When I am going through some aptitude questions I have got this problem

How many times the hours minutes and seconds hand will make equilateral triangle in 12 hours of clock

I can't understand how they form the equilateral triangle as they can't be the sides of triangle, may be they should be median. If so I am not able to solve when it happens

Can Anyone help me

Edit: Assuming the hands are of equal length may make the problem easier

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    $\begingroup$ Do you mean the triangle determined by the extreme points of the three clock-hands? $\endgroup$ – DonAntonio Nov 20 '13 at 17:04
  • $\begingroup$ @DonAntonio May be as to my knowledge that is the only case where we can form a equilateral triangle using the three hands $\endgroup$ – sai kiran grandhi Nov 20 '13 at 17:08
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    $\begingroup$ @sai Maybe it means that the tips of the hands form the vertices of the triangle? At first I thought the medians, as you did, but this would seem to assume the arms are all the same length. If they are not all the same length, then it would seem like an answer would heavily depend on the lengths, and upon what they mean by the hands forming an equilateral triangle. $\endgroup$ – rschwieb Nov 20 '13 at 17:08
  • $\begingroup$ @rschwieb To my knowledge if the hands are of different length, then an equilateral triangle is not possible. Is there any other case of forming the triangle rather than by tips $\endgroup$ – sai kiran grandhi Nov 21 '13 at 7:03
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    $\begingroup$ @saikirangrandhi But I'm guessing the question wants the arms to be all the same length... otherwise the question is too vague. $\endgroup$ – rschwieb Nov 21 '13 at 11:16

This kind of problem can be easier to handle if you rotate the clock backwards just fast enough to stop the hour hand. Then the minute hand will be seen to rotate at $330^{\circ}$ per hour, and the second hand at $60 \times 360 - 30 = 21570^{\circ}$ per hour.

The minute hand is at the $120^{\circ}$ position at $\frac{360m+120}{330}$ hours, and the second hand is at the $240^{\circ}$ position at $\frac{360s+240}{21570}$ hours.

For these times to coincide, we need integers $m$ and $s$ such that

$$\frac{360m+120}{330} = \frac{360s+240}{21570}$$

But this simplifies to

$$2157m + 703 = 33s$$

which is impossible because the rhs is divisible by 3 but the lhs isn't.

By symmetry (i.e. you run the film backwards in a mirror), the reflected position (minute hand at $240^{\circ}$, second hand at $120^{\circ}$) is also impossible.

  • $\begingroup$ +1 I did it Lubin's way, but this is way nicer, with the change of coordinates. A couple of notes. If one considers a point to be a "degenerate triangle" then there would be a solution at 0h00m00s (but I wouldn't say so). If the hands are not equal then there will be solutions, for some sets of time and lengths - I checked with Maxima but the general solution seems to be a set of very large equations. $\endgroup$ – Rolazaro Azeveires Nov 23 '13 at 9:45
  • $\begingroup$ "If the hands are not equal then there will be solutions..." That is certainly not guaranteed. $\endgroup$ – TonyK Nov 23 '13 at 14:21
  • $\begingroup$ Yes, it is not guaranteed. I said, there will be "for some sets of time and lengths" $\endgroup$ – Rolazaro Azeveires Dec 14 '13 at 1:04
  • $\begingroup$ TonyK: Nice proof. But one error: 60 360 - 30 is 21570, not 21750. No consequences though. I am told that this problem was known to college students in England in 1964, and probably earlier. $\endgroup$ – stan wagon Dec 22 '18 at 16:01
  • $\begingroup$ @stanwagon: Thanks for pointing that out. I have corrected it now. $\endgroup$ – TonyK Dec 22 '18 at 16:46

A reasonable reading of the question is whether there are any times where the hands are spaced at angles of $120^\circ$. All angles will be in degrees. The hour hand moves $\frac 1{120}$ per second, the minute hand moves $\frac 1{10}$ per second and the second hand moves 6 per second. Starting at noon, the minute hand gains $\frac {11}{120}$ per second on the hour hand, so will be $120$ ahead in $1309\frac 1{11}$ seconds. The hour hand has moved $10\frac {10}{11}$ in that time, the minute hand $130\frac {10}{11}$ and the second hand $21$ revolutions plus $294 \frac 6{11}$, so an equilateral triangle is not formed. There will be two times every hour when the hour and minute hands are $120$ apart. You can check the rest of them, but I would be surprised if it works.

  • $\begingroup$ We’re in agreement on this! $\endgroup$ – Lubin Nov 22 '13 at 18:37
  • $\begingroup$ You're just a pair of lazybones, is what I say :-) $\endgroup$ – TonyK Nov 22 '13 at 19:17

Here’s a strategy, not a solution, always assuming that the hands are of equal length and the “triangle” in question is formed by the tips. The desired condition happens when the angles between different hands are $120^\circ$. Now it’s a standard high-school algebra problem to determine when there’s a $120^\circ$ angle from the hour to the minute hand, it happens $11$ times in each $12$-hour period if the angle is positive, $11$ more if the angle is negative. I’d find these $22$ times and see whether the second hand is properly positioned at each. My wager is that it never happens.


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