,
,
,
,
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 
When does equality hold?
such that 
and 
for any real numbers 
and 
.
![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
.
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
.
have the digit 
and are divisible by 
?
and 
intersect at the point 
.
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
Y by Akatsuki1010, talkon, rmtf1111, 62861, tenplusten, Irrational_phi, BartSimpsons, TLP.39, MeineMeinung, bluephoenix, don2001, Einstein314, Tumon2001, pinetree1, enhanced, tapir1729, Tawan, MNJ2357, fatant, math_pi_rate, JasperL, Gems98, AlastorMoody, tangent-7, psi241, Ifhml, Aryan-23, AmirKhusrau, Naruto.D.Luffy, char2539, lilavati_2005, Kagebaka, A_Math_Lover, riadok, hellomath010118, itslumi, fjm30, amar_04, Gaussian_cyber, amuthup, parmenides51, hsiangshen, EmilXM, Bumblebee60, A-Thought-Of-God, Pluto1708, SnowPanda, Kanep, Aritra12, myh2910, spartacle, kirillnaval, Rg230403, samrocksnature, agwwtl03, edfearay123, 554183, Flying-Man, MatBoy-123, Miku_, VZH, guptaamitu1, Quidditch, kamatadu, CyclicISLscelesTrapezoid, EpicBird08, Adventure10, dkshield, Sedro, aidan0626, centslordm, ZVFrozel
Here is my first article about Desargues Involution Theorem. Any suggestions or ideas for the next articles would be appreciated.
Enjoy Reading!
EDIT : Added synthetic proof of Theorem 1.4 by TinaSprout.
EDIT2: Attached the v2 version.
Enjoy Reading!
EDIT : Added synthetic proof of Theorem 1.4 by TinaSprout.
EDIT2: Attached the v2 version.
This post has been edited 2 times. Last edited by MarkBcc168, Oct 10, 2020, 10:34 AM
is an involution on a line 
and 
then for any two points 
we have 
to the circle points, but I'm not sure what to do from there either.
.
.
.
also. Done.
be tangent to 
at 
.
be the 
at 
,
be the intersection of the tangents to 
.
we know 
are collinear, so 
are isogonal. Now it's clear that the sides of quadrilateral 
are tangent to 
are tangents to 
we know that 
form an involution. But the first two pairs of lines are images under reflection across the 
is not identity. Otherwise statement no longer true of course.
.
"
be six points lying on a conic 
.
and 
share the common orthocenter 
.
and 
.
)
be a triangle with orthocenter 
be any point on 
be a line passing through 
.
be inversion at 
.
concur at 
of 
.
be line passing through 
is reflection of 
in 
)
and 
be projection of 
So 
.
is isoseles with 
.
(which is 
)
![[asy]
size(250);
real r=0.35;
pair A=dir(110),B=dir(-150),C=dir(-30),H=A+B+C,X=r*C+(1-r)*B,Y=extension(H,X,A,C),Z=extension(H,X,A,B),A0=foot(A,X,H),Xp=2*A0-H,Yp=2*foot(B,H,X)-H,D=foot(H,B,C),Kp=extension(A,Xp,B,Yp);
draw(circumcircle(A,B,C),blue);
draw(A--D);
draw(Yp--Xp,green);
draw(A--B--C--A,red);
draw(A--A0,dotted);
draw(B--Kp--Xp,brown);
dot("$A$",A,dir(80));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$H$",H,dir(H));
dot("$X$",X,dir(50));
dot("$X'$",Xp,dir(Xp));
dot("$Y$",Y,dir(160));
dot("$Z$",Z,dir(-140));
dot("$A_0$",A0,dir(180));
dot("$H_A$",D,dir(140));
dot("$Y'$",Yp,dir(Yp));
dot(Kp);
[/asy]](http://latex.artofproblemsolving.com/4/8/d/48dee678d9b60aa9f653845a91b386c67215358f.png)
This means 
concur on 
concur on 
.
![[asy]
size(250);
real xmin = -11.56, xmax = 11.4, ymin = -5.99, ymax = 7.21; /* image dimensions */
pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen ffqqff = rgb(1,0,1); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902);
pair A=(-3.38,4.03), B=(-4.7,-2.17), C=(3.02,-2.45), K=(0.9058454679978847,4.546489276046068), H=(-3.5641836396205133,-1.0482060638227293);
/* draw figures */
draw(circle((-0.7479081801897431,0.2291030319113646), 4.6232807734267505), dtsfsf); draw(shift((0.8455388191156685,0.9618583318481063))*rotate(-177.60278509389428)*xscale(9.655743491396057)*yscale(3.579819715306276)*unitcircle, dbwrru);
draw((xmin, -0.9196368531884037*xmin-4.32596069034893)--(xmax, -0.9196368531884037*xmax-4.32596069034893),linewidth(1)); /* line */
draw((-3.38,4.03)--(-4.7,-2.17), wvvxds);
draw((-4.7,-2.17)--(3.02,-2.45), ffqqff);
draw((3.02,-2.45)--(-3.38,4.03), sexdts);
draw((-4.7,-2.17)--(0.9058454679978847,4.546489276046068), sexdts);
draw((3.02,-2.45)--(0.9058454679978847,4.546489276046068), wvvxds);
draw((-3.38,4.03)--(0.9058454679978847,4.546489276046068), ffqqff);
/* dots and labels */
dot("$A$",A,dir(150));
dot("$B$",B,dir(-150));
dot("$C$",C,dir(-60));
dot("$K$",K,dir(90));
dot("$H$",H,dir(30));
clip((-10,-5)--(11,-5)--(11,8)--(-10,8)--cycle);
[/asy]](http://latex.artofproblemsolving.com/1/2/2/122c34ea90bee4770984e3755853f0f37cbcf3e1.png)
and 
wrt 
intersect 
and 
intersect 
.
