AoPS Academy
, 
,
, 
, why 
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.. It is usually regarded that 
, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.? The issue here is that 
isn't even rigorously defined in this expression. What exactly do we mean by ? Unless the example in question is put in context in a formal manner, then we say that is meaningless.? Suppose that 
. Then we would have 
, absurd. A more rigorous treatment of the idea is that 
does not exist in the first place, although you will see why in a calculus course. So the point is that is undefined.
? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
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 is an indeterminate form.
, 
or , obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](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 
. On the other hand, 
is undefined because dividing by is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context. is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](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 it is perfectly OK to say that,![\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 is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](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 is allowed here! In fact, we have![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and are left undefined. We also have
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 , by defining them as![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by also being allowed with the same indeterminate forms (in addition to some other ones).
be real numbers such that 
Prove that 
Let be real numbers such that 
Prove that 
be real numbers such that Prove that Let be real numbers such that Prove that
, it is true that 
.
, it is true that
. When we have equality ?
rest on a plane. A sphere with radius 
is tangent to all of them, but does not intersect nor lie on the plane. A sphere with radius 
lies on the plane, and is tangent to all three spheres with radius 
Compute the shortest distance between the sphere with radius and the sphere with radius 
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.
be ny two points in the plane, take 
s.t. 
is equilateral, and take 
on the segments 
respectively s.t. 
. Now let 
be the points where the perpendiculars through to 
cut 
, and, finally, let 
.
is the same as that inscribed in 
, we get 
(for a point 
, the letter also denotes the value we assign to it). In the same way, we get 
, from which we find 
. Since were chosen arbitrarily, we're done.
, let 
be the number assigned to point .
be any two points in the plane; we want 
. Construct regular hexagon 
with center 
. Then, 
. be the intersection of lines 
and 
. By reflection through the angle bisector of 
, it is easy to see that 
and 
have the same incenter 
, and we have 
. Subtracting the equation 
, we get the desired result .
and construct triangles 
and 
so that is the medial triangle of which in turn is the medial triangle of (
are lines and 
share incenter ). Assume for the sake of contradiction that some two points in the plane don't have the same number; WLOG let and be these two points.
Also, because 
and 
share incircles and thus have a common incenter (this can be seen for example by reflecting about 
or noting that quadrilateral 
is tangential), we have![\[\frac{f(A)+f(B)+f(E)}{3}=\frac{f(A)+f(C)+f(F)}{3}\implies f(B)+f(E)=f(C)+f(F).\]](http://latex.artofproblemsolving.com/f/2/0/f20de6921748a2070970fec844ed852bbac13d9d.png)
Symmetry and (1) tell us that![\[f(A)+f(D)=f(B)+f(E)=f(C)+f(F)=2f(I).\]](http://latex.artofproblemsolving.com/5/6/f/56f218fb22aa8776d314f91a793d57084b10eb73.png)
But we clearly must also have![\[f(D)+f(X)=f(E)+f(Y)=f(F)+f(Z)=2f(I)\]](http://latex.artofproblemsolving.com/f/b/7/fb7754acee83d307b4a8672577310fe4dba05072.png)
by applying the same argument to figure 
as we did to 
, so 
and 
, contradiction.
and 
, with values 
and 
respectively. Draw the perpendicular bisector, and pick an arbitrary point on it, with value 
. Then, let 
be the incenter of 
, 
of 
, and so on. is 
, is 
, is 
, and so on, with 
where there are 
s, having a value of 
.
, we see that the value of goes to 
. As our function is necessarily continuous (slight perturbations of vertices slightly perturb incenter location), the midpoint 
of has a value equal to (the sequence 
converges to ). into successively smaller intervals, to find the value of a point dividing it in a certain ratio. The midpoint of 
has value 
, the midpoint of that point and again is 
, and so on. Remembering that the function is continuous, and as all fractions can be written as an infinite sum/difference of fractions with denominators powers of , it is clear that a point's value is the weighted average of the values of the endpoints of the segment (with the ratio of weights being equal to the ratio of the lengths, and the higher weight given to the endpoint it is nearest to).
