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I am really curious about ways to tackle this since it is such an open ended problem.

I thought about it purely visually and after an embarrassingly long time figured out the answer to be

\begin{align*} 44 \end{align*}

I would love to hear how others think about it! I have also found a neat "rigorized" solution I would be happy to share.

operatorerror
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  • that is not embarrassing! to most people, these kinds of specific problems tend to be easily but slowly solvable – M. Van Aug 28 '16 at 03:45
  • I think that's what makes them frustrating. I think it's easy to hide behind lots of machinery in higher math – operatorerror Aug 28 '16 at 03:46
  • You could sit in front of a clock for 24 hours and count. – Will Jagy Aug 28 '16 at 03:48
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    not the most elegant, but hey, that's just like, my opinion – operatorerror Aug 28 '16 at 03:50
  • Yes, 'higher math' (like category theoretic constructions, abstract group theory etc.) tends to be free of conventions, and so, in the end, when all the fear of abstraction settles, is easier than this kind of problems. This kind of problems ask you: given this and this particular convention, what can you say about that? – M. Van Aug 28 '16 at 03:50
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    @Will Jagy this is the most awesome answer I have see ever ..oll – Sathasivam K Aug 28 '16 at 04:58
  • Answer may be 48 – Sathasivam K Aug 28 '16 at 04:59
  • @SathasivamK it is not – operatorerror Aug 28 '16 at 05:40
  • Well,its 44...but how? – Sathasivam K Aug 28 '16 at 05:42
  • @SathasivamK check the linked answer, some of those are quite good – operatorerror Aug 28 '16 at 05:42
  • I check it already.but for every hours you have 2 right angle then totally 48 .but they give answer as 44 . I'm confused.@qbert – Sathasivam K Aug 28 '16 at 05:44
  • what about for 3 oclock or 9 oclock? – operatorerror Aug 28 '16 at 05:46
  • (1/2) Although this seems quite simple at first, it's deceptively tricky to conceptualise. Try thinking about how many times the hands cross.

    As was stated in the other question, you can reason that since the hour hand, $H$, rotates once in a $12$ hour period but the minute hand, $M$, rotates $12$ times, you can say that from the perspective of $H$, there are just $12-1=11$ rotations of $M$. So $H$ and $M$ cross $11$ times...

     – Jam Sep 03 '16 at 00:18
  • (2/2) You could also reason this as finding the places where the tips of $H$ and $M$ cross on a clockface on the unit circle centre $(0,0)$. i.e. solving $(\sin(t),\cos(t))=(\sin(12t),\cos(12t))$, which gives you $11$ solutions: $t\in{0,\frac{2\pi}{11},\dots,\frac{11\cdot2\pi}{11}}$.

    For each of our solutions where the hands cross, $M$ is perpendicular to $H$ exactly twice, when it is on the left and when on the right. Hence, our total number of solutions is $2\cdot2\cdot11=44$.

     – Jam Sep 03 '16 at 00:21

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