Two circles, one of radius 5 inches, the other of radius 2 inches, are tangent at point P. Two bugs start crawling at the same time from point P, one crawling along the larger circle at 3 inches per minute, the other crawling along the smaller circle at 2:5 inches per minute. How many minutes is it before their next meeting at point P?
Source: MathMovesU MATH CHALLENGE PROBLEM - 2012/01/10
The circumferences of the circles are 2*pi*5 = 10*pi and 2*pi*2 = 4*pi, so the time in minutes taken the bugs to crawl one
circumference of their respective circles are 10*pi/3 nd 4*pi/2.5. We are to find the least ordered pair of integers (m, n)
such that (10*pi/3)m = (4*pi/2.5)n. Multiplying the last equality by 7.5/pi, we obtain 25m = 12n. Clearly
25m = 12n = lcm(25, 12) = lcm(5^2, (2^2)*3) = (2^2)*3*(5^2) = 300, and so m = 300/25 = 12. Hence the bugs next meeting at
point P is (10*pi/3)*12 = 40*pi = 125.66 minutes later.
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