Tricky abc

Algebra Level 3

Let a a , b b and c c be distinct real numbers such that

a + 2 b = b + 2 c = c + 2 a \large a+\frac{2}{b}=b+\frac{2}{c}=c+\frac{2}{a}

Find the value of a b c |abc| .

Hint: You don't need to find the exact values of a a , b b and c c . Try to find the answer in another way.

4 2√2 √2 8

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1 solution

Ong Zi Qian
Oct 20, 2017

Due to a + 2 b = b + 2 c a+\frac{2}{b}=b+\frac{2}{c} ,

a b = 2 c 2 b a-b=\frac{2}{c}-\frac{2}{b}

a b = 2 ( b c ) b c a-b=\frac{2(b-c)}{bc}

b c = 2 ( b c ) a b bc=\frac{2(b-c)}{a-b} ---------------------------------(*)

With the same way, we get

b + 2 c = c + 2 a b+\frac{2}{c}=c+\frac{2}{a} ,

b c = 2 a 2 c b-c=\frac{2}{a}-\frac{2}{c}

b c = 2 ( c a ) a c b-c=\frac{2(c-a)}{ac}

a c = 2 ( c a ) b c ac=\frac{2(c-a)}{b-c} ---------------------------------(*)

c + 2 a = a + 2 b c+\frac{2}{a}=a+\frac{2}{b} ,

c a = 2 b 2 a c-a=\frac{2}{b}-\frac{2}{a}

c a = 2 ( a b ) a b c-a=\frac{2(a-b)}{ab}

a b = 2 ( a b ) c a ab=\frac{2(a-b)}{c-a} ---------------------------------(*)

Multiply the three (*) equations, we get

b c a c a b = 2 ( b c ) a b 2 ( c a ) b c 2 ( a b ) c a bc \cdot ac \cdot ab=\frac{2(b-c)}{a-b} \cdot \frac{2(c-a)}{b-c} \cdot \frac{2(a-b)}{c-a}

( a b c ) 2 = 8 (abc)^2=8

a b c = 2 2 |abc|=\boxed{2\sqrt2}

Nice solution, but to be complete, can you give an example of 3 distinct positive real numbers a , b , c a,b,c that satisfy the original system of equations?

Brian Charlesworth - 3 years, 7 months ago

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Your suggestion is good. I will did my best to try it. Thank you.

Ong Zi Qian - 3 years, 7 months ago

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