1) There are three race horses, whose respective names are Red, Orange and Yellow. They are owned by three breeders after which they are named. But no horse has the same name as his owner, and each horse has been trained by the breeder who is neither his owner nor namesake. One day, each breeder rode one of the three horses. John Red rode the horse trained by the namesake of his own horse. Stan Orange rode the horse which he himself had trained. The question is: which horse did Peter Yellow ride?
2) A female mouse is at the SE corner of a square. She is in love with a male mouse at the NE corner, who loves a female mouse at the NW corner, who loves a male mouse at the SW corner, who loves the mouse at the SE corner. At a Pavlovian signal, each mouse instantly moves directly towards his/her beloved. What happens at the centre of the square, when everyone meets up, isn’t within the parameters of our problem. The question is: if each side of the square is ten feet long and each mouse travels at the same constant speed and we ignore the length/width of the mice, how far does each mouse travel before they meet up?
The answers to June’s puzzles were supplied in the June issue.
Here are the answers to July’s puzzles: 1) Work out a table with three columns. The first row is headed Mr Red, Mr Orange, Mr Yellow. The next row is for the horse owned by that person. The third row is for the horse trained by the person. The final row is for the horse ridden on the day we are considering. There is only one way to complete the table and it shows that Mr Yellow rode the horse named Orange. 2) Ten feet. The path of each mouse is always at right angles to the path of the pursuing mouse, and at any moment each mouse is at the corner of a rotating and gradually shrinking square. Undoing the rotation, it becomes clear that the motion of the pursued mouse doesn’t alter the distance that the pursuer has to travel.