. Let be the midpoints of respectively, and let 
have values 
respectively. 
then has a value of and 
has 
. The midpoint of is at 
, and thus has value 
of is located at 
and has a value of 
. is also on the line connecting and . The distance 
is 
, and the distance 
is 
, and 
is 
's value of 
is also equal to 
After some effort, this (should!) simplify to 
. of 
as well, all points between and have value 
. Reflecting over to 
, we get the same result for segment 
. We can keep doing this infinitely, so all point on the line have the same value. As there is nothing special about , we have that all points on a given line have the same value. As any two points are collinear, we have that all points in the plane have an equal value.
be the value at point . be the center of . Let be the center of 
, be the center of 
, and 
, and let be the center of . Then it is easy to see that 
such that 
. Now using rotation and scaling, any two points and 
will be and hence have the same value. such that . This is not too
and 
. Let the two incenters be 
and 
respectively. By appropriate dilation and rotation of each triangle about its respective incenter, we can match up any two corresponding sides. It follows that for point 
, and any point 
, we have that 
. It follows that 
, 
, 
, 
, 
, 
, and all have the same weight, hence we can see that all other points must also have this weight since we can perform the same scaling and rotating on any other pair of triangles with at least one vertice from the original set that we worked with. 
would still be continuous, but there are noncontinuous solutions in this case: any "affine-linear" function satisfies the conditions.
) of the Motivation.
Given two points and 
of distance . Then, if we have proven that![\[ f(P) = f(P'); \]](http://latex.artofproblemsolving.com/f/f/c/ffcf5a06e0e497f08ec46e8b77fd3f4ce9c41deb.png)
we are finished with the problem.
There are two ways to finish here.
for any two points of the plane. Let the distance between those two points be 
. Then, apply the same argument on a plane magnified with homothety/dilation of scale .
for any two 
with distance at most . To do so, form an isosceles with sides 
. To prove the general statement, let we wish to prove 
Then find a 
in the segment 
so that 
and induct upwards to translate to 
.![\[\begin{tabular}{p{12cm}}
If we have enforced a way to force some two general points to be uniformly in order, then regardless of the size of the distance of any $P_1$ and $P_2$, the operation then can be magnified by $d$.
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/d/4/5/d45d2e11c271abf2a8a6aec5aef6e8309b00be8a.png)
Well, and the second way also puts it into perspective that![\[\begin{tabular}{p{12cm}}
No magnification is required: by translating a valid operation that forces uniform values on two points, we can actually match any two points to be equal!
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/e/9/a/e9a1972e9aad73a7e0eb8e2f411c304bb47a9b80.png)
This ends this Section. 
This might be the most absurd-ly worded Claim I've made in my Solutions, but here goes:![\[\begin{tabular}{p{12cm}}
Given $P,P'$ with distance $1$. Draw the segment $\overline{PP'}$ and draw two perpendicular lines to that segment that passes through $P$ and $P'$. Call the strip bounded by those two lines to be the $\textit{parallel range}$ of $P$ and $P'$. Moreover, divide the $\textit{parallel range}$ equally, and call the region closer to $P$ (and $P'$, respectively) to be the $\textit{P-half-range}$ (and $\textit{P'-half-range}$, respectively.) \\
\\
For any $Q$ which lies inside $\textit{P-half-range}$, we can prove that \\
\\
$\quad \quad \quad \quad \quad \quad \quad \quad \quad f(Q) - f(Q') = \dfrac{f(P)-f(P')}{3}.$ \\
\\
where $Q'$ is the reflection of $Q$ towards the perpendicular bisector of $PP'$.
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/f/6/3/f63090a16f654f08729f46cf83fd57cc7c6b6f7d.png)
We will proof that we can find two points and so that is the incenter of 
and 
is the incenter of 
.
be . Then, draw a circle 
with center and radius 
. Pick two points and so that![\[ (PP'X) \cap \omega = \{A,B\} \]](http://latex.artofproblemsolving.com/2/b/2/2b208e9b8645e26edc8ada7b68f0f8dd84b62504.png)
Since 
and 
is the internal bisector of angles 
, this validates the reverse construction --- 
is forced to be the respective incenters of the two triangle. 
