Equable ellipses I

This is not an exact solution but is a quick way to get close enough to the correct answer.

Since the major axis is twice the length of its minor axis, $a = 2b$ .

The area of the ellipse is $A = \pi ab$ , and substituting $a = 2b$ gives $A = 2 \pi b^2$ .

A Ramanujan approximation of the perimeter of an ellipse is $P \approx \pi(3(a + b) - \sqrt{(3a + b)(a + 3b)})$ , and substituting $a = 2b$ gives $P \approx \pi b(9 - \sqrt{35})$ .

Therefore, $2 \pi b^2 \approx \pi b(9 - \sqrt{35})$ , which solves to $b \approx \frac{1}{2}(9 - \sqrt{35})$ , which means that $A \approx 2 \pi (\frac{1}{2}(9 - \sqrt{35}))^2 \approx \boxed{14.939}$ .

Since there's no closed-form expression for the perimeter of an ellipse, numerical methods or approximations are the only way to go. @David Vreken posted a solution using an approximation, so here are a couple of numerical methods.

In both, I'll use a scaling argument, so let's start with that. Let the ellipse with semi-axes $a,\frac12 a$ be $E$ , and let its area be $A(a)$ and its perimeter $P(a)$ .

Then by similarity, $A(a)=a^2 \cdot A(1)$ and $P(a)=a \cdot P(1)$ .

For the ellipse to be equable, $a^2 \cdot A(1) = a\cdot P(1)$

or, rearranging, $a = \frac{P(1)}{A(1)}$

Of course, $A(1)=\frac12 \pi$ , so $a=\frac{2P(1)}{\pi}$

This means we just need to find $P(1)$ .

All the methods below give $P(1)\approx 4.84422 \ldots$ , which leads to $a=\frac{2P(1)}{\pi}=3.08392\ldots$ and $A(a)=\boxed{14.93924\ldots}$ .

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Method $1$ : non-calculus

We can draw the ellipse $E$ with the equations $x(t)=\cos t$ and $y(t)=\frac12 \sin t$ . Letting $t_k=\frac{2\pi k}{n}$ ,

$P(1) \approx \sum_{k=0}^{n-1} \sqrt{\left(x \left(t_{k+1} \right)-x \left(t_{k+1} \right) \right)^2+\left(y \left(t_{k+1} \right)-y \left(t_{k+1} \right) \right)^2}$

For various $n$ values, we get:

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Method $2$ : calculus

From the same parameterisation, $\begin{aligned} P(1)&=4\int_0^{\frac{\pi}{2}} \sqrt{\dot{x}^2+\dot{y}^2} \;dt \\ &=4\int_0^{\frac{\pi}{2}} \sqrt{\cos^2 t+\frac14 \sin^2 t} \;dt \\ &=2\int_0^{\frac{\pi}{2}} \sqrt{4\cos^2 t+\sin^2 t} \;dt \\ &=2\int_0^{\frac{\pi}{2}} \sqrt{1+3\cos^2 t} \;dt \end{aligned}$

and we can use numerical integration here, or Wolfram|Alpha. Incidentally, this is a complete elliptic integral of the second kind.

Method $3$ : AGM

See, for instance, here for an explanation of a rapidly converging algorithm using the arithmetic-geometric mean to evaluate $P(1)$ .