HMMT November

Algebra Difficulty 4.8 AIME Find the answer

Find all triples of real numbers $(a, b, c)$ such that $a^{2}+2 b^{2}-2 b c=16$ and $2 a b-c^{2}=16$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

$a^{2}+2 b^{2}-2 b c$ and $2 a b-c^{2}$ are both homogeneous degree 2 polynomials in $a, b, c$ , so we focus on the homogeneous equation $a^{2}+2 b^{2}-2 b c=2 a b-c^{2}$ , or $(a-b)^{2}+(b-c)^{2}=0$ . So $a=b=c$ , and $a^{2}=2 a b-c^{2}=16$ gives the solutions $(4,4,4)$ and $(-4,-4,-4)$ .

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