The reason I put three blacksquares above was that the conclusion should be in sight after the 
: pick in the 
of 
--- and pick 
to be in the of 
. It should be immediately clear that must be in the parallel range of .
lies in the parallel range of , then![\[ f(R)-f(R') = \dfrac{f(P)-f(P')}{3} \]](http://latex.artofproblemsolving.com/c/3/a/c3ad83addf3d3cad325708c95bcdb8a76b8d273c.png)
but since lies in the parallel range of and lies in the parallel range of ,![\[ f(R)-f(R') = \dfrac{f(Q)-f(Q')}{3} = \dfrac{\frac{f(P)-f(P')}{3}}{3} \]](http://latex.artofproblemsolving.com/a/4/b/a4b840e2b616c08e2297f09a84035bc8de48680e.png)
rendering 
be an arbitrary circle and let 
be arbitrary tangent lines to no two of which are parallel. Let the function mapping points to real numbers be . Then we have![\[f(\ell_i\cap\ell_j)+f(\ell_i\cap\ell_k)+f(\ell_j\cap\ell_k)=c\]](http://latex.artofproblemsolving.com/1/0/2/102e27537a1bbc9e2f43cb8dd00cf294c7502c7a.png)
for all distinct 
and some fixed . Thus we have equations in variables. To solve these equations, note![\[f(\ell_i\cap\ell_j)=c-f(\ell_1\cap\ell_i)-f(\ell_1\cap\ell_j)\forall i,j\ne 1,i\ne j.\]](http://latex.artofproblemsolving.com/3/f/a/3fadcbc911852e107909f1689e35dd22501282ed.png)
Let 
so ![$c=c-a_i-a_j+c-a_j-a_k+c-a_k-a_i=3c-2[a_i+a_j+a_k]$](http://latex.artofproblemsolving.com/6/7/3/6737fb8f43816352ef34378dac8c02c04e01c530.png)
. This implies 
is fixed at , and so the desired follows: each of the 
is equal to 
. But this implies that any non-degenerate triangle has all its real numbers of vertices. To finish, note that for fixed and , all points 
not on line must have 
, so for particular 
on line and not on line , looking at 
finishes. denote the map from the plane 
to 
.
with 
which is not a rectangle, 
.
. By Pitot's Theorem, whichever one of 
or 
is concyclic has an incircle. Thus 
, finishing.
finishes, as 
so 
.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(20cm);
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 = -7.7799430983503575, xmax = 10.220716282714264, ymin = -1.8198421720682916, ymax = 9.297129470225183; /* image dimensions */
pen qqwwzz = rgb(0,0.4,0.6); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqttzz = rgb(0,0.2,0.6); pen ffqqtt = rgb(1,0,0.2);
draw((-2.48,0.34)--(1.56,0.36)--(3.562679491924312,3.8687426312891326)--(1.5253589838486228,7.357485262578265)--(-2.5146410161513773,7.337485262578268)--(-4.51732050807569,3.8287426312891357)--cycle, linewidth(0.4) + qqwwzz);
/* draw figures */
draw((-2.48,0.34)--(1.56,0.36), linewidth(0.4) + qqwwzz);
draw((1.56,0.36)--(3.562679491924312,3.8687426312891326), linewidth(0.4) + qqwwzz);
draw((3.562679491924312,3.8687426312891326)--(1.5253589838486228,7.357485262578265), linewidth(0.4) + qqwwzz);
draw((1.5253589838486228,7.357485262578265)--(-2.5146410161513773,7.337485262578268), linewidth(0.4) + qqwwzz);
draw((-2.5146410161513773,7.337485262578268)--(-4.51732050807569,3.8287426312891357), linewidth(0.4) + qqwwzz);
draw((-4.51732050807569,3.8287426312891357)--(-2.48,0.34), linewidth(0.4) + qqwwzz);
draw((xmin, 0.004950495049504941*xmin + 0.35227722772277226)--(xmax, 0.004950495049504941*xmax + 0.35227722772277226), linewidth(0.4) + qqwwzz); /* line */
draw((xmin, -1.7124171761194096*xmin + 9.969536186268696)--(xmax, -1.7124171761194096*xmax + 9.969536186268696), linewidth(0.4) + qqwwzz); /* line */
draw(circle((3.031422123213444,1.846063139364821), 1.4787607512476095), linewidth(0.4) + yqqqyq);
draw((1.56,0.36)--(1.5253589838486228,7.357485262578265), linewidth(0.4) + qqttzz);
