,
,
,
,
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 
coins numbered from 
.
,
.
people and 
colors, where each person has one ball of each color. For each person, their 
and 
.
Equality holds when 
Equality holds when 
:o
so 
and since 
is just 
therefore 
so 
?
apples in the pile.
Y by Adventure10, ImSh95, Mango247
Let 
and 
be concentric circles, with in the interior of . From a point 
on one draws the tangent 
to (
). Let 
be the second point of intersection of and , and let 
be the midpoint of . A line passing through intersects at 
and 
in such a way that the perpendicular bisectors of 
and 
intersect at a point 
on . Find, with proof, the ratio 
.
and 
be concentric circles, with in the interior of . From a point 
on one draws the tangent 
to (
). Let 
be the second point of intersection of and , and let 
be the midpoint of . A line passing through intersects at 
and 
in such a way that the perpendicular bisectors of 
and 
intersect at a point 
on . Find, with proof, the ratio 
.
be the common center of the concentric circles 
.
,
.
.
as well. Hence, the quadrilateral CDEF is cyclic. The intersection M of the perpendicular bisectors of its diagonals CE, DF is its circumcenter. If this circumcenter is to be on one of its side CD, it must be the midpoint of this side, 
.
,
and 
and 
.
of the outer circle touches the inner circle at 
.
is the midpoint of 
.
and 
.
and 
meet at 
on the line 
?
,
.
is cyclic. Hence 
.
.
![[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 = -4.300000000000004, xmax = 15.96000000000002, ymin = -5.740000000000006, ymax = 6.300000000000007; /* image dimensions */
/* draw figures */
draw(circle((2.120000000000002,0.5600000000000007), 2.308246087400564));
draw(circle((2.120000000000002,0.5600000000000007), 3.548182633405449));
draw((-1.290876635526356,1.537507328474018)--(3.747148166480568,2.197189312302931));
draw((-1.290876635526356,1.537507328474018)--(3.609775926061166,3.780274474346656));
draw((0.1261857316220342,1.723058323220483)--(-0.06571349512947573,2.098199114942177));
draw((3.609775926061166,3.780274474346656)--(3.747148166480568,2.197189312302931));
/* dots and labels */
dot((2.120000000000002,0.5600000000000007),dotstyle);
label("$O$", (2.200000000000004,0.6800000000000008), NE * labelscalefactor);
dot((-1.290876635526356,1.537507328474018),dotstyle);
label("$A$", (-1.220000000000001,1.660000000000002), NE * labelscalefactor);
dot((0.1261857316220342,1.723058323220483),dotstyle);
label("$E$", (0.2000000000000008,1.840000000000002), NE * labelscalefactor);
dot((3.747148166480568,2.197189312302931),dotstyle);
label("$F$", (3.820000000000006,2.320000000000002), NE * labelscalefactor);
dot((1.159449645267405,2.658890901410337),dotstyle);
label("$B$", (1.240000000000002,2.780000000000003), NE * labelscalefactor);
dot((-1.290876635526356,1.537507328474018),dotstyle);
dot((3.609775926061166,3.780274474346656),dotstyle);
label("$C$", (3.680000000000006,3.900000000000004), NE * labelscalefactor);
dot((-0.06571349512947573,2.098199114942177),dotstyle);
label("$D$", (0.02000000000000059,2.220000000000003), NE * labelscalefactor);
dot((1.772031215465845,2.939236794644417),dotstyle);
label("$M$", (1.860000000000003,3.060000000000004), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */ [/asy]](http://latex.artofproblemsolving.com/8/e/b/8eba8505f680d4b79ba1283760e0f107a9631329.png)
so 
is the midpoint of 
We then have 
and hence 
.
so 
so by converse of PoP, points 
are concyclic. Hence, the point of intersection of the perpendicular bisectors of 
and 
implying 
Now, we have that 
so 
and 
Hence, our desired answer is 
I tried both complex and bary bash but the calculations are getting huge(alas!)
we actually get ![\[\dfrac{AM}{MC}=\dfrac{AC-MC}{MC}=\dfrac{AC}{MC}-1=\dfrac83-1=\dfrac53,\]](http://latex.artofproblemsolving.com/b/3/c/b3c2879ec9a8babad72a1726e5f9a06899aaafec.png)
not 
as he suggests.
.
and 
,
is cyclic. However notice that 
is the radical axis of 
and 
,
,
.
