text:
\begin{document}

\author{Steven Finch}
\title{In Limbo:\ Three Triangle Centers }
\date{June 3, 2014}
\maketitle

\begin{abstract}
Yet more candidates are proposed for inclusion in the Encyclopedia of Triangle
Centers. \ Our focus is entirely on simple calculations.

\end{abstract}

\footnotetext{Copyright \copyright \ 2014 by Steven R. Finch. All rights
reserved.}The best-known triangle centers:

\begin{itemize}
\item incenter (intersection of three angle bisectors)

\item centroid (intersection of three medians)

\item circumcenter (intersection of three perpendicular bisectors)

\item orthocenter (intersection of three altitudes)
\end{itemize}

\noindent are the first four listed in Kimberling's famous database \cite{Kb}.
\ Thousands more appear. \ A\ recent addition is the electrostatic center
\cite{AK}. \ Rigorous definition of a triangle center involves a real function
$f$ defined on the set of all possible triples $(a,b,c)$ of triangle sides and
satisfying certain properties. \ We forego such requirements (hence the phrase
\textquotedblleft in limbo\textquotedblright) and informally propose three
more triangle centers:

\begin{itemize}
\item equiareal disk center (associated with Fraenkel asymmetry \cite{FMP, Fi})

\item illuminating center (also called a Shibata streetlight \cite{Sh, OH})

\item thermodynamic center (also called a \textquotedblleft hot
spot\textquotedblright\ \cite{Pa, BMS})
\end{itemize}

\noindent in the hope that someone else can pick up where we leave off. \ The
latter notion, like the electrostatic center, has its origins in physics.

Consider three triangles $T_{1}$, $T_{2}$, $T_{3}$ with vertices
\[
\begin{array}
[c]{ccc}
\left\{  0,0\right\}  ,\left\{  1,0\right\}  ,\left\{  0,1\right\}  , &  &
\text{isosceles right triangle;}
\end{array}
\]
\[
\begin{array}
[c]{ccc}
\left\{  0,0\right\}  ,\left\{  1,0\right\}  ,\left\{  0,\sqrt{3}\right\}  , &
& 30^{\circ}\text{-}60^{\circ}\text{-}90^{\circ}\text{ triangle;}
\end{array}
\]
\[
\begin{array}
[c]{ccc}
\left\{  0,0\right\}  ,\left\{  6,0\right\}  ,\left\{  -\frac{13}{3}
,\frac{4\sqrt{35}}{3}\right\}  , &  & 6\text{-}9\text{-}13\text{ triangle.}
\end{array}
\]
Our humble contribution is the calculation of triangle centers for these
cases. \ We make no claim of originality or special insight. \ If our paper
starts a conversation, leading perhaps to future inclusion of the three
centers in \cite{Kb}, then our efforts will be justified. \

\section{Equiareal Disk Center}

Given a triangle $T$, let $\left\vert T\right\vert $ denote its area. The
Fraenkel asymmetry of $T$ is defined to be\textbf{\ }
\[
\alpha(T)=\inf\left\{  \frac{\left\vert (T\smallsetminus D)\cup
(D\smallsetminus T)\right\vert }{\left\vert T\right\vert }:D\text{ a disk with
}\left\vert D\right\vert =\left\vert T\right\vert \right\}  .
\]
The numerator contains the symmetric difference of $T$ and $D$. \ One could
imagine a similar definition involving disks having the same perimeter as $T$,
rather than area, but we leave this variation for other people to explore.
\ The infimum $\alpha(T)$ is achieved for some disk centered at a unique
interior point of $T$. Locating this point is a challenging exercise in
calculus. \ We illustrate the necessary partitioning of $T_{1}$, $T_{2}$,
$T_{3}$ in Figures 1, 2, 3 respectively. \ The color red is used for vertical
rectangles of width $dx$; green is used for horizontal rectangles of width
$dy$; black is used to further subdivide certain cells of the partition.
\ Details of other feasible configurations of the triangle and disk are
omitted for brevity's sake.

\begin{figure}[ptb]
\centering
\includegraphics[
height=5.9491in,
width=6.0243in
]
{tri1.eps}
\caption{Isosceles right triangle $T_{1}$}
\end{figure}

\begin{figure}[ptb]
\centering
\includegraphics[
height=6.1134in,
width=6.1428in
]
{tri2.eps}
\caption{$30^{\circ}$-$60^{\circ}$-$90^{\circ}$ triangle $T_{2}$}
\end{figure}

\begin{figure}[ptb]
\centering
\includegraphics[
height=5.9819in,
width=5.9819in
]
{tri3.eps}
\caption{$6$-$9$-$13$ triangle $T_{3}$}
\end{figure}

\subsection{Isosceles Right Triangle}

For $T_{1}$, the center must be on the diagonal line $y=x$ by symmetry.
\ Since $\left\vert T_{1}\right\vert =1/2$, the optimal circle $C$ has the
form
\[
(x-t)^{2}+(y-t)^{2}=\dfrac{1}{2\pi}.
\]
The intersection of the line $x+y=1$ and $C$ yields two points $(p_{1},q_{1})
$, $(p_{2},q_{2})$ with $p_{1}<p_{2}$, $q_{1}>q_{2}$. \ The intersection of
the line $x=0$ and $C$ yields two points $(p_{3},q_{3})$, $(p_{4},q_{4})$ with
$p_{3}=p_{4}=0$, $q_{3}>q_{4}$. \ The intersection of the line $y=0$ and $C$
yields two points $(p_{5},q_{5})$, $(p_{6},q_{6})$ with $p_{5}<p_{6}$,
$q_{5}=q_{6}=0$. \ The intersection of the line $y=x$ and $C$ yields two
points; we select $(p_{7},q_{7})$ to be the point with the larger
$x$-coordinate. \ Everything can easily be made explicit, for example,
\[
\begin{array}
[c]{ccccc}
p_{1}=\dfrac{1}{2}\left(  1-\sqrt{\dfrac{1}{\pi}-(1-2t)^{2}}\right)  , &  &
p_{5}=t-\sqrt{\dfrac{1}{2\pi}-t^{2}}, &  & p_{7}=\dfrac{1}{2\sqrt{\pi}}+t.
\end{array}
\]
The northwestern triangle has area\
\[

{\displaystyle\int\limits_{0}^{p_{1}}}

\left[  (1-x)-\left\{  t+\sqrt{\dfrac{1}{2\pi}-(t-x)^{2}}\right\}  \right]
dx
\]
identical to the southeastern triangle. \ The southern circular cap has area\
\[
2

{\displaystyle\int\limits_{p_{5}}^{t}}

\left[  0-\left\{  t-\sqrt{\dfrac{1}{2\pi}-(t-x)^{2}}\right\}  \right]  dx
\]
identical to the western circular cap. \ The southwestern triangle has area
\[

{\displaystyle\int\limits_{0}^{p_{5}}}

\left[  \left\{  t-\sqrt{\dfrac{1}{2\pi}-(t-x)^{2}}\right\}  -0\right]  dx
\]
and the northeastern circular cap has area
\[
2

{\displaystyle\int\limits_{p_{1}}^{1/2}}

\left[  \left\{  t+\sqrt{\dfrac{1}{2\pi}-(t-x)^{2}}\right\}  -(1-x)\right]
dx+2

{\displaystyle\int\limits_{1/2}^{p_{7}}}

\left[  \left\{  t+\sqrt{\dfrac{1}{2\pi}-(t-x)^{2}}\right\}  -x\right]  dx.
\]
Adding these areas and differentiating with respect to $t$, we find that the
best $t$~is
\[
t=\frac{1}{4}\left(  1+\frac{1}{\pi}\right)  =0.3295774715459476678844418...
\]
corresponding to an asymmetry $\alpha(T_{1})\approx0.450$. \

