Confusing x , y , z x, y, z

Algebra Level 5

x 2 y + y 2 z + z 2 x = x y 2 + y z 2 + z x 2 = k x y z x^2y+y^2z+z^2x=xy^2+yz^2+zx^2=kxyz

Find the range of k k for which there is at least one triplet ( x , y , z ) (x,y,z) of non-zero real numbers such that the equation above holds true.

If it is in the form ( − ∞ , a ] ∪ [ b , ∞ ) (-\infty, a]\cup [b,\infty) , find a + b a+b .


The answer is 2.

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Roman Marcarelli
Jul 21, 2016

Begin by subtracting the middle equation: x 2 ( y − z ) + y 2 ( z − x ) + z 2 ( x − y ) = 0 Note that this is the same as: ( x − y ) ( y − z ) ( x − z ) = 0 So either x = y , y = z , or z = x . Without loss of generality, choose x = y : x 3 + x 2 z + z 2 x = k x 2 z The problem requires non-zero real solutions, so we can divide by x and rearrange to obtain: x 2 + ( 1 − k ) z x + z 2 = 0 The discriminant must be positive: ( 1 − k ) 2 z 2 − 4 z 2 ≥ 0 → ( 1 − k ) 2 ≥ 4 → 1 − k ≤ − 2 or 1 − k ≥ 2 So k ≤ − 1 or k ≥ 3 , yielding ( − ∞ , − 1 ] ∪ [ 3 , ∞ ) . Note that substituting y = z or z = x would yield the same result, so this is the final range of k . Therefore, a + b = − 1 + 3 = 2 . QED \textrm{Begin by subtracting the middle equation:}\\ x^2(y - z) + y^2(z - x) + z^2(x - y) = 0\\\textrm{ }\\ \textrm{Note that this is the same as:}\\ (x - y)(y - z)(x - z) = 0\\\textrm{ }\\ \textrm{So either } x = y \textrm{, } y = z \textrm{, or }z = x \textrm{.}\\ \textrm{ }\\ \textrm{Without loss of generality, choose } x = y \textrm{:}\\ x^3 + x^2z + z^2x = k x^2z\\\textrm{ }\\ \textrm{The problem requires non-zero real solutions, so we can divide by } x \textrm{ and rearrange to obtain: }\\ x^2 + (1-k)zx + z^2 = 0\\ \textrm{ }\\ \textrm{The discriminant must be positive:}\\ (1-k)^2z^2 - 4z^2\geq0 \rightarrow (1-k)^2\geq4 \rightarrow 1 - k\leq-2 \textrm{ or } 1 - k \geq2\\ \textrm{ }\\ \textrm{So } k\leq-1 \textrm{ or } k\geq3 \textrm{, yielding } (-\infty, -1] \cup [3, \infty)\textrm{. Note that substituting } y = z \textrm{ or } z = x \textrm{ would yield}\\\textrm{the same result, so this is the final range of } k \textrm{.} \\ \textrm{Therefore, } a + b = -1 + 3 = 2 \textrm{. QED}

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