Spain | A Nice Algebra Problem | Math Olympiad

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[yukigo]

(abc)² = 100 * 200 * 300 =
abc = ±1000√6
a + b + c = abc/bc + abc/ac + abc/ab
= abc(1/200 + 1/300 + 1/100)
= abc/100(1/2 + 1/3 + 1)
= ±(55√6)/3

[MarcoPolo-xute]

It's a nice problem, but I guess you could have solved it faster if you had first isolated b, this way: b = 100/a; b = 200/c; c = 2a. I found a = 5 sqrt 6; c = 10 sqrt 6; b = (10/3) sqrt 6; a+b+c = (55/3) sqrt 6; but I did not have to deal with big numbers, like root of 6050. I wonder if my solution has something wrong, but I think it's easier. Anyway, thanks for the problem.

[ericrobin]

AB=100
B=100/A so from BC=200 we have 100C/A=200 ie C=2A
Then from CA=300 we have 2A^2=300 ie A=sqrt(150)
We can deduce that:
C=2sqrt(150) since C=2A and

[nenedillats]

After 3:14 (the value of a^2, b^2 and c^2), you can continue by :
ab > 0, bc > 0 and ac > 0 so a, b and c are positives.
And now you know the value of a, b and c
i.e the value of a + b + c.
It's faster.
Your way after 3:14 is beautiful, I appreciate your idea.

[Yuri_Kravets]

Исходные уравнения:
1. a *b = 100
2. b *c = 200
3. c *a = 300
Шаг 1: Найдем b через a
b = 100/a
Шаг 2: Подставим b в уравнение (2)
100/a *c = 200
100c = 200a
c = 2a
Шаг 3: Подставим c = 2a в уравнение (3)
c *a = 300
(2a) *a = 300
2a^2 = 300
a^2 = 150
a = √150= 5√6
Шаг 4: Найдем b через a
b = 100/a = 100/5√6= 20√6
Шаг 5: Найдем c через a
c = 2a = 2 *5√6= 10√6
Итак, решение:
a = 5√6
b = 20√6
c = 10√6

[tonytsang]

(1) ab=100
(2) bc=200
(3) ca=300
Obviously, a=/=0, b=/=0, c=/=0, just simply (1)/(2) get c=2a, then substitute to (3) get 2a^2=300, no need for that complex solution

[ebEliminator]

I started with, [ab = 100, bc = 200], [200 = 2 * 100], [bc = 2ab], cancel out the b's so that [c = 2a], then plug it into ca = 300 so that 2a^2 = 300 and solve for a.

[ЧувакИзКосмоса]

But if you solve it isolating every value, you'll get only the positive answer.

[joenorsworthy]

I was imagining a longer possible approach, but I could not think of one.

[meurdesoifphilippe]

Or you multiply everything to get (abc)^2 then you deduce abc and a by dividing by bc and so on.

[philipbao]

a^2=150
b^2=200/3
a=5*sqrt(6)
b=10/3*sqrt(6)
c=10*sqrt(6)

[emmanueltanguy]

bc/ab=200 /100 <=> c=2a <=> ac=a*2a <=> 2a^2=300 <=> a=5√6 <=> c=10√6
ca/ab=300/100 <=> c=3b <=> b=(10/3)√6

[El.dabl]

ab=100, bc=200 ca=300. a=300/c b=200/c ab= 300/c * 200/c =60000/c2.... 60000/c2=100 c=10 √6
b*10 √6=200 b=20/ √6 a*20/ √6 =100 a=5/ √6

[alestee]

a=±5*sqrt(6); b=±(10/3)*sqrt(6); c=±10*sqrt(6).

[SidneiMV]

(bc)/(ab) = 2 => c = 2a
ac = 300 = 2a² => a = ± √150
=> c = ± 2√150
ab = 100 => b = ± 100/√150

a + b + c = ± (3√150 + 100/√150)
a + b + c = ± (550/√150)
a + b + c = ± [550/(5√6)]
a + b + c = ± (110/√6)
a + b + c = ± (110√6/6)
*a + b + c = ± (55√6/3)*

[XfeeXg]

ab=100 ; bc=200
So bc=2ab or
c=2a
ca=300=2a.a=2a*2
a*2=150
a= 5.(6)^2
b=100/(5×6^2)
b=(10×6^2)/3
c=10×6^2
a+b+c=(5+10/3+10)×6^2
a+b+c=(55×6^2)/3

[pashafedas]

b = 100/a; c = 300/a;
100/a * 300/a = 200; a² = 150; a = √150;
a * b = 100; √150 * b = 100; b = 100/√150;
c * a = 300; c * √150 = 300; c = 300/√150;
a + b + c = √150 + 100/√150 + 300/√150 = 150/√150 + 100/√150 + 300/√150 = 550/√150 = 550/5√6 = 110/√6

[lecombustor]

A, b, c are positive, so sum can't be negative.

[steveneagan]

I get
a= (3b)/2
b=√66.67
c=3b
I had to use a calculator to check the answer and it came surprisingly close.
ab=100.001
bc=200.001
ca=300.001

[НиколайП-шн]

A= 150/ кор.150
B= 100/ кор.150
С= 2× кор.150
Сумма 550/кор.150