Symmetric Time
At some time close to a quarter past two, the hour and minute hands of the clock are mirror images of each other, with respect to the horizontal line through 3 o'clock.
What is the precise time that this happens? Round off to the nearest second; give the answer in the form HMMSS . (For instance, if the answer were 2:12:41, type 21241.)
Extra challenge : How many times does a symmetric situation like this occur in a 12-hour period?
The answer is 21828.
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4 solutions
Can you post the "extra challenge" separately? I believe that the answer would surprise many people.
I can't understand. Do you have any other nice solution.
Another solution for @Utku Demircil :
Let the hour hand be ϕ degrees above "3 o'clock" and the minute hand ϕ degrees below "3 o'clock".
Also, suppose the time is t hours after 2 o'clock.
Because every hour corresponds to 3 6 0 ∘ / 1 2 = 3 0 ∘ , the position of the hour hand must satisfy ϕ = 3 0 ∘ ⋅ ( 1 − t ) . Because every minute corresponds to 3 6 0 ∘ / 6 0 = 6 ∘ , the position of the minute hand must satisfy ϕ = 6 ∘ ⋅ ( 6 0 t − 1 5 ) . Equating the two yields 3 0 ∘ ⋅ ( 1 − t ) = 6 ∘ ⋅ ( 6 0 t − 1 5 ) ; 1 2 0 ∘ = 3 9 0 ∘ ⋅ t ; t = 3 9 0 ∘ 1 2 0 ∘ = 0 . 3 0 7 6 9 2 . . . hours after 2 o’clock ; multiplying by 60 gives 1 8 . 4 6 1 5 . . . minutes; multiplying the 0 . 4 6 1 5 . . . minutes by 60 gives 2 7 . 6 9 2 . . . seconds, showing that the time is 2 : 1 8 : 2 8 .
Hope that helps...
2 0 − x = 1 2 x
x = 1 8 1 3 6
Thus, 2 : 1 8 : 2 8
At 2:15 the HH(hour hand) will be 22.5 degrees behind horizontal.
MH will be horizontal.
In m minutes, HH will be m/2 degrees from its position, that is 22.5- m/2 degrees behind horizontal.
In m minutes, MH will be 6m degrees from horizontal.
So both are inclined at same angle when 22.5 - m/2= 6m.
So m=3.462 minutes. implies 3m and 28s.
Answer 2:15:00+0:03:28=21828
The hour hand moves 360o in 12hr - to convert to seconds we get (360o/12hr) (1 hr/60min) (1 min/60 sec) = 1/120 of a degree per second The min hand moves 360o in 60 min - or (360o/60min)*(1 min/60 sec) = 1/10 of a degree per second
At 2:15, the hour hand is at 67.5o (1/4 of the way between 2 and 3, plus the 60o between 12 and 2, = 60 + 30/4; The second hand is at 90o (evenly on the 3) Thereafter the position of the hour hand after x seconds is: 67.5 + (1/120)x The position of the minute hand after x seconds is: 90 + (1/10)x
The number of degrees past 90 for the minute hand is purely x/10 The number of degrees before 90 for the hour hand is 90-(67.5+x/120) = 22.5-x/120
If we set those two equations equal, then all we need to do is solve for x: x/10 = 22.5-x/120 rearrange: x/120 + x/10 = 22.5 multiply by 120: x + 12x = 2700 simplify: 13x = 2700 divide by 13: x = 207.692308 ~ 208 seconds.
208 seconds = 180 + 28 seconds, or 3min 28sec Since we started at 2:15, we need to add 3:28 to that time = 2:18:28.
Let x be the time, in hours. This is also the position of the hour hand h = x .
Because the minute hand travels 12 times faster than the hour hand, and pointed to zero (12 o'clock) at 2 o'clock, the position of the minute hand is m = 1 2 ( x − 2 ) .
Since the two hands are mirror images around 3 o'clock, we have h + m = 2 ⋅ 3 = 6 . We are now ready to solve: x + 1 2 ( x − 2 ) = 6 ∴ 1 3 x − 2 4 = 6 ∴ x = 3 0 / 1 3 = 2 1 3 4 .
Converting the fraction to minutes: 6 0 × 1 3 4 = 1 8 1 3 6 ; and the remaining fraction to seconds: 6 0 × 1 3 6 = 2 7 1 3 9 . Thus the time is h : m m : s s = 2 : 1 8 : 2 8 .
Extra challenge
In general, a symmetric situation of this kind happens 13 times in a 12-hour period, evenly spread out. The most obvious case is precisely six o'clock. The other 12 solutions are obtained by adding or subtracting multiples of 55 minutes, 2 3 1 3 1 seconds.