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).
that is not isosceles at 
, let 
be its incircle, which is tangent to 
at 
, respectively. The lines 
and 
intersect the line passing through and parallel to 
at 
and 
, respectively. The lines passing through 
and perpendicular to 
intersect 
and 
at 
and 
, respectively.
are collinear and that 
are concurrent.
be a prime number. Prove that there exists an integer 
with 
such that neither 
nor 
is divisible by 
.
such that: 
be an odd prime number. Prove that there exist sequence such that: 
MathJuice bottles and Bob starts with 
MathJuice bottles. An unfair coin is then flipped, with probability 
of landing heads. If the coin lands heads, Alice gives Bob a bottle; otherwise, Bob gives Alice a bottle. This process repeats until someone runs out of bottles.
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 such that 
is equilateral.Then 
means 
is isosceles. Also since 
bisects 
is cyclic. Also easily seen 
. Hence 
.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/1/e/1/1e13f33eae83aba0191f80efd34d3eb7e7d1a1fd.png)
onto 
so that 
and 
reflection of 
across ; 
is equilateral. Clearly 
is a kite, so 
, consequently, as 
, we get 
, i.e. is the circumcenter of 
. Additionally, as 
is cyclic. From 
, we infer 
.
. If 
is midpoint of 
and 
projection of 
onto 
, then 
, hence 
, but 
, done.
-gon in this case with vertices 
. Let 
and 
. Then we observe that 
, 
lies on 
and 
lies on . Wishfully thinking, we want to prove 
, which is equivalent to 
lying on 
. Hence the problem is equivalent to
Which is straightforward by complex numbers. You might need the fact that 36th cyclotomic polynomial is 
. 
-gon in this case with vertices . Let and . Then we observe that , lies on and lies on . Wishfully thinking, we want to prove , which is equivalent to lying on . Hence the problem is equivalent toWhich is straightforward by complex numbers. You might need the fact that 36th cyclotomic polynomial is . 
be a triangle with 
and 
.
be a point on the ray 
but outside of segment ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
such that 
. Show that 
.
. Since 
is cyclic, 
. So this will take us to the above well-known problem.
be a quadrilateral such that 
, 
, 
and 
. Show that 
.
, it is the same quadrilateral in the original problem.
and 
.
. It is easy to see through cosine rule that 
(1). Consider 
. Through angle chasing we see that 
. Join 
with to get a cyclic quadrilateral 
. Since sum opposite sides of a cyclic quadrilateral sum up to 
, we have that, 
. Now, consider 
. Through angle chasing we see that 
. Now, consider 
. Using sine rule we see that 
. Now, consider 
. Using sine rule we see that, 
and 
we see that
.
. Using sine rule we see that, 
(Substituting the value of 
). And, 
comparing with (1) we have the desired claim.
be on such that 
and 
. Now we have 
and 
are isosceles so 
so is center of circle 
so 
so 
so 
is equilateral. 
so is center of 
. 
and 
so triangles 
and are congruent so .
be circumcenter of 
. We will show that lies on the line which is the main part of the problem.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/d/1/7/d1733997ec6892629582d694b4e13ed07743ea76.png)
is isosceles, we have 
. Finally apply angle sum property in 
to get 
. Call . We will use trigonometry now to prove 
. Apply Law of Sines in 
to get
Apply Law of Sines in 
using the fact that 
to get
Now eliminate 
and 
from 
and 
by dividing them. This would give![\[\frac{\sin(25^\circ)}{\sin(\theta)} = \frac{\cos(40^\circ)}{\sin(35^\circ + \theta)} = \frac{\sin(50^\circ)}{\sin(35^\circ + \theta)}\]](http://latex.artofproblemsolving.com/d/d/f/ddf3a69d056ef4ce1a1453997acb80ea6d95e50f.png)
From here on, it is just a brute force calculation. Write 
and expand 
to get![\[2\sin(\theta)\cos(25^\circ) = \sin(35^\circ)\cos(\theta) + \sin(\theta)\cos(35^\circ)\]](http://latex.artofproblemsolving.com/3/1/b/31bff53d145c5ec1420475317ac4340c6594ae2c.png)
Divide the equation by 
and solve for 
to finally get
Now we prove the following lemma.
.
and 
angles and force some cancellations.
and the lemma is proven.
to get 
since 
. Its about time we get back to geometry. to the intersection of perpendicular bisector of 
and 
. Since 
we get 
. By inscribed angle theorem, must be the circumcenter of 
restoring the original definition.
is equilateral. Due to the equilateral triangle, we would have![\[\overline{XE} = \overline{XC} = \overline{XD} = \overline{DC} = \overline{BC}\]](http://latex.artofproblemsolving.com/3/d/3/3d381985a6b679d8017631a7a11eb519306c1e6b.png)
This would even give 
. Due to all this, one can conclude 
with AAS congruency criteria. It will give 
and we are done. 
and therefore 
. By quadrilateral Ceva, we have that: 
. and now by sine rule, we have that: 
(on the opposite side of with respect to ) such that 
is equilateral. Because 
, is the perpendicular bisector of 
. 
