360 written examination-about the watch hands

Thinking about the algorithm of the watch pointer

we all know that the watch is three seconds, the hour hand can determine the position of the minute hand and the second hand, the minute hand can determine the position of the second hand. For the sake of simplicity, now assume only the hour and minute hands.

Now assume that a moment is h:m, (h=0,1,2 ... 11;m=0,1,2..59), then at that moment the angle of the minute hand (distance from the beginning 0 points) is:, the angle of the hour hand:,

Then the angle between the hour hand and the minute hand:

1. When is the hour and minute hand coincident?

Order, after simplification:

Thus, every hour there is a moment to meet the hour and minute coincidence, notice when h=11, m=60, that is, 12 points, that is, 0 points. All moments that meet the hour minute hand coincide are:

[0:0], [1: (05-06)]], [2: (10-11)] ... [11:60] that is [12:0] a total of 11 moments

Since the seconds are not considered, as in [1: (05-06)]), the hour hand and the second hand are completely coincident between 05 and 06 minutes.

Question 2: How long is the hour and minute swap positions?

Moment 1 is: h1:m1, Moment 2 is: h2:m2, make d1=d2, suppose h2>h1, then must have m1<m2. Simplification to:

Also, when the needle runs into minutes: the minute hand runs to the hour hand:

, simplifying the arrangement of:

There should be more than one answer to this moment, but when the h2-h1=2,360 written in a given case is two hours, how many minutes has it been? Ternary one-time equations should be better solved?

360 written examination-about the watch hands