536 Puzzles: Problem & Solution #282: Those Russian Cyclists Again

Published by Fudgy McFarlen on January 8, 2017January 8, 2017

The solutions giving in the book 536 Puzzles and Curious Problems are quite shorthanded and in order to understand them often you have to dig.

P^2=(a+b+c)^2

P^2=a^2+b^2+c^2+2ab+2ac+2bc

Since in a right triangle: c^2=a^2+b^2 and area A=ab=hc we can substitute and get

P^2=2c^2+2ac+2bc+2hc

P^2=2c(a+b+c)+2hc

c=P^2/(2P+2h) c=60^2/(2*60+2*12) = 25

Now since A=ab=hc=12*25=300 we can substitute in c^2=a^2+b^2

c^2=a^2+(300/a)^2 which is rearranged to:

a^4-a^2c^2+300^2=0

a^4-625a^2+300^2=0 giving

a=15 and b=20

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