# Sunday Times Teaser 2532

### by Danny Roth

#### Published: 3 April 2011 (link)

In hot pursuit George and Martha are jogging around a circular track. Martha starts at the most westerly point, George starts at the most southerly point, and they both jog clockwise at their own steady speeds. After a short while Martha is due north of George for the first time. Five minutes later she is due south of him for the first time. Then George catches up with her during their second laps at the most northeasterly point of the track.

What is Martha’s speed (in degrees turned per minute)?

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1. Considering the two speeds, George (speed $$g$$ degrees/minute) covers thirteen 45 degree segments in the same time that Martha (speed $$m$$ degrees/minute) covers eleven. Hence the relationship between their speeds is $$11g=13m$$.
When Martha is due north of George for the first time, she will be the same number of degrees ahead of due west as George is behind. Hence at time $$t_1$$ we have: $m t_1=90-g t_1$When Martha is subsequently due south of George, she is as far ahead of due east as he is behind. Hence at time $$t_2$$ we have: $m t_2-180=270 – g t_2$ Since these two times are 5 minutes apart; $t_2-t_1=\frac{450}{g + m}-\frac{90}{g+m}$This gives $$g+m$$ as 72 degrees/min and, since $$g=(13/11)m$$, we can now eliminaate $$g$$ to show that Martha’s speed is 33 degrees/minute.