# Five-section Of An Ellipse

In Pentallipse, Chandler West noticed that given an ellipse, if we were to choose 5 points which are equally spaced out around the perimeter, then the area of the pentagon is independent of the starting point.

He reached this conclusion by randomly testing starting points, but had difficulty evaluating the integrals.

How can we prove that this statement is true?

In the comments below, it is stated that this conjecture is actually not true. The ellipse was small enough, and too close to a circle, and the observation was within limits of experimental error.

Note by Calvin Lin
6 years, 9 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Since I started Brilliant, I think this is the best problem I've run across so far. I'm losing sleep over this.

- 6 years, 9 months ago

To be fair, Chandler tagged it with Computer Science, because he used numerical integration to calculate the area. There is no easy way to deal with the elliptical integration.

However, his observation that the area is independent of the starting point, is interesting mathematically. I'm wondering if this is always true, and if so, why.

Staff - 6 years, 9 months ago

Right, to my great surprise, when I work this out numerically, the "conjecture" seems to hold. But it's a mighty fishy conjecture. I'm trying to find out just what other conditions are required for this to work at all.

- 6 years, 9 months ago

??

- 6 years, 9 months ago

Look, Finn, for something as simple as a $1 x 2$ rectangle with a perimeter of $6$, this doesn't work. That rules out a "general law" for any closed figure. So, apparently it seems to work for a $3:4$ ellipse, and what else? And why?

- 6 years, 9 months ago

Oh I see. Also, just looking at your profile picture, are you blind? :P

- 6 years, 9 months ago

No, it makes it hard for the women to tell that I'm looking.

- 6 years, 9 months ago

HAHAHAHAHA risque. :D

- 6 years, 9 months ago

This is really interesting....I'm given to know that arc length is a calculus idea, so I'm wondering if there's an "Euclidean" way to look at this problem. i.e. using pure geometric methods...(projection maybe? just a crazy idea that doesn't seem likely to work).

- 6 years, 9 months ago

Okay, I tried this out with a $1:2$ ellipse, perimeter divided into fifths. Numerically, it's clear that it doesn't work. So, my suspicion is that with a $3:4$ ellipse, it's close enough to a circle where we could be fooled by this. I think this "conjecture" is false, however appealing. I'll try again with other sets of points on the $3:4$ to bury this one.

- 6 years, 9 months ago

This is obviously true if we are given a circle, since the pentagon will be a regular pentagon by symmetry, which has a fixed area.

However, we can't simply scale from the circle to the ellipse, since perimeters / lengths do not scale nicely, though area scales perfectly.

Staff - 6 years, 9 months ago

excuse me Calvin did you copy this problem or make it yourself

- 6 years, 9 months ago

You can click on the link to Chandlers problem. He made the above observation in his solution, which i found interesting mathematically, and felt that it was worth discussing.

The above comment deals with the special case where the ellipse is a circle. However, I do not easily see how to extend the argument.

Staff - 6 years, 9 months ago

@Chandler West Moving the discussion here.

Staff - 6 years, 9 months ago

This problem sounds similar to the slicing the pizza problem? Maybe it is the same concept in a way? When I try and do this for an ellipse funnily enough I get a lot of elliptic integrals which isn't making my day.

I am thinking of creating a question on MSE regarding this.

- 6 years, 9 months ago

Has anybody already tried to work this out with Green's Theorem? I'm talking about that method you use to calculate the area of a polygon whose coordinate points you know by 'working backwards' from Green's theorem.

Green's Theorem: $\int_{\partial D} P \,dx + Q \,dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \,dA$

So if we set $P = -\frac{1}{2}y$, and $Q = \frac{1}{2}x$, the double integral in Green's theorem just evaluates the area of whatever region $D$ you are dealing with (whose boundary is $\partial D$).

Working out the annoying integration and algebra for a polygon with known coordinates for its points, we have the formula:

$\iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})\,dA = \frac{1}{2}\sum\limits_{i=1}^n (x_iy_{i+1} - x_{i+1}y_i)$

$x_{n+1} = x_1 , y_{n+1} = y_1$

Which basically says the area of the polygon is equal to that summation on the right-hand side of the equation.

My guess is that if there is any pattern or funny thing going on here, you'll be sure to find it in that summation by exchanging the $x$'s or the $y$'s with their relationship to their coordinate counterpart (i.e $x_1$ is the coordinate counterpart of $y_1$, etc.) via the equation for whatever ellipse you're using.

- 6 years, 9 months ago

To be fair, Chandler tagged it with Computer Science, because he used numerical integration to calculate the area. There is no easy way to deal with the elliptical integration. However, his observation that the area is independent of the starting point, is interesting mathematically. I'm wondering if this is always true, and if so, why.

