On the shores of circle lake

Question

Eight trees $ABCDEFGH$ are standing (in this order) on the shores of circle lake.

• The four trees $ACEG$ form a square of area $500m^2$.
• The four trees $BDFH$ form a rectangle of area $400m^2$.

What is the largest possible area of the octagon $ABCDEFGH$ under these conditions?

Answer 1

I think the answer is

$300\sqrt{5}=670.8 m^2$.

Reasoning

(I'm going to leave out the meters unit on all forthcoming measurements.)

As others have figured out, the square $ACEG$ holds the key to the circle's measurements. With an area of 500, the sides are each of length $\sqrt{500}$, leading to a diagonal of $10\sqrt{10}$. Since the square is circumscribed along the circle, this is equal to the diameter. Thus, the radius, $r$, is $5\sqrt{10}$.

Let's create a Cartesian coordinate system with the center of the circle at $(0, 0)$. Next, let's place the square $ACEG$ rotated by 45° from its "normal" view. (It should look like a "diamond" in the standard, if not mathematical, sense.) With $A$ in the upper-left and working clockwise (I know it's breaking convention, but this is how I drew it and I don't want to confuse myself), the points of the square are $(-r, 0)$, $(0, r)$, $(r, 0)$, and $(0, -r)$.

Now let's place the rectangle down. Here is the part that I cannot prove: I propose that the maximal area occurs when a line drawn from the center of the circle to any point on the square $ACEG$ intersects the midpoint of the edge of the rectangle $BDFH$. If you imagine the square as tilted 45°, then the rectangle is placed "flat" such that the edges are parallel with the edges of the page. Let's call the length of the long edges—i.e. segments $BD$ and $FH$—$a$ and the length of the short edges—i.e. segments $DF$ and $HB$—$b$. We have a system of equations to solve the positions of the corners:

• The area is 400, so $ab=400$.
• Either diagonal intersects the center of the circle and is therefore the diameter. Thus, $a^2+b^2=(2r)^2$.

Solving this system for $a, b >0$ yields $a=20\sqrt{2}$ and $b=10\sqrt{2}$. Now we know the points of rectangle $BDFH$, in order: $(-a/2, b/2)$, $(a/2, b/2)$, $(a/2, -b/2)$, $(-a/2, -b/2)$.

The total area of the octagon is that of the rectangle $BDFH$ plus four triangles: $BCD$, $DEF$, $FGH$, and $HAB$. The rectangle has area $ab=400$. The triangles $BCD$ and $FGH$ each have an area of $a(r-b/2)/2$, and the triangles $DEF$ and $HAB$ each have an area of $b(r-a/2)/2$. Combining these results in a total area of $ab+b(r-a/2)+a(r-b/2)$. Plugging in the values found for $a$ and $b$ yields the number I stated above.

Reference Figure

Answer 2

Total area of 670.8

We know the square sits evenly on the circle, with the diagonal equal to the diameter. That is $10\sqrt{10}$.
The diagonal of the rectangle is also equal to the diameter, and the area of that is $D^2 * \sin{x} * \cos{x}$ where X is the angle between the diagonal and a side of the rectangle. That works out to 26.57° or 63.43°.

So, our octagon can be made by having the rectangle rotated around the square. That in turn creates 8 slices that added up make the octagon's area. The area of each slice is $\frac{r^2 * \sin{x}}{2}$. As the only variable there is the sin(x), that is what we have to maximize.

The 4 (mirrored) slices (if the square has its diagonal vertical, and the rectangle's longer centerline is horizontal-ish) are $\sin{x}, \sin{(90-x)}, \sin{(x+36.86)}, \sin{(53.14-x)}$. Maximizing the sum of those gives $x = 26.57°$ or having the points be symmetrical.

So, now we have an octagon made up of slices either 26.57° or 63.43°. Those work out to areas of 55.9 and 111.8. Four of each gives a total area of $670.8$.

Answer 3

Stating Point / Not Even a Partial Answer

I'm putting this in an answer just to use the spoiler tag for those that want to do all the math themselves.

A square of area $500m^2$ has sides of $\sqrt{500}=10\sqrt{5}$.
A square of side $\sqrt{500}$ has a diagonal of $\sqrt{1000}=10\sqrt{10}$
As the square is inscribed within the circle, that means the circle has a diameter of $10\sqrt{10}$ and a radius of $5\sqrt{10}$

The rectangle BDFH has an area of $\mathbf{BD}\times\mathbf{BF}=400m^2$
The length of a chord is given by $2R\sin{\frac{\theta}{2}}$ which, in this case, becomes $10\sqrt{10}sin{\frac{\theta}{2}}$ where $\theta$ is the angle between the two points as taken from the center of the circle. Reference

Answer 4

The side length $s$ of the square is:

$$ s = \sqrt{500} = 10\sqrt{5} $$

The radius of the circle $r$ is half the diagonal of this square:

$$ r = \frac{s}{\sqrt{2}} = 5\sqrt{10} $$

The diagonal of the rectangle (of width $w$ and height $h$) is also equal to twice the radius, and its area is equal to the product of its dimensions:

$$ w^2 + h^2 = (2r)^2 = 1000\\ wh = 400 $$

Solving for $h$ and $w$, we have:

$$ \begin{align} w &= 20\sqrt{2} \\ h &= 10\sqrt{2} \\ \end{align} $$

The angle $\phi$ between the long side ($w$) of the rectangle and its diagonal is:

$$ \phi = \arctan\frac{h}{w} = \arctan\frac{1}{2} $$

Consider a vertex of the square adjacent to the endpoints of the short side of the rectangle. The angle between the two corners of the rectangle is $2\phi$, so if the rectangle is oriented so that there is an angle $\theta$ between one of its corners and the vertex of the square, there will be an angle $2\phi-\theta$ between that vertex and the other corner.

Since each of the sides of the square occupies an angle of $\pi/2$ measured from the circle's center, the angles between the corners of the rectangle and the next two corners of the squares are $\pi/2-\theta$ and $\pi/2-(2\phi-\theta)$.

The final four angles are repeats of these, since the arrangement has twofold rotational symmetry.

Thus the octagon is made of eight sectors. Each has two side lengths of $r$ and the angles between those sides are (two each of):

$$ \theta, \\ 2\phi-\theta, \\ \pi/2 - \theta, \\ \pi/2 - 2\phi + \theta $$

The total area of the octagon is then just:

$$ r^2\left(\sin\theta + \sin(2\phi-\theta) + \sin(\pi/2 - \theta) + \sin(\pi/2 - 2\phi + \theta)\right) \\ =300(\sin\theta + 2\cos\theta) $$

Plotting this quantity, we can see that it is maximized when $\theta=\phi$, and its maximum value is:

$$ 300\sqrt{5} $$