This week the #puzzle is: Will You Be Right on Time? #angles #clock #RightAngles #geometry
| A standard analog clock includes an hour hand, a minute hand, and 60 minute markers, 12 of which are also hour markers. |
| At a certain time, both the hour hand and minute hand are both pointing directly at minute markers (either of which could also be an hour marker). The hour hand is 13 markers ahead (i.e., clockwise) of the minute hand. |
| At what time does this occur? |
And for extra credit:
| At various times of day, the minute and hour hands form a right angle. But is there a time of day when the hour hand, minute hand, and second hand together form two right angles, as illustrated below? If you can find such a time or times, what are they? |
![]() |
| If you cannot find any such times, suppose the measures of the two nearly right angles formed by the three hands measure A and B degrees. What time or times of day minimize the square error function f(A, B) = (A − 90)2 + (B − 90)2? |
| Either way, your answer should be precise to at least a thousandth of a second. |

Solution, possibly incorrect:
We can have any time from 00:00 to 11:59.
The time:
So many minutes have passed since 00:00:
It takes 720 minutes for the hour hand to go all the way around the clock.
When the hour hand points directly at the nth minute marker:
We also have some extra knowledge about the distance between the hands:
There are 2 ways to compute where the hour hand is.
Method 1, example:
Method 2, example:
As these 2 methods don’t result in the same n for the example, this isn’t the time we’re looking for.
As we have, from method 1:
n is an integer, so m is a multiple of 12. Further, from method 1:
This leads to MM being a multiple of 12. So:
Then we use method 2 to find n:
| MM | 00 | 12 | 24 | 36 | 48 |
| n | 13 | 25 | 37 | 49 | 1 |
And method 1, backwards, to find m:
| MM | 00 | 12 | 24 | 36 | 48 |
| n | 13 | 25 | 37 | 49 | 1 |
| m | 156 | 300 | 444 | 588 | 12 |
And then find HH:
| MM | 00 | 12 | 24 | 36 | 48 |
| m | 156 | 300 | 444 | 588 | 12 |
| HH | 2.6 | 4.8 | 7 | 9.2 | 11.4 |
Only the time 07:24 gives an integer HH. So this is the solution.
And for extra credit:
I write a program to go through a lot of different options. I fine tune the program to find better and better solutions, with better and better error functions. The results are:
Good time, in seconds..: 13744.07855
-- Good time..........: 03:49:04.07855
-- Error function.....: 0.0079550
Good time, in seconds..: 29455.92145
-- Good time..........: 08:10:55.92145
-- Error function.....: 0.0079550
So the best possible times are 03:49:04.079 and 08:10:55.921.