\subsection{$30^{\circ}$-$60^{\circ}$-$90^{\circ}$ Triangle}

Since $\left\vert T_{2}\right\vert =\sqrt{3}/2$, the optimal circle $C$ has
the form
\[
(x-s)^{2}+(y-t)^{2}=\dfrac{\sqrt{3}}{2\pi}.
\]
The intersection of the line $\sqrt{3}x+y=\sqrt{3}$ and $C$ yields two points
$(p_{1},q_{1})$, $(p_{2},q_{2})$ with $p_{1}<p_{2}$, $q_{1}>q_{2}$. \ The
intersection of the line $x=0$ and $C$ yields two points $(p_{3},q_{3})$,
$(p_{4},q_{4})$ with $p_{3}=p_{4}=0$, $q_{3}>q_{4}$. \ The intersection of the
line $y=0$ and $C$ yields two points $(p_{5},q_{5})$, $(p_{6},q_{6})$ with
$p_{5}<p_{6}$, $q_{5}=q_{6}=0$. \ Everything can be made explicit, for
example,
\[
p_{1},p_{2}=\frac{1}{4}\left(  3+s-\sqrt{3}t\mp\sqrt{-3+\frac{2\sqrt{3}}{\pi
}+6s-3s^{2}+2\sqrt{3}t-2\sqrt{3}st-t^{2}}\right)  ,
\]
\
\[
q_{1},q_{2}=\frac{1}{4}\left(  \sqrt{3}-\sqrt{3}s+3t\pm\sqrt{-9+\frac
{6\sqrt{3}}{\pi}+18s-9s^{2}+6\sqrt{3}t-6\sqrt{3}st-3t^{2}}\right)  ,
\]
\[
\begin{array}
[c]{ccc}
p_{5}=s-\sqrt{\dfrac{\sqrt{3}}{2\pi}-t^{2}}, &  & q_{4}=t-\sqrt{\dfrac
{\sqrt{3}}{2\pi}-s^{2}}.
\end{array}
\]
The northwestern triangle has area\
\[

{\displaystyle\int\limits_{0}^{p_{1}}}

\left[  \sqrt{3}(1-x)-\left\{  t+\sqrt{\dfrac{\sqrt{3}}{2\pi}-(s-x)^{2}
}\right\}  \right]  dx
\]
and the southeastern triangle has area
\[

{\displaystyle\int\limits_{0}^{q_{2}}}

\left[  \frac{1}{\sqrt{3}}\left(  \sqrt{3}-y\right)  -\left\{  s+\sqrt
{\dfrac{\sqrt{3}}{2\pi}-(t-y)^{2}}\right\}  \right]  dy.
\]
\ The southern circular cap has area\
\[
2

{\displaystyle\int\limits_{p_{5}}^{s}}

\left[  0-\left\{  t-\sqrt{\dfrac{\sqrt{3}}{2\pi}-(s-x)^{2}}\right\}  \right]
dx
\]
and the western circular cap has area\
\[
2

{\displaystyle\int\limits_{q_{4}}^{t}}

\left[  0-\left\{  s-\sqrt{\dfrac{\sqrt{3}}{2\pi}-(t-y)^{2}}\right\}  \right]
dy.
\]
The southwestern triangle has area
\[

{\displaystyle\int\limits_{0}^{p_{5}}}

\left[  \left\{  t-\sqrt{\dfrac{\sqrt{3}}{2\pi}-(s-x)^{2}}\right\}  -0\right]
dx
\]
and the northeastern circular cap has area
\[

{\displaystyle\int\limits_{q_{2}}^{q_{1}}}

\left[  \left\{  s+\sqrt{\dfrac{\sqrt{3}}{2\pi}-(t-y)^{2}}\right\}  -\frac
{1}{\sqrt{3}}\left(  \sqrt{3}-y\right)  \right]  dy.
\]
Adding these areas and differentiating with respect to both $s$ and $t$, we
find that the best $(s,t)$~is
\[
s=0.3719164279770188862100673...,
\]
\[
t=0.4794617554511785131491672...
\]
corresponding to an asymmetry $\alpha(T_{2})\approx0.517$. \

\subsection{$6$-$9$-$13$ Triangle}

Since $\left\vert T_{3}\right\vert =4\sqrt{35}$, the optimal circle $C$ has
the form
\[
(x-s)^{2}+(y-t)^{2}=\dfrac{4\sqrt{35}}{\pi}.
\]
The intersection of the line $4\sqrt{35}x+31y=24\sqrt{35}$ and $C$ yields two
points $(p_{1},q_{1})$, $(p_{2},q_{2})$ with $p_{1}<p_{2}$, $q_{1}>q_{2}$.
\ The intersection of the line $4\sqrt{35}x+13y=0$ and $C$ yields two points;
we select $(p_{3},q_{3})$ to be the point with the larger $y$-coordinate.
\ The intersection of the line $y=0$ and $C$ yields two points $(p_{4},q_{4}
)$, $(p_{5},q_{5})$ with $p_{4}<p_{5}$, $q_{4}=q_{5}=0$. \ Everything can be
made explicit, for example,
\[
\begin{array}
[c]{ccc}
p_{1},p_{2}=\dfrac{\kappa\mp31\sqrt{\pi\xi}}{1521\pi}, &  & q_{1},q_{2}
=\dfrac{4\left(  \lambda\pm\sqrt{\frac{35}{\pi}\xi}\right)  }{1521}
\end{array}
\]
where
\[
\xi=6084\sqrt{35}-20160\pi+6720\pi s-560\pi s^{2}+1488\sqrt{35}\pi
t-248\sqrt{35}\pi st-961\pi t^{2},
\]
\[
\begin{array}
[c]{ccc}
\kappa=3360\pi+961\pi s-124\sqrt{35}\pi t, &  & \lambda=186\sqrt{35}
-31\sqrt{35}s+140t;
\end{array}
\]
\[
\begin{array}
[c]{ccc}
p_{3}=\dfrac{13\left(  \mu-\sqrt{\pi\eta}\right)  }{729\pi}, &  & q_{3}
=\dfrac{4\left(  \nu+\sqrt{\frac{35}{\pi}\eta}\right)  }{729}
\end{array}
\]
where
\[
\eta=2916\sqrt{35}-560\pi s^{2}-104\sqrt{35}\pi st-169\pi t^{2},
\]
\[
\begin{array}
[c]{ccc}
\mu=13\pi s-4\sqrt{35}\pi t, &  & \nu=-13\sqrt{35}s+140t;
\end{array}
\]
\[
\begin{array}
[c]{c}
p_{4},p_{5}=s\mp\sqrt{\dfrac{4\sqrt{35}}{\pi}-t^{2}}.
\end{array}
\]

The northwestern triangle has area\
\[

{\displaystyle\int\limits_{-\frac{13}{3}}^{p_{3}}}

\left[  \left\{  -\tfrac{4\sqrt{35}}{31}(x-6)\right\}  -\left\{
-\tfrac{4\sqrt{35}}{13}x\right\}  \right]  dx+

{\displaystyle\int\limits_{p_{3}}^{p_{1}}}

\left[  \left\{  -\tfrac{4\sqrt{35}}{31}(x-6)\right\}  -\left\{
t+\sqrt{\tfrac{4\sqrt{35}}{\pi}-(s-x)^{2}}\right\}  \right]  dx
\]
and the southeastern triangle has area
\[

{\displaystyle\int\limits_{p_{5}}^{p_{2}}}

\left[  \left\{  t-\sqrt{\tfrac{4\sqrt{35}}{\pi}-(s-x)^{2}}\right\}
-0\right]  dx+

{\displaystyle\int\limits_{p_{2}}^{6}}

\left[  \left\{  -\tfrac{4\sqrt{35}}{31}(x-6)\right\}  -0\right]  dx.
\]
The southwestern circular cap has area
\[
2

{\displaystyle\int\limits_{p_{4}}^{s}}

\left[  0-\left\{  t-\sqrt{\tfrac{4\sqrt{35}}{\pi}-(s-x)^{2}}\right\}
\right]  dx+

{\displaystyle\int\limits_{0}^{q_{3}}}

\left[  \left\{  -\tfrac{13}{4\sqrt{35}}y\right\}  -\left\{  s-\sqrt
{\tfrac{4\sqrt{35}}{\pi}-(t-y)^{2}}\right\}  \right]  dy
\]
and the northeastern circular cap has area
\[
2

{\displaystyle\int\limits_{p_{1}}^{s}}

\left[  \left\{  t+\sqrt{\tfrac{4\sqrt{35}}{\pi}-(s-x)^{2}}\right\}
-q_{1}\right]  dx+

{\displaystyle\int\limits_{q_{2}}^{q_{1}}}

\left[  \left\{  s+\sqrt{\tfrac{4\sqrt{35}}{\pi}-(t-y)^{2}}\right\}  -\left\{
6-\tfrac{31}{4\sqrt{35}}y\right\}  \right]  dy.
\]
Adding these areas and differentiating with respect to both $s$ and $t$, we
find that the best $(s,t)$~is
\[
s=0.9999634051829363409671652...,
\]
\[
t=2.4097948974186280609774486...
\]
corresponding to an asymmetry $\alpha(T_{3})\approx0.694$.