draw((-2.48,0.34)--(3.562679491924312,3.8687426312891326), linewidth(0.4) + qqttzz);
draw((xmin, -0.5707684211618897*xmin + 3.5763031585065823)--(xmax, -0.5707684211618897*xmax + 3.5763031585065823), linewidth(0.4) + linetype("4 4") + ffqqtt); /* line */
/* dots and labels */
dot((-2.48,0.34),linewidth(4pt) + dotstyle);
label("$A$", (-2.350296352528207,0.020716046854468963), NE * labelscalefactor);
dot((1.56,0.36),linewidth(4pt) + dotstyle);
label("$B$", (1.5700926537772786,-0.05290628190244146), NE * labelscalefactor);
dot((3.562679491924312,3.8687426312891326),dotstyle);
label("$C$", (3.6315178589707733,4.051538546295315), NE * labelscalefactor);
dot((1.5253589838486228,7.357485262578265),dotstyle);
label("$D$", (1.6069038181557338,7.54859916224856), NE * labelscalefactor);
dot((-2.5146410161513773,7.337485262578268),dotstyle);
label("$E$", (-2.4423242634743447,7.530193580059332), NE * labelscalefactor);
dot((-4.51732050807569,3.8287426312891357),dotstyle);
label("$F$", (-4.632588543992433,4.08834971067377), NE * labelscalefactor);
dot((5.6,0.38),linewidth(4pt) + dotstyle);
label("$K$", (5.67453748197504,0.5360723481528419), NE * labelscalefactor);
dot((1.5484529946162078,2.692495087526089),linewidth(4pt) + dotstyle);
label("$L$", (1.6621205647234167,2.9472036149416585), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/a/0/4/a04c1551fc18cad1f589bdbbeab283babd0fb1c3.png)
and in the plane and let 
denote the number written at point . Let be a regular hexagon and let 
. We see that 
and 
have same incenter (center of hexagon) and so we have that 
. Now, we see that if 
is the incircle of 
, then the reflection of about perpendicular bisector of segment 
must be the incircle of 
, but we see that is also the incircle of 
and 
implying that center of lies on line 
but this is exactly the perpendicular bisector of segment , which means that is preserved under reflection through perpendicular bisector of due to symmetry, which means that is also the incircle of , therefore 
, subtracting 
from 
we get that 
, but since and were arbitrarily chosen, we must have that all of the points in the plane must have same number written on them/assigned to them.
be a scalene triangle with incenter 
. Let 
, 
, 
be the incenters of 
, 
, 
, and let 
be the incenter of 
. Then 
. is the incenter of . Let 
, 
, 
, and let 
, 
, 
touch 
, 
, 
at 
, 
, 
.![[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,IA,IB,IC,X,Y,Z,P,Q,R; A=dir(130); B=dir(200); C=dir(340); I=incenter(A,B,C); IA=incenter(I,B,C); IB=incenter(I,C,A); IC=incenter(I,A,B); X=extension(A,I,IB,IC); Y=extension(B,I,IC,IA); Z=extension(C,I,IA,IB); P=foot(IA,B,C); Q=foot(IB,C,A); R=foot(IC,A,B);
draw(incircle(A,B,C),linewidth(.4)); draw(A--I,gray); draw(B--I,gray); draw(C--I,gray); draw(IA--IB--IC--cycle,linewidth(.5)); draw(A--B--C--cycle,linewidth(.5)); draw(incircle(I,B,C),linewidth(.3)); draw(incircle(I,C,A),linewidth(.3)); draw(incircle(I,A,B),linewidth(.3));
dot("\(A\)",A,N); dot("\(B\)",B,SW); dot("\(C\)",C,SE); dot("\(I\)",I,NE); dot("\(I_A\)",IA,S); dot("\(I_B\)",IB,NE); dot("\(I_C\)",IC,W); dot("\(X\)",X,dir(60)); dot("\(Y\)",Y,S); dot("\(Z\)",Z,dir(290)); dot("\(P\)",P,S); dot("\(Q\)",Q,NE); dot("\(R\)",R,NW); [/asy]](http://latex.artofproblemsolving.com/a/c/e/ace8b896d11a11193c527f861d4342aa70714135.png)
and 
, so 
, Then 
and 
, so 
. and are tangent at 
, so 
lies on their radical axis, i.e.\ 
. Analgoously 
and 
, so , , are the points where the incircle touches the sides of the triangle.
