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.

## Comments

Sort by:

TopNewestSince I started Brilliant, I think this is the best problem I've run across so far. I'm losing sleep over this. – Michael Mendrin · 3 years, 3 months ago

Log in to reply

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. – Calvin Lin Staff · 3 years, 3 months ago

Log in to reply

– Michael Mendrin · 3 years, 3 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.Log in to reply

– Finn Hulse · 3 years, 3 months ago

??Log in to reply

– Michael Mendrin · 3 years, 3 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?Log in to reply

– Finn Hulse · 3 years, 3 months ago

Oh I see. Also, just looking at your profile picture, are you blind? :PLog in to reply

– Michael Mendrin · 3 years, 3 months ago

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

– Finn Hulse · 3 years, 3 months ago

HAHAHAHAHA risque. :DLog in to reply

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). – Xuming Liang · 3 years, 3 months ago

Log in to reply

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. – Michael Mendrin · 3 years, 3 months ago

Log in to reply

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. – Soham Ghosh · 3 years, 3 months ago

Log in to reply

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. – Milly Choochoo · 3 years, 3 months ago

Log in to reply

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.

math.stackexchange question – Ali Caglayan · 3 years, 3 months ago

Log in to reply

@Chandler West Moving the discussion here. – Calvin Lin Staff · 3 years, 3 months ago

Log in to reply

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. – Calvin Lin Staff · 3 years, 3 months ago

Log in to reply

– Jovin Joseph · 3 years, 3 months ago

excuse me Calvin did you copy this problem or make it yourselfLog in to reply

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. – Calvin Lin Staff · 3 years, 3 months ago

Log in to reply

Integrate tan x^(1/3) and please share the solution – Manas Kumar Das · 3 years, 3 months ago

Log in to reply

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. – Hosam Hajjir · 3 years, 3 months ago

Log in to reply

Note that the perimeter of the ellipse is not easily known. We only have approximations to this 'elliptical integral'z – Calvin Lin Staff · 3 years, 3 months ago

Log in to reply

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

Dim major

axis, minoraxis As DoublePublic 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 > target

length And xlength < targetlength Thent1 = t - 0.1

t2 = t

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((major

axis * 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. – Hosam Hajjir · 3 years, 3 months ago

Log in to reply

– Hosam Hajjir · 3 years, 3 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.Log in to reply

– Calvin Lin Staff · 3 years, 3 months ago

Thanks! Can you post your code for reference?Log in to reply