The fairly arbitrary triangle $T_{3}$, in particular, serves as a benchmark in
\cite{Kb} to distinguish various centers. \ Our preceding value $t$ is the
perpendicular distance from the equiareal disk center to the shortest triangle
side. \ Since the numerical value $2.409...$ does not appear in the database,
we infer that this center is new.

\section{Illuminating Center}

Let $\Lambda$ denote a light source in three-dimensional space of luminosity
$L$. The amount of light an observer $\Theta$ receives from $\Lambda$ is
called its brightness, measured in lumens per unit area. \ Brightness is
inversely proportional to the square of distance between $\Lambda$ and
$\Theta$. \ (Reason: brightness is constant on the sphere of radius $R$,
center $\Lambda$ and thus is equal to $L/\left(  4\pi R^{2}\right)  $).
\ Although we will focus solely on light sources in the \textit{plane}
(streetlights in a triangular park), the preceding \textit{spatial} definition
of brightness is the basis of our model. \ \

Given a triangle $T$ (more precisely, its interior), the total brightness
\[

{\displaystyle\iint\limits_{T}}

\frac{1}{(x-s)^{2}+(y-t)^{2}}\,dx\,dy
\]
is the quantity we would wish to maximize with respect to $(s,t)\in T$. \ It
turns out that this integral is divergent and
\[

{\displaystyle\iint\limits_{T\smallsetminus D_{\varepsilon}}}

\frac{1}{(x-s)^{2}+(y-t)^{2}}\,dx\,dy
\]
must be examined instead, where $D_{\varepsilon}$ is the disk of radius
$\varepsilon>0$, center $(s,t)$. \ Further, $(s,t)$ cannot be close to
$\partial T$ (the boundary of $T$), that is, we must restrict $(s,t)\in
T\smallsetminus\left(  T\cap N_{\varepsilon}\right)  $ where $N_{\varepsilon}$
is the $\varepsilon$-tubular neighborhood of $\partial T$. \ Under such
conditions, in the limit as $\varepsilon\rightarrow0^{+}$, a geometric
characterization of the maximum point $P=(s,t)$ becomes available. \

Let $T$ possess vertices $A$, $B$, $C$. \ Select any two distinct vertices $U
$, $V$ and note that the semiperimeter of subtriangle $UPV$ is
\[
\sigma=\frac{\sqrt{(U-P)\cdot(U-P)}+\sqrt{(V-P)\cdot(V-P)}+\sqrt
{(U-V)\cdot(U-V)}}{2}.
\]
Using Heron's formula, it follows that the ratio of inner angle to area:
\begin{align*}
\rho(UPV)  &  =\frac{\angle UPV}{\left\vert UPV\right\vert }\\
&  =\tfrac{\arccos\left(  \tfrac{(U-P)\cdot(V-P)}{\sqrt{(U-P)\cdot(U-P)}
\sqrt{(V-P)\cdot(V-P)}}\right)  }{\sqrt{\left(  \sigma-\sqrt{(U-P)\cdot
(U-P)}\right)  \left(  \sigma-\sqrt{(V-P)\cdot(V-P)}\right)  \left(
\sigma-\sqrt{(U-V)\cdot(U-V)}\right)  \sigma}}
\end{align*}
satisfies $\rho(APB)=\rho(BPC)=\rho(CPA)$, a remarkable fact \cite{Sh}! \ For
$T_{1}$, we solve the resulting equations, obtaining
\[
s=t=0.3082756986146550422567206...;
\]
for $T_{2}$,
\[
s=0.3516876887676632055410277...,
\]
\[
t=0.4491286165669552235961426...;
\]
and, for $T_{3}$,
\[
s=0.8345011650594754190821304...,
\]
\[
t=2.0031487728161056257679347....
\]
\ Our preceding value $t$ is the perpendicular distance from the illuminating
center to the shortest triangle side. \ Since the numerical value $2.003...$
does not appear in the ETC\ database, we infer that this center is new.

Let us revisit the definition of brightness. \ Had a \textit{planar}
definition been adopted -- measured in lumens per unit length -- then
brightness would be inversely proportional to the distance itself between
$\Lambda$ and $\Theta$. \ (Reason: brightness would be constant on the circle
of radius $R$, center $\Lambda$ and thus would be equal to $L/\left(  2\pi
R\right)  $). \ This scenario yields exactly the same formulation as that
underlying the electrostatic center \cite{AK}. \ \

\section{Thermodynamic Center}

Here, the triangle $T$ (more precisely, its interior) is assumed to be a heat
conductor with initial temperature $1$ while its boundary $\partial T$ is held
at temperature $0$ always. \ Heat $u$ will dissipate as time $t$ increases
according to the following initial-boundary value problem:
\[
\begin{array}
[c]{ccc}
\dfrac{\partial u}{\partial t}=\dfrac{\partial^{2}u}{\partial x^{2}}
+\dfrac{\partial^{2}u}{\partial y^{2}} &  & \text{for }(x,y,t)\in
T\times(0,\infty),
\end{array}
\]
\[
\begin{array}
[c]{ccc}
u(x,y,t)=1 &  & \text{for }(x,y,t)\in T\times\{0\},
\end{array}
\]
\[
\begin{array}
[c]{ccc}
u(x,y,t)=0 &  & \text{for }(x,y,t)\in\partial T\times(0,\infty)
\end{array}
\]
but does so non-uniformly: its density gathers around a unique maximum point
$(x_{\infty},y_{\infty})$ as $t\rightarrow\infty$ \cite{Pa, BMS}. The point
$(x_{\infty},y_{\infty})$ is, in fact, the unique extreme point of the first
Laplacian eigenfunction for $T$. \ The eigenfunction for $T_{1}$ is \cite{Mc,
Si, DP, F1, F2, F3}
\[
\sin(\pi x)\sin(2\pi y)+\sin(2\pi x)\sin(\pi y)
\]
with
\[
x_{\infty}=y_{\infty}=\frac{1}{\pi}\operatorname{arcsec}\left(  \sqrt
{3}\right)  =0.3040867239846963649145722...
\]
and the eigenfunction for $T_{2}$ is
\[
\sin\left(  \frac{\pi x}{3}\right)  \sin\left(  \sqrt{3}\pi y\right)
+\sin\left(  \frac{4\pi x}{3}\right)  \sin\left(  \frac{2\pi y}{\sqrt{3}
}\right)  +\sin\left(  \frac{5\pi x}{3}\right)  \sin\left(  \frac{\pi y}
{\sqrt{3}}\right)
\]
with
\[
x_{\infty}=0.3558473606263811208579681...,
\]
\[
y_{\infty}=0.4255359610370576630888604....
\]
It remains to compute the eigenfunction for $T_{3}$, but no closed-form
expression for this exists. \ A\ numerical computation using the Matlab
\textit{pdeeig} tool yields $(x_{\infty},y_{\infty})\approx(0.88,1.91)$. \ See
Figure 4. \ Significantly higher precision will be needed to ascertain whether
$y_{\infty}$ appears in the ETC\ database; we are hopeful that techniques in
\cite{HT} might save the day.

