# What is the Maximum Area of an Inscribed Pentagon with Perpendicular Diagonals?

• MHB
• anemone
In summary, the formula for finding the maximum area of a pentagon is A = (5/4) x s x h, and this maximum area can be achieved when the pentagon is a regular pentagon with equal sides and angles. If the side length is not given, the formula A = (5/4) x s^2 x cot(π/5) can be used, where s is the apothem. The minimum area of a pentagon is achieved when it is a degenerate pentagon with all 5 points in a straight line, and the maximum area is always larger than the minimum area. A pentagon cannot have a negative area, even for a degenerate pentagon where the area is
anemone
Gold Member
MHB
POTW Director
Find the maximum area of a pentagon $ABCDE$ inscribed in a unit circle such that the diagonal $AC$ is perpendicular to the diagonal $BD$.

[TIKZ]\draw circle (3) ;
\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 $1 + \frac{\sqrt3}2\bigl(1+ \frac12\bigr) = 1 + \frac{3\sqrt3}4$.

## 1. What is the formula for finding the maximum area of a pentagon?

The formula for finding the maximum area of a pentagon is A = (1/4)√(5(5+2√5))s^2, where s is the length of one side of the pentagon.

## 2. How do you determine the length of the sides of a pentagon to maximize its area?

To determine the length of the sides of a pentagon to maximize its area, you can use the formula s = (2√5A)/(5+√5), where A is the desired maximum area. This will give you the length of one side, and you can then use this value to find the length of the other sides using the same formula.

## 3. Can the maximum area of a pentagon be found using trigonometry?

Yes, the maximum area of a pentagon can be found using trigonometry. You can use the formula A = (1/4)√(5(5+2√5))s^2, where s is the length of one side of the pentagon, and the trigonometric functions sin and cos to find the maximum area.

## 4. Is there a specific shape or type of pentagon that will always have the maximum area?

No, there is not a specific shape or type of pentagon that will always have the maximum area. The maximum area of a pentagon depends on the length of its sides, so any pentagon with the correct side lengths can have the maximum area.

## 5. Can the maximum area of a pentagon be found using calculus?

Yes, the maximum area of a pentagon can be found using calculus. By taking the derivative of the area formula with respect to the length of one side, setting it equal to 0, and solving for the side length, you can find the maximum area of the pentagon.

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