, implying 
, and so 
, contradiction. 
For any two points 
and 
, it is possible to rotate and scale the above diagram so that 
and 
. Hence 
, so 
is constant.
define to be the number it was assigned. We would like to show 
for all 
. Indeed, consider points 
where 
is a regular hexagon. Triangles 
and 
share an incenter so ![\[ f(X)+f(A)+f(C) = f(Y)+f(B)+f(D) \]](http://latex.artofproblemsolving.com/a/7/2/a72729cf2576e7ec67e306d73ee5b16de9a96377.png)
So, it is sufficient to show that 
. Indeed, if we let and intersect at 
then we notice that triangles 
and 
share an incenter. So, ![\[ f(A) + f(C) + f(T) = f(B) + f(D) + f(T) \Leftrightarrow f(A) + f(C) = f(B) + f(D) \]](http://latex.artofproblemsolving.com/f/e/a/feac9a404b00581c2c0c65cd042f747abd84a0fd.png)
as desired.
be the number assigned to a point . be (fixed) points with on segment 
. and 
be a line perpendicular to and passing through segment 
. Let be a variable point on . Let be the point such that is the incenter of 
, and let be the point such that is the incenter of 
. The latter two points don't always exist, but they will when lies within a certain distance of . Observe that we have
which is fixed. varies along some projective algebraic plane curve as varies. That is, there exist real rational functions 
such that (on some subset of where it exists) can be parameterized as 
where 
varies in some interval. This is due to the theory of moving points: if we move linearly, we can obtain 
and 
by reflecting over 
and 
, so has a well-defined degree. Then we can obtain 
and 
by reflecting 
over 
and 
, so has a well-defined degree too. thus varies along some projective algebraic plane curve as desired.
at is now a real rational function in , so there exist finitely many points where its sign changes. Take some finite contiguous section 
of the locus of where the curve is always concave or always convex (WLOG concave) and where no two points on this curve have the same 
-coordinate. Fix a tiny open disc 
lying close to but entirely above (so there exists a point on lying directly below any point in , and does not intersect ) and let be a point directly above the center of which lies very far above . be any point in : I claim that there exist points 
such that is the incenter of 
. To do this, we vary a circle 
of radius centered at and let 
be the points where the tangents from to first intersect , which exist by our definition of . Consider the expression 
, where denotes signed distance and is positive if lies above line 
and negative otherwise ( being vertical never happens). If this expression is positive, then lies entirely in the interior of 
, and if it's negative, then lies partially outside . The expression is clearly continuous in and positive as 
. Moreover, when is chosen minimally so that intersects (so they're tangent), because is convex, the intersection point lies nonstrictly and thus outside , so the expression is nonpositive. Thus this expression is for some value of by the intermediate value theorem, at which point will be the incenter of as desired (by choosing 
for this value of ).
is fixed and 
is fixed, should also be fixed, so we find that is constant on 
. Now, consider two points 
such that the circle with diameter is contained entirely in . Varying a point along the perpendicular bisector of lying inside this circle with diameter and letting be the point such that 
has incircle , we find that the locus of is the entire perpendicular bisector minus its intersection with , so is also constant (and equal to its constant value on ) along the entire perpendicular bisector. Now rotating about its midpoint and continually applying this fact implies is constant everywhere, as desired. 
be the real number assigned to . The key idea is the following picture:
. Clearly, and 
share an incenter, so 
. Notice that 
is a kite, so it has an incircle. This incircle is the incircle of 
and 
, so they also share an incenter: 
. Therefore, 
.
