Greatest Product Circle Point
Let $ A = (2, 0) $, $ B = (0, 2) $, $ C = (-2, 0) $, and $ D = (0, -2) $. Compute the greatest possible value of the product $ PA \cdot PB \cdot PC \cdot PD $, where $ P $ is a point on the circle $ x^2 + y^2 = 9 $.
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- 9
- $\frac{a}{b}$
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- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
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- $\sin{}$
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- $\cap$
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- $\infty$
Solution
We use complex numbers. Let $ a = 2 $, $ b = 2i $, $ c = -2 $, and $ d = -2 $ be the complex numbers corresponding to $ A $, $ B $, $ C $, and $ D $, respectively. Let $ p $ be the complex number corresponding to $ P $, so that $ |p| = \sqrt{9} = 3 $. Then we have \[\begin{aligned} PA \cdot PB \cdot PC \cdot PD &= |p-2| \cdot |p-2i| \cdot |p+2| \cdot |p+2i| \\ &= |(p-2)(p+2)| \cdot |(p-2i)(p+2i)| \\ &= |p^2-4| \cdot |p^2+4| \\ &= |p^4-16|.\end{aligned}\]
Since $ |p| = 3 $, we have $ |p^4| = 3^4= 81 $, so by the triangle inequality, \[|p^4-16| \le |p^4| + |-16| = 81 + 16 = 97.\]Equality holds if and only if $ p^4 = -81 $, which occurs when $ p = 3\left(\frac{\sqrt2}{2} + \frac{\sqrt2}{2}i\right) $. Therefore, the answer is $ \boxed{97} $.
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