I am working through some old questions from the Swiss Mathematics Games. Specifically, I am trying to solve problem 18 from the 35th Quarter Finals (from 2021). The question and solutions can be found on the FSJM website, but no working is provided. I have recreated the problem below (reworded a little).
We have three cedar trees at the vertices of an isoceles right-angled triangle, which has two sides of length 51m. A point is located a non-zero distance less than 30m from the nearest tree, and the three distances between the trees and the point are all integer values. What is the sum of these three distances?
It is intended that there are multiple solutions: in particular there are three (solutions for the sum of the distances, obviously for any valid point there is another one by symmetry).
However, I have no idea how to do this efficiently, much less how to do this without any calculator or computer access (as the question was originallly intended). If we orient the triangle so that the right-angle is at $(0,0)$, then the other two vertices are at $(51,0)$ and $(0,51)$. Then, if we let $X = (x,y)$ be a general point, we can get expressions for the three integer distances from each vertex to the point $X$:
$$\sqrt{x^2 + y^2}, \quad \sqrt{x^2 + |51 - y|^2}, \quad \sqrt{|51 - x|^2 + y^2}$$
As these must be integers, by rearranging these expressions I am able to show that $102x$ and $102y$ must be integers.
Some solutions are easily found, for example by setting either $x$ or $y$ to be $0$. By very thorough checking of a long list of Pythagorean triples I was able to find a second solution myself, but the third solution (which has non-integer co-ordinates) seems completely unobtainable by hand without blind guessing (or writing out Pythagorean triples up to and including $1632^2 + 6426^2 = 6630^2$).
Is there a neat, efficient way to find these solutions?
For convenience, the solutions are:
$122 = 20 + 37 + 65$
$130 = 25 + 52 + 53$
$170 = 17 + 68 + 85$
I. One solution is easy. Extend side $CB$ to $D$ such that $BD=17$, and join $AD$. Triangle $ACD$ is thus an unreduced $3-4-5$ right triangle with integer sides $51-68-85$, and $DB,\,DC,\,DA=17,\,68,\,85$.
II. The problem is of course equivalent to that of finding a point which is an integer distance from three vertices of a square. In his Unsolved Problems in Number Theory (3rd ed., p. 283), R. K. Guy notes that J. A. H. Hunter in correspondence with Leslie J. Upton gives an infinity of solutions for a square of side $s$ where the distances $a, b, c$ of point $P$ from three vertices are in arithmetic progression: $$a=m^2-2mn+2n^2,\; b=m^2+2n^2,\; c=m^2+2mn+2n^2$$ and $$s^2=2m^2(m^2+4n^2)$$where $s$ is an integer if$$m=2(u^2+2uv-v^2),\; n=u^2-2uv-v^2$$ A brief search yields, for $u=1$, $v=2$, that $s=40$ and $a,\,b,\,c=130,\,102,\,74$, in arithmetic progression. These four even distances reduce to $20,\,65,\,51,\,37$.
Further, Guy notes that each solution has its "inverse", in which side $s$ and distance $b$ switch roles. In the figure below, the smaller square has side $EB=20$, and $CE,\,CB,\,CG=a,\,b,\,c=65,\,51,\,37$.
On side $BC=51$ build square $BCDA$ and join $EA$. Since $\triangle BEA\cong\triangle BGC$, then $EB,\,EC,\,EA=20,\,65,\,37$, and point $E$ is a second solution to the problem.
III. Finally, Guy notes that John Conway and Michael Guy found an infinity of integer solutions of$$(s^2+b^2-a^2)^2+(s^2+b^2-c^2)^2=(2bd)^2$$This equation is satisfied for $s=51,\,a=25,\,b=52,\,c=53$. Solving the equation knowing only that $s=51$ may not seem easy, but since it is given that $EA=a<30$, and as Guy notes "one of $s,\,a,\,b,\,c$ is divisible by $3$, one by $4$, and one by $5$" (and here $51$ is divisible by $3$), a fairly short search reveals that $a=25$, $b=52$ (divisible by $4$), and $c=53$ (not divisible by $3,\,4$ or $5$).
In the figure below, $EA,\,ED,\,EC=a,\,b,\,c=25,\,52,\,53$.
Addendum: As for the unsolved problem of finding a point at integer distances from four vertices of a square, Guy says it "remains as an astonishingly hard nut to crack" (p. 284). But note how closely this third solution of the three-distance problem for $s=51$ approximates a solution of the four-distance problem as well: four lines $a,\,b,\,c,\,d$ of integer length radiating from a point to the vertices of a square, such that $a^2+c^2=b^2+d^2$. For here $$25^2+53^2=52^2+d^2$$and$$d^2=EB^2=625+2809-2704=730=27^2+1$$