hours minutes and seconds hand will make equilateral triangle 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
 A: 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.
A: 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.
A: 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.