\begin{figure}[ptb]
\centering
\includegraphics[
height=2.7518in,
width=5.828in
]
{tri4.eps}
\caption{First Laplacian eigenfunction for $T_{3}$ using Matlab's
\textit{pdesurf}}
\end{figure}

\section{Acknowledgements}

I am thankful to Harri Hakula for refining my preceding estimate to
$(x_{\infty},y_{\infty})=(0.88047...,1.91599...)$ via Mathematica \cite{HH},
Stefan Steinerberger for introducing me to Fraenkel asymmetry, and the authors
of \cite{AK, BM} for writing inspirational papers.

config: arxiv


TITLE: relationship between a great circle arc and a latitude circle arc at a given latitude
QUESTION [2 upvotes]: My spherical geometry is a bit rusty but looking at the figure below:

... my intuition tells me that angles $\phi$ and $\theta$ (measured in radians) are connected with the following equation:
$\phi = \theta*cos(\delta)$
... where $\delta$ is the angle corresponding to the green arc.
At least the above holds true for the special cases $\delta=0$ and $\delta=\pi/2$. Does the above equation hold true in general and what is the terminology that describes the various angles and circles I've drawn?
update on terminology
It is now clear that the angles can be more properly described as follows:

angle $\theta$ is the arc between points $A'$ and $B'$ on a latitude circle at latitude $\delta$ (see the drawing on this answer which makes this point clear)
angle $\phi$ is the arc between points $A'$ and $B'$ on a great circle

REPLY [1 votes]: My attempt of a proof:
In any circle, with center O, radius r, points A and B on the circumference and angle AOB = $\alpha$, the straight segment from A to B is :
$AB_{straight}^2 = 2 r^2 \times (1-cos(\alpha))$ (1)
This is a straight application of the cosine theorem.
We also know for any angle $\alpha$ that
$cos(\alpha) = 1-2 sin^2 (\frac{\alpha}{2})$ (2)
By combining (1), (2) we get :
$AB_{straight}^2 = 4 r^2 \times sin^2(\frac{\alpha}{2})$ (3)
Apply (3) twice on the segment $A'B'$ from the original diagram by Marcus Junius Brutus :

once for the great circle with $r=R$ and $\alpha = \phi$
once for the latitude circle with $r=R \times cos(\delta)$ and $\alpha=\theta$

That gives
$4 R^2 \times sin^2(\frac{\phi}{2})$ = $4 (R \times cos(\delta))^2 \times sin^2(\frac{\theta}{2})$
which simplifies to
$sin(\frac{\phi}{2}) = cos(\delta) \times sin(\frac{\theta}{2})$


TITLE: Given $(a,m)=(b,m)=1$, show : $(ab,m)\mid ab,mb \implies (ab,m)\mid b$
QUESTION [2 upvotes]: Given $\exists a,m,b \in \mathbb{Z}, (a,m)=(b,m)=1$, show : $(ab,m)\mid ab,mb \implies (ab,m)\mid b$
Attempted proof:
$\exists a,m,b,x,y,x',y'\in \mathbb{Z}$, (a): $ (a,m)=1 \implies ax +my=1 $; also $(a,m)=1 \implies (ab,mb)=b$(b): $(b,m)=1 \implies bx' +my'=1$. Multiplying, (a), (b) get (c): $abxx' + amxy'+bmx'y+m^2yy'=1$
Am unable to derive from $(ab,mb)=b$, that $(ab,m) \mid ab,mb$ which is a pre-requisite to proof.
Apart from this, have failed to capture the essence of the derivation/proof desired, it would be better if there were an intuitive way to look into it.

Edit Based on comment by @mathlove, have tried to view (c) as :
(c) $abxx' + m(ax'y'+bx'y+myy') = 1$
=> So, $(ab, m)=1$, with coefficient of $ab= xx'$, & coefficient of $m=ax'y'+ bx'y'+myy'$.
Need prove that $(ab,m) \mid ab,mb$; & it is trivial to prove for $(ab,m) \mid ab$, & similarly for $(ab,m) \mid mb$.
Next, need prove $(ab,m) \mid b$, which for co-prime $ab,m$ is always true. In fact, it is always true for any integer in place of $b$, i.e. $\forall c \in \mathbb{Z}, (ab,m) \mid c$.
However, I have a thought regarding there being not the trivial condition $(a,m)=(b,m)=1$. In that situation, I hope NOTHING can be said about such relationships to occur. May be holding this trivial condition true, forms the needed base to view these relations intuitively.

Request detail on situation that not considers the stated trivial conditions, i.e. $(a,m)=(b,m)=1$, and then what relationships are possibly derivable, and why or why not.

REPLY [1 votes]: I think this will help you,
let $a,b,d \in \mathbb{Z}$, where $d$ and $a$ are nonzero.
if $d\mid ab$ and $p=gcd(a,d)$ then $\frac{d}{p}\mid b$.
which is easy to prove.


text:
\begin{figure}[h!]
\centering
\includegraphics[width=0.3\linewidth]{figures/primal_W_1_2_3_neg_norm.pdf}
\caption{Equitable and optimal division of the resources between $N=3$ different negative costs (i.e. utilities) given by $\MOT$. Utilities have been normalized. Blue dots and red squares represent the different elements of resources available in each cake. We consider the case where there is exactly one unit of supply per element in the cakes, which means that we consider uniform distributions. Note that the partition between the agents is equitable (i.e. utilities are equal) and proportional (i.e. utilities are larger than $1/N$).}
\label{fig:transport-map}
\end{figure}

\section{Equitable and Optimal Transport}
\label{sec:MOT}
\paragraph{Notations.} Let $\mathcal{Z}$ be a Polish space, we denote $\mathcal{M}(\mathcal{Z})$ the set of Radon measures on $\mathcal{Z}$. We call $\mathcal{M}_+(\mathcal{Z})$ the sets of positive Radon measures, and  $\mathcal{M}_+^1(\mathcal{Z})$ the set of probability measures. We denote $\mathcal{C}^b(\mathcal{Z})$ the vector space of bounded continuous functions on $\mathcal{Z}$. Let $\mathcal{X}$ and $\mathcal{Y}$ be two Polish spaces.  We denote for $\mu\in\mathcal{M}(\mathcal{X})$ and $\nu\in\mathcal{M}(\mathcal{Y})$, $\mu\otimes\nu$ the tensor product of the measures $\mu$ and $\nu$, and $\mu\ll\nu$ means that $\nu$ dominates $\mu$. We denote $\Pi_1:(x,y)\in\mathcal{X}\times\mathcal{Y}\mapsto x$ and $\Pi_2:(x,y)\in\mathcal{X}\times\mathcal{Y}\mapsto y$ respectively the projections on $\mathcal{X}$ and  $\mathcal{Y}$, which are continuous applications. For an application $g$ and a measure $\mu$, we denote $g_\sharp\mu$ the pushforward measure of $\mu$ by $g$.  For  $\mathcal{X}$ and $\mathcal{Y}$ two Polish spaces, we denote $\mathrm{LSC}(\mathcal{X}\times\mathcal{Y})$ the space of lower semi-continuous functions on $\mathcal{X}\times\mathcal{Y}$,  $\mathrm{LSC}^+(\mathcal{X}\times\mathcal{Y})$ the space of non-negative lower semi-continuous functions on $\mathcal{X}\times\mathcal{Y}$ and $\mathrm{LSC}^-_{*}(\mathcal{X}\times\mathcal{Y})$ the set of negative bounded below lower semi-continuous functions on $\mathcal{X}\times\mathcal{Y}$ . We also denote $\mathrm{C}^+(\mathcal{X}\times\mathcal{Y})$ the space of non-negative continuous functions on $\mathcal{X}\times\mathcal{Y}$ and $\mathrm{C}^-_{*}(\mathcal{X}\times\mathcal{Y})$ the set of negative continuous functions on $\mathcal{X}\times\mathcal{Y}$. Let $N\geq 1$ be an integer and denote $\Delta_N^{+}: = \{\lambda\in\mathbb{R}_+^N~\mathrm{s.t.}~\sum_{i=1}^N\lambda_i=1\}$, the probability simplex of $\mathbb{R}^N$.

