Interplanetary blips and bleeps

Question

Things were quite different in 3000 AD. We'd discovered other planets with sentient life for instance. Five to be precise. Adam, Bill, Carl, Dave, and Eric we called them, and we used the lately discovered instantaneon beams to communicate without delay across the vast distances between us. Light-speed communication was so last millenum. Of course there were the recreational 'radio' operators on each planet as well... they'd set up an instantaneon channel (Channel A) on which any signals on one of the light-speed channels (Channel B) were recorded by stations on each planet and broadcast instantaneously across the universe. Channel B stations on each planet recorded the signals on Channel A and broadcast them into the universe at light-speed.

One time we had a solar flare, and our silly Channel A station broadcast a blip accidentally across the universe. Some time later (I don't recall how long), the Channel B recordings of that blip started coming in as five bleeps from our five friends, at intervals (surprisingly enough) of exactly a year.

Later, Eric's star supernovaed. It kind of sucked knowing that an entire planet got wiped out like that, but life continued on. Adam, Bill, Carl and Dave became our best friends in the sky; from anywhere on earth you could look out and see at least one of them.

Anyway, later we had another solar flare and we sent out a blip on Channel A again, but this time we noticed something interesting. The blip of course came back to us on Channel A as well, from each station on Adam, Bill, Carl and Dave recording the resulting Channel B bleep from the other three. A certain period of time after the flare we heard a blip, and less than half that time later another one, but those 2 blips were all we got on Channel A. Some of the planets' stations must have received some of the Channel B bleeps at the same time.

Question: How far apart in time did we receive these two blips?

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Assume no relative motion of the planets. Assume the galaxy is flat - all planets within it lie in a plane. Ignore any other blips or bleeps that might be generated in this situation - the only ones we are considering are the Channel B bleeps from the five planets in response to our Channel A blip, and for the second flare, the channel A blips resulting from the four planets' receipt of each of the other three's Channel B bleeps in response to our blip. Assume the stations' recording and broadcasting is instantaneous.

This is a mathematical puzzle, not a lateral thinking puzzle. There is nothing hidden in the text; interpreting the situation described in the puzzle at face value leads to a single unique solution. I believe this mathematical problem is interesting as a puzzle because it seems (I think) quite surprising that a unique solution exists given the limited data. While it is entirely possible to solve this puzzle by hand (I did - the underlying mathematical problem is one I once formulated to myself as a puzzle and solved by hand; I thought it might be interesting dressed up in the form of this story), there is some work involved, for which computer assistance is allowable. An exact expression for the answer is preferred.

Answer 1

Let's start with the information in the third paragraph. Essentially it tells us that the mutual distances between the planets Adam, Bill, Carl and Dave have exactly two distinct values. Since Adam, Bill, Carl and Dave are in a single common plane, this means that they must be arranged in one of the following six constellations:



The ratio between these lengths is, in reading order: $\sqrt{2} \approx 1.41$, $\frac{\sqrt{5}+1}{2} \approx 1.62$, twice $\sqrt{3} \approx 1.73$, and twice $\frac{\sqrt{6}+\sqrt{2}}{2} \approx 1.93$. Now since the second signal arrived within half again the time of the first signal, the larger of the two distances must be less than $1.5$ times the shorter distance, so the only permitted constellation is the square.

Let's say Adam, Bill, Carl and Dave are at points $A,B,C,D$, and Earth is at $X$. Because of the information in the first paragraph, we know that $a = |XA|$, $b = |XB|$, $c = |XC|$, $d = |XD|$ are four of five consecutive terms of an arithmetic progression (the remaining one would be the distance between Earth and Eric). By trial and error (repeat the following steps until we find a solution) it turns out that $|XE|$ must be the second term. Suppose $a$ is the lowest term, and $b$ is the third lowest. Then it is true that $2b = a + d$ and $2c = b + d$.

We can use coordinate geometry to figure this out. Let $ACDB$ be the standard unit square with $A = (0,0), B = (0,1), C = (1,0), D = (1,1)$, and suppose that $X = (x,y)$. We find $a = \sqrt{x^2+y^2}$, $b = \sqrt{x^2+(y-1)^2}$, and so on. Substituting these into the previous two equations and squaring the radicals away yields a system of two quartic equations in $x$ and $y$ that Wolfram Alpha can solve. There are three real solutions to this system of equations. One is $(1/2, 1/2)$, which would make $X$ the center of the square, but we can exclude that since we know that $a,b,c,d$ are not all the same. Of the two remaining solutions, one lies outside of the square, but paragraph 2 excludes that possibility. So only one possibility for $X$ remains, at coordinates $(\frac{63\sqrt{17}-161}{848}, \frac{35\sqrt{17}+99}{848})$.

Finally, we can compute the solution to the problem. We know that $|XD|-|XC|$ is exactly one lightyear, and we want to know $|BC|-|BD|$ in lightyears, so the answer is $$\frac{|BC|-|BD|}{|XD|-|XC|} = \frac{\sqrt{2}-1}{2}\sqrt{65+7\sqrt{17}} \approx 2.01.$$