored with an integer from 
.
prime?
Y by Adventure10, Mango247, Rounak_iitr, and 1 other user
Let 
, 
and 
be midpoints of sides 
and 
, respectively, and let 
be circumcenter of acute-angled triangle 
. Circumcircles of triangles 
and 
intersect at two different points 
and 
inside of triangle . Prove that ![\[\angle BAX=\angle CAY.\]](//latex.artofproblemsolving.com/3/9/9/39910fdd000cbed820b33ac08a8200ca06ed8193.png)
, 
and 
be midpoints of sides 
and 
, respectively, and let 
be circumcenter of acute-angled triangle 
. Circumcircles of triangles 
and 
intersect at two different points 
and 
inside of triangle . Prove that ![\[\angle BAX=\angle CAY.\]](http://latex.artofproblemsolving.com/3/9/9/39910fdd000cbed820b33ac08a8200ca06ed8193.png)
![$[BC] = a, [CA] = b, [AB] = c$](http://latex.artofproblemsolving.com/7/5/0/75011d7d392ed255c5334be27b558f15cf10f5dc.png)
and 
.
,
and 
with radii ![$[OA]=R$](http://latex.artofproblemsolving.com/5/f/7/5f7b00297950d0dc4251e654ae96cc8f7c459b0f.png)
,![$[KM]=\frac{R}{2}$](http://latex.artofproblemsolving.com/a/9/c/a9c0e4971fc5a2a1b42ab323c777227681724054.png)
and ![$[QO]=\frac{R}{2 \cos \widehat A}$](http://latex.artofproblemsolving.com/a/1/9/a190b24d0325af751c927d89997e66b268b13334.png)
,
cut 
.
cut 
.
is W-isosceles and 
.
.![$\frac{p(A, (K))}{p(A, (Q))} = \cos^2 \widehat A = \frac{[KM]^2}{[QO]^2}$](http://latex.artofproblemsolving.com/9/1/9/9196884d466e324c3750decaf895ffe755a35f6b.png)
![$\frac{[AK]^2}{[AQ]^2} = \frac{p(A, (K)) + [KM]^2}{p(A, (Q)) + [QO]^2} = \frac{[KM]^2}{[QO]^2} $](http://latex.artofproblemsolving.com/d/1/3/d13d124db66ee252c863b4722bdd71e4e0b69cde.png)
![$\frac{[AK]}{[AQ]} = \frac{[KM]}{[QO]}$](http://latex.artofproblemsolving.com/0/e/2/0e295760298214c2e3a916bbf4dd181b26d849da.png)
.
cut center line 
at 
,
are isogonals WRT 
bisect 
![$-\frac{\overline{IK}}{\overline{IQ}} = \frac{\overline{JK}}{\overline{JQ}} = \frac{[AK]}{[AQ]} = \frac{[KM]}{[QO]}$](http://latex.artofproblemsolving.com/9/0/6/906db1ae59879f15bcdb8f37ea9e7347a98b066a.png)
.![$\frac{p(I, (K))}{p(I, (Q))} = \frac{[IK] ^2 - [KM^2]}{[IQ]^2 - [QO]^2} = \frac{[KM^2]}{[QO]^2} = \frac{p(A, (K))}{p(A, (Q))}$](http://latex.artofproblemsolving.com/0/a/8/0a8058be6c25a25141c5e5402a829525b5de92da.png)
and likewise, 
is coaxal with 
.
.
be feet of altitudes from 
to 
then consider inversion of pole 
it pictures the Euler's circle of 
into circle 
and circle 
is the point where circle 
what circle 
and the symmetry wrt the internal bisector of 
we picture 
intersect 
than 
so 
and 
so 
are not collinear. also if 
or 
.
which implies 
a contradiction.
must be bisectors of 
together with the reflection w.r.t. the angle bisector of 
.
is circumcircle 
,
is Euler's circle of 
is Euler's line 
.
intersection 
second times.)
circumcenters triangle 
is circumcircle of triangle 
and 
(*)
is radical axis circles 
- cyclic and circumcircle is 
and 
,
concurrent;
,
concurrent;
,
concurrent;
,
concurrent;
Such that 
is cyclic.
and 
similar and 
and ![\[ AX.AY_1 = AY.AX_1.\]](http://latex.artofproblemsolving.com/3/7/8/3787d45a147c77a161c53eefd5c5177693514c11.png)
and 
collinear and 
so the isogonal conjugate of S
an
.
with circle 
are not on the bisector of 
such that 
and 
it's easy to prove 
,
,
and 
so ow 
so 
so 
and 
and 
followed by reflection in bisector maps these two circles two one another. Since the intersection points aren't fixed points under this operation, 
.
.
.
and 
by 
since I like this label. Let 
be the feet of altitude from 
of 
.
.
gives 
are concurrent at 
gives 
are concurrent at 
.
cut 
at 
.
cut 
.
.
as desired.
and quadrilateral 
,
By above claim, this involution must be isogonality w.r.t. 
be the transformation for which we first invert around 
and reflect the whole image along the angle bisector of 
From this we have that this transformation sends 
to 
,
followed by symmetry wrt bisector of angle 
are swapped, 
are swapped. Also note that 
so 
,
be the Dumpty point of vertice 
, reflect through the bisector of angle 
,
.
.
so 
inversion, which swaps 
,
.
,
by symmedian properties. Hence 
and 
and 
are isogonal.
-![\[ \odot(BOC)\leftrightarrow \odot(MNP). \]](http://latex.artofproblemsolving.com/0/d/3/0d38d04543819865870cdad28ae319f480d1134b.png)
which clearly finishes. 
,
.
and 
swap under 
,
inversion, so their two intersections swap. 
inversion, so their intersections swap as well, implying 
.
)
and so 
inversion centered at 
b
and 
inversion centered at 
Then:
so 
Furthermore, the foot from 
Hence, by taking a 2 times homothety centered at 
we get that the nine point circle and 
inversion at 
,
,
.
is the nine point circle of 
