Sylvester's four point problem

Key takeaways

Sylvester's four point problem belongs to mathematics; place it in that map before memorising details.
Learn the definition first, then one example that makes the definition concrete.
Connect Sylvester's four point problem to a quantity you can measure, compute or draw — that is where exam questions come from.
Reproduce the core statement of Sylvester's four point problem from memory before moving on to harder problems.

Reference excerpt

Sylvester's four point problem in geometric probability asks for the probability that four randomly chosen points in the Euclidean plane form a convex quadrilateral. Together with Buffon's needle problem, it has been called "one of the prime paradigms in geometric probability theory". The answer depends on the probability distribution from which the points are drawn, and finding a distribution for which this probability is small is closely connected to the crossing number of complete graphs. The problem was posed in 1864 by J. J. Sylvester, who asserted (with fallacious reasoning) that if the points are drawn from the entire plane then the probability is 3 4 {\displaystyle {\tfrac {3}{4}}} but that if they are drawn from a bounded convex set then the probability is bounded below 3 4 {\displaystyle {\tfrac {3}{4}}} . (Sylvester originally posed the problem in a complementary form, asking for the probability that four points do not form a convex quadrilateral, and giving answers 1 4 {\displaystyle {\tfrac {1}{4}}} and bounded above 1 4 {\displaystyle {\tfrac {1}{4}}} .) Among continuous uniform distributions over bounded convex sets the probability of a convex quadrilateral is maximized by any circle or ellipse (probability approximately 0.704) and minimized by any triangle (approximately 0.667). For uniform distributions over bounded open sets the minimum probability that can be achieved is unknown, but has upper and lower bounds that are both near 0.380.

Specific solutions

Sylvester, and Arthur Cayley, presumed that three of the points could be taken as forming the largest-area triangle of the four triangles determined by the points. This assumption limits the fourth point to a larger similar triangle, the anticomplementary triangle of the first three points, with four times the area. The four points form a convex quadrilateral if the fourth point does not also lie within the triangle of the first three points. The fraction of area of the anticomplementary triangle in which a convex quadrilateral is formed is 3 4 {\displaystyle {\tfrac {3}{4}}} . However, it is fallacious to use this reasoning to justify Sylvester's claim that the probability of obtaining a convex quadrilateral is 3 4 {\displaystyle {\tfrac {3}{4}}} : there is no uniform distribution on the entire plane from which one can draw four points and apply this sort of area argument. For four points chosen uniformly at random within a unit disk, the probability that they are in convex position is

1 − 35 12 π 2 ≈ 0.7045. {\displaystyle 1-{\frac {35}{12\pi ^{2}}}\approx 0.7045.}

This was calculated soon after Sylvester posed his problem by Wesley S. B. Woolhouse; Woolhouse then assumed (again, fallaciously) that the solution for the entire plane could be taken as a limit of disks with arbitrarily large radii, all having the same probability. The same probability applies to an ellipse. For points chosen uniformly at random from within a triangle, square, or regular hexagon (or their affine transformations, including the rectangles and parallelograms), the probabilities are respectively 2 3 ≈ 0.6667 {\displaystyle {\tfrac {2}{3}}\approx 0.6667} , 11 36 ≈ 0.6944 {\displaystyle {\tfrac {11}{36}}\approx 0.6944} , and 289 972 ≈ 0.7027 {\displaystyle {\tfrac {289}{972}}\approx 0.7027} . Among uniform distributions over convex shapes, the maximum probability of obtaining a convex quadrilateral is given by choosing random points in a disk, while the minimum probability is obtained by choosing points in a triangle. For points drawn from a two-dimensional Gaussian distribution, the probability of obtaining four points in convex position is

6 π arcsin ⁡ 1 3 ≈ 0.6490. {\displaystyle {\frac {6}{\pi }}\arcsin {\frac {1}{3}}\approx 0.6490.}

… excerpt ends here. Continue reading the full article.

Worked examples

Example 1 — a first encounter with Sylvester's four point problem

Start with the simplest possible case. Write down what Sylvester's four point problem claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Sylvester's four point problem before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Sylvester's four point problem ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Sylvester's four point problem

In research: Sylvester's four point problem appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry: Engineering practice reuses Sylvester's four point problem in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom: Sylvester's four point problem is common in secondary-school and first-year university syllabi. It links to neighbouring topics Discrete geometry, Geometric probability, Graph drawing, so understanding it makes those chapters shorter.
In everyday life: Look for Sylvester's four point problem outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

How to study Sylvester's four point problem in 20 minutes

Read the reference excerpt below once, without taking notes.
Close the page and write down what Sylvester's four point problem means in your own words.
Compare your version with the excerpt and mark what you missed.
Work through the three examples above with pen and paper.
Explain Sylvester's four point problem out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Sylvester's four point problem in simple terms?

Sylvester's four point problem in geometric probability asks for the probability that four randomly chosen points in the Euclidean plane form a convex quadrilateral. Together with Buffon's needle problem, it has been called "one of the prime paradigms in geometric probability theory".

Why does Sylvester's four point problem matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Sylvester's four point problem?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Sylvester's four point problem.