,
,
,
,
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 
such that 
satisfying 
where 
.
cannot be a perfect cube of a positive integer.
has an incenter 
Suppose 
.
be the midpoint of 
.
meets 
again at point 
and 
be such that 
and 
.
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
at 
and 
.
.
![\[(2+0+2+4)\left((2\cdot2\cdot4)^2-\dfrac{2^2+0^2+2^2+4^2}{2+0+2+4}\right)=2024.\]](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)
have the digit 
and are divisible by 
?
is a quadratic polynomial such that 
and 
.
.
and 
intersect at the point 
.
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
Y by mathematicsy, CoolJupiter, Adventure10, Mango247
Let the incircle of a nonisosceles triangle 
touch 
, 
and 
at points 
, 
and 
respectively. The corresponding excircle touches the side at point 
. Let 
be the common point of 
and the incircle, closest to , and 
be the common point of 
and 
. Prove that 
.
touch 
, 
and 
at points 
, 
and 
respectively. The corresponding excircle touches the side at point 
. Let 
be the common point of 
and the incircle, closest to , and 
be the common point of 
and 
. Prove that 
.
be the incircle of 
.
,
is the point diametrically opposite 
is 
.
,
and 
be the line through 
parallel to 
median then the rest is trivial
refers to directed angles.![[asy]
import geometry; import olympiad;
unitsize(4cm);pair A = dir(135), B = dir(210), C = -1/B;
pair I = incenter(A, B, C);
pair D = foot(I, B, C), E = foot(I, C, A), F = foot(I, A, B);
pair T_ = 2I - D, N = B + C - D, T = intersectionpoints(incircle(A,B,C), A -- N)[1];
pair K = extension(E, F, D, T);
pair L = extension(D, I, A, K);
draw(circumcircle(A,E,F)^^circumcircle(D, T, A), linetype("0 2")); draw(D--L, linetype("2 4"));
draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(A--N); draw(D--K--F); draw(A--K, dashed);
draw(rightanglemark(A, L, D, 1.5)^^rightanglemark(I, D, B, 1.5));
draw(rightanglemark(A, F, I, 1.5)^^rightanglemark(A, E, I, 1.5));
draw(rightanglemark(A, T, D, 1.5));
draw(I--F^^I--E, linetype("2 5"));
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(T); dot(T_); dot(N); dot(K); dot(I); dot(L);
label("A", A, dir(90));
label("B", B, dir(-90));
label("C", C, dir(-90));
label("D", D, dir(-90));
label("E", E, dir(90));
label("F", F, -dir(30));
label("K", K, dir(90));
label("T$'$", T_, dir(-135));
label("T", T, dir(-7));
label("I", I, dir(-45));
label("N", N, dir(-90));
label("L", L, dir(90));[/asy]](http://latex.artofproblemsolving.com/1/7/6/17683b1851b292373700bf0479d24b6665b21dbd.png)
are concyclic with diameter 
.![\[\measuredangle AEI = 90^{\circ} = \measuredangle AFI\]](http://latex.artofproblemsolving.com/8/7/7/877120eed02179c1a9c7530890b852803c843be4.png)
![\[\measuredangle ATD = 90^{\circ}\]](http://latex.artofproblemsolving.com/9/b/3/9b3bf32dba5f84cd72b1d481b6e00e45c5f75712.png)
,
.
is a diameter of 
.
are collinear and 
.
and 
,
is the radical axis of 
is the radical axis of 
and 
.
and 
,
.
meets 
and ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.176743075134857, xmax = 11.428752031839883, ymin = -3.7969421960496423, ymax = 7.034290346656071; /* image dimensions */
pair A = (0.,5.), B = (-2.,0.), C = (5.,0.), D = (0.6570484595545034,0.), X = (2.3429513665770023,0.);
draw(A--B--C--cycle, linewidth(1.));
/* draw figures */
draw(A--B, linewidth(1.));
draw(B--C, linewidth(1.));
draw(C--A, linewidth(1.));
draw(circle((0.6570484976345142,1.798909413008393), 1.7989094130083934), linewidth(1.));
draw((0.6570484595545034,3.597818826016786)--D, linewidth(1.));
draw(X--A, linewidth(1.));
/* dots and labels */
dot(A,dotstyle);
label("$A$", A,dir(90));
dot(B,dotstyle);
label("$B$", B,dir(180));
dot(C,dotstyle);
label("$C$", C,dir(0));
dot(D,linewidth(4.pt) + dotstyle);
label("$D$", D,dir(-90));
dot((0.6570484595545034,3.597818826016786),linewidth(4.pt) + dotstyle);
label("$E$", (0.6570484595545034,3.597818826016786),dir(-135));
dot(X,linewidth(4.pt) + dotstyle);
label("$X$", X,dir(-90));
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/5/b/d/5bd0babd042ab4b6bb32a68205f401ed7e5f4d37.png)
and 
respectively. As 
.
