302 // Professor Stewart’s Superlative Sneaky Solutions
Remember, the avatars ‘watch over’ every other cushion. Otherwise
........................................... you could make the mat bigger.
Target Practice Tuck hit the outer ring (light grey), while Robin hit the three inner rings (dark grey).
The rings Robin and Tuck hit.
The areas of successive circles are pr2, where r ¼ 1, 2, 3, 4, 5, respectively: p,
4p, 9p, 16p,
25p
The areas of the rings are the differences between these: p,
3p, 5p, 7p, 9p
These are p multiplied by consecutive odd integers. Robin’s integers are less than or equal to Tuck’s, since Robin’s arrows are closer to the centre. The two sets of odd integers must have the same sum. The only possibility is 1 þ 3 þ 5 ¼ 9. Bonus point: Six rings. The sixth ring has area 11p, so 1 þ 3 þ 7 ¼ 11 is a second solution. Further bonus point: Eight rings. Robin’s odd numbers must be consecutive, and so must Tuck’s. The next two rings have areas 13p, 15p, and 3 þ 5 þ 7 ¼ 15 is a second solution with consecutive odd
........................................... numbers.