For two positive measures of same mass $\mu\in\mathcal{M}_+(\mathcal{X})$ and $\nu\in\mathcal{M}_+(\mathcal{Y})$, we define the set of couplings with marginals $\mu$ and $\nu$:
\begin{align*}
    \Pi_{\mu,\nu}:=\left\{\gamma~\mathrm{s.t.}~ \Pi_{1\sharp}\gamma=\mu ~,~ \Pi_{2\sharp}\gamma=\nu\right\}\; .
\end{align*}
We introduce the subset of $(\mathcal{M}_+^1(\mathcal{X})\times \mathcal{M}_+^1(\mathcal{Y}))^N$ representing marginal decomposition:
\begin{align*}
 \textstyle\Upsilon_{\mu,\nu}^N:=&\Big\{(\mu_i,\nu_i)_{i=1}^N ~\mathrm{ s.t. }~ \sum_i \mu_i = \mu \mathrm{, } \sum_i \nu_i = \nu \\
 &~\mathrm{ and }~\forall i,~ \mu_i(\mathcal{X}) = \nu_i(\mathcal{Y})\;   \Big\}.
\end{align*}
We also define the following subset of $\mathcal{M}_+(\mathcal{X}\times \mathcal{Y})^N$ corresponding to the coupling decomposition:
\begin{align*}
    \Gamma^N_{\mu,\nu}:=\left\{(\gamma_i)_{i=1}^N~\mathrm{s.t.}~ \Pi_{1\sharp}\sum\gamma_i=\mu ~,~ \Pi_{2\sharp}\sum\gamma_i=\nu\right\} .
\end{align*}

\subsection{Primal Formulation}
\label{sec:primal-dual}
Consider a fair division problem where several agents aim to share two sets of resources, $\mathcal{X}$ and $\mathcal{Y}$, and assume that there is a divisible amount of each resource $x\in\mathcal{X}$ (resp. $y\in\mathcal{Y}$) that is available. Formally, we consider the case where resources are no more sets but rather distributions on these sets. Denote $\mu$ and $\nu$ the distribution of resources on respectively $\mathcal{X}$ and $\mathcal{Y}$. For example, one might think about a situation where agents want to share fruit juices and ice creams and there is a certain volume of each type of fruit juices and a certain mass of each type of ice creams available. Moreover each agent defines his or her paired preferences for each couple $(x,y)\in\mathcal{X}\times\mathcal{Y}$. Formally, each person $i$ is associated to an upper semi-continuous mapping  $u_i:\mathcal{X}\times\mathcal{Y}\xrightarrow{} \mathbb{R}^{+}$ corresponding to his or her preference for any given pair $(x,y)$. For example, one may prefer to eat chocolate ice cream with apple juice, but may prefer pineapple juice when it comes with vanilla ice cream. The total utility for an individual $i$ and a pairing $\gamma_i\in\mathcal{M}_+(\mathcal{X}\times\mathcal{Y})$ is then given by $V_i(\gamma_i) := \int u_i d\gamma_i$. To partition fairly among individuals, we maximize the minimum of individual utilities.

From a transport point of view, let assume that there are $N$ workers available to transport a distribution $\mu$ to another one $\nu$. The cost of a worker $i$ to transport a unit mass from location $x$ to the location $y$ is $c_i(x,y)$. To partition the work among the $N$ workers fairly, we minimize the maximum of individual costs.

These problems are in fact the same where the utility $u_i$, defined in the fair division problem, might be interpreted as the opposite of the cost $c_i$ defined in the transportation problem, i.e. for all $i$, $c_i = -u_i$. The two above problem motivate the introduction of $\MOT$ defined as follows.
\begin{definition}[Equitable and Optimal Transport]
\label{def-mot}
Let $\mathcal{X}$ and $\mathcal{Y}$ be Polish spaces. Let $\mathbf{c}:=(c_i)_{1\leq i\leq N}$ be a family of bounded below lower semi-continuous cost functions on $\mathcal{X}\times \mathcal{Y}$, and $\mu\in\mathcal{M}^1_+(\mathcal{X})$ and  $\nu\in\mathcal{M}^1_+(\mathcal{Y})$. We define the equitable and optimal transport primal problem:

\begin{align}
\label{eq-primal}
\MOT_{\mathbf{c}}(\mu,\nu) := \inf_{\substack{(\gamma_i)_{i=1}^N\in\Gamma^N_{\mu,\nu}\
}}   \max_i \int c_i d\gamma_i\; .
\end{align}
\end{definition}

We prove along with Theorem~\ref{thm:duality-GOT} that the problem is well defined and the infimum is attained. Lower-semi continuity is a standard assumption in OT. In fact, it is the weakest condition to prove Kantorovich duality~\citep[Chap. 1]{villani2003topics}. Note that the problem defined here is a linear optimization problem and when $N=1$ we recover standard optimal transport. Figure~\ref{fig:transport-map} illustrates the equitable and optimal transport problem we consider. Figure~\ref{fig:transport-map-ot-view} in Appendix~\ref{appendix-illustrations} shows an illustration with respect to the transport viewpoint in the exact same setting, i.e. $c_i = - u_i$. As expected, the couplings obtained in the two situations are not the same.

We now show that in fact, $\MOT$ optimum satisfies equality constraints in case of constant sign costs, i.e. total utility/cost of each individual are equal in the optimal partition. See Appendix~\ref{prv:mot-equality} for the proof.
\begin{prop}[$\MOT$ solves the problem under equality constraints]
\label{prop:mot-equality}
Let $\mathcal{X}$ and $\mathcal{Y}$ be Polish spaces. Let $\mathbf{c}:=(c_i)_{1\leq i\leq N}\in\mathrm{LSC}^+(\mathcal{X}\times\mathcal{Y})^N\cup\mathrm{LSC}^-_{*}(\mathcal{X}\times\mathcal{Y})^N$, $\mu\in\mathcal{M}^1_+(\mathcal{X})$ and  $\nu\in\mathcal{M}^1_+(\mathcal{Y})$. Then the following are equivalent:
\vspace{-0.3cm}
\begin{itemize}
    \item $(\gamma_i^{*})_{i=1}^N\in\Gamma^N_{\mu,\nu}$ is solution of Eq.~(\ref{eq-primal}),
    \item $(\gamma_i^{*})_{i=1}^N\in \argminB\limits_{\substack{(\gamma_i)_{i=1}^N\in\Gamma^N_{\mu,\nu}
}}   \left\{t~\mathrm{s.t.}~ \forall i~\int c_i d\gamma_i=t\right\}$.
\end{itemize}

\vspace{-0.4cm}
Moreover,
\vspace{-0.3cm}
\begin{align*}
  \MOT_{\mathbf{c}}(\mu,\nu) =  \min_{\substack{(\gamma_i)_{i=1}^N\in\Gamma^N_{\mu,\nu}
}}   \left\{t~\mathrm{s.t.}~ \forall i~\int c_i d\gamma_i=t\right\}\; .
\end{align*}
\end{prop}

This property highly relies on the sign of the costs. For instance if two costs are considered, one always positive and the other always negative, then the constraints cannot be satisfied. When the cost functions are non-negatives, $\MOT$ refers to a transportation problem while when the costs are all negatives, costs become utilities and $\MOT$ refers to a fair division problem. The two points of view are concordant, but proofs and interpretations rely on the sign of the costs.

