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Three clocks which show the times for three different time zones are hanging on a wall. The length of their minute hands is the same - but not the lengths of their hour hands. The lengths of the hour hands of clock 1, clock 2 and clock 3 are 7 cm, 14 cm and 21 cm respectively.

The clocks are arranged on the wall in such a way that the tips of their minute hands form an equilateral triangle at any point in time.

Clock 3 shows UTC. Which time zones clock 1 and clock 2 show such that under this condition the tips of the three hour hands also form an equilateral triangle at any point in time? Hint: There may be more than one solution.

enter image description here

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As I have already noted elsewhere combining two regular triangles by cornerwise vector addition (or subtraction) yields another regular triangle. This is easy to see by subtracting the barycentres and using 120° symmetry.

It remains to show that a regular triangle can be formed with the given lengths. This is also easy to establish using the following

triangular grid:enter image description here

So if clock 3 has UTC then clocks 1 and 2 could show

UTC +2h and UTC -2h. (Order doesn't matter.) or +/- 10 or some combination

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    $\begingroup$ Nice solution using this triangular grid! I could remove the "calculation-puzzle" tag from the description! "Order does not matter" may be a bit ambiguous though. For a given clock configuration you cannot exchange the + and - signs. $\endgroup$
    – Herbert Kociemba
    Jun 4 at 11:44
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    $\begingroup$ Very nice indeed! Made a visualization in GeoGebra (adjustable g and t), looks correct 🙂 $\endgroup$
    – ThePuzzler- or rather APuzzler
    Jun 4 at 12:00
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    $\begingroup$ @HerbertKociemba I'm not sure I understand this. The longest must be in the middle, yes, But why should the other two not be interchangeable? $\endgroup$
    – loopy walt
    Jun 4 at 12:04
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    $\begingroup$ @HerbertKociemba I think I see what you mean; one has to interchange them correctly meaning that the location and the angle/time difference are tied together but one still can swap the lengths. $\endgroup$
    – loopy walt
    Jun 4 at 12:53
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    $\begingroup$ @justhalf: This is exactly my point. If you keep the clocks in place this will not work. In my picture above the indices 1, 2, 3 of the clocks go counterclockwise. Here clock 1 has UTC-2. and clock 2 has UTC+2. If you want the other solution you must arrange the indices 1,2 and 3 clockwise. Else this does not work. The two solutions are mirrors of each other. If you look into the mirror and let the time run backwards, you see the second solutions. $\endgroup$
    – Herbert Kociemba
    Jun 5 at 9:16
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Partial for now

Let us ignore the clock labels to begin with. Without loss of generality, let all clocks have the same centre. (else the answer(s) would not be independent of clock arrangement).
Now you can see the shape formed by the hour hand tips is rotated around the centre as time varies, so as soon as we find timezones which give an equilateral, we are done.
Draw 3 concentric circles with radii 7, 14, 21, and attempt to draw an equilateral with a vertex on each.

I'm not good enough at Euclidean geometry to see how one goes forward, and I'll stop here for now rather than using GeoGebra for an answer.

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    $\begingroup$ You may not assume in full generality that the solution is completely independent of the clock arrangement. If you switch clock 1 and 2 in the picture above the red triangle is not equilateral any more though the green one still is. $\endgroup$
    – Herbert Kociemba
    Jun 4 at 10:04
  • $\begingroup$ Apart from differentiating clock numbers it still appears wlog can be used to help solve given how the question is posed. It would appear to me that each triple of timezones that solve the concentric case correspond to one solution when they are non-concentric $\endgroup$
    – Shuri2060
    Jun 4 at 10:22

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