I'm questioning the solution of a high-school mathematics competition problem involving the geometric mean of distances from a point to a circle. The specific question is as follows:
Let O be the center of a circle with radius 4. Let A be a point outside the circle such that the distance from A to O is 14. Find the geometric mean of the distances from point A to every point on circle O.
For convenience, let O be centered at the origin, and let A be the point (-14, 0). The intended solution uses power of a point on point A to derive the result that the geometric mean of the lengths will be equal to the length of the tangent line from A to circle O.
https://i.sstatic.net/id1MJ.png (AD (blue line) is a tangent line from A to circle O)
The solution reasons as follows: For distances AB and AC, by power of a point, AB * AC = AD^2. The line can have greater or lesser slope arbitrarily. Thus, all the non-tangent points of the circle can be paired up in such a way that the geometric mean of the lengths from A to each point in the pair is AD. Clearly, the distance from A to a point of tangency is also AD. From common sense, the geometric means of all the relevant lengths will then be AD. By the Pythagorean Theorem, the length AD is sqrt(14^2 – 4^2) = sqrt(196 – 16) = sqrt(180) = 6 sqrt(5).
The reason I believe this is incorrect is because it seems to assign a heavier weight to points on circle O that are on the "A side". Specifically, the smaller arc length between the two point of tangency will be less than πr, while the larger arc length is greater than πr. But, each arc length contains an equal number of points (one from each pair) in the pairing solution above. This makes points more concentrated on the smaller arc (closer to A) for any finite number of lines drawn. I see no reason why this should not extend to an infinite number of lines including all the points of the circle. As an analogy, drawing many lines passing through the origin all with different slopes, for example with a angular separation of 1 degree between each line, will result in a much higher concentration of "line" near the origin than far away from the origin.
I have tried to think of a more rigorous solution for this, especially applying calculus-based techniques, but was unable to think of one.
Is the solution given really correct?
If not, what is the correct solution?
Best Answer
You are correct. To see why, suppose that $A$ is much closer to the circumference of $O$ than is illustrated in the original question; e.g., suppose $AO = 4.1$. Then the major and minor arcs subtended by the angle formed between the two external tangents from $A$ will have drastically different measures, one being almost the full circumference. While it is true that points can be mapped in one-to-one correspondence between the two arcs, this is like saying that the lengths of all finite intervals on the real line are equal simply because there is a similar one-to-one correspondence--and we know that this is not true: the length of $[0,1]$ is not equal to the length of $[0,100]$, even though the mapping $$f : [0,1] \to [0,100], \quad f(x) = 100x$$ is a bijection.
In order to compute the geometric mean of the distances from $A$ to points on the circle, we have to specify a measure with respect to the set of those distances. For instance, we could choose a measure that is uniform on the circumference of the circle, and in the absence of any other information, this would be the most natural interpretation. If we do, then the given solution is clearly wrong for the reasons we have elucidated.
Since the logarithm of the geometric mean of a sample is $$\log \left(\prod_{i=1}^n x_i \right)^{1/n} = \frac{1}{n} \sum_{i=1}^n \log x_i,$$ we can see the geometric mean is the antilogarithm of the arithmetic mean of the log-transformed sample. Thus, the computation of interest with respect to the uniform measure on the circumference is $$\exp \int_{\theta = 0}^{2\pi} \frac{1}{2\pi} \log \sqrt{(4 \cos \theta - 14)^2 + 16 \sin^2 \theta} \, d\theta = \exp \log 14 = 14.$$
This computation is supported by the following numeric simulation with Mathematica:
the output of which is very close to $14$.