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How often, and when, are the minute & hour hands at right-angles?
This is a bit more complicated, since there are two possible internal right-angle configurations to be considered, i.e. at angular differences of 90° and 270°.

But, if having worked through puzzles 2 & 3, it should be clear how to answer the following:

  • when are the hands first at right-angles after the start at midnight?
  • at what subsequent time interval will right-angles occur?
  • when would the 17th occasion be?
  • how many occasions does that make during the day?
    If you predict twice as many occurrences compared to overlapping or opposite hands, how confident are you of that prediction?
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Answers
  • The first occurrence is at 0hr 16 4/11 mins (00:16:22).
  • The time interval thereafter is 32 8/11 mins.
  • The 17th occasion is at 09:00:00.
  • The total occurrences per 24hrs is 44.
Times that the hands are at right-angles:

Notes:
  • See the Notes to puzzle 2 regarding the potential errors involved when quoting times to the nearest second.
  • It is geometrically impossible for all three hands to be mutually 90° apart, but there are several configurations when any two are coincident or opposite while the third is at (or nearly at) right-angles to at least one of the others, such as at 03:00:00 and 05:10:55 for example.
  • If the question is instead:  "When, after any chosen time, will the chosen angle between them be 90°?", see puzzle 9.
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Maths
Method 1 (see Basics):
When at right-angles (A = 90°) the first minute/hour angular difference D is 90°, but the next is when D = 270°.
Hence, D = |5.5M - 30H| = 90 & 270 alternately, and M = (30H + 90) × 2/11 on odd-numbered occasions, but (30H + 270) × 2/11 on even-numbered occasions.
Modular arithmetic can be used to allow for this alternation. Since (1 - Nmod2) = (1 - 1) = 0 for odd N, and (1 - 0) = 1 for even N, the Nth occurrence can be found from M = (30H + 90 + (180 × (1 - Nmod2))) × 2/11 mins. Hence:
  • during the first hour (H = 0):  M = (0 + 90 + (180 × 0)) × 2/11 = 180/11 = 16 4/11, or 16mins 22secs†.
    The time is thus 0hr + (16mins 22secs) = 00:16:22.
  • next, but still during the first hour (H = 0):  M = (0 + 90 + (180 × 1)) × 2/11 = 540/11 = 49 1/11, or 49mins 5secs†.
    The time is thus 0hr + (49mins 5secs) = 00:49:05.
  • during the second hour (H = 1):  M = (30 + 90 + (180 × 0)) × 2/11 = 240/11 = 21 9/11, or 21mins 49secs†.
    The time is thus 1hr + (21mins 49secs) = 01:21:49.
As for puzzle 3, the Nth occurrence can be calculated from I1 + ((N - 1) × IR). In this case, however, I1 = 16 4/11 mins (from H = 0), and IR (from the next pair) is 32 8/11 mins.
So, for N = 17:  the time is 16 4/11 + (16 × 32 8/11) = 540 mins, or 9hrs precisely (09:00:00).

Two right-angles form during each hour until the 2nd occurrence after H = 2, when M = (60 + 90 + (180 × 1)) × 2/11 = 660/11 = 60 mins, a whole hour, bringing the time to 2 + 1 hrs or 03:00:00. This pattern continues such that only one right-angle can occur during the 2nd & 8th hours of either half day, totalling 44 (so if you were confident that there really were 2 × 22 occurrences per day, you were correct).
The total occurrences is again found from mins per day / IR: total N = 1440 / 32 8/11, = 44.

NB:  Again, take care to use the precise figures of minutes and fractions of a minute, rather than times rounded to the nearest second.

Method 2 (see Basics):
The first occasion is when the minute/hour hand angular difference reaches 90°. Subsequent occurrences will be after the cumulative angular difference C amounts to a further 180° and multiples thereof. Hence:
  • for N = 1:  C = 90.
  • for N = 2:  C = 90 + (180 × 1) = 270.
  • for N = 3:  C = 90 + (180 × 2) = 450 etc...
  • it follows that C = 90 + 180(N - 1), and since C = T × 11/120, T = (90 + 180(N - 1)) × 120/11.
So, for N = 17:  T = (90 + (180 × 16)) × 120/11 = 32400 secs, = 09:00:00 precisely.

Total occurrences per day can again be predicted: total N = ((86400 × 11/120) - 90) / 180, = 44.

† = to the nearest second

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