,
,
,
,
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 
,
intersects the extension of 
at point 
.
is the midpoint of 
,
intersects the circumcircle of 
.
.
be positive real numbers such that 
.![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
has how many digits?
with integer coefficients such that for all positive number 
and prime 
satisfying 
,
.
and 
.
.
have the digit 
and are divisible by 
?
![\[(2+0+2+4)\left((2\cdot2\cdot4)^2-\dfrac{2^2+0^2+2^2+4^2}{2+0+2+4}\right)=2024.\]](http://latex.artofproblemsolving.com/0/0/7/0074cd2c9c2f8327db52e113ef2beefe98ceca90.png)
![\[(2+0+2+4)(2\cdot2\cdot4)^2-\left(2^2+0^2+2^2+4^2\right)=2024.\]](http://latex.artofproblemsolving.com/d/5/4/d54152fae9e6e259dc464dbef3ed39272eb9e599.png)
Y by buratinogigle, Davi-8191, Adventure10, Mango247
Given an acute-angled triangle ABC, let 
be the orthocenter. A cirlcle passing through the points 
and a cirlcle with a diameter 
intersect at two distinct points 
. Let 
be the foot of the perpendicular drawn from 
to line 
, and let 
be the foot of the perpendicular drawn from to line 
. Show that 
.
be the orthocenter. A cirlcle passing through the points 
and a cirlcle with a diameter 
intersect at two distinct points 
. Let 
be the foot of the perpendicular drawn from 
to line 
, and let 
be the foot of the perpendicular drawn from to line 
. Show that 
.This post has been edited 1 time. Last edited by Kunihiko_Chikaya, Feb 11, 2013, 4:54 PM
and 
be the feet of the B- and C- altitudes. 
and 
with diameter 
the circle 
with diameter 
and the circle 
concur at their radical center 
From the complete quadrangle 
the cross ratio 
is harmonic. Since 
then it follows that 
bisect 
internally and externally.
be a triangle and 
is point such that 
.
are projection of 
,
cuts 
at 
is circle diameter 
.
is a circle passing through 
.
is radical axis of 
.
is projection of 
.
.
are cyclic, so 
,
is cyclic, now, all you have to do is to apply the same argument as in the initial problem.
be altitudes in triangle 
is cyclic. Note that 
,
is radical axis of circle passing through points 
.
and 
is perpendicular to 
and we are done.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/7/a/3/7a340f64b32c32be984340e6d19feb791d959fcf.png)
and 
be the feet of the perpendiculars from 
and 
,
be the intersection of 
and 
to get that 
and 
are colinear.
and since 
.
be the feet of perpendiculars and let 
be the circle passing through 
,
are concurrent at a point 
.
,
are harmonic points, 
.
.
and if we grab the 3 radical axes it intersects at one point 
and 
form a harmonic point. therefore We are done.
