,
,
,
,
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 the orthocenter of acute triangle 
to 
be the reflection of 
intersects line 
.
:
positive integers have no single-digit prime factors?
,
such that 
,
in terms of 
and 
,
.
?
Y by awesomehuman, aansc1729, mathmax12, peelybonehead, GeoKing, Rounak_iitr, GoodMorning, Spiritpalm, Mogmog8, Tastymooncake2, centslordm
Isosceles triangle 
, with 
, is inscribed in circle 
. Let 
be an arbitrary point inside 
such that 
. Ray 
intersects again at 
(other than 
). Point 
(other than ) is chosen on such that 
. Line 
intersects rays 
and 
at points 
and 
, respectively. Prove that 
.
Proposed by Anton Trygub
, with 
, is inscribed in circle 
. Let 
be an arbitrary point inside 
such that 
. Ray 
intersects again at 
(other than 
). Point 
(other than ) is chosen on such that 
. Line 
intersects rays 
and 
at points 
and 
, respectively. Prove that 
.Proposed by Anton Trygub
This post has been edited 2 times. Last edited by brianzjk, Mar 23, 2023, 11:12 PM
a typo?
be projections of 
correspondingly. Clearly, points 
lie on a single circle with diameter 
.
,
lie on a single circle. Similarly, points 
lie on a single circle.
.
,
.
be the harmonic conjugate of 
be the point on 
.
with respect to 
if and only if there exists a point ![\[\measuredangle AEB=90^{\circ},\, \measuredangle CEB=\measuredangle BED.\]](http://latex.artofproblemsolving.com/c/8/1/c81264bcef52fc0df1ec43d896a23f6fe12c1eeb.png)
Since 
we find that 
.
through 
are collinear. Draw 
we only need to show that 
are collinear. But now![\[\measuredangle HFE=\measuredangle HDE=\measuredangle BDE=\measuredangle ACE=\measuredangle AFE,\]](http://latex.artofproblemsolving.com/7/3/5/735d11de2f7229873f6f7385650da1c93fbbb681.png)
done. 
and 
.
is tangent to 
,
.
passes through the minor arc midpoint 
of 
,![\[-1=(AM;BC) \stackrel D= (KE;BC) \stackrel A=(PE;XY).\]](http://latex.artofproblemsolving.com/2/6/7/26720c9856febdd47806ee22113779b6707c7e7d.png)
Bisector bundle lemma finishes.
?
and 
intersect 
at 
respectively not equal to 
,
.
,
are cyclic quadrilaterals. However 
therefore the radiuses of 
,
are equal and are thus reflections over 
so we can conclude.
and 
are congruent since 
by LoS. So it then follows that 
be 
,
.
so 
and similarly 
,
is perpendicular to 
.
.
.
we get 
and by PoP 
and by similar triangles again since 
,
is the projection of 
.
where 
is the 
as desired.
.
from right angles.
through 
at 
.
are collinear, as then 
and Appolonian circles finishes.
is divided by 
.
,![\[\frac{DQ\times \sin\theta}{AD\times\sin 90^\circ}\]](http://latex.artofproblemsolving.com/3/3/b/33bd3558c8918b852397012a2c88189cb305b7b8.png)
and the second ratio, by ratio lemma on 
,![\[\frac{EQ\times \sin\alpha }{EA\times \sin{90^\circ-\alpha}}\]](http://latex.artofproblemsolving.com/f/9/f/f9f075323fb2813d98301e097043fa97247b4683.png)
We equate the two, if equality holds working backwards we would be done.
This is true if 
,![\[\angle ADC=\angle EA'A=\angle ECA\]](http://latex.artofproblemsolving.com/2/2/1/221bc19bb04f170279747ca38597a8e72d2d2cf2.png)
But this is true clearly, by Shooting Lemma.
and 
intersect 
at 
,
,
are concurrent at 
is cyclic. Similarly, 
is as well. Now note that![\[ \text{radius}[(ADPB)] = \dfrac{AD}{2 \sin(\angle ABC)} = \dfrac{AD}{2 \sin(\angle ACB)} = \text{radius}[(ADQC)]\]](http://latex.artofproblemsolving.com/d/f/5/df5dd4612ed9d16da821a6d3bfca54ae700c0c26.png)
so 
and 
are symmetric about 
.
.![\[\angle XDE = \angle ABP = \angle ABC + \angle PBD = \angle ACB + \angle QCD = \angle ACQ = \angle EDY \]](http://latex.artofproblemsolving.com/a/1/1/a11296c059c29869dfe560a5c8f91cd1234cecac.png)
as desired.
.
.
. Notice that 
is tangent to the circle with diameter 
.
is right angled at 
so 
passes through midpoint of arc 
which is what we needed to show.
at 
,
.
.![\[(XY;KE) \overset{A}{=} (BC;EK') \overset{D}{=} (C,B;A,DK' \cap (ABC)),\]](http://latex.artofproblemsolving.com/b/4/2/b42e147a12738c40430edbdfd135de0e61963059.png)
to be the midpoint of arc 
.
,
,![\[KD^2 = KE \cdot KF = KA \cdot KK'. \quad \blacksquare\]](http://latex.artofproblemsolving.com/9/8/4/9840d4f37738a10a347c3f3f3bc255744e992a92.png)
.
and 
are cyclic, as
,
,
,
meet 
and 
meet 
in 
respectively.
.
.
and 
imply 
and 
.
.
gives that 
.
,
.
to 
(where 
.![\[\angle AES^\ast=\angle AEB=\angle AEC=\angle AET^\ast\]](http://latex.artofproblemsolving.com/b/1/a/b1aa82de5d28aae0de0ae8250f2606b39d49074f.png)
and 
,
and 
are reflections over 
meet 
.
,
.
So 
.
as the "main triangle". The new problem is:
not containing 
.
.
harmonic.
,
we have 
and 
,
and 
and 
are symmetric about 
so we are done.
so 
and 
are tangent.
we have 
passes through the antipode of 
which is the arc midpoint.
tangent to 
Do the same for 
.
are concurrent. Furthermore, this implies 
Simply project through 
