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{ a = b z + c x x = b y + c z y = b = c z + a x z = a x + b y c = a z + b x \large \left\{ \begin{aligned} a &= bz + cx\\ x &= by + cz\\ y = b &= cz + ax\\ z &= ax + by\\ c &= az + bx\\ \end{aligned} \right.

What is the value of ( a + b + c ) ( x + y + z ) 0 (a + b + c)(x+ y +z) \ne 0 ?


The answer is 2.25.

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1 solution

Michael Mendrin
May 5, 2018

First, by adding equations 1, 3, 5, we have

( a + b + c ) = ( b + c + a ) z + ( c + a + b ) x (a+b+c)=(b+c+a)z+(c+a+b)x

so that we know z = 1 x z=1-x

From this, we solve for x x in equations 1, 3, and then solve for c c from the two

c = 1 2 ( a + b 3 a 2 + 2 a b b 2 ) c = \dfrac{1}{2}\left( a + b - \sqrt{3} \sqrt{ -a^2 +2ab -b^2 }\right)

so that c c is real only if a = b a=b . This is the crux step. Immediately from this, we see that a = b = c = y a=b=c=y

From this, we then solve for x x and a a from equations 2, 4, and find that x = a = 1 2 x = a = \dfrac{1}{2} . Hence

( a + b + c ) ( x + y + z ) = 9 4 (a+b+c)(x+y+z) = \dfrac{9}{4}

Another possible solution would be

( a , b , c , x , y , z ) = ( 1 , 1 , 1 , 1 2 , 1 , 1 2 ) (a, b, c, x, y, z)=(-1, -1, -1, \frac{1}{2}, -1, \frac{1}{2}) which would have yielded 0 0

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