2023 Christmas Mock AIME #8 3x3 complex non linear system

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2023 Christmas Mock AIME #8 3x3 complex non linear system

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#1 h Jan 16, 2024, 2:26 PM

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Let $a$ , $b$ , and $c$ be complex numbers such that they satisfy these equations:
$abc + 4a^4 = 3$ $abc + 2d^4 = 2$ $abc +\frac{3c^4}{4}=-1$ If the maximum of $\left|\frac{a^2b^2c^2+1}{abc}\right|$ can be expressed as $\frac{p+\sqrt{q}}{r}$ for positive integers $p$ , $q$ , and $r$ , find the minimum possible value of $p + q + r$ .

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#2 h Mar 25, 2024, 9:06 PM

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parmenides51 wrote:

Let $a$ , $b$ , and $c$ be complex numbers such that they satisfy these equations:
$abc + 4a^4 = 3$ $abc + 2b^4 = 2$ $abc +\frac{3c^4}{4}=-1$ If the maximum of $\left|\frac{a^2b^2c^2+1}{abc}\right|$ can be expressed as $\frac{p+\sqrt{q}}{r}$ for positive integers $p$ , $q$ , and $r$ , find the minimum possible value of $p + q + r$ .

We find: $4a^{4}=3-abc$ and $2b^{4}=2-abc$ and $\frac{3}{4}c^{4}=-1-abc$ .

Let $abc=t$ ; multiplying: $6t^{5}+(3-t)(2-t)(1+t)=0$
with one of the solutions: $t=-\frac{1+\sqrt{385}}{24}+ i \cdot \sqrt{\frac{95-\sqrt{385}}{288}}$ .

Then $\left|t+\frac{1}{t}\right|=\frac{1+\sqrt{385}}{12}$ .

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#3 h Feb 5, 2025, 10:13 PM

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Doing the same $x=abc$ substitution as @above, we get $6x^4 + x^3 - 4x^2 + x + 6 = 0$ . Seeing that the polynomial is symmetric, we divide by $x^2$ and let $y = x + \frac{1}{x}$ . From here, one gets the quadratic $6y^2+y-16=0$ , which gives $y = \frac{-1 \pm \sqrt{385}}{12}$ . Since we want the maximum absolute value, we have $|\frac{a^2b^2c^2+1}{abc}| = |x+\frac{1}{x}| = |y| = \frac{1+\sqrt{385}}{12}$ giving the answer of $\boxed{398}$ .

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