Nice Algebra Problem || Tricky Maths Olympiad Question

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[musicsubicandcebu]

a² = ab*ca/bc = 3000/2
b² = ab*bc/ ca = 2000/3
c² = bc*ca/ab = 6000

[himo]

c=2a 2a²=3000 a²=1500
c=3b 3b²=2000　　 b²=2000/3　　　　
c²=4a²=6000
a²+b²+c²=24500/3

[walterwen]

Nice Algebra Problem: ab = 1000, bc = 2000, ca = 3000; a² + b² + c² =?
(ab)(bc)(ca) = (1000)(2000)(3000), (abc)² = 6(10⁹)
a² = [(abc)²]/[(bc)²] = [6(10⁹)]/(2000²) = (3/2)(10³)
b² = [(abc)²]/[(ca)²] = [6(10⁹)]/(3000²) = (2/3)(10³)
c² = [(abc)²]/[(ab)²] = [6(10⁹)]/(1000²) = 6(10³)
a² + b² + c² = (3/2)(10³) + (2/3)(10³) + 6(10³) = (3/2 + 2/3 + 6)(10³)
= (49/6)(10³) = 49000/6 = 24500/3

[cyruschang]

ab = 1000
bc = 2000
ça = 3000
a^2 + b^2 + c^2 = 3000/2 + 2000/3 + 6000 = 8166 + 2/3

[CNunez-vnhr]

Let me suggest a different approach. First, solve for b in terms of a from the first equation, and solve for c in terms of a from the third equation. Next, substitute b and c in terms of a in the second equation to get the value for a. Last step, use the value of a to get the values of b and c, substitute these values in the sum of squares, and voila!:). Hope this helps.

[Nuncamasusado]

Muy buena la form de abordar el ejercicio y mejor la conclución, gracias

[риколай-щю]

зачем так сложно сразу найти a²=1500 b²= c² =

[jerryhopfe]

Simpler to start with abc =1000c, abc =2000a, abc=3000b, then subtract 1000c-2000a to find that c=2a. Similiarly we find that b = 2/3 a Then since ca= 3000, then a = the square root of 1500. Of course the rest of the problem is trivial.