with any points 
, we get the desired result.
be the corresponding number for some point in the plane. be a regular hexagon with the incenter of both and being . Then let 
. Then notice that 
is a reflection of 
over the angle bisector of 
so the angle bisectors of 
, 
and all concur. This implies that 
and 
have a shared incenter, so 
. However from earlier we know that and share an incenter so 
, so combining the two equations gives . Varying hexagon gives us that all points are equal, done.
such that 
and 
share an incenter , such that if the incenters of and are and respectively, then 
is obtuse but not straight.
equilateral and cocentric such that 
is small. Details are left to the reader. . is obtuse but not straight, there exists a point such that is the incenter of 
. Note however that the arithmetic mean of 
and 
is the arithmetic mean of 
, which is 
. But this is also the arithmetic mean of 
, so we get 
. Now a composition of a rotation, dilation, and translation maps the pair 
onto any pair of points in the plane, as desired.
with incenter let 
be the incenters of 
. We claim there exists such that is not the incenter of . and line and vary such that is isosceles. As go to infinity, we can check 
goes to 
so it is impossible for to have always been the incenter of . be the actual incenter of and notice we must have 
. Thus for any 
take the spiral similarity sending 
to 
and do the same thing on the image of to get 
as desired.
.. Let be a variable point in the left halfplane determined by , and be the reflection of in . Then 
is invariant. and be two arbitrary points on the same side of , and let 
. Let 
and 
be the reflections of and in .
and 
have a common incenter, so ![\[ f(B) + f(C') = f(B') + f(C) \implies f(B) - f(B') = f(C) - f(C'). \]](http://latex.artofproblemsolving.com/6/5/b/65bf2969ca62b70c078fb3a928b30117b3a4564b.png)
This proves the desired claim. 
.
, 
, and 
be three concurrent directed lines whose angles are in this order. Then ![\[ \kappa(\ell_y) = \kappa(\ell_x) + \kappa(\ell_z). \]](http://latex.artofproblemsolving.com/5/e/d/5edb5be3628e0bd8a934a288c80112236d17d54e.png)
Proof. Let the lines concur at . Consider a triangle with circumcenter , and sides 
, 
, and 
.![[asy]
unitsize(1cm);
pair O = (0, 0); real r = 4; pair A = r*dir(50), pA = dir(140); pair B = r*dir(80), pB = dir(170); pair C = r*dir(100), pC = dir(190);
draw(O -- 1.5A, black+1, arrow=Arrow(TeXHead)); draw(O -- 1.5B, black+1, arrow=Arrow(TeXHead)); draw(O -- 1.5C, black+1, arrow=Arrow(TeXHead));
label("$\ell_x$", 1.4A, SE); label("$\ell_y$", 1.4B, ESE); label("$\ell_z$", 1.4C, E);
pair Z = r*dir(30); pair Y = 2 * extension(Z, Z+pA, O, A) - Z; pair X = 2 * extension(Y, Y+pC, O, C) - Y;
draw(X -- Y -- Z -- cycle, blue); dot("$X$", X, S, blue); dot("$Y$", Y, N, blue); dot("$Z$", Z, S, blue); dot("$O$", O, S); [/asy]](http://latex.artofproblemsolving.com/2/2/c/22c94dbbf6951d7f8e50e69aa65fd05ded546f11.png)
Then we have 
Adding the first two equations implies 
, as desired. 
for all . and , while varying in between. The above claim means 
is the same for all these lines. This is valid as long as 
.
denote the 
rotation of any line . Note that 
. Select 
and 
so that 
. In particular we have ![\[ \kappa(\ell_a) + \kappa(\ell_b) = \kappa(\ell_x) + \kappa(\ell_z) + \kappa(\ell_x') + \kappa(\ell_z') = 0. \]](http://latex.artofproblemsolving.com/3/a/5/3a5aa54418d292cd44065d80f10ef73abefd3567.png)
Then by above, any line 
has 
. In particular this also hits some 
. Thus for any line through , and the same reasoning applies to every line. is constant. 
such that 
The problem I was talking about is
A first thought I had was to consider regular shapes, among which were equilateral triangles and squares. If 
,
will imply![\[ f(A)+f(B)+f(C) = f(X)+f(Y)+f(Z) \]](http://latex.artofproblemsolving.com/e/2/1/e21a1aee70e517d3fd5f46cf8ad537e8c4459691.png)
and I can only extract similar information from considering the incenters of 
(those three incenters have the same incenter as 
.
involving both 
;
(some fixed circumcircle), then 
up to 
,
was bad already. However, while this may not explicitly cater to cartesian distance due to 
play a huge role in the Solution.
