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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?

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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?

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Thanks. Doesn't minimizing the distance to point $D$ however imply that you will pick $D$ to be arbitrarily close to $AB$ (implying a degenerate triangle)? –  Dreamer Oct 30 '13 at 9:39
@rbm Now this depends on how line $AB$ is situated with respect to the circle. If $AB$ intersects or touches the circle, then yes, we would be able to pick $D$ arbitrarily close to $AB$, so there would be no minimum. But if $AB$ doesn't meet the circle, then the minimum does exist. –  Dan Shved Oct 30 '13 at 9:40
Ah ok, I see. I understand now that this minimum exists through pictures, however, how can I algebraically find this minimum? –  Dreamer Oct 30 '13 at 9:46
Where $AB$ does not cut or touch the circle, the point $D$ lies on the perpendicular from the centre of the circle to line $AB$. Note that there are two diametrically opposite points, and you need to choose the right one. The other point gives a maximum. You can find the perpendicular and then the point on the perpendicular distance from centre of circle = radius. –  Mark Bennet Oct 30 '13 at 9:58
@rbm I'm not even sure what you mean by "algebraically". Mark Bennet's comment explains how to find the optimal $D$ in geometric terms. This is the solution I was hinting at. I don't see a reason to try and rewrite it in any significant way. –  Dan Shved Oct 30 '13 at 10:01
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

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}

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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?

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Thanks. Is there an algebraic way to find this minimum? –  Dreamer Oct 30 '13 at 9:49
Channelling Mark Bennet, drop a perpendicular from $D$ onto $AB$ extended. You want to minimise this distance. Clearly this can be zero if $AB$ extended cuts or touches the circle. Otherwise it is minimised when this perpendicular passes through $C$ and $D$ is between $C$ and $AB$ extended. –  Henry Oct 30 '13 at 10:09
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