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anemone

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- MHB
- Thread starter anemone
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- #1

anemone

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- #2

Opalg

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\draw [help lines, ->] (-3.5,0) -- (3.5,0) ;

\draw [help lines, ->] (0,-3.5) -- (0,3.5) ;

\coordinate [label=below left:$O$] (O) at (0,0) ;

\coordinate [label=above:$A$] (A) at (110:3) ;

\coordinate [label=left:$B$] (B) at (210:3) ;

\coordinate [label=below:$C$] (C) at (250:3) ;

\coordinate [label=right:$D$] (D) at (330:3) ;

\coordinate [label=right:$E\ $] (E) at (40:3) ;

\draw [thick] (A) -- (B) -- (C) -- (D) -- (E) -- cycle ;

\draw (C) -- (A) -- (D) -- (B) ;

\draw [thin] (A) -- (0,0) -- (E) ;

\draw [thin] (0,0) -- (D) ;

\draw (0.6,0.25) node{$\delta$} ;

\draw (0.1,0.35) node{$\gamma$} ;

\draw (-1,3.5) node{$(\cos(\delta + \gamma), \sin(\delta + \gamma))$} ;

\draw (-4.5,-1.9) node{$(-\cos(\delta - \gamma), \sin(\delta - \gamma)$} ;

\draw (-1,-3.5) node{$(\cos(\delta + \gamma), -\sin(\delta + \gamma)$} ;

\draw (4.4,-1.9) node{$(\cos(\delta - \gamma), \sin(\delta - \gamma)$} ;

\draw (3.8,1.9) node{$(\cos\delta, \sin\delta)$} ;[/TIKZ]

Choose a coordinate system with the unit circle centred at the origin $O$, the $x$-axis parallel to $BD$ and the $y$-axis parallel to $AC$, as in the diagram. Split the pentagon into the quadrilateral $ABCD$ and the triangle $ADE$. If $ABCD$ is kept fixed then the area of $ADE$ is maximised when $E$ is midway between $A$ and $D$ on the arc $AD$. Suppose that $OE$ then makes an angle $\delta$ with the $x$-axis, and let $2\gamma$ be the angle $AOD$, so that the angles $AOE$ and $EOD$ are both $\gamma$. The coordinates of $A$, $B$, $C$, $D$ and $E$ are then as shown in the diagram.

The area of $ABCD$ is the sum of the areas of triangles $BAD$ and $CAD$, with base $BD$ and combined height $AC$. So (using a product-to-sum identity) $$\text{Area}(ABCD) = \tfrac12AC\cdot BD = 2\sin(\delta+\gamma)\cos(\delta-\gamma) = \sin(2\delta) + \sin(2\gamma).$$ The triangle $ADE$ has base $AD = 2\sin\gamma$ and height $1-\cos\gamma$, so its area is $\sin\gamma(1-\cos\gamma)$.

Thus the area of the pentagon is $\sin(2\delta) + \sin(2\gamma) + \sin\gamma(1-\cos\gamma) = \sin(2\delta) + \sin\gamma(1+\cos\gamma)$. As far as $\delta$ is concerned, this is maximised when $\sin(2\delta) = 1$, or $\delta = 45^\circ$. To maximise the $\gamma$-function, differentiate it, getting $\cos\gamma(1+\cos\gamma) - \sin^2\gamma = 0$. That gives $2\cos^2\gamma + \cos\gamma - 1 = 0$, so that $(2\cos\gamma - 1)(\cos\gamma + 1) = 0$. The maximum occurs when $\cos \gamma = \frac12$, or $\gamma = 60^\circ$.

The maximum area of the pentagon is therefore $1 + \frac{\sqrt3}2\bigl(1+ \frac12\bigr) = 1 + \frac{3\sqrt3}4$.

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