,
,
,
,
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 
,
is rational for 
.
is called a rational point in 
-
.
and 
in 
and 
as 
.
,
,
be reals such that 
.Prove that
Where 
the equation 
and integers 
,
satisfying 
for 
(with the convention 
)![\[
\sum_{j=1}^n a_j^{2m-1} \leq \frac{mt}{2} \left( \sum_{j=1}^n a_j^{m-1} \right)^2.
\]](http://latex.artofproblemsolving.com/5/c/d/5cd52e9ce307563bde96f697a75079229d4bfcc2.png)
red marbles, 
yellow marbles, 
green marbles, and 
blue marbles. If Amanda takes the marbles one by one without replacing them until the 
marble. Then the probability that the 
marble is red is?
of occurring, Person A wins.
of occurring, Person A wins.
of occurring, Person A wins.
of occurring. Now, note that there is no possible way for someone to win with less than 
points at this point, so we can ignore that condition.
chance of obtaining a 
-
Solving yields 
.
(as a percent) as a blind guess.
Y by farhad.fritl, Frd_19_Hsnzde, ehuseyinyigit, pomodor_ap, Nuran2010, Rounak_iitr, user4747
In an acute-angled triangle 
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
Proposed by Theoklitos Parayiou, Cyprus
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.Proposed by Theoklitos Parayiou, Cyprus
This post has been edited 1 time. Last edited by falantrng, Apr 27, 2025, 4:38 PM
.
we have
Suppose 
.
,
is the foot of the altitude, we have
All segments on the left-hand sides are effectively computable (Sine Law/Ratio Lemma seems enough) via 
and the variable 
,
from the tangency to 
.
are all cyclic. Let 
be reflection of 
in 
.
.
so 
are collinear. Now Pascal on 
solves.
.
.
.
.
gives 
and similarly 
as well.
is the external angle bisector of 
.
.
,
are collinear, and 
are concyclic.
be the intersection of 
and the tangent from 
to 
and 
be the intersection of 
and 
,
are collinear, qed.
and 
.
be the second intersection of 
then 
and 
are reflections over 
,
is cyclic, implying that 
,
now.
,
,
,
,
,
,
denote 
,
,
,
,
,
,
,
respectively. (I know that's a lot but they're just common notation anyway)
.
.
so 
.
.
.
,
and 
.
so we're done.
,
are cyclic and 
,
is cyclic. Now, we claim that 
lies on the radical axis of 
and 
,
due to 
.
and 
.
and 
,
,
are collinear. Then 
,
is cyclic, i.e. 
and thus 
lie on the same line is equivalent to saying 
are concurrent lines.
and 
.
be the feet of the heights from 
are collinear,
,
is parallel to these two lines as well.
.
.
,
and the conclusion follows.
,
Let lines 
intersect the unit circle again at 
respectively. Now since 
are concyclic, we have
So
and similarly
Then
(So 
,
And since 
,
Now we find the coordinate of 
.
Since 
,
Now we find the vectors
and
Then
which is real. 
are concyclic, 
.
hence projecting DDIT at 
.
thus, 
are collinear. Pascal at 
gives the result as desired
,
.
.
.
