Minimize the area of a triangle Let $A \neq B$ be ﬁxed points outside a ﬁxed circle with centre
$C$. The point $D$ can be chosen freely on the circle. The goal is to minimise
the area of triangle $ABD$. Degenerate triangles (triangles that are merely line
segments) are excluded. In which conﬁgurations of $A, B, C$ and the circle does
this problem have a solution and how can one construct its solution?

I expressed the area as $A = \frac{1}{2}ab \sin(\gamma)$ and then derived this expression with respect to $\gamma$ but this gives a maximum, not a minimum. I think a minimum would occur for $\gamma \rightarrow 0$ but this would resulte in a degenerate triangle. So I'm inclined to say that a solution does not exist. Does anyone have any thoughts on this?
 A: Forget about $\frac{1}{2}ab\sin \gamma$. The area of $ABD$ equals $\frac{1}{2}|AB| \cdot h$, where $h$ is the distance from line $AB$ to point $D$. Points $A$ and $B$ are fixed, so the problem is to minimize the distance from point $D$ to the fixed line $AB$. The nondegenerate condition means that you cannot pick $D$ on line $AB$. Can you solve the problem when formulated like that?
A: For any point $D$ on the circle, there is a line through $D$ parallel to $AB$.
The area of $ABD$ is half the product of the length of $AB$ and the perpendicular distance between the parallel line through $D$ and $AB$.
So the area is has a local minimum or maximum when $D$ lies on a tangent of the circle parallel to $AB$.
To solve the question, you might need to look at examples (sketchs will do).  What happens if $AB$ (extended) is a tangent to the circle, or misses the circle, or cuts the circle in two points? 
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
Let's set the circle center as the coordinates origin. Let's call $\vec{A}$, $\vec{B}$ and $\vec{D}$ the 'position vectors' of the points $A$, $B$ and $D$. Also, $\vec{s} \equiv \vec{B} - \vec{A}$. The triangle area ${\cal A}$ is given by the magnitude of a vector $\vec{\cal A}$:
$$
\vec{\cal A}
=
{1 \over 2}\vec{D}\times\vec{A} + {1 \over 2}\vec{A}\times\vec{B}
+
{1 \over 2}\vec{B}\times\vec{D}
=
{1 \over 2}\vec{A}\times\vec{B} - {1 \over 2}\vec{D}\times\vec{s}
$$
Our goal is to minimize ${\cal A}^{2} = \vec{\cal A}\cdot\vec{\cal A}$ given the constraint $\vec{D}\cdot\vec{D} = a^{2} = \mbox{constant}$. $a > 0$ is the circle radius. We'll use Lagrange multipliers technique. Let's define
${\cal F} \equiv \vec{\cal A}\cdot\vec{\cal A} - \mu\vec{D}\cdot\vec{D}/2$ where $\mu/2$ is a Lagrange multiplier.
$$
\delta{\cal F} = 2\delta\vec{\cal A}\cdot\vec{\cal A} - \mu\delta\vec{D}\cdot\vec{D}
\,,\qquad
2\delta\vec{\cal A}\cdot\vec{\cal A}
=
-\delta\vec{D}\times\vec{s}\cdot\vec{\cal A}
=
\delta\vec{D}\cdot\vec{\cal A}\times\vec{s}
$$
$$
