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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?

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  • $\begingroup$ Are the trees all equidistant from the shore, or is that irrelevant? $\endgroup$ – Portali5t Sep 29 '15 at 12:52
  • $\begingroup$ The trees are points on the perimeter of circle lake. (The puzzle is tagged "math" and "geometry".) $\endgroup$ – Gamow Sep 29 '15 at 12:54
  • $\begingroup$ Just to be clear, All Points are concylic, right?? $\endgroup$ – Rohcana Sep 29 '15 at 15:47
  • $\begingroup$ Yes, all points are concylic. $\endgroup$ – Gamow Sep 29 '15 at 15:52
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    $\begingroup$ I'll be that guy. $\endgroup$ – user1717828 Sep 29 '15 at 18:53
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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

enter image description here

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  • $\begingroup$ Looks good. But your final area simplifies to $r(a + b) = 5 \sqrt{10}(10 \sqrt{2} + 20 \sqrt{2}) = 300 \sqrt{5}$, right? $\endgroup$ – Tyler Seacrest Sep 29 '15 at 18:48
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    $\begingroup$ @TylerSeacrest: Yes, you're correct. I did a cursory calculation in MATLAB, and I just found a typo. The answer was so pretty that I didn't check it. I have corrected my answer accordingly. $\endgroup$ – dpwilson Sep 29 '15 at 19:38
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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 $670.8$.

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    $\begingroup$ I don't get your last paragraph. It's true you can get the exact dimensions of the rectangle but you're not finished then. The two figures can be rotated as you wish so you still need to maximize the area that way I think $\endgroup$ – Ivo Beckers Sep 29 '15 at 17:18
  • $\begingroup$ @IvoBeckers Added in the reasoning for the angle chosen $\endgroup$ – JonTheMon Sep 29 '15 at 18:42
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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

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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:

$$ 300\sqrt{5} $$

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