,
,
,
,
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 
be a nonzero polynomial such that 
for every real 
,
.
,
and 
are relatively prime positive integers. Find 
.
be positive real numbers. Prove that 
.
Y by samrocksnature, megarnie, HWenslawski, son7, Mango247
Let 
be the incircle of a fixed equilateral triangle 
. Let 
be a variable line that is tangent to and meets the interior of segments 
and 
at points 
and 
, respectively. A point 
is chosen such that 
and 
. Find all possible locations of the point , over all choices of .
Proposed by Titu Andreescu and Waldemar Pompe
be the incircle of a fixed equilateral triangle 
. Let 
be a variable line that is tangent to and meets the interior of segments 
and 
at points 
and 
, respectively. A point 
is chosen such that 
and 
. Find all possible locations of the point , over all choices of .Proposed by Titu Andreescu and Waldemar Pompe
This post has been edited 2 times. Last edited by djmathman, Jun 22, 2020, 5:29 AM
.
and 
by the Pythagorean Theorem, or the reflection of that point over 
and 
instead of 
and 
.
by perpendicularity lemma. Then proceed to length chase.
.
.
Which is trivially true 
too.
However, 
because 
.
,
.
sends 
to 
,
and the arc made by the homothety we mentioned before.
be the center of the equilateral triangle 
and 
intersect each other at 
(this is the circumcircle of 
.
lies on ![[asy]
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.
.
and 
are congruent. Because 
and there have a right angle. Therefore 
and hence done. 
,
and 
are congruent. Since the length 
also does not change. Since the line 
,
,
be the tangency point of 
,
intersect 
.
is cyclic. Indeed, since 
and 
it follows that 
,
,
where 
.
.
by the midpoint property so let us show 
.
so by SAS it is sufficient to show 
.
and 
.
.![\[ \angle QIL = \angle EIL + \angle EIQ = \angle TIB + \angle TIQ = \angle QIB \]](http://latex.artofproblemsolving.com/d/6/b/d6bc862f70eefbe3dfc0a40330bf84d8d9c482c1.png)
Proving the claim. Using the fact that 
we have 
.
so 
.
.
is the arc 
on 
.
be tangent to the incircle at 
such that 
and 
where 
is the sidelength of 
.
and 
.
be the tangency points on 
respectively. Then![\[PA^2 = PX^2 + AX^2 = PT^2 + (\frac{\sqrt3}{2} s)^2 = PT^2 + TR^2 = PR^2.\]](http://latex.artofproblemsolving.com/3/5/d/35d8e0f2e8129b7a99ef002d1cc516bdd61b632b.png)
A similar argument shows that 
.
.
,
in the direction opposite 
in the direction of 
arc 
.
centered at 
be the contact points, 
the circumradius, 
and 
so 
and 
and the second part follows as 
so 
Also, 
or 
Notice 
on 
and so by homothety, 
degree arcs bounded by 
and 
That is, 
of 
or minor arc 
of 
where 
and 
I would appreciate it if anyone could look over my writeup and give me some feedback.
be the circle with center 
,
be the circle with center 
.
as the tangency points of 
,
,
.
Therefore,
as desired.
with respect to 
collinear. By the Pythagorean Theorem, we have
But as 
(inradius), we have 
.
,
and 
denote the images of 
,
circular arcs 
and 
centered at 
be the contact triangle, with 
It is clear that 
are collinear.
and 
Notice that 
Thus 
Since 
it suffices to show that 
By Law of Cosines we have 
so 
Similarly 
so we finish.
Letting 
so we finish.
are on the same side of 
If 
which is the aforementioned image of 
with ratio 
satisfies the given conditions. Set 
and 
the incircle touchpoint on 
.
and 
,
is the midpoint of 
.
.
and 
have radical axis 
.
is a point with 
and hence 
fixed.
be the point on minor arc 
,
,
,
.
and 
,
,![\[\triangle PMA \cong \triangle PXR_1 \cong \triangle PXR_2, \quad \triangle QNB \cong \triangle QXR_1 \cong \triangle QXR_2,\]](http://latex.artofproblemsolving.com/f/5/3/f53ed83a13d2c44456b3497030d811ea85d7af76.png)
,
on ![[asy]
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,
meets 
and from
and its 
homothety centered at 
be the intersection of 
