A golden ellipse is an ellipse whose axes are in golden proportion. That is, the ratio of the major axis length to the minor axis length is the golden ratio φ = (1 + √5)/2.

Draw a golden ellipse and its inscribed and circumscribed circles. In other words draw the largest circle that can fit inside and the smallest circle outside that contains the ellipse.

Then the area of the ellipse equals the area of the annulus bounded by the two circles. That is, the area of the green region

equals the area of the orange region.

The proof is straight forward. Let a be the semimajor axis and b the semiminor axis, with a = φb.

Then the area of the annulus is

π(a² − b²) = πb²(φ² − 1).

The area of the ellipse is

πab = πφb².

The result follows because the golden ratio satisfies

φ² − 1 = φ.

Related posts

• Posts on ellipses
• Gold, Silver, and Bronze ratios

Barycentric coordinates are sometimes called area coordinates or areal coordinates in the context of triangle geometry. This is because the barycentric coordinates of a point P inside a triangle ABC correspond to areas of the three triangles PBC, PCA and PAB.

(This assumes ABC has unit area. Otherwise divide the area of each of the three triangles by the area of ABC. We will assume for the rest of this post that the triangle ABC has unit area.)

Areal coordinates take three numbers two describe a point in two dimensional space. Why would you do that? It’s often useful to use an overdetermined coordinate system. The benefit of adding one more coordinate is that you get a coordinate system matched to the geometry of the triangle. For example, the vertices of the triangle have coordinates (1, 0, 0), (0, 1, 0), and (0, 0, 1), regardless of the shape of the triangle.

Here is an example of a theorem [1] that is convenient to state in terms of areal coordinates but that would be more complicated in Cartesian coordinates.

First we need to define the midpoint triangle, also called the medial triangle. This is the triangle whose vertices are the midpoints of each side of ABC. In terms of areal coordinates, the vertices of this triangle are (0, ½, ½), (½, 0, ½), and (½, ½, 0).

Now let P be any point inside the midpoint triangle of ABC. Then there is a unique ellipse E inscribed in ABC and centered at P.

Let (α, β, γ) be the areal coordinates of P. Then the area of E is

Because P is inside the medial triangle, each of the areal coordinates are less than ½ and so the quantity under the square root is positive.

Finding the equation of the inscribed ellipse is a bit complicated, but that’s not necessary in order to find its area.

Related posts

• Trilinear coordinates
• Converting between areal and trilinear coordinates
• Triangle centers

[1] Ross Honsberger. Mathematical Plumbs. 1979

I keep running into Edward John Routh (1831–1907). He is best known for the Routh-Hurwitz stability criterion but he pops up occasionally elsewhere. The previous post discussed Routh’s mnemonic for moments of inertia and his “stretch” theorem. This post will discuss his triangle theorem.

Before stating Routh’s theorem, we need to say what a cevian is. Giovanni Ceva (1647–1734) was an Italian geometer, best known for Ceva’s theorem, and for a construction in that theorem now known as a cevian.

A cevian is a line from the vertex of a triangle to the opposite side. Draw three cevians by connecting each vertex of a triangle to a point on its opposite side. If the cevians intersect at a point, Ceva’s theorem says something about how the lines divide the sides. If the cevians form a triangle, Routh’s theorem find the area of that triangle.

Routh’s theorem is a generalization of Ceva’s theorem because if the cevians intersect at a common point, the area of the triangle formed is zero, and then Routh’s area equation implies Ceva’s theorem.

Let A, B, and C be the vertices of a triangle and let D, E, and F be the points where their cevians intersect the opposite sides.

Let x, y, and z be the ratios into which each side is divided by the cevians. Specifically let x = FB/AF, y = DC/BD, and z = EA/CE.

Then Routh’s theorem says the relative area of the green triangle formed by the cevians is

If the cevians intersect at a point, the area of the triangle is 0, which implies xyz = 1, which is Ceva’s theorem.

Related posts

• Routh’s mnemonic
• Kepler triangle
• Triangle centers and Page Rank

Take a compass and draw a circle on a globe. Then take the same compass, opened to the same width, and draw a circle on a flat piece of paper. Which circle has more area?

If the circle is small compared to the radius of the globe, then the two circles will be approximately equal because a small area on a globe is approximately flat.

To get an idea what happens for larger circles, let’s a circle on the globe as large as possible, i.e. the equator. If the globe has radius r, then to draw the equator we need our compass to be opened a width of √2 r, the distance from the north pole to the equator along a straight line cutting through the globe.

The area of a hemisphere is 2πr². If we take our compass and draw a circle of radius √2 r on a flat surface we also get an area of 2πr². And by continuity we should expect that if we draw a circle that is nearly as big as the equator then the corresponding circle on a flat surface should have approximately the same area.

Interesting. This says that our compass will draw a circle with the same area whether on a globe or on a flat surface, at least approximately, if the width of the compass sufficiently small or sufficiently large. In fact, we get exactly the same area, regardless of how wide the compass is opened up. We haven’t proven this, only given a plausibility argument, but you can find a proof in [1].

Note that the width w of the compass is the radius of the circle drawn on a flat surface, but it is not the radius of the circle drawn on the globe. The width w is greater than the radius of the circle, but less than the distance along the sphere from the center of the circle. In the case of the equator, the radius of the circle is r, the width of the compass is √2 r , and the distance along the sphere from the north pole to the equator is πr/2.

Related posts

• Area and volume of hypersphere cap
• How Dürer drew an 11-sided figure
• Duality in spherical trig
• How faithful can a map be?

