Nerd Sniping

In a unit square (side length=1), is there any point inside the square with rational distance to all four corners?

Representable as a rational number, you mean? I'd naïvely think they all are, because they're almost all less than one and greater than zero. That's all the thought I'm willing to put into this tonight.

Note that the diagonal is , which is not rational.

rational distance to all four corners?
If it's not the same it's easy, say 0.1;0.9
But you want it to be all lengths are equal, right?

Perhaps a drawing will help:
@swayde Placing P at (0.1, 0.1) fails because then:
$$ \ell_1: \sqrt{(0.1)^2 + (0.1)^2} = \sqrt{\frac{2}{100}} = \frac{\sqrt{2}}{10} \notin \mathbb{Q} $$

It depends. Does "on a border" count as "inside the square"?

Ah, like that.

It wouldn't normally, but out of interest, do you have a point on the border that works?

... Can't we just try points until we find one?

We can, but I had hoped to get something constructive done today...
This has me looking up generation properties of Pythagorean triples to try to get something out of even the simplest possible cases... ugh

Which I remembered before you replied ...

I was thinking of (0,0), but I forgot the opposite corner...

In a unit square (side length=1), is there any point
inside theon the same plane as the square with rational distance to all four corners?Unknown: http://unsolvedproblems.org/index_files/RationalDistance.htm
Though someone had a go at saying "no": http://www.rxiv.org/pdf/1403.0009vA.pdf

Unknown
Motherfucker, I spent a better part of the day figuring this out.
Though with the "inside the square" constraint, there might be a proof that involves showing that all such points necessarily lie outside of the square  but I haven't found that either.
Seems like Heron's formula might be a way to go there, but there's no way I'm doing this on paper.

Though someone had a go at saying "no"
Now you've sniped me into trying to check this proof. I hope you're happy.It looks OK up to section C.4. That part I'm not sure about yet (though the fact that one of the references is "an answer someone posted on math.stackexchange.com" is a bit worrying).

You're assuming that any of these angles are going to be right angles, and that's not true at all.
If P were .5,.5, then every angle is 45degrees, which makes all of the lengths .5*sqrt(2).
I think you have to use cosin law to find the lengths.

No, the right angles arise naturally (but they're not the angles at point P). If the point is at (x, y) inside the unit square, then by dropping a perpendicular line to each edge in turn, you get right angled triangles yielding that the four distances are
√(x² + y²)
√((1x)² + y²)
√(x² + (1y)²)
√((1x)² + (1y)²
In the case P = (0.5, 0.5) then all of these become √(0.5² + 0.5²) = 0.5√2 as you said. In the case (0.1, 0.1) you get √(0.1² + 0.1²) = 0.1√2, √(0.9² + 0.9²) = 0.9√2, and (twice) √(0.1² + 0.9²) = √(41/50).It should of course be manifestly obvious that no point on either diagonal can work (because then two of the rational lengths would add up to that diagonal, which has an irrational length).

... Can't we just try points until we find one?
Brute force. I like it. Did I teach you how to code?

Frack, I missed the obvious.
Well, that means that the length squared has to be the sum of two rational numbers who are squares over squares
So let's say
A/B is the rational length.
C/D and E/F must be the rational coordinates.````A/B
is the second rational length by 1C/D,E/F
A/B``` and so on C/D,1E/F ```
A/`B``` 1C/D,1E/FA^2 / B^2 = C^2 / D^2 + E^2 / F^2
But of course, D and F must share a common denominator, so
A^2 / B^2 = (C^2 * F^2) / (D^2 * F^2) + (E^2 * D^2) / (D^2 * F^2)
So
B^2 = D^2 * F^2
and
A^2 = (C^2 * F^2) + (E^2 * D^2)
But then it has to also be true for 1  C/D, and 1  E/F, substituted independently.
Each producing
````A/B and A/B and
A/`B```

Brute force. I like it.
Better yet, the part of the alleged proof linked above that says that any solutions must have rational coordinates seems pretty solid, so you only have a countably infinite set to search!... What? It's better than an uncountably infinite set, almost all of whose members are not representable in any meaningful way, isn't it?


I'm not the one planning to do a brute force search for solutions.

The mean time to a solution is 1/2 of countably infinite. It is doable.

Less than that.
The center point isn't a solution and squares can be flipped and rotated  which means we only have to look for points in one half of one quadrant of the square, because they're all the same  which means our space of possible solutions is only an eighth as big!

I like the way you think, with Germanlike efficiency.

The mean time to a solution is 1/2 of countably infinite.
That's presuming there's exactly one solution. There might be (most likely) none, or some finite number, or possibly (though extremely unlikely) a countably infinite number.
<a Of course you'll still get infinite mean time to a solution in all these cases, except if there are infinitely many solutions with a nonzero density in the limit (I can define this phrase rigorously if you really want me to, but I assume you get the general idea), which is even more extremely unlikely.>

Now you've sniped me into trying to check this proof.
Looks like we run into problems when he derives equation (E.10) on page 9. For a start, the last three equations in (E.9) are missing terms of (respectively) +m², +m², and +2m² on the right hand side, but that's not immediately fatal.When we subtract the fourth equation from the first and divide through by 2^{g+1}, one of the terms is 2^{g3} (q_{1}²  q_{4}²). He's subsuming this into the 2T term, but since he's only proved that g is at least 2, there's no guarantee that this is even an integer.
Now in the g=2 case, we can take all the q_{i} to be zero (the form of solutions is simpler in this case), so we can still get to the claim that q_{1} and q_{4} have to be equal. But in the g=3 case... hmmm, OK, for g>2 we can first note that 2T is an integer to prove that q_{1} = q_{4}, and then this term vanishes and 2T really is even. So it looks like this hole can be patched, guess I'll have to keep reading.

Now in the g=2 case, we can take all the qi to be zero (the form of solutions is simpler in this case)
Thinking about it, that's not right; this works when the residue is 1 mod 4 (1 or 1=3 work), but not when the residue is 0 mod 4 (0 and 2 are not negatives of each other mod 4). A better generalisation is that we can always take the sign of r to be positive, with q_{I} varying. So it looks like there's a hole for the g=2 case still. Haven't looked any further at the g>2 case yet.