A350141 -id:A350141 - cp's OEIS frontend

A347040 Times on a 12-hour digital clock with 6 digits at which the three continuously moving hands of an analog clock, in the best approximation, enclose the same angles with one another, i.e., have the smallest sum of squares of the deviations from 120 degrees. When interpreting the terms as times of the day in the form hh:mm:ss, padding to the left with zeros is assumed.

2142, 4324, 12647, 14929, 23253, 25435, 33758, 35940, 44404, 50546, 54909, 61051, 65414, 71556, 80020, 82202, 90525, 92707, 101031, 103313, 111636, 113818, 122142, 124324

Offset: 1

Hugo Pfoertner, Aug 13 2021

An exact hit in which all angles are exactly 120 degrees is impossible. The smallest possible deviation occurs at two points in each display cycle, namely at 02:54:34.5617..., and at 09:05:25.4383... . With rounding to the nearest integer second, this corresponds to the terms a(6)=25435 and a(17)=90525.

The least squares clock solution actually occurs at an exact rational time, namely 5333364000/509173 seconds after 00:00:00, or 285998/509173 seconds after 02:54:34; and the exact least squares sum (in units of squared rotations) is 1/3055038 or (in units of squared clock-second-ticks) 3600/3055038 = 600/509173. - Robert B Fowler, Oct 29 2021

Henry Ernest Dudeney, Amusements in Mathematics, London, New York, Nelson, 1917. (Problem #63: The Stop-Watch)

Cf. A120500, A347039, A348637, A350141.

a(8) corrected by Robert B Fowler, Oct 29 2021

A348637 Largest clock triangle area.

Original entry on oeis.org

1, 2, 9, 9, 0, 3, 5, 3, 0, 7, 1, 0, 7, 3, 3, 2, 6, 3, 2, 7, 3, 9, 7, 2, 0, 1, 1, 3, 8, 8, 2, 4, 9, 6, 6, 9, 5, 5, 5, 6, 7, 5, 6, 0, 6, 6, 5, 9, 5, 3, 3, 3, 9, 3, 6, 9, 4, 2, 9, 1, 3, 7, 3, 1, 6, 8, 5, 7, 8, 8, 4, 1, 6, 8, 5, 6, 0, 5, 6, 7, 7, 7, 5, 7, 1, 2, 4, 8, 2, 5, 5, 1, 9, 4, 6, 4, 3, 1, 3, 6, 3, 7, 8, 6, 8

Offset: 1

Robert B Fowler, Oct 26 2021

Consider an analog clock face to be a unit circle, with unit-length clock hands; the endpoints of the hands lie on the unit circle and form the vertices of a Clock Triangle inscribed within the circle.
The area within this Clock Triangle has maximum value 1.2990353071..., which occurs around 02:54:35 and at its mirror image around 09:05:25.
At time T seconds after 00:00:00, the clock hands are at angles
      S (seconds hand) = T/60 * 360,      (degrees)
      M (minutes hand) = T/60/60 * 360,
      H (hours hand) = T/60/60/12 * 360.
The clock cycle repeats every 12 hours = 43200 seconds.
The second 6 hours of the cycle is a mirror image of the first 6 hours.
The area within the Clock Triangle at any time is equal to
      F(T) = abs(sin(H-M) + sin(M-S) + sin(S-H))/2.
(The derivation of this equation is not overly-complicated.)
The hour and minute hands are exactly 120 degrees apart at times
      T = 14400/11*(3k+1)   and   T = 14400/11*(3k+2)  for integer k.
There are 22 such times during every 12-hour cycle.
Empirically examining the relative extrema of F(T) near these 22 times, it is found that the largest F(T) occurs near T = 10475 (02:54:35), and near its mirror image T = 32725 (09:05:25).
Using Newton's iterative method to solve for Tmax in F'(Tmax) = 0,
      Tmax = 10474.561690797181984...
      F(Tmax) = 1.299035307107332...
Note: an equilateral triangle has area sqrt(3)*3/4 = 1.2990381056...

H. E. Dudeney, Amusements in Mathematics, Dover, 1958, pages 11 and 154; Problem #63, "The Stop-Watch", notes that at times 02:54:35 and 09:05:25 the clock hands are "nearly equidistant" and that "exact equidistance for the three hands is not possible", but does not point out that these two times are the most nearly equidistant times in the 12-hour clock cycle. Of the 430 puzzles in the book, 4 are represented on the front cover illustration; the stop-watch appears in the very center.

Henry Ernest Dudeney, Amusements in Mathematics, London, New York, Nelson, 1917.
Henry Ernest Dudeney, Cover illustration of Amusements in Mathematics with clock, New York, Dover, 1958.

Cf. A104954, A347040, A350141.

f[x_] := (Abs[Sin[2*Pi*x*(1/43200 - 1/3600)] + Sin[2*Pi*x*(1/3600 - 1/60)] + Sin[2*Pi*x*(1/60 - 1/43200)]])/2; RealDigits[FindMaximum[f[x], {x, 10475}, WorkingPrecision -> 110][[1]], 10, 100][[1]] (* Amiram Eldar, Oct 27 2021 *)

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