[1] Nick Lord. On an alternative formula for the area of a spherical cap. The Mathematical Gazette, Vol. 102, No. 554 (July 2018), pp. 314–316

Draw a quadrilateral by pick four arbitrary points on a circle and connecting them cyclically.

Now multiply the lengths of the pairs of opposite sides. In the diagram below this means multiplying the lengths of the two horizontal-ish blue sides and the two vertical-ish orange sides.

Ptolemy’s theorem says that the sum of the two products described above equals the product of the diagonals.

To put it in colorful terms, the product of the blue sides plus the product of the orange sides equals the product of the green diagonals.

The converse of Ptolemy’s theorem also holds. If the relationship above holds for a quadrilateral, then the quadrilateral can be inscribed in a circle.

Note that if the quadrilateral in Ptolemy’s theorem is a rectangle, then the theorem reduces to the Pythagorean theorem.

Related posts

• Area of a quadrilateral as a determinant
• Van Aubel’s theorem
• Ptolemy and trig tables

Suppose you’re in a boring meeting and you start doodling. You draw a circle, and then you draw a triangle on the outside of that circle.

Next you draw a circle through the vertices of the triangle, and draw a square outside that.

Then you draw a circle through the vertices of the square, and draw a pentagon outside that.

Then you think “Will this ever stop?!”, meaning the meeting, but you could ask a similar question about your doodling: does your sequence of doodles converge to a circle of finite radius, or do they grow without bound?

An n-gon circumscribed on the outside of a circle of radius r is inscribed in a circle of radius

So if you start with a unit circle, the radius of the circle through the vertices of the N-gon is

and the limit as N → ∞ exists. The limit is known as the polygon circumscribing constant, and it equals 8.7000366252….

We can visualize the limit by making a plot with large N. The plot is less cluttered if we leave out the circles and just show the polygons. N = 30 in the plot below.

The reciprocal of the polygon circumscribing constant is known as the Kepler-Bouwkamp constant. The Kepler-Bouwkamp constant is the limiting radius if you were to reverse the process above, inscribing polygons at each step rather than circumscribing them. It would make sense to call the Kepler-Bouwkamp constant the polygon inscribing constant, but for historical reasons it is named after Johannes Kepler and Christoffel Bouwkamp.

In 1609, Kepler remarked that the perimeter of an ellipse with semiaxes a and b could be approximated either as

P ≈ 2π(ab)^½

or

P ≈ π(a + b).

In other words, you can approximate the perimeter of an ellipse by the circumference of a circle of radius r where r is either the geometric mean or arithmetic mean of the semi-major and semi-minor axes.

How good are these approximations, particularly when a and b are roughly equal? Which one is better?

When can choose our unit of measurement so that the semi-minor axis b equals 1, then plot the error in the two approximations as a increases.

We see from this plot that both approximations give lower bounds, and that arithmetic mean is more accurate than geometric mean.

Incidentally, if we used the geometric mean of the semi-axes as the radius of a circle when approximating the area then the results would be exactly correct. But for perimeter, the arithmetic mean is better.

Next, if we just consider ellipses in which the semi-major axis is no more than twice as long as the semi-minor axis, the arithmetic approximation is within 2% of the exact value and the geometric approximation is within 8%. Both approximations are good when a ≈ b.

The next post goes into more mathematical detail, explaining why Kepler’s approximation behaves as it does and giving ways to improve on it.

More ellipse posts

• Eccentricity, flattening, and aspect ratio
• Conformal map of rectangle to ellipse
• Ellipses and elliptic integrals

An Archimedean spiral has the polar equation

r = b θ^1/n

This post will look at the case n = 1. I may look at more general values of n in a future post. (Update: See here.) The case n = 1 is the simplest case, and it’s the case I needed for the client project that motivated this post.

In this case the spacing between points where the spiral crosses an axis is constant. Call this constant h. Then

h = 2πb.

For example, when rolling up a carpet, h corresponds to the thickness of the carpet.

Suppose θ runs from 0 to 2πm, wrapping around the origin m times. We could approximate the spiral by m concentric circles of radius h, 2h, 3h, …, mh. To visualize this, we’re approximating the length of the red spiral on the left with that of the blue circles on the right.

We could approximate this further by saying we have m/2 circles whose average radius is πmb. This suggests the length of the spiral should be approximately

2π²m²b

How good is this approximation? What happens to the relative error as θ increases? Intuitively, each wrap around the origin is more like a circle as θ increases, so we’d expect the approximation to improve for large θ.

According to Mathworld, the exact length of the spiral is

πbm √(1 + (2πm)²) + b arcsinh(2πm) /2

When m is so large that we can ignore the 1 in √(1 + (2πm)²) then the first term is the same as the circle approximation, and all that’s left is the arcsinh term, which is on the order of log m because

arcsinh(x) = log(x + (1 + x²)^1/2).

So for large m, the arc length is on the order of m² while the error is on the order of log m. This means the relative error is O( log(m) / m² ). [1]

We’ve assumed m was an integer because that makes it easier to visual approximating the spiral by circles, but that assumption is not necessary. We could restate the problem in terms of the final value of θ. Say θ runs from 0 to T. Then we could solve

T = 2πm

for m and say that the approximate arc length is

½ bT²

and the exact length is

½ bT(1 + T²)^1/2 + ½ b arcsinh(T).

The relative approximation error is O( log(T) / T² ).

Here’s a plot of the error as a function of T assuming b = 1.

Related posts

• Logarithmic spiral
• Spiral of Theodorus
• Approximating a golden spiral

[1] The error in approximating √(1 + (2πm)²) with 2πm is on the order of 1/(4πm) and so is smaller than the logarithmic term.