A clock with twin hands
A clock has an hour hand and a minute hand that look exactly alike, and both move continuously. Usually you can still tell the time, because only one of the two ways of reading the hands gives a valid position.
In a 12-hour period, at how many moments is it impossible to tell the time, because reading the hands the other way round also gives a valid, different time?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Measure positions in full turns. At time hours, the hour hand is at and the minute hand at , both modulo 1.
The hour hand alone determines the time. So if the hands are swapped, the new time is fixed by the old minute hand's position. When does the new minute hand land on the old hour hand?
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s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. The hour hand's position determines the time completely. So a swap of the hands gives a valid time only when a simple equation holds, and we count its solutions in 12 hours, leaving out the moments when the two hands are on top of each other.
Step 1: positions. Measure angles in full turns. At time hours, the hour hand is at and the minute hand at , both taken modulo 1.
Step 2: the swap. Reading the minute hand as the hour hand means a time with , that is . That reading is valid if the minute hand at time , which is at , coincides with the old hour hand : So for : 143 moments.
Step 3: overlaps are not ambiguous. When the two hands are on top of each other, swapping them gives the same time, so the time can be read. The hands overlap when , that is when is a whole number: 11 moments in 12 hours. They are among the 143, for a multiple of 13.
Step 4. That leaves moments when the time cannot be told.
Example. The first one comes just after 12:05: at hours, about 12:05:02, the hands could also be read as about 1:00:25.