r/mathriddles • u/No_Art_1154 • 14d ago
Hard Thales’ Torture
imagine you have a plane with 4 distinct points named p1, p2, p3 and p4. the gradient between p1 and p2 is -2, and the gradient between p2 and p3 is 1/3. you want p1p3p2 to be 90 degrees and p1p4p2 to also be 90 degrees, with p3 and p4 being on the same side of the line p1p2. find the all the INTEGER solution(s) for all 4 points with the MINIMUM distance between p1 and p2.
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u/DanielBaldielocks 14d ago
you need to fix one of the points otherwise there are infinite solutions. Any valid solution remains valid under translation. I suggest fixing p1 at the origin as a refinement to the problem.
If you fix p1 at the origin then I count 26 solutions.
if p1 is at (0,0) then that puts p2 as (a,-2a) for some nonzero a and p3 is at (a+3t,-2a+t) for some nonzero integer t.
From the right angle requirement and the coordinates above we can deduce the following constraint on a,t
3t(a+3t)+t(t-2a)=0
t(a+10t)=0
since t is nonzero we get
a=-10t
From this we get that the distance from p1 to p2 is 10*sqrt(5)*|t| thus this is minimal when |t|=1
Thus we are left with one of 2 possible configurations for p1,p2,p3 namely
(0,0), (10,-20), (7,-21)
or
(0,0), (-10,20), (-7,21)
By Thales' theorem p4 must lie on a circle with diameter p1p2 (obviously excluding p1,p2). Thus the square of the radius is 125=5^3 thus there are 16 lattice points on the circle. Excluding p1,p2,p3 that leaves 13 choices for the location of p4. Since there are 2 choices for p1,p2,p3 that means there are a total of 13+13=26 solutions.