random? chord

Given a circle with inscribed equilateral triangle, what is the probability that a randomly chosen chord is longer than the side of the triangle.

Originally posted by luskin
Given a circle with inscribed equilateral triangle, what is the probability that a randomly chosen chord is longer than the side of the triangle.

I'm not normally great with geometrical problems but I'll give this a go. I apologise if it's unclear or messy.

Take a semi-circle with one corner of the equilateral triangle as a corner of the semi-circle. The line of the triangle then creates an angle of 30 degrees from the base of the semi-circle. Any randomly chosen chord from the corner of the semi-circle must make an angle between 0 and 90 degrees from the base. If the angle is >0 and 30 and

Originally posted by PaddyB
I'm not normally great with geometrical problems but I'll give this a go. I apologise if it's unclear or messy.

Take a semi-circle with one corner of the equilateral triangle as a corner of the semi-circle. The line of the triangle then creates an angle of 30 degrees from the base of the semi-circle. Any randomly chosen chord from the corner of the semi-ci ...[text shortened]... cle must make an angle between 0 and 90 degrees from the base. If the angle is >0 and 30 and

Looks like you got cut off, but I think you're saying the same thing I was thinking. Call the points of the triangle A, B and C. If you start your chord on A, then only if the endpoint lies inside the arc formed by B and C will the chord be longer than the side of the triangle. Since the length of arc BC is 1/3 of the total circumference, the probability that a random chord will be longer than the side of the triangle is also 1/3.

Originally posted by PBE6
Looks like you got cut off, but I think you're saying the same thing I was thinking. Call the points of the triangle A, B and C. If you start your chord on A, then only if the endpoint lies inside the arc formed by B and C will the chord be longer than the side of the triangle. Since the length of arc BC is 1/3 of the total circumference, the probability that a random chord will be longer than the side of the triangle is also 1/3.

That looks right. But what about this? Every chord has a unique mid-point, so you could choose a random point to determine your chord, and that mid-point would need to be within the inner concentric circle, which has half the radius of the original. This suggests probability of 1/4.

If you start at one of the vertices, then 60 degrees are within the triangle, and there are a total of 180 degrees (with 0 degrees and 180 degrees producing chords with length 0).

The chords that are "inside" the triangle will be longer than the sides of the triangle, while the others will not be. So I agree - 1/3

Originally posted by luskin
That looks right. But what about this? Every chord has a unique mid-point, so you could choose a random point to determine your chord, and that mid-point would need to be within the inner concentric circle, which has half the radius of the original. This suggests probability of 1/4.

The midpoint does not need to be inside an inner concentric circle with half the radius of the original. For example, a chord at 1 degree will not have such a midpoint.

Originally posted by AThousandYoung
The midpoint does not need to be inside an inner concentric circle with half the radius of the original. For example, a chord at 1 degree will not have such a midpoint.

What I meant is if the chord is to be longer than the`side of the triangle, then the mid-point must be within the inner circle.

Originally posted by luskin
That looks right. But what about this? Every chord has a unique mid-point, so you could choose a random point to determine your chord, and that mid-point would need to be within the inner concentric circle, which has half the radius of the original. This suggests probability of 1/4.

No, as there is more than one chord going through the mid point. In fact, if you take the area of the centre of the circle to be infinite (that is, there are infinitly many lengths of chords) then there is infinitly many chords of length=the diameter of the circle, and the infinity mucks everything up.

What you have outlined, I think, is the probabiliy of a chord of length x where (size of side of triangle

It does look like the answer is going to depend on the probability distribution.

Do you consider the distribution to be even with respect to the angle of the chord, the end point around the circumference of the circle, the midpoint of the chord within the circle, or something else? They'll each give you a different answer.

Originally posted by genius
No, as there is more than one chord going through the mid point. In fact, if you take the area of the centre of the circle to be infinite (that is, there are infinitly many lengths of chords) then there is infinitly many chords of length=the diameter of the circle, and the infinity mucks everything up.

What you have outlined, I think, is the probabiliy of a chord of length x where (size of side of triangle

ah-I was cut off. You have calculated the probability of chords of length greater than or equal to the length of the side of the triangle and strictly less than the diameter of the circle.

Originally posted by genius
ah-I was cut off. You have calculated the probability of chords of length greater than or equal to the length of the side of the triangle and strictly less than the diameter of the circle.

But what is the probability of a random chord going through the centre? Or any other randomly chosen point? It's going to be zero, isn't it?

Or to use a bit of jargon, the chord will almost surely not go through the centre.

Originally posted by mtthw
But what is the probability of a random chord going through the centre? Or any other randomly chosen point? It's going to be zero, isn't it?

Or to use a bit of jargon, the chord will almost surely not go through the centre.

That same argument applies to any point you care to choose. However, it doesn't matter where the mid-point of a random chord may lie when describing all points where the mid-points of all possible chords do lie. The problem arises when each point corresponds to a different number of chords.

As genius pointed out, there are an infinite number of chords that have their mid-points at the centre so there is a problem with double counting. Fixing one point of the chord on the edge, and then sweeping out all possible chords in order, alleviates this problem.

After doing a Monte Carlo simulation on Excel, by placing 5000 random pairs of points on a circle with radius 1 and calculating the distance between them, the result was that about 36% of the time the chord was longer than the side of the triangle. This is a little higher than 1/3, which makes me think the random number generator on Excel is a little wonky. Either that or there's more to this problem than meets the eye.

Originally posted by PBE6
That same argument applies to any point you care to choose. However, it doesn't matter where the mid-point of a random chord may lie when describing all points where the mid-points of all possible chords do lie. The problem arises when each point corresponds to a different number of chords.

As genius pointed out, there are an infinite number ...[text shortened]... r on Excel is a little wonky. Either that or there's more to this problem than meets the eye.

Yes, it does apply to any point. But it still suggests you can except any single point without affecting the final answer.

For the second bit: as I said before, there are several ways of defining your distribution. You've chosen a sensible one, but there are others.

Originally posted by mtthw
Yes, it does apply to any point. But it still suggests you can except any single point without affecting the final answer.

For the second bit: as I said before, there are several ways of defining your distribution. You've chosen a sensible one, but there are others.

I agree you can normally leave out any single point without affecting the final answer in these types of questions, but you can't leave out a singularity with infinite weighting while every other point has finite weighting!

Before everyone gets all het-up over this beware of mtthw's warning!

I vaguelly remember a similar problem at school (a long time ago!) when the Maths teacher proved three different answers!

No tricks. All were valid!