Puzzle page

Issue 53
in
December 2009

Clock confusion

We all know that homemade presents are supposed to be the best, but the clock your Aunty Mabel made you is a little hard to get used to — the hour and minute hands are exactly the same! You can muddle through most of the time but sometimes, say 26 minutes past 2 or just after 12 minutes past 5, you can't tell which of the two times it is. In the 12-hour period between noon and midnight, how many moments are there when it is not possible to tell the time on this clock?

This puzzle was taken from Norman Do's regular Puzzle Corner column for the Gazette of the Australian Mathematical Society. Why not try your hand at the problems in the latest Puzzle Corner?

If you are stumped by last issue's puzzle, here is the solution.

For some challenging mathematical puzzles, see the NRICH puzzles from this month or last month.

Clock confusing

The confusing clock.

Aunt Mabels clock shows 132 situations you can’t see what time it is. How do we calculate these 132 possibilities?
Take two normal clocks. Let the angle between the hour hand and the vertical be φ1 and between the minute hand and the vertical be α1 for the first clock and for the second clock φ2 and α2 respectively. The hour hand covers in 1 minute an angle of 6°, the minute hand covers in that time ½°. Let the time clock1 shows be a1 minutes past n1 o’clock and let clock 2 point to a2 minutes past n2 o’clock. So we can write:
φ1 = 6a1 and
α1 = 30n1 + ½ a1
(the minute hand covers 30° in 1 hour, so 30n1° in n1 hour and ½ a1° in a1 minutes). We obtain for the second clock:
φ2 = 6a2 and
α2 = 30n2 + ½ a2.
Aunt Mabels clock will confuse us if
φ1 = α2 and
φ2 = α1.
This gives
6a1 = 30n2 + ½ a2 and
6a2 = 30n1 + ½ a1.
From the first equation:
a1 = 5n2 + (1/12)a2 (i).
Fill in into the second equation:
6a2 = 30n1 + (5/2)n2 + (1/24)a2 or
a2 = (720/143)n1 + (60/143)n2 (ii).

With i and ii we can calculate all possibilities.
An example: take n1 = 4 and n2 = 8. ii gives a2 = 23,50 and i gives a1 = 41,96. Aunt Mabels clock can’t decide between 41,96 minutes past 4 and 23,50 minutes past 8.

Remark 1: You can easily find the time the hands are on top of each other:
a = (60/11)n.
Remark 2: You can uniquely determine φ when you know α.
φ = 12(α – 30n), n such that 0 ≤ α – 30n < 30 .

Hub Boreas

Clock confusing

I inadvertently swapped the names of the hands: minute hand <-----> hour hand

Hub Boreas