2025-09-12 21:00:31
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].
[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
The post More triangle inequalities first appeared on John D. Cook.2025-09-12 18:59:40
Let D be the unit disk in the complex plane and let f be a univalent function on D, meaning it is analytic and one-to-one on D.
There is a simple way to compute the area of f(D) from the coefficients in its power series.
If
then
The first equality follows from the change of variables theorem for functions of two variables and applying the Cauchy-Riemann equations to simplify the Jacobian. The second equality is a straight-forward calculation that you can work out in polar coordinates.
Let’s apply this to what I called the minimalist Mandelbrot set the other day.
The orange disk has radius 1/4 and so its area is simply π/16.
Finding the area of the blue cardioid takes more work, but the theorem above makes it easy. The cardioid is the image of the set {α : |α| < ½} under the map f(z) = z − z². To apply the theorem above we need the domain to be the unit disk, not the unit disk times ½, so we define
as a function on the unit disk. Now c1 = ½ and c2 = −¼ and so the area of f(D) = 3π/8.
I said in the earlier post that the minimalist Mandelbrot set makes up most of the Mandelbrot set. Now we can quantify that. The area of the minimalist Mandelbrot set is 7π/16 = 1.3744. The area of the Mandelbrot set is 1.5065, so the minimalist set shown above makes up over 91% of the total area.
2025-09-12 09:08:41
Ted Dunning left a comment on my post on random sampling from a triangle saying you could extend this to sampling from a polygon by dividing the polygon into triangles, and selecting a triangle each time with probability proportional to the triangle’s area.
To illustrate this, let’s start with a irregular pentagon.
To pick a point inside, I used the centroid, the average of the vertices. Connecting the centroid to each of the vertices splits the pentagon into triangles. (Here I implicitly used the fact that this pentagon is convex. The centroid of a non-convex polygon could be outside the polygon.)
We can find the area of the triangles using Heron’s rule.
Here’s what we get for random samples.
2025-09-12 05:41:02
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.
The post A triangle inequality by Erdős first appeared on John D. Cook.2025-09-11 21:48:58
If you have a triangle with vertices A, B, and C, how would you generate random points inside the triangle ABC?
One idea would be to use barycentric coordinates.
This generates points inside the triangle, but not uniformly.
Another idea is to use an accept-reject method. Draw a rectangle around the triangle, generate points in the square, and throw them away if they land outside the triangle.
An advantage to this method is that it obviously works because it doesn’t rely on anything subtle. Three cheers for brute force!
The method is fairly efficient; on average only half the points will be rejected.
A disadvantage to all accept-reject methods is that they have variable runtime, though this only matters in some applications.
There is a clever variation on the accept-reject method. Create a parallelogram by attaching a flipped copy of the triangle. Now randomly sample from the parallelogram. Every sample point will either land inside the original triangle or in its flipped twin. If it landed in the original triangle, keep it. If it landed in the twin, flip it back over and use that point.
This is like an accept-reject method, except there’s no waste. Every point is kept, possibly after flipping.
You can find code for implementing this method on Stack Overflow.
Felix left a note in the comments that the barycentric approach would work if you draw the random weights from an exponential distribution rather than a uniform distribution. That goes against my intuition. I thought about using a different distribution on the weights, but I didn’t work out what it would need to be, but I did not expect it to be exponential. Apparently it works. Here’s a proof by plot:
2025-09-10 19:55:53
George Marsaglia was a big name in random number generation. I’ve referred to his work multiple times here, most recently in this article from March on randomly generating points on a sphere. He is best remembered for his DIEHARD battery of tests for RNG quality. See, for example, this post.
I recently learned about a mental RNG that Marsaglia developed. It’s not great, but it’s pretty good for something that’s easy to mentally compute. The output is probably better than if you simply tried to make up random numbers; people are notoriously bad at doing that. Hillel Wayne explores Marsaglia’s method in detail here.
Here’s a summary the generator in Python.
state = 42 # Set the seed to any two digit number
def random_digit():
tens = lambda n: n // 10
ones = lambda n: n % 10
global state
state = tens(state) + 6*ones(state)
return ones(state)
