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 triangle. Let 
be the feet of the altitudes from 
respectively. Let 
be the projections of onto line 
. Show that 
.
be real numbers such that ![\[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \]](http://latex.artofproblemsolving.com/8/f/5/8f5d49c9b44e3db3cd85fea7b6521e1dc02d4269.png)
Prove that ![\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]](http://latex.artofproblemsolving.com/c/0/4/c0438d6eba686d0a11fd4decb4afa9a14aa50bac.png)
are non-negative integers.
rest on a plane. A sphere with radius 
is tangent to all of them, but does not intersect nor lie on the plane. A sphere with radius 
lies on the plane, and is tangent to all three spheres with radius 
Compute the shortest distance between the sphere with radius and the sphere with radius 
be a triangle with orthocentre 
. The tangent lines from 
to the circle with diameter 
touch this circle at 
and 
. Prove that 
and are collinear.
be a triangle with orthocenter . The tangent lines from to the circle 
with diameter ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
touch it on and . Prove that 
.
of 
, 
, 
for which 
and 
. Observe that 
, i.e. 
what is a characterization of 
- harmonical division, i.e. is harmonical conjugate of w.r.t. 
. In conclusion, belongs to polar of w.r.t. , i.e. .
separates , 
. Denote 
so that 
and 
so that 
. Observe that 
and 
. Thus 
, i.e. 
, 
, 
are concurrently .
be a cyclic hexagon. Prove that 
.
as the circle of reciprocation.
are collinear its the same as proving that their polars are concurrent. Because both belong to the circle, their polars are the tangents to the circle that meet at . passes through .
be the perpendicular to 
(
is the midpoint of ). If 
is the inverse of , then is the polar of and we are done.
.
~
we have :
. 
the altitudes of 
, and 
the midpoint of . Because 
, the points 
are concyclic. Let 
the circumcircle of 
. and radius 
. Because 
, we obtain that is the image of 
under this inversion, and 
remains invariant. For the other hand, is transformed into the line containing . Because 
, we can conclude that 
.
and are the intersections of the circles 
and 
, and 
being the midpoints of 
and 
respectively.
are the feet of the altitudes on 
and respectively and these altitudes concur at , then we know that 
, i.e. has equal power w.r.t. to both circles, consequently it belongs to their radical axis .
are switched.
are switched.
from Virgil's lemma, we can also assume that it's wrong i.e there is three intersection point, if I remembre well it's we get that 
with 
, 
and 
which obviously wrong, and then deduct the desired result.
and 
(X is on the same side with A from BC)
is collinear and perpendicular to 
and 
is harmonic
is harmonic
is collinear since 
is collinear
.Let me inform you that this problem is from China 1997 and i have three solutions but two of them have already posted so i'll post the third one:
the altitudes of , and the midpoint of .Since 
then the points are concyclic.
and 
be a triangle and its altitude 
.The tangent lines from to the circle with diametre touch this circle on and .If 
then prove that is the orthocenter of 
be the midpoint of .Extend 
to meet 
at 
where and are on the same side of .Now is the pole of wrt the circle 
.Again note that 
is the polar of .So polar of passes through 
lies on .Hence proved.
, , 
. Let the foot of the perpendicular from to be 
. By http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2173258&sid=fd7168f527cac736647f7961645922f0#p2173258, 
so the polar of with respect to the semicircle with diameter passes through , and therefore . So , , are collinear.
and .
lie on the circle 
with diameter .
cut at 
. 
is the polar of with respect to 
is also the polar of . It follows that 
are collinear.
be the midpoint of , 
be the foot of onto , 
be the foot of 
onto 
. We just want to show that is on the radical axis of the circles with diameters and (which is ). But this is obvious, since 
. is implied to be acute...
be the intersections of the -altitude with the circle.
which is obvious.
be the feet of perpendiculars from respectively 
as the center and radius 
. to themselves and it takes to feet of perpendicular from to (
)thus
and are concyclic hence the conclusion
is on the cicrcle such as the is perpendicular to and analogus the point F.
and 
and we have from theorem of Hire that the intersection point of BE and CF is on the polar of the point A which is PQ so P,Q and H are collinear.
be circumcircle of and 
be a circle so that 
and exists 
such that 
Let 
and the symmetric of 
w.r.t. 
The tangent lines from to the circle touch it on and 
Prove that 
(see PP10 from here).
. Then let 
be points on the unit circle, so 
. Then if 
is the orthic triangle, 
Now with the chord intersection formula, 
It remains to compute 
so 
are collinear, as required.
be a triangle with orthocenter . The tangent lines from to the circle with diameter touch it on and . Prove that .
denote the mid of segment 
Clearly, 
are concyclic in that order. Also, 
..Therefore 
must lie on the radical axes of the circles 
and 
which clearly is the claimed line the 
.
be a cyclic hexagon. Prove that . lies outside triangle in given problem.
wrt 
both pass through , so by La Hire it suffices to prove that the polar of does as well. Let the reflection of over the midpoint of be 
, and 
; we can see that
by orthocenter reflection, and since is the -queue point, we know that 
, which implies that 
is the polar of .
and radius . Then the problem is equivalent to showing that lies on the radical axis of this circle and the circle with diameter . We can do this by showing has equal powers with respect to both circles.
or 
. Notice that we have 
, and 
, giving 
. Then let the foot from to be . We can then write 
. We then want to show 
, which we can reduce to 
. Then by Law of Cosines, we have 
where is the foot from to . We can then write 
. By Power of a Point on 
, we have 
. We then just need to show 
. Multiply everything by two to get the equivalent 
and rearranging gives the equivalent 
, but 
, so we are done by LOC.
be the altitude from to 
and the circumcenter of 
. Also denote by 
the angle 
modulo 
. Then we make the following claim:
is cyclic.![\[\measuredangle APO = \measuredangle AQO = \measuredangle ADO = 90^{\circ}. \blacksquare\]](http://latex.artofproblemsolving.com/6/7/b/67b7b13bcc6f410eb5ce219358d90f2ee5ea159e.png)
and 
, that is the radical center of all of our present circles so that ![\[AP^2 = AQ^2 = AH \cdot AD \implies \triangle AHP \sim \triangle APD; \phantom{c} \triangle AHQ \sim \triangle AQD.\]](http://latex.artofproblemsolving.com/d/2/a/d2a48806b7ca4c79586873d9f66fa476062228a5.png)
so that are collinear, as desired. 
be the orthic triangle, be the midpoint of .