\delta{\cal F} = \delta\vec{D}\cdot\pars{\vec{\cal A}\times\vec{s} - \mu\vec{D}}\,,
\qquad
\pars{~\delta{\cal F} = 0\quad\imp\quad\mu\vec{D} = \vec{\cal A}\times\vec{s}~}
$$
$$
\mu\vec{D} = \vec{\cal A}\times\vec{s}
=
\vec{C} - {1 \over 2}\pars{\vec{D}\times\vec{s}}\times\vec{s}
\quad\mbox{where}\quad
\color{#ff0000}{\large\vec{C} = {1 \over 2}\pars{\vec{A}\times\vec{B}}\times\vec{s}}
\quad\mbox{is a known vector}
$$
$$
\mu\vec{D} = \vec{C} + {1 \over 2}\pars{s^{2}\vec{D} - \vec{D}\cdot\vec{s}\,\vec{s}}
\quad\imp\quad
\mu\vec{D}\cdot\vec{s} = \vec{C}\cdot\vec{s} = 0
\tag{1}
$$
Then
$$
\vec{D}
=
{2\vec{C} \over 2\mu - s^{2}}
\quad\imp\quad
a^{2} = {4C^{2} \over \pars{2\mu - s^{2}}^{2}}
\quad\imp\quad
{1 \over 2\mu - s^{2}} = \pm {a \over 2C}
$$
$$
\vec{D} = \pm\, a\,{\vec{C} \over C}
=
\pm\, a\,
{\pars{\vec{A}\times\vec{B}}\times\pars{\vec{B} - \vec{A}}
 \over
 \verts{\pars{\vec{A}\times\vec{B}}\times\pars{\vec{B} - \vec{A}}}}
\quad\mbox{and}\quad
\vec{\cal A}
=
{1 \over 2}\,\vec{A}\times\vec{B}
\mp
{1 \over 2}\,{a \over C}\,\vec{C}\times\vec{s}\,,
\quad
\vec{A} \not\parallel \vec{B}
$$
Notice that $\vec{D} \perp \pars{\vec{B} - \vec{A}}$. Geometrically, we look for a vector $\vec{D}$ which is perpendicular to the segment that joins point $A$ and $B$.
When $\vec{A} \parallel \vec{B}$, we have $\pars{~\mbox{from Eq.}\ \pars{1}~}$ $\vec{D} \perp \pars{\vec{B} - \vec{A}}$ and the solution is a vector
$\vec{D}$ which is perpendicular to the segment which joins $\vec{A}$ and $\vec{B}$:
$\sum_{i\ =\ x, y, z}D_{i}s_{i} = 0$.
Let's check an example $\pars{~\mbox{with a 'simple circle'}\ x^{2} + y^{2} = 1~}$:
$$
\vec{A} = \pars{3,3}\,,\quad \vec{B} = \pars{-4,0}\quad\mbox{and}\quad a = 1
$$
Then
$$
\vec{A}\times\vec{B} =\pars{A_{x}B_{y} - A_{y}B_{x}}\hat{z} = 12\hat{z}\,,
\quad
\vec{s} = \vec{B} - \vec{A} = \pars{-7,-3}
$$
$$
\pars{\vec{A}\times\vec{B}}\times\pars{\vec{B} - \vec{A}}
=
12\hat{z}\times\pars{-7\hat{x} - 3\hat{y}} = -84\hat{y} + 36\hat{x}
$$
$$
\vec{D} = \pm\pars{{9 \over \root{522}}\,\hat{x} - {21 \over \root{522}}\,\hat{y}}
$$
\begin{align}
\vec{\cal A}_{\pm}
&=
6\hat{z}
-
\pars{1 \over 2}\bracks{\pm\pars{{9 \over \root{522}}\,\hat{x} - {21 \over \root{522}}\,\hat{y}}}\times\pars{-7\hat{x} - 3\hat{y}}
\\[3mm]&=
6\hat{z}
\pm
{1 \over 2\root{522}}\pars{9\hat{x} - 21\hat{y}}\times
\pars{7\hat{x} + 3\hat{y}}
=
\pars{6 \pm {87 \over \root{522}}}\hat{z}
\approx\left\lbrace%
\begin{array}{lcl}
9.8079\,\hat{z} & \mbox{if} & +
\\[2mm]
2.1920\,\hat{z} & \mbox{if} & -
\end{array}\right.
\end{align}
\begin{align}
\vec{D}_{+} &= \phantom{-\,}
{9 \over \root{522}}\,\hat{x} - {21 \over \root{522}}\,\hat{y}
\quad\imp\quad \mbox{max. Area} = 9.8079
\\[3mm]
\vec{D}_{-}
&=
-\,{9 \over \root{522}}\,\hat{x} + {21 \over \root{522}}\,\hat{y}
\quad\imp\quad \mbox{min. Area} = 2.1920
\end{align}
