Source: Original.
From Claudio Baiocchi:
With respect to a polar-coordinates system with origin O at the center of the track, the velocity of each car is the angular speed times the radius of the lane; but the geometric restrictions force the values of angular speeds to be all identical; thus we can start by working with a still picture, and the solution will be given by a rotation of this picture at a constant angular speed.
Let P(j) denote the position of the j-th car; the distances |OP(j)| must increase with j and (possibly watching the picture in a mirror) the line OP(j+1) must form with OP(j) a positive acute angle, say A(j), whose cosinus must be given by |OP(j)|/|OP(j+1)|.
In particular, if we ask for constant ratio between the speeds (which means constant ratio between lane's radius) all the angles A(j) must be equal; the condition on the last car says that the origin O lies on the straight line P(N)P(0); which means that the sum of the angles A(j) is a multiple of pi. Thus the general solution of the problem in case 2 can be described by:
fix any q >= 3; set c=cos(pi/q), k=1/c; and choose N of the form q, 2q, 3q, ...
The case 1 of the problem seems more delicate. We must solve in (s,k,N) the problem:
SUM (j from 0 to N-1) ACOS { (s+jk)/[s+(j+1)k] } = q*pi (q integer)
Let us call F(s,k,N) the left hand member; for any fixed s > 0 one has F(s,0,N)=0 and F(s,infty,N)=N*pi/2; thus we can solve in k the equation with any q with q < N/2.
Call the angle between the radii to the positions of car i and i + 1, Ai.
For any values of t (an integer), s > 0, and N (greater than a minimum
depending on t):
Case 1. a value of k can be found such that the N angles A0 to A(N-1) sum
to 180 * t (ie fulfilling the condition that cars 0 and N and the track
centre lie in line).
Case 2. all the angles Ai are equal to 180*t/N and k = sec(180*t)/N
Further explanations and examples.
The radius, Ri, of the track of each car is proportional to its speed.
All the angles Ai remain constant while the cars move. Car N, car 0 and
the centre of the circular tracks lie in a straight line. That is A0 + A1
+.......+A(N-1) = 180 * t degrees where t is a positive integer; for
odd/even t, cars 0 and N are on opposite sides/same side of the centre of
the tracks. The straight ahead direction for each car is the direction of
the tangent to the track and that is (of course) perpendicular to the
radius to the car. Then cos ( Ai) = Ri/R(i+1).
Case 1.
Cos (Ai) = (s + ki)/(s + k(i + 1)), and A0 + A1 +.......+A(N-1) = 180 *
t. The angles Ai depend only on the ratio of s and k and it is convenient
to take s = 1. For other s, k should be multiplied by s and the angles
left unchanged.
Ai > A(i + 1), and the larger the value of k the smaller the value of
cos(Ai) and the larger the value of Ai. The upper limit of Ai is computed
by supposing k to be infinite so that cos(Ai) = (as i increases) 0, 1/2,
2/3, 3/4, 4/5, 5/6, 6/7, etc. The corresponding values of Ai are 90, 60,
48.19, 41.41, 36.87, 33.56, 31.00, 28.96, 27.27, 25.84, 24.62, 23.56,
22.62, 21.79, 21.04, 20.36, ....... . The upper limits for the cumulative
sums A0 + A1 + .......... are: 90, 150, 198.19, 239.60, 276.47, 310.02,
341.03, 369.98, 397.25, 423.09, 447.71, 471.27, 493.89, 515.68, 536.71,
557.08, ......... . The minimum value of N for t = 1, 2, 3 are
respectively 3, 8, 16, these being the positions in the list where the
cumulative sums first exceed 180, 360, 540.
In fact for t = 1, N = 3, s = 1, k = 3.7661, A0 + A1 + A2 = 180.0 where A0
= 77.89, A1 = 56.04, A2 = 46.07.
For t = 1, N =4, s = 1, k = 0.83315, A0 + A1 + A2 + A3 = 180.0 where A0,
A1, A2, A3 = 56.94, 46.57, 40.37, 36.13 respectively.
For all higher values of N and any value of s, k can be determined such
that A0, A1, A2, ...., A(N-1) sum to 180 degrees.
For t = 2, N = 8, s = 1, k = 9.201, the 8 angles A0, A1, ....... ,A7 sum to
360; the angles are, in order, 84.37, 58.28, 47.29, 40.83, 36.46, 33.25,
30.76, 28.76.
For t = 2, N = 9, s = 1, k = 2.079, the 9 angles A0, A1, ...... , A8 sum to
360; the angles are 71.05, 53.35, 44. 54, 39.03, 35.16, 32.25, 29.97,
28.10, 26.55.
Generally for any values of t, s and N (greater than a minimum depending
on t), there is a value of k such that the N angles, A0 to A(N-1) sum to
180 * t.
Case 2.
cos Ai = s*(k^i)/s*(k^(i + 1)) = 1/k, and, for any t, s and N (greater than
a minimum depending on t), all the Ai are equal, their common value being
(180 * t)/N = Ac (say).
For t = 1 the smallest value of N that gives an acceptable value of Ac
(less than 90) is N = 3. Then Ac = 60, cos Ac = 1/2 and k = 2.
For t = 1, N = 4, Ac = 45, cos 45 = 1/sqr 2, and k = sqr 2.
For t = 2, 3, etc the smallest values of N that give acceptable values of
Ac (less than 90) are N = 5, 7, etc, generally N = 2 * t + 1.
For t = 2, N = 5, Ac = 72, cos Ac = 0.309, k = 3.236.
For t = 2, N = 6, Ac = 60, cos Ac = 1/2, k = 2 - the example given in
the puzzle.
For t = 2, N = 7, Ac = 51.43, cos Ac =0.623, k = 1.605.
Generally for any values of s and t and of N (greater than a minimum
depending on t) Ac = 180 * t/N and k = sec Ac.