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

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

Additional Links

Categories: MathPuzzles