# one more clock related challenge!

Find the exact time between $2.25$ pm and $2.26$ pm when the second hand bisects the angle between the hour hand and the minute hand.

I mention below my solution to this question. I would like to know whether my solution is correct and also are there any other ways to solve this question.

As the second hand bisects the angle formed by the hour and minute hand: angle between hour and second hand $=$ angle between second and minute hand.

Let $x$ be the number of seconds from $2.25$ to $2.26$ when angle between hour and minute hand is bisected.

Hour hand rotates through $\frac{1}{120}$ degrees per minute. As angle between hour and second hand $=$ angle between second and minute hand.

After $x$ seconds hour hand from the beginning of the hour will rotate through- $60$ degrees $+$ $\frac{x}{120}$ degrees.

After $x$ seconds minute hand from the beginning of the hour will rotate through- $150$ degrees + $\frac{x}{10}$ degrees.

As angle between hour and second hand = angle between second and minute hand: $6x$ degrees $-$ $(60 + \frac{x}{120})$ degrees $=$ $150$ degrees $+$ $\frac{x}{10}$ degrees $-$ $6x$ degrees

....Simplifying we get: $x= \frac{25200}{1427} \approx 17.659$ seconds.

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This looks fine to me. The only error is when you say $\frac{25200}{1427} = 17.659$, which is manifestly false. – TonyK May 16 '12 at 11:15
@TonyK sorry but I did not get what you said. – mgh May 16 '12 at 12:35
TonyK is pointing out that it is not an equality because $\frac {25200}{1427}$ is not a terminating decimal. Your last = should be $\approx$ – Ross Millikan May 16 '12 at 12:55
@RossMillikan thanks for the clarification! – mgh May 16 '12 at 13:33

You are right except for the the conversion to decimal, which is not exact. Artin has fixed that by changing the $=$ to $\approx$. It depends on the department-math insists that $=$ is exact equality, while others accept it as "reasonably close".