![[asy]
size(250,250);
draw(circle((0,0),5)); draw((-4,3)--(4,3)); draw(circle((0,0),3));
label("$A$", (-4,3), NW); filldraw(circle((-4,3),0.05),black);
label("$B$", (0,3), N); filldraw(circle((0,3),0.05),black);
label("$C$", (4,3), NE); filldraw(circle((4,3),0.05),black);
label("$D$", (-2,3), NW); filldraw(circle((-2,3),0.05),black);
label("$M$", (1,3), N); filldraw(circle((1,3),0.05),black);
label("$E$", (-1.918677,2.306225775), W); filldraw(circle((-1.918677,2.306225775),0.05),black);
label("$F$", (2.918677,0.6937742252), SE); filldraw(circle((2.918677,0.6937742252),0.05),black);
draw(circle((1,3),3)); draw((-4,3)--(2.918677,0.6937742252));
[/asy]](http://latex.artofproblemsolving.com/c/3/b/c3bb4ea9d7b06bb2157855c236d1bf93dc47bb14.png)
is always 
.
,
.
,
is cyclic.![\[AB^2=BC^2=AD(AC),\]](http://latex.artofproblemsolving.com/8/9/4/894e2da109bc5b3e491303655de4e07bc103ac17.png)
since by \textbf{(1)}, ![\[AB^2=AE(AF) \iff AE(AF)=AD(AC).\]](http://latex.artofproblemsolving.com/6/e/3/6e334bf0360091753c404de963325ae653efed67.png)
Finally, using the converse of Power of a Point, we have that ![\[AE(AF)=AC(AD),\]](http://latex.artofproblemsolving.com/7/9/8/7983022f5db584497b143842d8fe150c0a04a92a.png)
which actually gives us that![\[\frac{AD}{AF}=\frac{AE}{AC},\]](http://latex.artofproblemsolving.com/1/9/0/190fa92cdad8862f81b171086d0e0fcd7546336f.png)
meaning that by SAS similarity, 
.![\[\angle AFC + \angle EDC = \angle AFC + (180-\angle ADE) = \angle AFC + (180-\angle AFC) = 180,\]](http://latex.artofproblemsolving.com/8/6/9/86986de0214a459eb92652a48154c1816913222c.png)
meaning that 
.
.
,
and of 
.
.
.
.
is cyclic.
,
and because of 
,
.
is cyclic. This is because
. 
![\[AE \cdot AF = AB^2=AB \cdot BC = AD \cdot AC\]](http://latex.artofproblemsolving.com/d/9/c/d9cb5c0cd29b68cc25d2f5ec659ad728cc906560.png)
,
.
hence 
.
![\[AF\cdot AE=AB^2=(\tfrac{1}{2}AB)(2AB)=AD\cdot AC\]](http://latex.artofproblemsolving.com/a/7/b/a7bdd1b5d781d75b493d297b0829de401dc2030a.png)
and 
is cyclic. Hence, 
.
,
so 
so 
so 
.
,
,
,
,
.
we get 
it follows that 
this tells us that the circumcenter of 
we then get
![[asy]
draw(circle((2,1),1));
draw(circle((2,1), sqrt(5)));
pair O=(2,1), A=(0,0), B=(2,0), C=(4,0), D=(1,0), M=(5/2,0);
draw(A--C);
dot(D);
dot(B);
dot(M);
dot(O);
dot(A);
dot(C);
label("$A$", A, SW);
label("$B$", B, S);
label("$C$", C, SE);
label("$D$", D, S);
label("$M$", M, S);
label("$O$", O, S);
pair E=(2-2sqrt(5)/5, 1-sqrt(5)/5), F=(2+2sqrt(5)/5, 1+sqrt(5)/5);
dot(E);
dot(F);
label("$E$", E, NW);
label("$F$", F, NE);
draw(A--F);
pair G=D/2+E/2, H=F/2+C/2;
draw(G--M--H);
draw(D--E);
draw(F--C);
[/asy]](http://latex.artofproblemsolving.com/1/4/2/142fde05ae1370460612ec340f4376af831b0392.png)
Since 
so 
and 
By power of a point, 
so 
intersect at the circumcenter of 
so 
As such, 
as desired. 
.
,
and 
.
.
.![\[ AD \cdot AC = x \cdot 4x = 4x^2.\]](http://latex.artofproblemsolving.com/2/0/2/20201dd7508e35ca5d1a2825fd5b390b1ff578ad.png)
Since ![\[ AF * FE = (2x)^2 = 4x^2 \]](http://latex.artofproblemsolving.com/f/0/e/f0e8aa97e9578855ad87779e2650381ce7b275ab.png)
from before, we have 
and quadrilateral 
be the circumcircle of quadrilateral 
.![\[\frac{AM}{MC} = \frac{x+\frac{3x}{2}}{\frac{3x}{2}} = \frac{\frac{5x}{2}}{\frac{3x}{2}} = \boxed{\frac 53} \]](http://latex.artofproblemsolving.com/a/3/c/a3c894d9b61cc855a204e74f593453bf78be8820.png)
.
so 
so the ratio is 
.
using the fact that 
.
say by extending 
and noting that 
.
by Power of Point. Therefore by Power of Point 
is cyclic, so 
and thus 
so 
.
,
as shown previously, so 
, so 
.
.
.
.
.
.