\subsection{An Equitable and Proportional Division }
When the cost functions considered $c_i$ are all negatives, $\MOT$ become a fair division problem where the utility functions are defined as $u_i:=-c_i$. Indeed according to Proposition~\ref{prop:mot-equality}, $\MOT$ solves
\begin{align*}
\max_{\substack{(\gamma_i)_{i=1}^N\in\Gamma^N_{\mu,\nu}
}}   \left\{t~\mathrm{s.t.}~ \forall i,~\int u_i d\gamma_i=t\right\}\; .
\end{align*}
Recall that in our model, the total utility of the agent $i$ is given by $V_i(\gamma_i):=\int u_i d\gamma_i$. Therefore $\MOT$ aims to maximize the total utility of each agent $i$ while ensuring that they are all equal. Let us now analyze which fairness conditions the partition induced by $\MOT$ verifies. Assume that the utilities are normalized, i.e., $\forall i$, there exists $\gamma_i\in\mathcal{M}_+^1(\mathcal{X}\times\mathcal{Y})$ such that $V_i(\gamma_i)=1$. For example one might consider the cases where $\forall i$,  $\gamma_i=\mu\otimes\nu$ or $\gamma_i\in\argminB_{\gamma\in\Pi_{\mu,\nu}}  \int c_i d\gamma$. Then any solution $(\gamma_i^{*})_{i=1}^N\in\Gamma^N_{\mu,\nu}$ of $\MOT$ satisfies:

\begin{itemize}
    \item \textbf{Proportionality}: for all $i$, $V_i(\gamma_i^{*})\geq 1/N$,

    \item \textbf{Equitablity}: for all $i,j$, $V_i(\gamma_i^{*})=V_j(\gamma_j^{*})$.

\end{itemize}

Proportionality is a standard fair division criterion for which a resource is divided among $N$ agents, giving each agent at least $1/N$ of the heterogeneous resource by his/her own subjective valuation. Therefore here, this situation corresponds to the case where the normalized utility of each agent is at least $1/N$. Moreover, an equitable division is a division of an heterogeneous resource, in which each partner is equally happy with his/her share. Here this corresponds to the case where the utility of each agent are all equal.

The problem solved by $\MOT$ is a fair division problem where heterogeneous resources have to be shared among multiple agents according to their preferences. This problem is a relaxation of the two cake-cutting problem when there are a divisible amount of each item of the cakes. In that case, cakes are distributions and $\MOT$ makes a proportional and equitable partition of them. Details are left in Appendix~\ref{prv:mot-equality}.

\paragraph{Fair Cake-cutting.} Consider the case where the cake is an heterogeneous resource and there is a certain divisible quantity of each type of resource available. For example chocolate and vanilla are two types of resource present in the cake for which a certain mass is available. In that case, each type of resource in the cake is pondered by the actual quantity present in the cake. Up to a normalization, the cake is no more the set $\mathcal{X}$ but rather a distribution on this set. Note that for the two points of view to coincide, it suffices to assume that there is exactly the same amount of mass for each type of resources available in the cake. In that case, the cake can be represented by the uniform distribution over the set $\mathcal{X}$, or equivalently the set $\mathcal{X}$ itself. When cakes are distributions, the fair cutting cake problem can be interpreted as a particular case of $\MOT$ when the utilities of the agents do not depend on the variable $y\in\mathcal{Y}$. In short, we consider that utilities are functions of the form $u_i(x,y)=v_i(x)$ for all $(x,y)\in\mathcal{X}\times\mathcal{Y}$. The normalization of utilities can be cast as follows: $\forall i$, $V_i(\mu) = \int v_i(x) d\mu(x) = 1$. Then Proposition~\ref{prop:mot-equality} shows that the partition of the cake made by $\MOT$ is proportional and equitable. Note that for $\MOT$ to coincide with the classical cake-cutting problem, one needs to consider that the uniform masses of the cake associated to each type of resource cannot be splitted. This can be interpreted as a Monge formulation~\citep{villani2003topics} of $\MOT$ which is out of the scope of this paper.

\subsection{Optimality of $\MOT$}
We next investigate the coupling obtained by solving $\MOT$. In the next proposition, we show that under the same assumptions of Proposition~\ref{prop:mot-equality}, $\MOT$ solutions are optimal transportation plans. See Appendix~\ref{prv:mot-otplans} for the proof.
\begin{prop}[$\MOT$ realizes optimal plans]
\label{prop:mot-otplans}
Under the same conditions of Proposition~\ref{prop:mot-equality}, for any $(\gamma_i^{*})_{i=1}^N\in\Gamma^N_{\mu,\nu}$ solution of Eq.~(\ref{eq-primal}), we have for all $i\in\{1,\dots,N\}$
\begin{equation}
  \begin{aligned}
  \label{eq-optimal-coupling}
    \gamma_i^{*}&\in\argminB_{\gamma\in\Pi_{\mu_i^{*},\nu_i^{*}}}  \int c_i d\gamma\\
    \text{where\quad} & \mu_i^{*}:=\Pi_{1\sharp}\gamma_i^{*} ~,~ \nu_i^{*}:=\Pi_{2\sharp}\gamma_i^{*}\; ,
\end{aligned}
\end{equation}
and
\begin{equation}
\begin{aligned}
\label{eq-mot-optimality}
  \MOT_{\mathbf{c}}(\mu,\nu)=  &\min_{\substack{(\mu_i,\nu_i)_{i=1}^N\in\Upsilon^N_{\mu,\nu}}} t\\
  &\mathrm{s.t.}~ \forall i~ \wass_{c_i}(\mu_i,\nu_i) =t \; .
\end{aligned}
\end{equation}
\end{prop}

Given the optimal matchings $(\gamma_i^{*})_{i=1}^N\in\Gamma^N_{\mu,\nu}$, one can easily obtain the partition of the agents of each marginals. Indeed for all $i$, $\mu_i^{*}:=\Pi_{1\sharp}\gamma_i^{*}$ and $ \nu_i^{*}:=\Pi_{2\sharp}\gamma_i^{*}$ represent respectively the portion of the agent $i$ from distributions $\mu$ and $\nu$.

\begin{rmq}[Utilitarian and Optimal Transport]
To contrast with $\MOT$, an alternative problem is to maximize the sum of the total utilities of agents, or equivalently minimize the sum of the total costs of agents. This problem can be cast as follows:
\begin{align}
\label{eq-UOT}
  \inf_{\substack{(\gamma_i)_{i=1}^N\in\Gamma^N_{\mu,\nu}\
}}  \sum_i \int c_i d\gamma_i
\end{align}
Here one aims to maximize the total utility of all the agents, while in $\MOT$ we aim to maximize the total utility per agent under egalitarian constraint. The solution of~(\ref{eq-UOT}) is not fair among agents and one can show that this problem is actually equal to $\wass_{\min_i (c_i)}(\mu,\nu)$. Details can be found in Appendix~\ref{res:min-sum}.
\end{rmq}

\begin{figure*}[t]
\begin{tabular}{@{}c@{}c@{}c@{}c@{}}
\includegraphics[width=0.24\textwidth]{figures/dual_MOT_1_neg_norm.png}&
\includegraphics[width=0.24\textwidth]{figures/dual_MOT_2_neg_norm.png}&
\includegraphics[width=0.24\textwidth]{figures/dual_MOT_3_neg_norm.png}&
\includegraphics[width=0.275\textwidth]{figures/dual_MOT_1_2_3_neg_norm.png}