![[asy]
size(200);
real xmin = -11.56, xmax = 11.4, ymin = -6.71, ymax = 6.49; /* image dimensions */
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196);
draw((-5.046763269188339,0.4619919977155256)--(-5.2967122506540925,0.8048119310102981)--(-5.639532183948865,0.5548629495445441)--(-5.389583202483111,0.21204301624977182)--cycle);
/* draw figures */
draw(circle((-4.146300247797356,0.6906136426211456), 3.716456689608643),rvwvcq);
draw((-1.36,3.15)--(-7.34,-1.21), wrwrwr); draw(shift((-2.622027210432846,0.20515263701486836))*rotate(171.095677480916)*xscale(6.055749201148707)*yscale(3.1315285909720743)*unitcircle, dbwrru);
draw((-7.24,2.75)--(-3.1317669005609376,-2.8846866822581645), wrwrwr);
/* dots and labels */
dot("$W_1$",(-7.24,2.75),dir(130));
dot("$X_1$",(-1.36,3.15),dir(70));
dot("$Z_1$",(-7.34,-1.21),dir(-120));
dot("$Y_1$",(-3.1317669005609376,-2.8846866822581645),dir(-50));
dot("$H$",(-5.389583202483111,0.21204301624977182),dir(-20));
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/d/8/7/d87154e591aa1e150e23c056e5501f48e0632091.png)
passing through 
and 
.
So it suffices to prove the following Lemma:
pass through 
.
Suppose that
Then 
.![[asy]
size(250);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10);
real xmin = -11.56, xmax = 11.4, ymin = -6.71, ymax = 6.49; /* image dimensions */
pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902);
draw((-5.046763269188339,0.4619919977155256)--(-5.2967122506540925,0.8048119310102981)--(-5.639532183948865,0.5548629495445441)--(-5.389583202483111,0.21204301624977182)--cycle, rvwvcq);
draw((-5.810310316701492,0.15737405465845028)--(-5.75564135511017,-0.2633530595599306)--(-5.33491424089179,-0.20868409796860912)--(-5.389583202483111,0.21204301624977182)--cycle, sexdts);
/* draw figures */
draw(circle((-4.146300247797356,0.6906136426211456), 3.716456689608643),dtsfsf);
draw((-1.36,3.15)--(-7.34,-1.21), rvwvcq); draw(shift((-2.622027210432846,0.20515263701486836))*rotate(171.095677480916)*xscale(6.055749201148707)*yscale(3.1315285909720743)*unitcircle, dbwrru);
draw((-7.24,2.75)--(-3.1317669005609376,-2.8846866822581645), rvwvcq);
draw((-8.303273643131877,-0.166559706051003)--(2.5278862881010964,1.2408329868827077), sexdts);
draw((-5.049183679559928,-2.407639245714383)--(-5.784472042338983,3.2510698642735227), sexdts);
/* dots and labels */
dot("$W_{1}$", (-7.24,2.75),dir(130));
dot("$X_{1}$", (-1.36,3.15), dir(40));
dot("$Z_{1}$",(-7.34,-1.21) , dir(-130));
dot("$Y_{1}$", (-3.1317669005609376,-2.8846866822581645), dir(-70));
dot("$H$", (-5.389583202483111,0.21204301624977182) , dir(-30));
dot("$W_{2}$", (-5.784472042338983,3.2510698642735227),dir(90));
dot("$Y_{2}$",(-5.049183679559928,-2.407639245714383) , dir(30));
dot("$X_{2}$", (2.5278862881010964,1.2408329868827077), dir(20));
dot("$Z_{2}$", (-8.303273643131877,-0.166559706051003) , dir(200));
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/a/f/5/af50917f925ee05ff87fee34c6efe65bc4672962.png)
be the circle passing through points 
.
or 
,
lie inside 
But since 
lie outside 
Hence we obtain a contradiction, as desired. 
become 
,
= asymptotes of the rectangular hyperbola. So this works well with the inversion at 
be any three reciprocal pairs of 
;
to a line 
and 
become reciprocal pairs of an involution 
on a line 
)
become reciprocal pairs of an involution 
is the inversion image of 
.
are reciprocal pair on 
with center 
.
.
,
.
.
