Decisions - Futility Closet

A puzzle by David Silverman:

Able, Baker, and Charlie are playing tag. Able is faster than Baker, who’s faster than Charlie. All three of them start at point P, and Able is “it.” At time -T, Baker runs north and Charlie runs south. After a count that takes time T, Able starts chasing one of the two quarries. The game ends when Able has tagged both Baker and Charlie. If Baker and Charlie maintain their speeds and directions, who should Able chase first to minimize the time required to make the second tag?

	Select Click for Answer
Let a, b, and c be the respective speeds of Able, Baker, and Charlie, and let x and y be the times required for Able to catch the first and second person. If Able chases Baker first, then $\displaystyle x=\frac{bT}{a-b} \ \mathrm{and} \ y=\frac{aT(b+c)}{\left ( a-c \right )\left ( a-b \right )}$ If Able chases Charlie first, then $\displaystyle x=\frac{cT}{a-c} \ \mathrm{and} \ y=\frac{aT(b+c)}{\left ( a-c \right )\left ( a-b \right )}$ “Since both y’s are equal, the x’s show that the least time will elapse when Able chases Charlie first.” (From Pi Mu Epsilon Journal 6:5 [Fall 1976], 317.)

Click for Answer