Measuring the Mystery Planet
Geometry, Algebra; Difficulty: medium
Planetary Measurement
The Riddler Classic from this week (Mar 25, 2022) is a geometry and algebra problem wrapped up as an astronomical measurements puzzler. As Star-vind I felt compelled to get into action (also my puzzle submission was adapted to be this week’s Golden Globes Express problem—Yay!). Here’s the Classic reproduced:
The astronomers of Planet Xiddler are back at it. They have developed a new technology for measuring the radius of a planet by analyzing its cross sections.
And so, they launch a satellite to study a newly discovered, spherical planet. The satellite sends back data about three parallel, equally spaced circular cross sections which have radii A, B and C megameters, with 0 < A < B < C. Based on these values, the scientists calculate the radius of the planet is R megameters. To their astonishment, they find that A, B, C and R are all whole numbers!
What is the smallest possible radius of the newly discovered planet?
*spoiler alert* solution and answer follow in the sections below.—
Solution
This problem obviously involves solving right triangles. It becomes more interesting once you think outside the triangle (or at least outside Pythagorean triplets).
In the figure above, AD, BE and CF represent the three parallel and equally spaced cross sections with DE = EF. We are also told that A, B, C and R are all whole nos with \( 0 < A < B < C \)
Let the equal spacing between the cross sectional planes be \(x\);
The distance from the center of the planet to the middle section be \(d\); i.e. OE = \(d\)
The radius of the planet is R
This particular choice of set up will become clear in the next section where we explore what values \(x\) and \(d\) can take on.
From the three right triangles in the fig, we have:
\begin{align*}
&R^2 = A^2 + (d+x)^2 &&= A^2 + d^2 + 2dx + x^2 \\ \\
& R^2 = B^2 + d^2 \\ \\
&R^2 = C^2 +(d-x)^2 &&= A^2 + d^2 – 2dx + x^2
\end{align*}
Are d and x necessarily integers?
It’s natural (but not necessarily correct) to assume that the inter-sectional distance and hence \(d\) and \(x\) are also integers, and thus restrict the search for an answer to Pythagorean triples meeting the above conditions. This might very well be what the puzzle creator intended.
But the puzzle as stated does not stipulate that the equal spacing between the cross sections be a whole number. We do know R, A, B, C are all positive integers. Consequently neither \(d \) or \(x\) can be fractional nos. If \(d\) was fractional, so would \(d^2\). In the second equation above, since B is a positive integer, then R could not be a positive integer. Similar reasoning for x.
So does that then imply that \(d\) and \(x\) have to be integers?
No! They could possibly both be irrational numbers such that their squares (\(d^2, x^2\)) and their product (\(dx\)) are integers. This would make all the quantities in the above equations positive integers. This opens the possibility that we find positive integer solutions for R, A, B, C. This is only possible if both \(d\) and \(x\) are irrational square roots of positive integers.
Solving for integer d and x
If we restrict our search to integer \(d\) and \(x\), we need to find Pythagorean triples (not necessarily primitives) with the same hypotenuse such that the three bases form an arithmetic progression. Writing a short program to search for these gives us the following three triples with the smallest common hypotenuse that satisfies the arithmetic progression of bases requirement:
(25,60,65) (33,56,65) and (39,52,65)
Hence the answer for the smallest R is 65 with A= 25, B=33, C = 39 and the inter sectional distance being 4 (assuming integer inter sectional distances). (All nos in megameters as the puzzle requires).
Relaxing integer requirements for d and x
We can find a much smaller R if we don’t require \(d\) and \(x\) to be integers. As noted before, \(x\) and \(d\) then have to be irrational square roots of positive integers. So we need to search for positive integer values of R, \(x^2\) and \(d^2\) that also result in positive integer values for A,B,C. Also, we only have to search for R < 65 to see if we can do better than the solution we have already have above. Writing another quick program identifies 39 values of R smaller than 65 meeting the conditions of the puzzle. The solution with the smallest R is:
R = 8, A = 2, B = 7, C = 8; inter sectional distance \( x = d = \sqrt{15}\). With C = R implies that the third section results in hemispheres—not explicitly prohibited by the question.
If we assume that R > C, then the required solution is:
R = 14, A = 7, B = 11, C = 13; inter sectional distance \(x = 2\sqrt{3}\) and \(d = 5\sqrt{3}\)
Third section in the 'other' hemisphere?
On Twitter, Spike Smith asks what if the third section is on a different hemisphere from the first two. I had explored this but did not include in my solution previously. Adding it now:
It is conceivable that not all the sections need be on the same ‘side’ or hemisphere: The third section with radius C could be on a different one from the first two. No other configuration would be feasible. Looking into this possibility, the solution with the smallest R is:
R = 63, A = 18, B = 58, C = 62; inter sectional distance \( = 16\sqrt{5}\) with the section with radius C being on the ‘other’ hemisphere from those with radii A, B.