# Random chord to a circle around a triangle

This puzzle is literally a math textbook question. Its only redeeming feature is that it also happens to be too broad in a surprisingly interesting way.

## The problem

Given the circle $\mathbf O$ passing through the vertices of an equilateral triangle $\triangle \mathbf{ABC}$, draw a randomly chosen chord $\mathbf S$ to the circle. What is the probability that the chord is longer than the side of the triangle?

## Only Very Convincing solutions allowed

A Very Convincing solution needs to contain at least the following:

1. The probability of $\mathbf S$ being longer than $\mathbf{AB}$
2. A description of the method used to calculate it
3. A convincing argument showing that the chord selection process is indeed random and distributes the chords evenly
4. An outline for a real world physical experiment that can be used to prove the result empirically.

## The Catch: Very Convincing is still ambiguous!

So far, this question is straight out of a math textbook; in fact, I seem to remember seeing this problem in more than one problem collection, always with the same answer.

However, it turns out that even with the requirement that the solution be Very Convincing, several solutions, each with different probabilities as the result, still exist.

So, with everyone working together as a team, we are going to find at least three distinct Very Convincing solutions. Every individual solution (with a new probability as the result) will receive fame and glory, but it will be the third distinct solution that will get the tick.

(My reasoning for this is that I personally know of three distinct Very Convincing solutions, and the later ones must surely be the more difficult ones to find, based on nothing else but the fact that the others were found earlier. I may change this criterion, if a more sensible way of choosing a "correct answer" occurs to me. If this happens, there'll be a bounty for the third solution.)

If you don't want to help others, you can of course post three Convincing Solutions in the same answer. Collaboration is likely to be more fun though, and will probably be appreciated more.

Have fun!

• So the question is basically: "Find three different distributions that distribute at least one parameter of the chord evenly"? – w l Apr 12 '18 at 11:13
• Well, yes, and each has to be achievable by simple enough physical means. But where is the fun in that formulation? :-) – Bass Apr 12 '18 at 11:16
• This seems like a bad fit for puzzling, in addition to being a easily googlable puzzle and having very broad conditions. – Quintec Apr 12 '18 at 15:48
• Nah, it’s a pretty good fit, I think. It’s just that until you try it, there’s no way to tell how much fun finding the alternate ”proofs” actually is (well at least if you don’t deliberately spoil it for yourself by googling), and you have to go by the other people’s evaluations. Once w l gave away the secret in his comment and framed his really clever answer as being somewhat boring busywork, it was all over for any fun, of course, but it’s not always mandatory to decide to not to have fun. Thanks for the comment by the way, much more appreciated than the anonymous silent downvotes. – Bass Apr 12 '18 at 19:14

1.

Probability: $\frac{1}{3} \approx 33\%$
Calculation: The length of the circumference inside the triangle compared to the whole circumference.
Distribution: We evenly distribute the endpoints around the circumference.
Experiment: Spin circle two times to randomly pick two points on the circumference, connect to get chord.

2.

Probability: $\frac{1}{3} + \frac{\sqrt{3}}{2\pi} \approx 61\%$
Calculation: The area of the triangle plus the arc segment compared to the whole circle.
Distribution: We evenly distribute the second point in the whole circle.
Experiment: Spin circle to randomly pick point on the circumference, throw dart to randomly pick point in circle, pick chord that goes through both points.

3.

Probability: $1 - \frac{\sqrt{3}}{2} \approx 13\%$
Calculation: The length of the side of the triangle compared to the diameter of the circle.
Distribution: We evenly distribute the length of the chord.
Experiment: Spin circle to randomly pick point on the circumference, randomly pick a length for the chord (can use darts again). If you want to remove the unimportant ambiguity, flip a coin to decide which of the two possibilities to draw.

4.

Probability: $\frac{1}{4} = 25\%$
Calculation: The area of the inscribed circle of the triangle compared to the area of the circle.
Distribution: We evenly distribute the perpendicular of the chord.
Experiment: Throw a dart to randomly pick a point on the circle. Pick chord that goes through that point and perpendicular to the center of the circle.

5.

Probability: $\frac{1}{2} = 50\%$
Calculation: The radius of the inscribed circle of the triangle compared to the radius of the circle.
Distribution: We evenly distribute the distance of the chord to the center of the circle.
Experiment: Spin circle to randomly pick a point on the circumference, randomly pick a distance to the center of the circle. Pick chord that goes perpendicular to the center of the circle with the given distance.

6.

Another possibility is to throw two darts to pick two points inside the circle and pick the chord that goes through both of those points, picking random points in an area larger than the circle and discarding chords that do not go through the circle is also possible.

Well, the simplest one (I believe). First, note that the position of the triangle doesn't matter, only its size. The question is not asking for the chance that the chord intersects the triangle.

[1., 2. and probably 3.] Fix one point of the chord (WLOG we can assume it's $A$ in the picture), and draw a half-line $T$ tangent to the circle through $A$ (i.e. a horizontal one to the right). Randomly choose an angle in $[0, \pi]$ (using the uniform distribution). Draw the chord, starting from $A$, such that the angle between $T$ and the chord is equal to that angle. It's easy to see that the chord is longer than $AB$ if the angle is between $\pi/3$ and $2\pi/3$, so the probability is 1 in 3.

[4.] Take a long, straight stick, at least twice as long as the diameter of the circle; fix its center at point $A$ and make sure it can rotate freely around it. Then, give it a sweep with random velocity, like a wheel of fortune. This produces the uniform distribution from the previous paragraph.

• What do you mean by "[1.,2. and probably 3.]" ? – Thomas Blue Apr 12 '18 at 12:04
• They correspond to parts needed for a single Very Convincing solution. I'm not sure about 3. but that's probably because there are more Very Convincing solutions out there which 'contradict' this one. – Glorfindel Apr 12 '18 at 12:05