In this puzzle, Flatland has set up three fixed GPS broadcast stations (A, B and C) on the vertices of an equilateral triangle with edges of length 100 kilometers. When Richard leaves his house in Flatland to visit Roger his GPS receiver gets a time signal of exactly 7 a.m. from C, but the time from B is 0.00001 seconds earlier and the time from A is 0.00004 seconds earlier. When Richard arrives at Roger’s house the time he receives from B is 0.00001 seconds later than the time he receives from C, and the time from A is 0.00004 seconds later than the time from C. How far is Richard’s house from Roger’s house?
A, B and C can be given two-dimensional kilometer coordinate values (0,0), (50, 50√3), and (100, 0), respectively. The speed of light is c ≈ 299,792 km/sec. Let (x, y) be the coordinates of Richard’s house and (u, v) be the coordinates of Roger’s house. So the time signals Richard receives when he leaves home result in the following equations:
√((x-100)2+y2)=√((x-50)2+(y-50 √3)2)+(0.00001 sec.)c and √((x-100)2+y2)=√(x2+y2)+(0.00004 sec.)c.
After wrestling with the numerical algebra this leads to (x, y) ≈ (43.1716, 27.052). Similarly the signals when Roger arrives at Richard’s house result in the equations following:
=√((u-50)2+(v-50 √3)2 )-
(0.00001 sec.)c and
=√(u2+v2 )-(0.00004 sec.)c.
After more rounds of numerical contortions this leads to (u, v) ≈ (57.0732,31.065). So the distance from Richard’s house to Roger’s house is about