AoPS Academy
, 
,
, 
, why 
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.. It is usually regarded that 
, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.? The issue here is that 
isn't even rigorously defined in this expression. What exactly do we mean by ? Unless the example in question is put in context in a formal manner, then we say that is meaningless.? Suppose that 
. Then we would have 
, absurd. A more rigorous treatment of the idea is that 
does not exist in the first place, although you will see why in a calculus course. So the point is that is undefined.
? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
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 is an indeterminate form.
, 
or , obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](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 
. On the other hand, 
is undefined because dividing by is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context. is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](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 it is perfectly OK to say that,![\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 is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](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 is allowed here! In fact, we have![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and are left undefined. We also have
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 , by defining them as![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by also being allowed with the same indeterminate forms (in addition to some other ones).
be a positive integer and 
a positive real number. Initially there are 
fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points 
and 
, with to the left of , and letting the flea from jump over the flea from to the point 
so that 
. such that, for any point 
on the line and for any initial position of the fleas, there exists a sequence of moves that will take them all to the position right of .
be a triangle with 
. Let the incenter and incircle of triangle be 
and 
, respectively. Let 
be the point on line 
different from 
such that the line through parallel to 
is tangent to . Similarly, let 
be the point on line different from 
such that the line through parallel to 
is tangent to . Let 
intersect the circumcircle of triangle at 
. Let 
and 
be the midpoints of and , respectively.
.
of positive integers such that for any pair of positive integers 
, the arithmetic and geometric means![\[ \frac{a_m + a_{m+1} + \cdots + a_n}{n-m+1}\quad\text{and}\quad (a_ma_{m+1}\cdots a_n)^{\frac{1}{n-m+1}}\]](http://latex.artofproblemsolving.com/2/0/b/20b6f3097b2a8fa8e3fbed54af1f7f892f4ca2fc.png)
are both integers.
, such that 
is not divisible by .
1.75, what is the average cost per ball if you buy 10 of ball A and 4 of ball B? Please write as an improper fraction
be triangle with incenter . A point 
in the interior of the triangle satisfies ![\[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\]](http://latex.artofproblemsolving.com/d/6/2/d622094b95a1f317063cf2f02a8b66b31751638f.png)
Show that 
, and that equality holds if and only if 
.
. Thus B, P, I, C lie on the same circle (Let's say with centre 
). But it is a well-known result that lies on line , since we have 
, so 
. Hence with equality holds iff 
.
, to both side of the equality we have:
, is fixed for each 
.
is also fixed, and it is clear that, the angle is equal to
is arc
.
at 
respectively.
goes through the center of circle .
is the nearest point of circle to . on the circle we have:
is , it is clear that 
.
of the perpendicular bisector of with satisfies 
.[/spam]
it is sufficient to prove 
.
Since P is inside AIB.
10 minute problem for me, too. must be a point on circle 
; 
. This is zero if (by the original form), so 
and 
is cyclic. is the closest point to 
on circle . Equivalently, the line passes through the center of circle . The center must lie on the perpendicular bisector of , which meets the bisector of angle 
at 
on the circumcircle of (the midpoint of arc ). We want to show that is the center of circle .
. Similarly, 
. The only point other than making these (signed) angles is the center of circle , so is that center and we are done.
are on a circle,we know that pass through circumcenter of therefore 
.
So, we have
From rectangles 
and 
,
and 
,
and 
are equal, as desired.
lies within rectangle 
.
,
,
,
?
i couldn't believe that this is an imo problem when i saw 
and the circle (since 
is the center), where 
Now let 
be the 
.
is cyclic, by the Incenter Excenter Lemma. Let the center of 
be 
.
it follows that 
is cyclic as well. So, 
.
and the equality obviously holds if and only if ![[asy]
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[/asy]](http://latex.artofproblemsolving.com/4/e/0/4e0293313576a6a1f5a3968cfd91a0d8fc12af17.png)
be the midpoint of arc 
of 
.
implying that 
with equality when 
.
But it is well-known that this is equal to 
therefore 
are concyclic with center 
But by the Incenter-Excenter Lemma, 
therefore 
QED
,
,
,
.
,
.
.
.
to both sides of our given equation gives us 
Furthermore, we know that 
by what we just found. Thus, 
Thus, we know that 
.
and the center of circle 
are collinear, so the shortest distance between 
and 
holds if 
and all other scenarios result in 
as desired.
and 
where 
.
so 
is cyclic. Extend 
Then 
are collinear, 
Thus, 
with equality occuring iff 
.
.![\[\angle{BIC} = 90^{\circ} + 90^{\circ} - \angle{PBA} - \angle{PCA} = \angle{BPC}\]](http://latex.artofproblemsolving.com/3/1/e/31e2f7c312e42b17729250907a9a087d4070bf7e.png)
,
,![\[LP + PA \geq LA\]](http://latex.artofproblemsolving.com/a/2/1/a21b9719f211be30ab44534164e94fd228bebe01.png)
as 
.![\[AP \geq AI\]](http://latex.artofproblemsolving.com/2/7/9/2798e076f62149aa935a403f8891e370653de5f2.png)
is a degenerate triangle; i.e., when 
is cyclic due to the angle chase. Use Chicken foot to conclude.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/5/e/a/5eafe259a04e97399aee35fb3acd4f7e9f532c75.png)
![\[\angle PBA+\angle PCA = \angle B - \angle PBA + \angle C - \angle PCA\]](http://latex.artofproblemsolving.com/f/6/f/f6f751345394d472e2a588ac45800dedc0d57917.png)
which simplifies to ![\[\angle PBA + \angle PCA = \frac{\angle B}{2}+\frac{\angle C}{2}.\]](http://latex.artofproblemsolving.com/1/a/f/1afbd6517fa7194cdd467bcf75cd57d886e64c45.png)
We can rearrange this to ![\[\angle PBA - \frac{\angle B}{2} = \frac{\angle C}{2}-\angle PCA.\]](http://latex.artofproblemsolving.com/9/6/0/9603fd64b0e39a93026abd3c3514fa4481aa2f54.png)
This implies that 
.![\[AP \ge AI\]](http://latex.artofproblemsolving.com/a/3/d/a3dc56739aebbe9d896fbf773a2f10a789327719.png)
for any 