centered at 
to 
which is the incircle to the excircle of ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.98, xmax = 16.42, ymin = -4.14, ymax = 9.34; /* image dimensions */
pair A = (0.,5.), B = (-2.,0.), C = (5.,0.), D = (-1.0131969613578624,2.4670075966053435), F = (0.6570484595545034,0.), T = (2.03941567143766,0.6477638648687565);
draw(A--B--C--cycle, linewidth(1.));
/* draw figures */
draw(A--B, linewidth(1.));
draw(B--C, linewidth(1.));
draw(C--A, linewidth(1.));
draw(circle((0.6570484976345142,1.798909413008393), 1.7989094130083934), linewidth(1.));
draw(A--(2.3429515404454966,0.), linewidth(1.));
draw(A--(11.327350454273194,5.), linewidth(1.));
draw(F--(11.327350454273194,5.), linewidth(1.));
draw((11.327350454273194,5.)--D, linewidth(1.));
/* dots and labels */
dot(A,dotstyle);
label("$A$", (0.08,5.2), NE * labelscalefactor);
dot(B,dotstyle);
label("$B$", (-2.3,-0.38), NE * labelscalefactor);
dot(C,dotstyle);
label("$C$", (5.08,0.2), NE * labelscalefactor);
dot(D,linewidth(4.pt) + dotstyle);
label("$D$", (-1.46,2.3), NE * labelscalefactor);
dot((1.9290695423130608,3.070930457686939),linewidth(4.pt) + dotstyle);
label("$E$", (2.,3.24), NE * labelscalefactor);
dot(F,linewidth(4.pt) + dotstyle);
label("$F$", (0.6,-0.38), NE * labelscalefactor);
dot((2.3429515404454966,0.),linewidth(4.pt) + dotstyle);
label("$N$", (2.34,-0.42), NE * labelscalefactor);
dot(T,linewidth(4.pt) + dotstyle);
label("$T$", (2.2,0.42), NE * labelscalefactor);
dot((11.327350454273194,5.),linewidth(4.pt) + dotstyle);
label("$K'$", (11.4,5.16), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/c/4/9c453ff80361494d0162a7a20827c6cbfaf8c915.png)
.
.
and 
.
and using the lemma, we'll be done.
,![\[\frac{s-b}{b}=\frac{\sin(\alpha - C)}{\sin \alpha}=\cos C - \sin C\cot \alpha.\]](http://latex.artofproblemsolving.com/8/0/1/801e922637cb8d60401a8e05f86e97ec011838bb.png)
So, 
.
,![\[\frac{AK'}{AD}=\frac{\sin\left(\frac{B+C}{2}\right)}{\sin\left(\frac{B-C}{2}\right)}\cdot \frac{\cos\left(\frac{B+C}{2}\right)}{\cos\left(\frac{B+C}{2}\right)}=\frac{\sin(B+C)}{\sin B-\sin C}=\frac{\sin A}{\sin B-\sin C}.\]](http://latex.artofproblemsolving.com/c/1/c/c1cf85916facfe3707b0fdf468dff3e358395cd2.png)
So, 
.
at 
,![\[AX=\frac{AC\cdot AK}{FC+AK}=\frac{b\cdot AK}{FC+AK}.\]](http://latex.artofproblemsolving.com/d/9/7/d97debfb0c900fdc9251f96dca83cbf5f4e50050.png)
.![\[\cot \beta = \frac{\frac{AK'}{AX}-\cos C}{\sin C}=\frac{FC+AK}{b\sin C}-\cot C = \frac{s-c+\frac{(s-a)a}{b-c}}{b\sin C}-\cot C.\]](http://latex.artofproblemsolving.com/4/e/4/4e4c308c54c51f5c748ebaaa9739e7e85b64344a.png)
or 
.
or ![\[b^2=ab\cos C - (s-b)(s-c)+\left(\frac{(s-a)a}{b-c}\right)\left(b\cos C - (s-b)\right).\]](http://latex.artofproblemsolving.com/e/5/e/e5ec295ee4a8ffe7761bf7f332f926f84293bad8.png)
,![\[0=\frac{a^2-b^2-c^2}{2}-\frac{(a+c-b)(a+b-c)}{4}+\left(\frac{a(s-a)}{b-c}\right)\left(\frac{a^2+b^2-c^2}{2a}-(s-b)\right).\]](http://latex.artofproblemsolving.com/6/0/0/600f7ae3623af9273b49851f7df96457f4449190.png)
.
or 
.
or 
which is clearly true.
be the point diametrically opposite 
.
colinear.
Also:
So 
as desired.
so 
and project from A we get thd desired result.
meet on the incircle at the antipode of 
are collinear. Now, let 
meet the line parallel to 
,
.
.
is cyclic from 
.
from 
is cyclic and 
,
are concurrent. Hence 
,
.
is diametrically op. to 
is a harmonic quad. Since 
collinear
.Where 
.
are harmonic pencils ! where 
is parallel to 
with the incircle of 
at 
,
are harmonic quadrilaterals, we have: 
= 
,
= 
= 
or 
meet the incircle again at a point 
,
and 
meet at a point 
is collinear, so 
,
,
.
collinear. Since 
is a diameter, 
.
