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The target is at T. Ike, Jay and Kay start out at I, J and K. Ike and Kay remain at their respective locations, but Jay moves to J'.
Since they all start equidistant from T, I, J and K are on a circle of unspecified radius centered on T. I is at the center of a circle of radius 4 km, whose intersections with the first circle mark the points J and K.
Triangles JTI and ITK are congruent isosceles triangles with base 4 km and sides 8 km. Triangle IJJ' shares a base angle with IJT at point (and angle) J, and so is similar. Side TK has to be an integer, and thus TI and TJ also, as the equal sides of two congruent isosceles triangles. Thus, for TJ' to be an integer as required, JJ' must be an integer. JI, at length 4 is the base of the larger triangle, JTI, and a side of the smaller triangle, JIJ', so IT/4 = 4/JJ', or (IT)=16/(JJ').
If IT were 1, then the circle around T would not intersect the 4 km circle around I. If IT were 2, J and K would have been at the same position initially. If IT were 4, Jay would have reached the target (J' located at T), but he is said to have stopped short of it.
With IT = 8 and JJ' = 2, isosceles triangle J'JI has base 2, and so consists of two back-to-back right triangles with leg 1 and hypotenuse 4, so angle J'JI (aka TJI) has measure arccos(1/4). Angle JTI, the apex of one of the large isosceles triangles, then has measure 180 degrees - 2 arccos(1/4). The two large isosceles triangles together, JTI and ITK, at apex T, form angle JTK with measure 360 deg - 4 arccos(1/4).
Then using the law of cosines in triangle J'TK,
(J'K)^2 = 6^2 + 8^2 - 2*6*8*cos(4 arccos(1/4)), as cos(360-x) = cos x.
Since cos(4a) = 8 cos^4(a) - 8 cos^2(a) + 1,
4 arccos(1/4) = 8/256 - 8/16 + 1 = 17/32
So (J'K)^2 = 6^2 + 8^2 - 2*6*8*17/32 = 49, and
J'K = 7 km.
From New Scientist, 16 September 2006, p.23, Enigma number 1409.
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