. Also, 
clearly lie on the circle with diameter . with radius 
, then and go to themselves, and goes to .
are concyclic, inverting back gives that collinear as desired. 
, 
, 
as the feet of the 
, 
, 
altitudes respectively and as the midpoint of 
.
and 
, thus allowing the circle with diameter to be the unit circle. and also lie on this circle. We set 
and 
as free variables.![\[A = \frac{2pq}{p+q}\]](http://latex.artofproblemsolving.com/1/5/6/1562090eb0402f0dba760e24a5a3f1d4dbee5e73.png)
We shall find , notice that it is the intersection of 
with the unit circle and is thus given by the formula![\[e = \frac{1-\frac{2pq}{p+q}}{\frac{2}{p+q}-1} = \frac{p+q-2pq}{2-p-q}\]](http://latex.artofproblemsolving.com/c/c/1/cc174555499b0bf36ff4183491aa8bbcec8775eb.png)
We similarly find![\[f = \frac{\frac{2pq}{p+q} + 1}{\frac{2}{p+q}+1} = \frac{2pq+p+q}{2+p+q}\]](http://latex.artofproblemsolving.com/c/e/a/cea3ff7bb2a70e507b7615b8cffd18ffcef05ab0.png)
To find we intersect 
and 
, we thus obtain![\[H = \frac{\frac{2pq-p-q}{2-p-q} \cdot (1 + \frac{2pq+p+q}{2+p+q}) - \frac{2pq+p+q}{2+p+q} \cdot (\frac{p+q-2pq}{2-p-q}-1)}{\frac{2pq-p-q}{2-p-q} - \frac{2pq+p+q}{2+p+q}}\]](http://latex.artofproblemsolving.com/7/3/a/73a4cdd6264e3f76a1c1cec9811c7029c31ce8cf.png)
Simplifying, we have![\[H = \frac{2(2pq-p-q)(pq+p+q+1) + 2(2pq+p+q)(pq-p-q+1)}{(2+p+q)(2pq-p-q) + (p+q-2)(2pq+p+q)}\]](http://latex.artofproblemsolving.com/7/c/6/7c6fa2016e4da7193377c76d0a73dc48d812990e.png)
Further simplifying, we obtain![\[H = 2 \cdot \frac{4pq(pq+1) - 2{(p+q)}^2}{4(p+q)(pq-1)} = \frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)}\]](http://latex.artofproblemsolving.com/2/5/1/251b49a39a403ab54910673e8b64ebe1f7ddd4f8.png)
Now we shall prove the colinearity condition, notice that![\[\frac{H-p}{H-q} = \frac{\frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)} - p}{\frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)} - q} = \frac{(2p^2q^2 - p^2 - q^2) - p(p+q)(pq-1)}{(2p^2q^2 - p^2 - q^2) - q(p+q)(pq-1)}\]](http://latex.artofproblemsolving.com/e/9/6/e96c6cc47757686914ce938a899f673f017dd322.png)
From which![\[\frac{H-p}{H-q} = \frac{q(q - p)(p^2-1)}{p(p-q)(q^2-1)} = -\frac{q(p^2-1)}{p(q^2-1)}\]](http://latex.artofproblemsolving.com/a/0/8/a0832234be68bc35b78b0f3ea61f30201004f1cc.png)
We can simply take the conjugate to obtain![\[\overline{\left(\frac{H-p}{H-q}\right)} = -\frac{\left(\frac{1}{q}\right) \cdot (1 - \frac{1}{p^2})}{\left(\frac{1}{p}\right) \cdot (1 - \frac{1}{q^2})} = -\frac{q(p^2-1)}{p(q^2-1)}\]](http://latex.artofproblemsolving.com/5/8/c/58c541677097ccd5d62e99e011862659446d81d5.png)
Thus 
.
. Then from Brokard's theorem, triangle 
is self-polar, leading to the required result.
.
min
.
then: 
and 
and since 
is cyclic, we have 
the equality becomes then: 
since 
thus: 
then 
from which follow the collinarity of 
are concyclic where 
is the midpoint of 
and 
circles with diamters 
and 
and 
then by radical axis theorem it follows immediatly that 
and hence the points 
.
by the definition of radical axis.
and 
,
after rearranging and multiplying by 4.
,
and 
we have
,
.
,
.
,
and 
.
.
,
.
so the result follows.
,
.
,
,
.
.
and 
,
.
,
,
with respect to 
,
,
,
be the inversion operator, ![\[\Psi(AD \cap PQ) = \Psi(AD) \cap \Psi(PQ) = AD \cap (APQ) = D \implies AD \cap PQ = H,\]](http://latex.artofproblemsolving.com/2/1/d/21df81fe7ef438f3a9ba23868558df7af227a6a7.png)
also.
defined similarly. It is clear that 
is a cyclic quadrilateral since
Thus notice that the power of 
,
is 
.
is Humpty, since 
,