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The watch in the picture below contains a square, around which the hours are marked. At 3:00 the area enclosed between the hour hand, the minute hand, and the square sides is ¼ of the total area of the square. At 9:00 it is also ¼ of the total area.

How many times during 12 hours does the area enclosed by the hour hand and minute hand equal exactly ¼ of the square?

enter image description here

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Answer

22 times

Explanation

Note that the area covered by by two hands is one fourth of the square if and only if they are perpendicular ie. make a $90°$ angle, which happens 22 times on 12 hours.

Proof

A) The area covered by two hands is one fourth of the square if and only if they are perpendicular.

enter image description here

Let $OF$ and $OG$ be the hands. Let $h$ be the distance from center to the sides.

Now,
$[OFAG]= {1 \over 4} [ABCD] = [AOB]$
$\implies [AOF]= [GOB]$
$\implies AF=GB$ (since the area and heights are equal)

Since, $\angle OBG =\angle OAF= 45°$ and $OA=OB$
the triangles $AOF, BOG$ congruent and

$\angle AOF =\angle GOB$
$\implies \angle GOF=\angle GOA + \angle AOF = \angle GOA +\angle GOB =\angle AOB =90°$


B) The clock hands are perpendicular 22 times.

The hour hand makes 1 revolution per 12 hours. The minute hand makes 12. So, if we observe the minute hand from the perspective of the hour hand, we would see that it makes 12-1=11 revolutions per 12 hours. In other words, the minute hand crosses or overlaps the hour hand 11 times. Between each two crossing, there are two cases where they are perpendicular which makes 22 times per 12 hours.

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    $\begingroup$ Please reconsider your answer. Remember that we have a square, so part of your claim with regard to proportionality to the angel between the hands is not right. Your claim would have been true for a circle shape. It is also not 22. $\endgroup$ – Moti Aug 19 '15 at 6:49
  • $\begingroup$ Brilliant edit. I was thinking of the same proof to insist that the area will be one-fourth of the square regardless of the geometrical shape. $\endgroup$ – CodeNewbie Aug 19 '15 at 7:10
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    $\begingroup$ I've gone over this and this should be correct (although you could get 23 if you start and end on a "hit"). Regardless of the shape of the watch face, there will be 2 times for each rotation of the minute hand around the hour hand that the enclosed area is hit. For a square, this is always perpendicular, which you can demonstrate by observing a 90* corner rotating from its centre; any area you "lose" by rotating, you gain the same amount on the leading edge. $\endgroup$ – Samthere Aug 19 '15 at 9:11
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    $\begingroup$ I suppose it would depend on whether the minute hand (and/or hour hand) moves smoothly around the clock, or "ticks" from one minute to the next. If it moves in discrete intervals, there are probably many less instances, because it might "tick" right past the location that would create a perfect 90° angle. $\endgroup$ – GentlePurpleRain Aug 19 '15 at 15:20
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    $\begingroup$ IT is 22 times. $\endgroup$ – Moti Aug 19 '15 at 19:00
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It's bound to happen once between two consecutive $(60k/11)*12$s (in mins) after (15/11)*12 mins (11 instances). In addition, "going backwards" from such an instance where both arrows overlap, it also happens (15/11)*12 mins ago (11 instances). It works the exact same way on a square as it does on a circle: Let's say that the two arrows are orthogonal and perpendicular to each other. We can choose the closest configuration to the one we have, so we don't need to rotate it all out of the quarter. To keep the area constant, we have to rotate both arrows an equal amount because of the symmetry, meaning the angle must always be 90 degrees (15 mins). Therefore it's 22, the total.

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I am counting 24 possibilities.

Like Anachor already said, both hands must be perpendicular, and his 22 are correct. But i think there are 2 more:
1. nearly before 3 o'clock, sth. like 02:58 or 02:59 where the hands are perpendicular
2. and the same nearly before 9 o'clock, so like 08:58 or 08:59.

So in both cases there are two appearances of the quarter within of two minutes.

I hope i've made no logical error. If so, please be kind in the comments :-)



Edit: Forget my answer, Roland is totally right.

For every minute the hour hand is moving by 1/720. So at 2:59 it will be at 179/720 and the minute hand will be at 708/720, that is a difference of 191. The difference must be exactly 180 for them to be perpendicular.

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    $\begingroup$ For 1: If the minute hand (pointing up) is approaching the hour hand (pointing right), the angle between the two is decreasing constantly between 2:50 and 3:15, for example. Doesn't this mean that the angle between the two can only be 90 for one time, 3:00? In other words, the angle must be larger than 90 at 2:59. $\endgroup$ – Roland Aug 19 '15 at 14:51

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