, is internal bisector of 
and 
, so 
are cyclic. 
, hence 
.
,
,
and 
.
to meet the circumcircle of triangle 
at 
.
.
.
and 
.
.
.
.
,
.
.
.
and 
.
and 
.
.
.
Therefore 
.
.
with 
as base, on the side of 
.
.
and 
.
,
.
is the angle bisector of the isosceles triangle 
.
.
and hence 
.
.
and hence 
.
.
,
.
)
.
.
.
such that 
.
and equilateral triangle and so now a bit of angle chase to show that 
and so 
and now some more angle chase to get that 
.
and since 
,
.
.
must be the external angle bisector of 
.
.
and so 
and hence 
.
.
and 
.
,
.
.
,
and the circle with center 
,
and 
.
and 
.
and 
,
.
,
.
,
.
,
and 
are congruent, so 
,
via trigonometry.
(Cosine Rule)
and combining above we get
Now, in 
we find that 
and we are done!
intersect the circumcircle of 
at 
,
.
,
is equilateral. It is now obvious that 
,
.
is equilateral
so 
so 
cut 
be a point on ![$[BD]$](http://latex.artofproblemsolving.com/3/2/6/3261e689901bce018ecdef47b9bc60a78ead3746.png)
such that 
.
and 
,
is a cyclic quadrilateral. So 
.
gives us 
since 
.
gives us 
since 
.
.
is equilateral.
is a rhombus since 
and 
is an angle bisector of 
implies 
.
,
and 
.
,
.
Pls someone verify, I was able to get to the 10^{\circ} part but i forgot it so after verifying on geogebra it was correct.![[asy]
import olympiad;
import geometry;
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pen dotstyle = black; /* point style */
real xmin = -5.660598523469371, xmax = 4.383194927003558, ymin = -5.995088517583327, ymax = 3.0490856793386323; /* image dimensions */
draw((0,2)--(2,0)--(1.418918607713596,-3.2954763357920323)--(-1.9273957624893574,-3.295476335792032)--cycle, linewidth(0.6));
/* draw figures */
draw(circle((-1.1445069205095593,-1.1445069205095593), 3.3463143702029523), linewidth(0.6));
draw((0,2)--(2,0), linewidth(0.6));
draw((2,0)--(1.418918607713596,-3.2954763357920323), linewidth(0.6) + blue);
draw((1.418918607713596,-3.2954763357920323)--(-1.9273957624893574,-3.295476335792032), linewidth(0.6) + blue);
draw((-3.707932448732714,-3.2954763357920323)--(-1.9273957624893574,-3.295476335792032), linewidth(0.6) + blue);
draw((0,2)--(1.418918607713596,-3.2954763357920323), linewidth(0.6));
draw((2,0)--(-1.9273957624893574,-3.295476335792032), linewidth(0.6) + blue);
draw((2,0)--(-3.707932448732714,-3.2954763357920323), linewidth(0.6));
draw((0,2)--(-3.707932448732714,-3.2954763357920323), linewidth(0.6));
draw((-1.1445069205095593,-1.1445069205095593)--(1.418918607713596,-3.2954763357920323), linewidth(0.6) + blue);
draw((-1.1445069205095593,-1.1445069205095593)--(2,0), linewidth(0.6));
draw((-1.1445069205095593,-1.1445069205095593)--(-3.707932448732714,-3.2954763357920323), linewidth(0.6) + blue);
/* dots and labels */
dot((0,2),linewidth(3pt) + dotstyle);
label("$A$", (-0.019889879483389942,2.1446682596464366), NE * labelscalefactor);
dot((2,0),linewidth(3pt) + dotstyle);
label("$D$", (2.0507500024861476,0.10972906533899539), NE * labelscalefactor);
dot((1.418918607713596,-3.2954763357920323),linewidth(3pt) + dotstyle);
label("$C$", (1.5509403758038454,-3.555541530255694), NE * labelscalefactor);
dot((-1.1445069205095593,-1.1445069205095593),linewidth(3pt) + dotstyle);
label("$O$", (-1.4598176134966891,-1.1635954597539648), NE * labelscalefactor);
dot((-1.9273957624893574,-3.295476335792032),linewidth(3pt) + dotstyle);
label("$B$", (-1.8763256357319409,-3.6388431347010277), NE * labelscalefactor);
dot((-3.707932448732714,-3.2954763357920323),linewidth(3pt) + dotstyle);
label("$E$", (-3.9707659761149214,-3.615042676288075), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/d/a/0/da0423b8b474bfba376606aed3d02934cea7595a.png)
is equilateral. It is easy to see that 
giving 
.
by angle sum property and 
because 
is circumcenter of 
.
,
and 
,
by cpct.