A Problem of time

528 /www/nrich/html/content/97/10/six1/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><blockquote> "It was the best butter," the March Hare meekly replied. "Yes, but some crumbs must have got in as well," the Hatter grumbled: "you shouldn't have put it in with the bread-knife." The March Hare took the watch and looked at it gloomily: then he dipped it into his cup of tea, and looked at it again: but he could think of nothing better to say than his first remark, "It was the best butter, you know." </blockquote> <table style="" border=""> <tbody> <tr valign="top"> <td style="" valign="top" width="380"><mdo:image align="left" alt="Tea Party" src="p72-300.gif"></mdo:image></td> <td style="" width="210"> Consider a watch face which has all the hours shown by the same mark, and both hands the same in length and shape. It is opposite to a mirror. Find the time between 6 and 7 when the time as read direct and in the mirror is exactly the same. </td> </tr> </tbody> </table> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Tim sent us his excellent answer to this problem: The time is approximately 27 minutes 42 seconds past 6 o'clock. At six o'clock the minute hand and the hour hand are exactly 180 degrees apart. Let the hour hand move through x degrees. Then, if the time is reversible in a mirror, the minute hand has moved through (180 - x) degrees. The hour hand moves at 30 degrees per hour. The minute hand moves at 360 degrees per hour. The time elapsed during which both hands move is identical. It follows that $\frac{(180 - x)}{360} = \frac{x}{30}$ x = $\frac{180}{13}$ Hence, this angle represents $\frac{180}{13}$ x $\frac{1}{360}$ x 60 minutes on the clock face. This reduces to 2$\frac{4}{13}$ minutes (approx. 2 minutes and 18 seconds). Therefore the time is 30 - 2$\frac{4}{13}$ minutes past 6 o'clock; i.e. 27$\frac{9}{13}$ minutes past 6 o'clock. Laura (West Flegg Middle School, Great Yarmouth, Norfolk) came very close to this answer. Her solution was 28 minutes past 6 o'clock. </mdoxml> 2 5 0 0 0 1 0 A Problem of time Consider a watch face which has identical hands and identical marks for the hours. It is opposite to a mirror. When is the time as read direct and in the mirror exactly the same between 6 and 7? Algebra Linear equations Using, Applying and Reasoning about Mathematics Visualising Transformations and their Properties Reflections 2D Geometry, Shape and Space Angles Measures and Mensuration Time Transformations and their Properties Symmetry