on triangle 
it is enough to prove: ![\[\frac{LH}{LH_A} \cdot\frac{H_AD}{DC} \cdot \frac{CX}{XH} = 1 \]](http://latex.artofproblemsolving.com/c/a/5/ca5a5fbc872bbbe5053a4c254dc952119b2c08ae.png)
Using the Ratio Lemma on triangle 
we get: ![\[\frac{LH}{LH_A} = \frac{\cos(\beta)}{\sin(\gamma)\sin(\alpha)}\]](http://latex.artofproblemsolving.com/c/6/b/c6b89a205981fb3614471caa45100f846abf9ede.png)
Using the Ratio Lemma on triangle 
we get: ![\[ \frac{CX}{CH} = \frac{\sin(\alpha)\sin(x)}{\cos(\beta)\cos(\gamma + x)} \]](http://latex.artofproblemsolving.com/f/b/b/fbb1710f7ba004243867ac4e5289368adb29ef17.png)
Using the Ratio Lemma on triangle 
we get: ![\[ \frac{H_AD}{DC} = \frac{\sin(\gamma)\cos(\gamma + x)}{\sin(x)} \]](http://latex.artofproblemsolving.com/1/c/1/1c12889bc42c4ae75f5e19990aa222c97ca0fc2d.png)
Multiplying the three expressions, we get:![\[ \frac{LH}{LH_A} \cdot \frac{H_AD}{DC} \cdot \frac{CX}{XH} = \frac{\cos(\beta)}{\sin(\gamma)\sin(\alpha)} \cdot \frac{\sin(\gamma)\cos(\gamma + x)}{\sin(x)} \cdot \frac{\sin(\alpha)\sin(x)}{\cos(\beta)\cos(\gamma + x)} = 1 \]](http://latex.artofproblemsolving.com/3/6/9/369a31e4ab0808652255bcd39573fe1fd1de4586.png)
cyclic and additional trivially from the angles we have 
,
is the tangent to 
at 
.
and 
.
and 
are concurrent at 
and 
which is 
,
and 
which is 
are collinear. However, since 
,
and 
also works.
.![\[\angle BDP = \angle AEC = \angle ADC = \angle CDQ.\]](http://latex.artofproblemsolving.com/1/0/7/10760fcdfeedf54b41023f3410dd8fcdbb095652.png)
Also, since 
,
is isosceles.
be the image of 
are concyclic,
are collinear. Note that 
,
.
,
,
,
and 
.
cases. Pick 
,
.
.
Let 
.
thus we're done 
which means that 
.
defined by 
is projective since 
,
is the tangent line to 
.
so by prism lemma (or Steiner conic) the line 
we get 
,
.
that map 
and 
.
are collinear has deg 
concyclic has a deg 
.
,
is tangent to 
are concyclic.
holds. 
and 
are concyclic.
One can similiarly prove that 
,
are collinear.
.
hold. This is enough to prove the claim. 
implies 
are collinear. 
inscribed 
.
,
and 
lie on the same straight line. This means that 
,
,
and 
.
at 
and connect 
then do ratio lemma with menelaus.
move on BC, similar to other solutions we know 
cyclic so 
if we let 
so 
since 
,
where 
is the foot from 
to 
(
are concylic points.
,
is collinear.
and sides 
and 
and therefore 
.
.
and so 
.
,
,
,
and 
.
.
and therefore 
by applying the Sine theorem to 
.
and 
we obtain 
and 
.
Q.E.D.
is the 
be the second intersection of the circles 
the intersection of 
and 
is a harmonic.
be on the circle 
.
with respect to 
.
be the midpoint of 
,
with 
.
be the center of the circle 
.
.
and 
,
colinear and 
,
,
colinear.
be the intersection of 
and 
.
and 
.
,
and 
is cyclic, by Power of a Point we have 
,
is the PoP of U with respect to 
is the PoP of U in respect to 
are colinear. Now, its easy to see that 
,
from 
are all cyclic. Let ![\[\measuredangle PDB = \measuredangle AEP = \measuredangle ADC = -\measuredangle ADB = -\measuredangle BDA'.\]](http://latex.artofproblemsolving.com/2/0/4/2047cf43305fe5cfb9de955727629a4293eef7fe.png)
are collinear. Finally, Pascal's on hexagon 
implies 
collinear, as needed.
and 
are cyclic.
and 
( Due to 
and 
being cyclic)
.
.
( due to 
being tangent to 
)
.
are not on the same line we get an obvious contraction )
.
and 
because 
and 
.
Which implies the claim. 
.
Which implies the claim. 
Now by applying Pascal's theorem on cyclic hexagon 
,
,
,
must be collinear, thus finishing the problem. 
