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IMO D
Let x/a+b−2c=y/b+c−2a=z/c+a−2b=k (any constant value)
Therefore we will get, x = k(a+b-2c); y =k(b+c-2a); z=k(c+a-2b)
Hence, X+Y+Z = ka + kb –2kc + kb+kc-2ka+kc+kc-2kb = 0
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Given: a, b, c are three different numbers. None of the numbers equals the average of the other two.

Asked: If \(\frac{x}{{a+b-2c}}=\frac{y}{{b+c-2a}}=\frac{z}{{c+a-2b}}\) , then what is the value of \(x+y+z\) ?

Let \(\frac{x}{{a+b-2c}}=\frac{y}{{b+c-2a}}=\frac{z}{{c+a-2b}} = k\)

x = k (a + b - 2c)
y = k (b + c - 2a)
z = k (c + a - 2b)

x + y + z = k (0) = 0

IMO D
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Kinshook
Given: a, b, c are three different numbers. None of the numbers equals the average of the other two.

Asked: If \(\frac{x}{{a+b-2c}}=\frac{y}{{b+c-2a}}=\frac{z}{{c+a-2b}}\) , then what is the value of \(x+y+z\) ?

Let \(\frac{x}{{a+b-2c}}=\frac{y}{{b+c-2a}}=\frac{z}{{c+a-2b}} = k\)

x = k (a + b - 2c)
y = k (b + c - 2a)
z = k (c + a - 2b)

x + y + z = k (0) = 0

IMO D

Can you explain the theory or rational behind assigning variable 'k' to the denominator and why it is okay to assign it a value of 0?
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Quote:
Can you explain the theory or rational behind assigning variable 'k' to the denominator and why it is okay to assign it a value of 0?
Solution:

  • The assigning of variable k is not done to the denominator but to the whole fractions
  • \(\frac{x}{a+b-2c}=k\) or \(x=ka+kb-2kc\)
  • \(\frac{y}{b+c-2a}=k\) or \(y=kb+kc-2ka\)
  • \(\frac{z}{c+a-2b}=k\) or \(z=kc+ka-2kb\)
  • Now we add this and get \(x+y+z=ka+kb-2kc+kb+kc-2ka+kc+ka-2kb\)
    \(⇒x+y+z=2ka-2ka+2kb-2kb+2kc-2kc\)
    \(⇒x+y+z=0\)

Hence the right answer is Option D
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