Yesterday I wrote about a triangle inequality discovered by Paul Erdős.

Let P be a point inside a triangle ABC. Let x, y, z be the distances from P to the vertices and let p, q, r, be the distances to the sides. Then Erdős’ inequality says

x + y + z ≥ 2(p + q + r).

Using the same notation, here are four more triangle inequalities discovered by Oppenheim [1].

• px + qy + rz ≥ 2(qr + rp + pq)
• yz + zx + xy ≥ 4(qr + rp + pq),
• xyz ≥ 8pqr
• 1/p+ 1/q + 1/r ≥ 2(1/x + 1/y + 1/z)

[1] A. Oppenheim. The Erdös Inequality and Other Inequalities for a Triangle. The American Mathematical Monthly. Vol. 68, No. 3 (Mar., 1961), pp. 226–230

Plane geometry has been studied since ancient times, and yet new results keep being discovered millennia later, including elegant results. It’s easy to come up with a new result by proving a complicated theorem that Euclid would not have cared about. It’s more impressive to come up with a new theorem that Euclid would have understood and found interesting.

Paul Erdős conjectured another triangle inequality in 1935 which was proved by Mordell and Barrow in 1937 [1].

Let P be a point inside a triangle ABC. Let x, y, z be the distances from P to the vertices and let p, q, r, be the distances to the sides. Then

x + y + z ≥ 2(p + q + r)

with equality only if P is the center of an equilateral triangle [2]. In the figure above, the theorem says the dashed blue lines together are more than twice as long as the solid red lines.

How far apart are the left and right sides of the inequality? This was the motivation for the previous post on selecting random points from a triangle. I wanted to generate random points and compare the two sides of the Erdős-Mordell-Barrow inequality.

We can visualize the inequality by generating random points inside the triangle and plotting the points with a color that indicates the inequality gap, darker blue corresponding to a larger gap.

This shows the inequality is sharper in the middle of the triangle than near the vertices.

[1] Problem 3740, American Mathematical Monthly, 44 (1937) 252-254.

[2] You could interpret this as a theorem comparing barycentric and trilinear coordinates.

Pick’s theorem is a surprising and useful to find the area of a region formed by connecting dots on a grid. The area is simply

A = i + p/2 − 1

where i is the number of dots in the interior and p is the number of dots on the perimeter.

Example

For example, the in the figure below there are 11 black dots on the perimeter and 17 blue dots in the interior.

Therefore Pick’s theorem says the area of the figure is 17 + 11/2 − 1 =  21½.

Intuition

It makes sense that the area would be approximately i if the figure is very large. But why do you need to add half the dots on the perimeter and why do you subtract 1?

Pick’s theorem is simple, but it isn’t obvious. If it were obvious, someone closer to the time of Descartes (1596–1650) would have discovered it. Instead, the theorem was discovered by Georg Alexander Pick in 1899. However, the theorem is fairly obvious for a rectangle, and this post will illustrate that case. For a rectangle, you can easily see why you need to add half the perimeter dots and subtract 1.

Rectangular case

Imagine an m by n rectangular array of dots. If you were to draw a frame around those dots with the corners of the rectangle being exactly between dots, the area would be mn, the number of dots inside the frame.

Pick’s theorem does not apply here because the corners of our frame do not lie on the grid. So let’s shift our grid down and to the left so that its corners do lie on the grid.

Now some of our original dots, marked in blue, are on the perimeter of the frame. We could add the number of dots on the perimeter to correct for this, but that would be too much because the blue dots on the perimeter are only on the left and top sides. Each has a reflection on the right and bottom, marked with red dots. So we shouldn’t add all the dots on the perimeter, but only half the dots on the perimeter.

But that still overcounts slightly. There are two dots on the perimeter that do not correspond to a blue dot or to the reflection of a blue dot. These are the hollow circles in the top right and bottom left corners. So when we take half the perimeter, we need to subtract 1 to account for half of this pair of dots.

This doesn’t prove Pick’s theorem in general, but it does prove it for rectangular regions. Even if we can’t see why the formula generalizes to more complicated regions, we can be grateful that it does.

Related posts

• The shoelace formula for area of a polygon
• Do perimeter and area determine a triangle?
• Triangulating a regular polygon

Iconjack posted on X today

(Possibly) surprising fact: a regular dodecahedron is rounder than a regular icosahedron.

You might reasonably suppose that having more sides would result in a rounder figure, so a figure with 20 sides should be rounder than a figure with 12 sides. But it’s the other way around.

Discrete curvature

Here’s one way to see the statement above. In a dodecahedron, three pentagons come together at each vertex. The interior angles of a pentagon have 108°. If you were to take the three pentagons, cut a slit long one slide, and push them flat on to a plane, there would be a gap of

360° − 3 × 108° = 36°.

This gap is known as the angle defect and is the discrete analog of curvature.

In an icosahedron you have five triangles coming together at a point. So if you were to push the five triangles at a vertex down to a point, there would be a gap (angle defect) of

360° − 5 × 60° = 60°.

The larger angle defect means the shape is less round at a vertex.

For completeness, the angle defect of a cube is 90°. For an octahedron it’s 120° and for a tetrahedron it’s 180°.

Dihedral angles

Angle defect is the standard way to measure how round a polyhedron is. It focuses on how faces fit together at a vertex. But we could instead focus on how faces fit together at an edge. The angle between two adjacent faces is the dihedral angle.

The faces of an icosahedron come meet at an angle of approximately 138.19°. For a dodecahedron, the angle is 116.57°. So the icosahedron is rounder in the sense that the faces come together at an angle closer to 180°.

The dihedral angles for the octahedron, cube, and tetrahedron are 109.47°, 90°, and 70.53°.

