,
,
,
,
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 
with no parallel sides is inscribed in a circle 
.
are inscribed in triangles 
,
and 
,
,
,
.
by 
and the number of distinct prime divisors of 
.
be a positive integer. Prove that there exist infinitely many positive integers 
such that 
and 
does not divide 
for any positive integers 
satisfying 
.
such that for any 
:
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
![$\sqrt[2]{2}$](http://latex.artofproblemsolving.com/1/8/6/1869ac018fb72a26a12ab14499cbf1b1c13b4a10.png)
,![$\sqrt[3]{3}$](http://latex.artofproblemsolving.com/1/7/a/17a6bb5dc11696b8650b6d3a11e770f861f4c326.png)
,![$\sqrt[5]{5}$](http://latex.artofproblemsolving.com/a/b/9/ab90f2d42dc24065e2a64988fe76f22eb9359799.png)
.
such that![\[
\sqrt{x - 2011} + \sqrt{2011 - x} + 10
\]](http://latex.artofproblemsolving.com/4/8/b/48b82d13b733947420a0d763c6af74636770835a.png)
is an integer.
,![\[ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 46 \qquad
\textbf{(D)}\ 66 \qquad
\textbf{(E)}\ 74
\]](http://latex.artofproblemsolving.com/8/2/f/82ffc15072ca99f6cd1abc312781eb4827817356.png)
:o
so 
and since 
is just 
therefore 
so 
Y by
Let 
be an acute triangle, 
fixed, 
moves on the big arc 
of 
. Let 
be the circumcenter of 
are not collinear, 
, 
is the incircle of triangle . tangents to at 
. Let 
be the -excenter of triangle . 
cuts 
at 
. Let 
lies on such that 
.
a)
cuts 
at 
. Prove that 
.
b) Let
lies on the circle 
go through 
such that 
. 
cuts again at 
. Prove that the midpoint 
of 
lies on a fixed circle.
be an acute triangle, 
fixed, 
moves on the big arc 
of 
. Let 
be the circumcenter of 
are not collinear, 
, 
is the incircle of triangle . tangents to at 
. Let 
be the -excenter of triangle . 
cuts 
at 
. Let 
lies on such that 
.a)
cuts 
at 
. Prove that 
.b) Let
lies on the circle 
go through 
such that 
. 
cuts again at 
. Prove that the midpoint 
of 
lies on a fixed circle.This post has been edited 1 time. Last edited by 799786, Mar 5, 2022, 6:48 AM
and 
.
be contact points of 
be Bevan point of 
.
,
.
be 
- excenter, 
- excenter of 
is orthocenter of 
so 
and 
.
and orthocenter 
.
be orthogonal projection of 
such as 
.
intersects 
again at 
be orthogonal projection of 
then it's easy to see that 
are homothetic. So 
which means 
.
we have 
is harmonic quadrilateral. Hence if we let 
be second intersection of 
with 
or 
are collinear
be midpoint of arc 
be intersection of 
.
which is fixed. We consider the following configuration
with 
are fixed, 
intersects 
intersects the line through 
tangents at 
be 
intersect 
.
then 
Then 
.
are fixed then 
are fixed, hence 
