,
,
,
,
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 
Prove that
denote the sum of the sum of the factors of all positive 
,
is the sum of the factors of 
added to the sum of the factors of 
and so on until the sum of the factors of 
.
denote 
.
or find a counterexample
such that
,
,
such that 
,
in terms of 
and 
,
.
are natural numbers satisfy that: 
and 
are prime numbers. Prove that 
is a perfect squared number
?
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.
and let 
and 
be tangency points of the incircle of 
,
and 
respectively. Let the circle with centre 
passing through 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
Y by A-Thought-Of-God, samrocksnature, Ya_pank, ohhh
Let 
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
. Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.
Proposed by Mahavir Gandhi
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
. Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.Proposed by Mahavir Gandhi
This post has been edited 4 times. Last edited by Rg230403, May 13, 2021, 11:41 AM
is the incircle of 
are the intouch points. Let 
be the foot of perpendicular from D onto EF. 
are defined analogously.
.
and define 
analogously. Let the tangent to 
meet AB, AC at 
and 
respectively. Now by Newton's Theorem on quadrilateral 
,
= 
and 
and 
= 
.
at 
and 
, 
and let the point of tangency between the A-Mixtilinear incircle and 
.
passes through U.
= 
athrough 
,
)
)
.
.
, 
are
.
=
.
= 
= 
.
is cyclic as required.
= 
.
and 
,
and let 
= 
.
and 
are tangent to 
and 
by Miquel's Theorem, 
.
lies on 
= 
.
.
.
is cyclic. So YZ is antiparallel to BC wrt 
.
.
.
.
.
and 
.
is similar to 
.
.
is similar to 
.
concur at T.
denote the incircle, 
the circle centered at 
,
the circumcircle; 
the respective intouch points opposite to 
respectively, and 
the midpoint of arc 
.![\[\measuredangle DIM_A = \measuredangle AM_AO = \measuredangle OAM_A = \measuredangle AIA'.\]](http://latex.artofproblemsolving.com/7/0/6/7069ff7ae7abe5dee411c0940a45e696deef1636.png)
As both 
are perpendicular to 
are perpendicular to 
.
and 
.
is tangent to 
,
and 
(as 
,
.
(as 
)
to 
with 
meets 
.
denote the 
be the intouch points of 
respectively.
,
intersect 
,
is tangent to both 
,
is the radical axis of 
)
and we get the desired
,
-
and 
to be the respective mixtilinear incircles.
,
.
and 
.
concur at 
are isogonals, where 
is the 
,
.
to 
,
.
goes through the exsimillicenter of the incircle and circumcircle.
be the incircle.
.
are equal. For this, observe that 
,
which is equivalent to the two inradii being equal.
mixtilinear incircle and the claim follows.
.
.
and 
be the center of 
since both lie on its perpendicular bisector. Thus, we also have 
.
on 
but this point is 
to 
and 
to ![[asy]
size(13cm);
defaultpen(fontsize(11pt));
import olympiad;
pair A = dir(116), B = dir(195), C = dir(-15), O = (0,0);
pair I = incenter(A,B,C);
pair D = foot(I,C,B), E = foot(I, A, C), F = foot(I, A, B);
pair Ma = 2*foot(O,A,I) - A;
path w = circumcircle(I, 2*foot(I,A,B)-I, 2*foot(I,A,C)-I);
pair[] P12 = intersectionpoints(w, circumcircle(A,B,C));
pair A1 = P12[0], A2 = P12[1];
w = circumcircle(I, 2*foot(I,C,B)-I, 2*foot(I,A,C)-I);
P12 = intersectionpoints(w, circumcircle(A,B,C));
pair C1 = P12[0], C2 = P12[1];
w = circumcircle(I, 2*foot(I,A,B)-I, 2*foot(I,B,C)-I);
P12 = intersectionpoints(w, circumcircle(A,B,C));
pair B1 = P12[0], B2 = P12[1];
w = incircle(A,B,C);
draw(A--B--C--cycle, royalblue);
pair A0 = foot(I, A1, A2), B0 = foot(I, B1, B2), C0 = foot(I, C1, C2);
pair X = extension(A, A0, B, B0);
pair Mb = 2*foot(O, I, B) - B, Mc = 2*foot(O, I, C) - C;
draw(circumcircle(A,B,C), royalblue);
draw(w, fuchsia);
draw(A1--Ma--A2, fuchsia);
draw(A1--A2, orange);
draw(B1--B2, orange + dashdotted);
draw(C1--C2, orange + dashdotted);
draw(A0--X--B0, brown);
draw(X--C0, brown);
draw(A--A0, brown + longdashed);
draw(C--C0, brown + longdashed);
draw(B--B0, brown + longdashed);
draw(O--I, black+dashed);
draw(X--I, black);
dot("$A$", A, dir(A));
dot("$A_1$", A1, dir(A1));
dot("$A_2$", A2, dir(A2));
dot("$B_1$", B1, dir(B1));
dot("$B_2$", B2, dir(B2));
dot("$C_1$", C1, dir(C1));
dot("$C_2$", C2, dir(C2));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M_A$", Ma, dir(Ma));
dot("$O$", O, dir(Ma));
dot("$I$", I, dir(Ma));
dot("$A'$", A0, dir(A0));
dot("$B'$", B0, dir(B0));
dot("$C'$", C0, dir(C0));
dot("$X$", X, dir(X));
[/asy]](http://latex.artofproblemsolving.com/3/e/b/3eb986c295c2f63c7edf08e04da8e1fc8a6695d2.png)
) .
,
similarly. The key claim of the problem is that 
.
are concyclic, we deduce that 
and 
and similarly.
.
and similarly 
.
so we have 
.
is isoceles and 
,
.
,
and 
are isoceles with two sides parallel, so 
,
not containing 
be points on 
such that 
and 
are tangent to incircle of 
is the angle bisector of 
.
is tangent to incircle of 
.
.
and the result follows from fact 5 
be the tangency points of incircle with 
respectively.
.
not containing 
be the 
but this would clearly imply 
are isogonal w.r.t 
and hence done 
similiarly for others.
and also 
so 
at 
.
.
,
is tangent to 
and so 
.
,
and so 
and so 
and now its easy enough to prove by angel chasing that 
.
are already tangent to the incircle, it must be the excircle. So, 
is perpendicular to 
,
.
is tangent to the incircle of 
,
is some permutation of 
,
and 
be the intouch triangle and circumcevian triangle of 
,
,
and 
at 
,![\[\measuredangle AYX=90^\circ-\measuredangle OAC=\measuredangle CBA\]](http://latex.artofproblemsolving.com/6/4/a/64a0305b41bc722fd9c7977660e83c0ee98322b6.png)
and thus quadrilateral 
is bicentric. It is then well-known that 
.
,
.
must also be homothetic as well. Therefore, 
and 
are concurrent at the homothetic center of two triangles, which lies on 
.
are constant and 
s.t. 
is the mixti touch point then 
concurr
be the arc midpoint of 
has both the same circumcircle and incircle as 
is the bisector of 
,
is 
is symmetric in 
.
is parallel to 
,
,
M
takes 
as 
,
and we get that the desired concurrence point is the exsimillicenter of 
become collinear since 
are concurrent and we are done 
concurrency and then you notice your configs and win.