Related posts

• Curvature at Cairo
• Dual polyhedra
• Planets and Platonic solids

If three circles are all tangent to each other, you can find two more circles that are tangent to all three, and the equation for finding these new circles is remarkably elegant. This is Descartes’ theorem.

Two tangent circles

To illustrate Descartes’ theorem, we first need three mutually tangent circles. Drawing two mutually tangent circles is easy. We choose our coordinate system so that the first circle is centered at the origin

O₁ = (x₁, y₁) = (0, 0)

and the center of the second circle is on the x-axis. If the two circles have radii r₁ and r₂, then the center of the second circle is at

O₂ = (x₂, y₂) = (r₁ + r₂, 0).

I’ve assumed that the second circle is outside the first. You could have a tangent circle inside another circle, but to illustrate Descartes’ theorem we need circles that are externally tangent.

Three tangent circles

Now we pick the radius of the third circle to be r₃. The center of this circle must belong to a circle of radius r₁ + r₃ centered at O₁ and to a circle of radius r₂ + r₃ centered at O₂. I went over finding the point(s) of intersection of two circles here.

For illustration I chose r₁ = 2, r₂ = 1, and r₃ = 2.8. Using the code from the aforementioned post I found

(x₃, y₃) = (2.93333333 3.79941516)

Here’s a plot.

Four tangent circles

Now we’re in a position to state and illustrate Descartes’ theorem. The theorem is most easily stated in terms of signed curvatures.

The curvature of a circle of radius r is 1/r. Big circles have small curvature and small circles have big curvature.

For Descartes’ theorem we need to use signed curvatures, taking the sign to be positive if circles are externally tangent and negative if the circles are internally tangent. The signed curvatures of our three circles are positive, that is

k_i = 1/r_i

for i = 1, 2, and 3. But Descartes’ theorem will give us two circles, one between the three given circles and one surrounding them, corresponding to a positive and negative solution for k₄.

Finding the radii

Now we can state Descartes’ theorem [1]:

(k₁ + k₂ + k₃ + k₄)² = 2(k₁² + k₂² + k₃² + k₄²)

This gives us a quadratic equation for k₄ with two solutions:

k₄ = k₁ + k₂ + k₃ ±2(k₁k₂ + k₂k₃ + k₁k₃)^½.

Finding the centers

Now we know the radii of our tangent circles, but not the centers. For this we view the circles as living in the complex plane with centers z_i. Then the centers satisfy a theorem analogous to the one satisfied by the curvatures.

(k₁z₁ + k₂z₂ + k₃z₃ + k₄z₄)² = 2(k₁²z₁² + k₂²z₂² + k₃²z₃² + k₄²z₄²)

In other words, the curvature-center products satisfy the same equation as the curvatures.

I’m unclear on the history here, but I don’t believe Descartes had a formula for the centers of the circles. He certainly did not have the formula above using complex numbers [2].

Here’s a plot of the two circles guaranteed by Descartes’ theorem.

[1] The Soddy-Gossett theorem generalizes Descartes theorem to the case of n + 2 spheres in ℝⁿ. The square of the sums of the curvatures equals n the sum of the curvatures squared.

[2] Jeffrey C. Lagarias, Colin L. Mallows, Allan R. Wilks. Beyond the Descartes circle theorem. https://arxiv.org/abs/math/0101066v1

There are several ways to define an ellipse. If you want to write software draw an ellipse, it’s most convenient to have a parametric form:

This gives an ellipse centered at (h, k), with semi-major axis a, semi-minor axis b, and with the major axis rotated by an angle θ from horizontal.

But if you’re going to physically draw an ellipse, there’s a more convenient definition: an ellipse is the set of points such that the sum of the distances to two foci is a constant s. This method is most often called the gardener’s ellipse. It’s also called the nails-and-string method, which is more descriptive.

To draw an ellipse on a sheet of plywood, drive a nails in the plywood at both foci, and attach an end of a string of length s to each nail. Then put a pencil in the middle of the string and pull it tight. Keep the string tight as you move the pencil. This will draw an ellipse [1].

Presumably the name gardener’s ellipse comes from the same idea on a larger scale, such as a gardener marking out an elliptical flower bed using two stakes in the ground and some rope.

Going back to drawing on a computer screen, there is a practical use for the gardener’s ellipse: to determine whether a point is inside an ellipse, add the distance from the point to each of the foci of the ellipse. If the sum is less than s then the point is inside the ellipse. If the sum is greater than s the point is outside the ellipse.

From parameters to nails and string

If you have the parametric form of an ellipse, how do you find the gardener’s method representation?

To make it easier to describe points, set θ = 0 and h = k = 0 for a moment. Then the foci are at (±c, 0) where c² = a² − b².

Since (a, 0) is a point on the string, you have to be able to draw it and so the string must stretch from the focus at (−c, 0) to the point (a, 0) and back to the focus at (c, 0). Therefore the length of the string is 2a.

The length of the string depends on the major axis and not on the minor axis.

When we shift and rotate our ellipse, letting θ, h, and k take on possibly non-zero values, we apply the same transformation to the foci.

Rotating and translating the ellipse doesn’t change the length of the major axis, so s stays the same.

From nails and string to parameters

Draw a line between the two “nails” (foci). The slope of this line is tan θ and it’s midpoint is (h, k). The parameter c is half the length of the line.

The semimajor axis a is half the string length s.

Once you know a and c you can solve for b = √(a² − c²).

Related posts

• An ellipse hidden in Pascal’s triangle
• Areal coordinates and ellipse area
• Rapidly convergent series for ellipse perimeter

[1] In practice, a carpenter might want to draw an ellipse a different way.