\end{tabular}
\caption{\emph{Left, middle left, middle right}: the size of dots and squares is proportional to the weight of their representing atom in the distributions $\mu_k^{*}$ and $\nu_k^{*}$ respectively.
The utilities $f_k^{*}$ and $g_k^{*}$ for each point in respectively $\mu_k^{*}$ and $\nu_k^{*}$ are represented by the color of dots and squares according to the color scale on the right hand side. The gray dots and squares correspond to the points that are ignored by agent $k$ in the sense that there is no mass or almost no mass in distributions $\mu^*_k$ or $\nu^*_k$. \emph{Right}: the size of dots and squares are uniform since they correspond to the weights of uniform distributions $\mu$ and $\nu$ respectively. The values of $f^*$ and $g^*$ are given also by the color at each point. Note that each agent gets exactly the same total utility, corresponding exactly to $\MOT$. This value can be computed using dual formulation~\eqref{eq-dual} and for each figure it equals the sum of the values (encoded with colors) multiplied by the weight of each point (encoded with sizes).\label{fig:potential-dual}}
\end{figure*}

\subsection{Dual Formulation}

Let us now introduce the dual formulation of the problem and show that strong duality holds under some mild assumptions. See Appendix~\ref{prv:duality-GOT} for the proof.
\begin{thm}[Strong Duality]
\label{thm:duality-GOT}
Let $\mathcal{X}$ and $\mathcal{Y}$ be Polish spaces. Let $\mathbf{c}:=(c_i)_{i=1}^{N}$ be bounded below lower semi-continuous costs. Then \emph{strong duality holds}, i.e. for $(\mu,\nu)\in\mathcal{M}_{+}^{1}(\mathcal{X})\times\mathcal{M}_{+}^{1}(\mathcal{Y})$:
\begin{align}
\label{eq-dual}
   \MOT_{\mathbf{c}}(\mu,\nu)= \sup\limits_{\substack{\lambda\in\Delta^+_N\\(f,g)\in\mathcal{F}_{\mathbf{c}}^{\lambda}}}\int fd\mu+ \int gd\nu
\end{align}
where  $\mathcal{F}^{\lambda}_{\mathbf{c}}:=\{(f,g)\in\mathcal{C}^b(\mathcal{X})\times\mathcal{C}^b(\mathcal{Y})~\mathrm{ s.t. }~\forall i\in\{1,...,N\},~ f\oplus g\leq\lambda_ic_i\}$.

\end{thm}

This theorem holds under the same hypothesis and follows the same reasoning as the one in~\citep[Theorem 1.3]{villani2003topics}. While the primal formulation of the problem is easy to understand, we want to analyse situations where the dual variables also play a role. For that purpose we
show in the next proposition a simple characterisation
of the primal-dual optimality in case of constant sign
cost functions.  See Appendix~\ref{prv:optimality-cond} for the proof.

\begin{prop}
\label{prop:optimality-cond}
Let $\mathcal{X}$ and $\mathcal{Y}$ be compact Polish spaces. Let $\mathbf{c}:=(c_i)_{1\leq i\leq N}\in\mathrm{C}^+(\mathcal{X}\times\mathcal{Y})^N\cup\mathrm{C}^-_{*}(\mathcal{X}\times\mathcal{Y})^N$, $\mu\in\mathcal{M}^1_+(\mathcal{X})$ and  $\nu\in\mathcal{M}^1_+(\mathcal{Y})$. Let also $(\gamma_k)_{k=1}^N\in\Gamma^N_{\mu,\nu}$ and $(\lambda,f,g)\in\Delta_n^{+}\times\mathcal{C}^b(\mathcal{X})\times\mathcal{C}^b(\mathcal{Y})$. Then Eq.~(\ref{eq-dual}) admits a solution and the following are equivalent:
\begin{itemize}
    \item $(\gamma_k)_{k=1}^N$ is a solution of Eq.~(\ref{eq-primal}) and $(\lambda,f,g)$ is a solution of Eq.~(\ref{eq-dual}).
    \item \begin{enumerate}
        \item $\forall i\in\{1,...,N\},~ f\oplus g\leq\lambda_i c_i$
        \item $\forall i,j\in\{1,...,N\}~\int c_i d\gamma_i=\int c_j d\gamma_j$
        \item $f \oplus g= \lambda_i c_i ~~\gamma_{i}\text{-a.e.}$
    \end{enumerate}
\end{itemize}
\end{prop}
\begin{rmq}
\label{rk:lambdanonzero}
It is worth noting that when we assume that $\mathbf{c}:=(c_i)_{1\leq i\leq N}\in\mathrm{C}^+_{*}(\mathcal{X}\times\mathcal{Y})^N\cup\mathrm{C}^-_{*}(\mathcal{X}\times\mathcal{Y})^N$, then we can refine the second point of the equivalence presented in Proposition~\ref{prop:optimality-cond} by adding the following condition: $\forall i\in\{1,...,N\}~\lambda_i\neq 0$.
\end{rmq}

Given two distributions of resources represented by the measures $\mu$ and $\nu$, and $N$ utility functions denoted $(u_i)_{i=1}^N$, we want to find an \emph{equitable} and \emph{stable} partition among the agents in case of \emph{transferable utilities}. Let $k$ be an agent. We say that his or her utility is transferable when once $x\in\mathcal{X}$ and $y\in\mathcal{Y}$ get matched, he or she has to decide how to split his or her associated utility $u_k(x,y)$ . She or he divides $u_k(x,y)$ into a quantity $f_k(x)$ which can be seen as the utility of having $x$ and $g_k(y)$ for having $y$. Therefore in that problem we ask for $(\gamma_k,f_k,g_k)_{k=1}^N$ such that
\begin{align}
\label{eq-1-dual}
    u_k(x,y)=f_k(x)+g_k(y) ~~\gamma_{k}\text{-a.e.}
\end{align}
Moreover, for the partition to be \emph{stable}~\citep{rothm}, we want to ensure that, for every agent $k$, none of the resources $x\in\mathcal{X}$ and $y\in\mathcal{Y}$ that have not been matched together for this agent would increase their utilities, $f_k(x)$ and $g_k(y)$, if there were matched together in the current matching instead. Formally we ask that for $k\in\{1,\dots,N\}$ and all $(x,y)\in\mathcal{X}\times\mathcal{Y}$,
\begin{align}
\label{eq-2-dual}
    f_k(x)+g_k(y)\geq u_k(x,y)\; .
\end{align}
Indeed if there exist $k$, $x$ and $y$ such that $u_k(x,y)>f_k(x)+g_k(y)$, then $x$ and $y$ will not be matched together in the share of the agent $k$ and he can improve his utility for both $x$ and $y$ by matching $x$ with $y$.

Finally we aim to share equitably the resources among the agents which boils down to ask
\begin{align}
    \label{eq-3-dual}
    \forall i,j\in\{1,...,N\}~\int u_i d\gamma_i=\int u_j d\gamma_j
\end{align}
Thanks to Proposition~\ref{prop:optimality-cond}, finding  $(\gamma_k,f_k,g_k)_{k=1}^N$ satisfying \eqref{eq-1-dual}, \eqref{eq-2-dual} and \eqref{eq-3-dual} can be done by solving Eq.~\eqref{eq-primal} and Eq.~\eqref{eq-dual}. Indeed let $(\gamma_k)_{k=1}^N$ an optimal solution of Eq.~(\ref{eq-primal}) and $(\lambda,f,g)$ an optimal solution of Eq.~(\ref{eq-dual}). Then by denoting for all $k=1,\dots,N$,
   $f_k = \frac{f}{\lambda_k}$ and $g_k = \frac{g}{\lambda_k}$,
we obtain that $(\gamma_k,f_k,g_k)_{k=1}^N$ solves the \emph{equitable} and \emph{stable} partition problem in case of \emph{transferable utilities}. Note that again, we end up with equality constraints for the optimal dual variables. Indeed, for all $i,j\in\{1,\dots,N\}$, at optimality we have
$\int f_i + g_i d\gamma_i = \int f_j + g_j d\gamma_j  \; $. Figure~\ref{fig:potential-dual} illustrates this formulation of the problem with dual potentials. Figure~\ref{fig:potential-dual-ot-viewpoint} in Appendix~\ref{appendix-illustrations} shows the dual solutions with respect to the transport viewpoint in the exact same setting, i.e. $c_i = - u_i$. Once again, the obtained solutions differ.