Since the statement present no changes if addition were replaced by multiplication, I thought that the values of the functions must be linear in some way.
as @above mentions, 
can be of value 
,
or doubly-linear: 
.
or 
,
= length of projection from 
through the origin will do.
then the statement is redundant in some way. So, I figured that it's more fruitful to consider individual points rather than whole structures; with the hopes that individual points will not lose the scarce degree of freedom available (that is, 
This final Section is mainly how I figured out to think more combinatorially rather than geometrically at first: suppose that I had a function (algebra comes out!) that isn't equal everywhere.
is a broad uncountable universal set, applying the same technique as if I were to prove 
:
for some two points (in contrast to injectivity, where we assumed that 
for two different nodes.)![\[\begin{tabular}{p{12cm}}
$\textbf{Poncelet's Porism, rephrased.}$ Let $\triangle ABC$ has incenter $I$, incircle $\omega$ and circumcenter $\Omega$. If $D,E$ is on $\Omega$ and $DE$ is tangent to $\omega$, then there exists exactly one $F$ so that $\triangle DEF$ has the same circumcircle and incircle with $\triangle ABC$. \\
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/f/9/4/f9430e009f35093b1a9f49e12f4354a06ea9ed74.png)
Proof: Construct 
is tangent to 
.
from 
,
incenter of 
,
,
and 
so that the circle with center 
intersect 
.
and 
be a triangle with incenter 
,![\[ |D'I'| = |D'E'| \ \text{or} \ |D'F'| = |D'I| \]](http://latex.artofproblemsolving.com/4/f/e/4fe43c2d9e007a39e845d1d50b9771165fe9a0ec.png)
so 
(this is now a triple phantom argument) has radius 
.![\[ R^2 = OI^2-2Rr; \quad R^2 = OI'^2-2Rr' = OI^2 - 2Rr' \]](http://latex.artofproblemsolving.com/b/2/f/b2f52fc64a2194825199f76c7c23930d10fdf1b7.png)
since the points 
,
and 
is automatically tangent to 
--- 
is very particular: you can't just pick 
and 
will be bisectors of 
and 
,
and 
can make it so that 
.
to be less than the difference of the initial points 
beforehand.
having a fixed height with respect to the perpendicular bisector of 
![\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/8/4/2/842413e566043c96b7a8f2300401b08af9d8ba48.png)
Final Comments: Take note that the green part 
has no technical purpose towards the Solution: if I were to write this in a contest, it'd be something like![\[\begin{tabular}{p{15cm}}
If there exists two points $P,P'$ so that $f(P) = f(P')$, construct $Q,Q'$ with distance to perp.bisector of $PP'$ less than of $P$'s and construct $R,R'$ in the same way with respect to $QQ'$. \\
Note that by (some construction algorithm) we can find $A,B$ for $P,P',Q,Q'$ so that $PAB$ and $P'AB$ have incenters $Q,Q'$. So, for any pair of pair of points $(PP') (QQ')$ that satisfies the distance construction, \\
\\
$ \quad \quad \quad f(Q)-f(Q') = \dfrac{f(A)+f(B)+f(P)}{3} - \dfrac{f(A)+f(B)+f(P')}{3} = \dfrac{f(P)-f(P')}{3}$ \\
\\
which upon telescoping with $RR'$ will yield a contradiction. \\
Then, all points have the same number. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/2/e/4/2e4ff6b841e8a979088ab0bf027d9f140d471c81.png)
But when explaining why the possibility of letting 
in the first place lies in the fact that 
,