,
at 
respectively.
be the line through 
be the antipode of 
.
,
are concurrent.
passes through 
by parallel lines.
and 
are cyclic.
are collinear, 
proving the first claim.
which proves our second claim. 
,
,
,
.
be a point such that 
so 
lies on the polar of 
.
so we are done.
concur ,at 
,
let 
and let 
It is well-known that 
Notice 
so 
is the point at infinity along 
,
then its known that if u let 
are colinear, now taking polars w.r.t. 
concurrent and that means 
and the line through 
where 
Projecting (with perspective at 
it suffices to show that 
where 
However, it is well-known that 
where 
where 
and 
.
.
which along with the fact that 
and as 
is tangent to the incircle, we have that 
is tangent to the incircle.
and 
.
lies on the polar of 
,
becomes that polar of 
is the line 
.
and we are done.
-
such that 
.
are collinear, so 
and 
can easily be seen to be cyclic, and by radical center on the incircle, 
,
,
,
,
concur, hence 
.
passes through 
It is well-known that 
is just the midpoint of 
which implies the desired result. 
and 
and the incircle. Then from EGMO chapter 4, we know that 
.
.
is tangent to the incircle, this is enough to imply that line 
But since 
,
,
,
.
.
onto 
and 
are concurrent. Let 
so that 
.
.
concur at some point 
is a diameter of the incircle.![\[ (DE;ST) \overset{F}= (DE;PK) \overset{A}= (B,C;M,\overline{AK} \cap \overline{BC}) = -1 \]](http://latex.artofproblemsolving.com/0/9/e/09ec0b588794b3e7e9863549622b8de0845af24c.png)
which implies that 
intersect at 
denote the incircle. Define 
.
is a diameter of 
.
.
Now projecting through 
so we indeed have 
through Diameter of the Incircle lemma.
is the pole of 
.
,
concur by an incircle concurrence lemma.
,
.
,
,
with 
,
.
.
is the polar of 
is tangent to 
,
is a right triangle.
are collinear.
,
,
,
.![\[(-a+b+c)x+(-a+b-c)y+(-a-b+c)z=0,\]](http://latex.artofproblemsolving.com/d/f/f/dfff949c49cae5f346f134ca66b3e5c21d134c08.png)
line ![\[(a-b-c)(c-b)x+a(a-b+c)y+a(-a-b+c)z=0,\]](http://latex.artofproblemsolving.com/c/a/6/ca674a37cc0b1d9d6c655bb8418be5358d7d485f.png)
and line ![\[(0)x+(1)y+(-1)z=0.\]](http://latex.artofproblemsolving.com/c/f/a/cfacd23ed3ee2b980cfeeb8bca9a79df33839a60.png)
Now we plug everything into the concurrence lemma:
as desired.![\[
-1=(Z,Y;P,T)\stackrel{X}{=}(Z,Y;Q,K).
\]](http://latex.artofproblemsolving.com/9/6/6/966390ec9e48635548387e793d367b9078d44b00.png)
However, since![\[
-1=(B,C;M,\infty_{BC})\stackrel{A}{=}(Z,Y;Q,(A\infty_{BC}\cap ZY)),
\]](http://latex.artofproblemsolving.com/f/f/8/ff89b604fe7859cb57c73454a1f4bb57a32bd84d.png)
we indeed have 
,
and 
.
and also 
.
.
.
and since FM = MN, then MT = MF and as MF is tangent to the incircle, we get that MT is also tangent to the incircle. Now its obvious that the polar of A is DE 
lies on the polar of A. Now by La Hire, A should lie on the polar of 
and since D and E lie on a circle, we have that X lies on the polar of 
is the line 
so 
,
so 
is cyclic. Thus 
is also cyclic, so 
(follows by taking inverses of lines 
concur.
Since (where 
and 
and 
we know 
so 
are collinear, and we are done.
denote a harmonic bundle.
.
.
be a point on 
.
On the other hand, using the ratio property of symmedians, we obtain 
We have therefore proven 
and thus 
is the antipode of 
.
.
is harmonic, we have![\[-1 = (Z,Y;T',T) \overset{X}{=} (Z,Y;P,K) \overset{A}{=} (B,C;M,Q).\]](http://latex.artofproblemsolving.com/6/f/e/6fea4d27d814f01286214cd0efe86b1d0dab7b26.png)
,
,
,
lies on the polar of 
,
is the midpoint of 
and 
be 
.
as per the OTIS problem. Let 
are collinear (proof by a homothety sending the incircle to the 
be the 
implies 
.
implies 
collinear.
.
says 
are collinear.
Clearly, 
We have that 
so 
so 
so 
so 
lie on the polar of 
Thus, 
which finishes
,
intersects 
at 
and 
intersects 
and 
implies that 
collinear. It is well-known that 
passes through 
implies that 
and since 