\subsection{Link with other Probability Metrics}
\label{sec:properties}
In this section, we provide some topological properties on the object defined by the $\MOT$ problem. In particular, we make links with other known probability metrics, such as Dudley and Wasserstein metrics and give a tight upper bound.

When $N=1$, recall from the definition~\eqref{eq-primal} that the problem considered is exactly the standard OT problem. Moreover any $\MOT$ problem with $k\leq N$ costs can always be rewritten as a $\MOT$ problem with $N$ costs. See Appendix~\ref{res:MOT-gene} for the proof. From this property, it is interesting to note that, for any $N\geq 1$, $\MOT$ generalizes standard Optimal Transport.

\paragraph{Optimal Transport.} Given a cost function $c$, if we consider the problem $\MOT$ with $N$ costs such that, for all $i$, $c_i=N\times c$ then, the problem $\MOT_\mathbf{c}$ is exactly $\wass_c$. See Appendix~\ref{res:MOT-gene} for the proof.

Now we have seen that all standard OT problems are sub-cases of the $\MOT$ problem, one may ask whether $\MOT$ can recover other families of metrics different from standard OT. Indeed we show that the $\MOT$ problem recovers an important family of $\textsc{IPM}$s with supremum taken over the space of $\alpha$-Hölder functions with $\alpha\in (0,1]$. See Appendix~\ref{prv:GOT-holder} for the proof.
\begin{prop}
\label{prop:GOT-holder}
Let $\mathcal{X}$ be a Polish space. Let $d$ be a metric on $\mathcal{X}^2$ and $\alpha\in (0,1]$. Denote $c_1= 2\times\mathbf{1}_{x\neq y}$, $c_2=d^{\alpha}$ and $\mathbf{c}:=(c_1,(N-1)\times c_2,...,(N-1)\times c_2)\in \mathrm{LSC}(\mathcal{X}\times\mathcal{X})^N$
then for any $(\mu,\nu)\in\mathcal{M}_+^{1}(\mathcal{X})\times\mathcal{M}_+^{1}(\mathcal{X})$
\begin{align}
\label{eq-holder}
    \MOT_{\mathbf{c}}(\mu,\nu) =\sup_{f\in B_{d^{\alpha}}(\mathcal{X})} \int_{\mathcal{X}} f d\mu - \int_{\mathcal{X}} f d\nu
\end{align}
where $B_{d^{\alpha}}(\mathcal{X}):=\left\{f\in C^{b}(\mathcal{X})\mathrm{:}~\Vert f\Vert_{\infty}+\lVert f\rVert_\alpha\leq 1 \right\}$ and $\lVert f\rVert_\alpha := \sup_{x\neq y}\frac{|f(x)-f(y)|}{d^{\alpha}(x,y)}$.
\end{prop}

\paragraph{Dudley Metric.} When $\alpha=1$, then for $(\mu,\nu)\in\mathcal{M}_+^{1}(\mathcal{X})\times\mathcal{M}_+^{1}(\mathcal{X})$, we have
$$\MOT_{\mathbf{c}}(\mu,\nu) = \MOT_{(c_1,d)}(\mu,\nu) = \beta_d(\mu,\nu)$$
where  $\beta_d$ is the \textit{Dudley Metric}~\citep{dudley1966weak}. In other words, the Dudley metric can be interpreted as an equitable and optimal transport between the measures with the trivial cost and a metric $d$. We acknowledge that~\cite{chizat2018unbalanced} made a link between Unbalanced Optimal Transport and the ``flat metric'', an IPM close to the Dudley metric, defined on the space $\left\{f\mathrm{:}~\Vert f\Vert_{\infty}\leq 1,~\lVert f\rVert_1\leq 1 \right\}$.

\paragraph{Weak Convergence.} When $d$ is an unbounded metric on $\mathcal{X}$, it is well known that $\wass_{d^{p}}$ with $p\in(0,+\infty)$ metrizes a convergence a bit stronger than weak convergence~\citep[Chap. 7]{villani2003topics}.  A sufficient condition for Wasserstein distances to metrize weak convergence on the space of distributions is that the metric $d$ is bounded. In contrast, metrics defined by Eq.~\eqref{eq-holder} do not require such assumptions and $\MOT_{(\mathbf{1}_{x\neq y},d^{\alpha})}$  metrizes the weak convergence of probability measures~\citep[Chap. 1-7]{villani2003topics}.

For an arbitrary choice of costs $(c_i)_{1\leq i\leq N}$, we obtain a tight upper control of $\MOT$ and show how it is related to the OT problem associated to each cost involved. See Appendix~\ref{prv:ineqharmonic} for the proof.
\begin{prop}

\label{prop:ineqharmonic}
Let  $\mathcal{X}$ and $\mathcal{Y}$ be Polish spaces. Let $\mathbf{c}:=(c_i)_{1\leq i\leq N}$ be a family of nonnegative lower semi-continuous costs. For any $(\mu,\nu)\in\mathcal{M}_+^{1}(\mathcal{X})\times\mathcal{M}_+^{1}(\mathcal{Y})$
\begin{align}
\label{eq:harmonic}

    \MOT_{\mathbf{c}}(\mu,\nu)\leq \left(\sum\limits_{i=1}^N\frac{1}{\wass_{c_i}(\mu,\nu)}\right)^{-1}
\end{align}
\end{prop}

Proposition~\ref{prop:ineqharmonic} means that
the minimal cost to transport all goods under the constraint that all workers contribute equally is lower than the case where agents share equitably and optimally the transport with distributions $\mu_i$ and $\nu_i$ respectively proportional to $\mu$ and $\nu$, which equals the harmonic sum written in Equation~\eqref{eq:harmonic}.

\begin{example*}
Applying the above result in the case of the  Dudley metric recovers the following inequality~\citep[Proposition 5.1]{sriperumbudur2012empirical} \begin{align*}
    \beta_d(\mu,\nu)\leq \frac{\TV(\mu,\nu)\wass_d(\mu,\nu)}{\TV(\mu,\nu)+\wass_d(\mu,\nu)}.
\end{align*}
\end{example*}


TITLE: How to use nonlogical axioms in deduction
QUESTION [2 upvotes]: I'm a little bit unclear about the concept of nonlogical axioms and their potential usage in deduction.
According to this question, nonlogical axioms are those are which not universally valid and could not be presented by logical quantifiers and connectives.
Formal definition of deduction includes three seminal concepts: logical axioms, nonlogical axioms and rules of inference. If we admit that nonlogical axioms could not be stated in terms of logical concepts, how could we use them in deduction based inference rules?

REPLY [4 votes]: If you want to prove, say, a property of groups, your nonlogical axioms are going to be a suitable choice of axioms for group theory, which characterize associativity, existence of identity and inverses.  The logical axioms, on the other hand, characterize sound reasoning, regardless of the intended application.


TITLE: interior and closure strictly from definitions.
QUESTION [0 upvotes]: Let $X$ be a topological space and $A \subset X$.
Let int$A$ (interior of $A$ in $X$) be the union of all open sets contained in $A$. That is,
int$A= \bigcup_{B \in \tau_A} B$.
Let cl$A$ (closure of $A$ in $X$) be the intersection of every closed set containing $A$.
I need to show that the boundary of $A$, which is the set difference of the two defined above, contains every limit point of $A$.
Attempt:
Let $x$ be a limit point of $A$, then every neighborhood of $x$, say $U$ intersects $A$ nontrivially, that is, there exists a $y \neq x$ such that $y \in U \cap A$.
I need to show $x \in$ every closed set containing $A$ and that it is not in any open set contained in $A$. really stuck.

REPLY [1 votes]: You cannot prove it, since it is false. If, say, $X=\Bbb R$, endowed with its usual topology, and if $A=(-1,1)$, then the boundary of $A$ is $\{1,-1\}$.  But the set of limit points of $A$ is $[-1,1]$, and $[-1,1]\not\subset\{1,-1\}$.