I Counting problem for fun

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Summary
interesting counting problem for fun
Summary: interesting counting problem for fun

Imagine we draw a circle with diameter d and mark off sixty equal intervals like minutes on a clock. Then we draw two diameters perpendicular to one another and divide each in sixty equal intervals. Using the intervals on the diagonals we lay out a Cartesian plane with corresponding grid. Then we start one of the diameters moving like the minute hand of a clock. And like a minute hand the diameter stops at each point we marked off on the circle. As the diameter makes one revolution what is the sum of corners of the grid it intersects with at all of the stops. If we divided each interval of our plane ten times how many? 100 times? Is there an expression to reflect the number of intersections each time we increase the intervals by one order of magnitude.
 

.Scott

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If I understand the problem, only on certain stops will the minute hand land on any corners at all. And when it does land on a corner point, it is likely to land on many at a time.

At all stops, it will touch the center point - so that's 1. When the minute hand is at the 12:00/6:00 position, it will touch another 60 corners. At 3:00/9:00 is will touch another 60. At 1:30/7:30 and 4:30/10:30 it will touch another 42 each.

But then we get into the problem of finding rational solutions to ##cos(\pi y)## where ##y## is rational. I don't believe there are any such solutions beyond what I have already listed.

So the tally will stand at 205.
 
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Thank you for replying, Scott. However, I don’t understand 1:30/7:30 designation as the countable corners are onlybat the stops of the minute hand and at these times the minute hand is not at a stop.
 

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