- 6 years, 8 months ago

We can think of the irregular pentagon within the ellipse as a stretched version of a regular pentagon within a circle. Whatever the orientation of the "original" regular pentagon, when we scaled it by the ratio of the major axis to the minor axis, its area will be multiplied by this scale ratio. Suppose the major axis of the ellipse is horizontal, and that the ratio of major to minor axes is R, then when calculating the area of the pentagon (whatever its orientation is) by adding horizontal strips (slices) of the pentagon we have the differential area dA = L(y) dy = R Lc(y) dy , where Lc(y) is the length of the horizontal strip of the original regular pentagon, within the circle. Hence when summing the differential areas of the horizontal strips, the total resulting area is R times the area of the unstretched regular pentagon contained in the circle.

- 6 years, 9 months ago

The error in your argument is that perimeter doesn't scale when stretched in one direction. As such, when stretching the regular pentagon the shape that we get will not have vertices that are equidistant along the perimeter of the ellipse, hence the first line is wrong.

Note that the perimeter of the ellipse is not easily known. We only have approximations to this 'elliptical integral'z

Staff - 6 years, 9 months ago

The code I wrote (in Visual Basic for Applications - Microsoft Excel ) follows.

Note that the semi-major axis I used here is 15 instead of 4, to exaggerate the eccentricity of the ellipse.

Dim majoraxis, minoraxis As Double

Public Sub pentagon()

Dim vertices(5, 3) As Double

major_axis = 15

minor_axis = 3

p = WorksheetFunction.Pi()

perimeter = integration(2 * p, 300)

' k is an parameter that varies the starting point

For k = 0 To 19

area = 0

For i = 0 To 4

target_length = ((i / 5) + (k / 20) * (1 / 5)) * perimeter

xlength = 0

For t = 0 To 2 * p Step 0.1

Length = integration(t, 300)

If target_length = 0 Then

vertices(i + 1, 1) = major_axis

vertices(i + 1, 2) = 0

ElseIf Length > targetlength And xlength < targetlength Then

t1 = t - 0.1

t2 = t

Do

 tm = (t1 + t2) / 2

midlength = integration(tm, 300)

If midlength > target_length Then

t2 = tm

Else

t1 = tm

End If

If Abs(t2 - t1) < 0.0000000001 Then Exit Do


Loop

t = (t1 + t2) / 2

vertices(i + 1, 1) = major_axis * Cos(t)

vertices(i + 1, 2) = minor_axis * Sin(t)

Exit For

End If

xlength = Length

Next t

Next i

For i = 1 To 5

vertices(i, 3) = Sqr(vertices(i, 1) ^ 2 + vertices(i, 2) ^ 2)

Next i

For i = 1 To 5

j = i Mod 5 + 1

anglecos = (vertices(i, 1) * vertices(j, 1) + vertices(i, 2) * vertices(j, 2)) / (vertices(i, 3) * vertices(j, 3))

Angle = WorksheetFunction.Acos(anglecos)

area = area + 0.5 * vertices(i, 3) * vertices(j, 3) * Sin(Angle)

Next i

ActiveSheet.Cells(k + 1, 1) = area

Next k

Exit Sub

End Sub

Public Function integration(ByVal t As Double, ByVal n As Integer) As Double

a = 0

b = t

h = (b - a) / n

s = f(a) + f(b)

For i = 1 To n - 1 Step 2

s = s + 4 * f(a + i * h)

Next i

For i = 2 To n - 2 Step 2

s = s + 2 * f(a + i * h)

Next i

s = s * h / 3

integration = s

End Function

Public Function f(x) As Double

f = Sqr((majoraxis * Sin(x)) ^ 2 + (minoraxis * Cos(x)) ^ 2)

End Function

The corresponding output is (The areas for increasing shift in the starting vertex)

100.9786729

100.5668878

99.95746287

99.48788261

99.20344447

99.10868612

99.20344447

99.4878826

99.95746286

100.5668878

100.9786729

100.5668878

99.95746284

99.48788258

99.20344445

99.10868609

99.20344443

99.48788254

99.95746273

100.5668875

A look at these areas reveals some interesting periodic patterns.

- 6 years, 9 months ago

I've just implemented a simulation of this problem. And it turns out the area of the pentagon depends on the starting vertex.

- 6 years, 9 months ago

Thanks! Can you post your code for reference?

Staff - 6 years, 9 months ago

Integrate tan x^(1/3) and please share the solution

- 6 years, 8 months ago