## I need an average

by The other day our 9-year old said to me, “A is 1 inch from B, and B is 1 inch from C.  How far apart are A and C?”   It turns out this was a joke, and the answer was supposed to be 1 inch (I think he got it from Calvin and Hobbes, though I don’t remember seeing it there), but of course I couldn’t help mentioning that the answer could be anywhere from 0 to 2 inches.

It got me thinking — what would the “average” distance be?  And I think the average would be the square root of 2. My logic goes something like this:  I think of the line segment AB being fixed, and then BC rotating around B in a circle.    There is symmetry between the angle ABC being acute and being obtuse, so I think the “average” distance AC would occur when ABC is a right angle.  In that case, the distance AC would be √2.

So here’s my question:

1) Are there any other reasonable ways of coming up with an answer (whether or not it is the same)?

2) I want the answer, √2, to somehow be an average of the two extremes 0 and 2.  But it’s not the arithmetic mean (1), the geometric mean (0), or the harmonic mean (0).  That bothers me.  Is it some other kind of average?  (Yes, I’m looking for a process that gives me the answer I want, which is kind of backwards but there it is.)

Number photo from gokoroko.

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### 12 Responses to “I need an average”

1. Jonathan Vos Post Says:

Depends. The 9-year old was echoing someone who assumed that the points were colinear. You assume that they’re in a plane. Any other hidden assumption here, such as Metric Space?

2. jonah Says:

Unless I’m missing something, isn’t the harmonic mean undefined because it would require dividing by 0?

3. jonah Says:

Also, it’s the RMS average.

4. Dave Richeson Says:

How about $pi/4$ for an average (approx. 1.27324…)?

Here’s my reasoning. If the angle between the segments is $\theta$, then the distance between A and C is $2\sin(\theta/2)$. Then the average value of this function on the interval $[0,\pi]$ is $\displaystyle\frac{1}{\pi}\int_0^{\pi}2\sin(\theta/2)d\theta=\frac{\pi}{4}$.

5. Dave Richeson Says:

Oops. It should have been $4/\pi$, not $\pi/4$.

6. Fadam Says:

I agree with Dave, \pi/4 would be the mean. What you calculated was the median length.

A simpler way of asking the same question would be what is the average length of a chord in a circle with a radius of 1.

7. Mitchell Says:

The problem with the reasoning in the article is that he seems to have considered the problem discretely, rather than continuously. Thinking continuously, you easily get Dave’s answer above.

8. Sue Says:

@Jonathan: If the points were colinear, the answer would be 2, not 1. I was guessing the joke was that the A and C in the snetence were one inch apart. Also, any 3 points are in a plane, so that wasn’t an assumtion.

@Mitchell: Can you explain what you see as the difference between a discrete and a continuous perspective on this problem?

Interesting question.

9. jedward706 Says:

perhaps the “discrete” view could be considered like “taxi-cab” geometry — the only allowed positions for the points are on a Cartesian grid ?

10. Ξ Says:

Thanks for all the ways to think about this [and sorry for the delay in response — our internet access was really limited over the past week, so I’m just getting a chance to read them].

I think I was viewing the problem continuously, not discretely, but I’m still not wrapping my brain around why it would be a median but not a mean — I’m seeing symmetry, so the two should be the same. (But I think it’s quite possible I’m making some other assumptions, as jedwards points out. I was certainly assuming Euclidean, but being in a plane follows from that and the fact that there are three points. [Wait, is that right? Maybe the mean would change if C varies over a sphere rather than a circle. OK, I think I’m seeing how the mean and median might be different.]

11. TwoPi Says:

How does one define the median in the continuous context? What is the median of an uncountable set of data?

12. Veky Says:

Well, of course. Look at it this way: there is only one mean, and it is usually called “arithmetical mean”. It can be calculated in the continuous case using integrals, as shown above (result 4/pi).

What about other “means”? It turns out these are also instances of the above mean, just in different codomains. To calculate f-mean (mean pulled by f), we map the values by function f into some other space, calculate the mean there, and pull back by f^{-1}.

For example, geometric mean is log-mean, and harmonic mean is reciprocal-mean. (Square mean is literally squaring-mean.) And these are useful in many applications, so they have special names.

Now I suppose it is clear what happened: you have calculated the angle-mean of all the chords, where angle is the function that maps the length of the chord to size of its central angle. You’ve mapped chord lengths to angle sizes, calculated the mean (very easily, by symmetry) in the angle-space, and pulled that mean (=0) to chord-space, yielding sqrt2.

So you can get almost any value in domain space (at least in the interval spanned by actual values) as f-mean for suitably chosen f. You have just chosen f so that the codomain mean can be easily calculated, not as some intrinsic property of chord lengths. But you don’t have to be too disappointed, since great mathematicians had almost the same problems in a slightly more difficult setting. See http://en.wikipedia.org/wiki/Bertrand_paradox_(probability)