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, 2022

Algebra Difficulty 5.7 AIME, harder Prove it United States

Problem:

Vijay chooses three distinct integers $a, b, c$ from the set $\{1,2,3,4,5,6,7,8,9,10,11\}$ . If $k$ is the minimum value taken on by the polynomial $a(x-b)(x-c)$ over all real numbers $x$ , and $l$ is the minimum value taken on by the polynomial $a(x-b)(x+c)$ over all real numbers $x$ , compute the maximum possible value of $k-l$ .

Solution

Solution:

Quadratics are minimized at the average of their roots, so
$\begin{aligned} & k = a\left(\frac{b+c}{2} - b\right)\left(\frac{b+c}{2} - c\right) \\ & l = a\left(\frac{b-c}{2}\right)\left(\frac{c-b}{2}\right) = -\frac{a(b-c)^2}{4}, \text{ and } \\ & l = a\left(\frac{b-c}{2} - b\right)\left(\frac{b-c}{2} + c\right) = a\left(\frac{-b-c}{2}\right)\left(\frac{b+c}{2}\right) = -\frac{a(b+c)^2}{4} . \end{aligned}$
Therefore,
$k-l = -\frac{a}{4}\left((b-c)^2 - (b+c)^2\right) = a b c$
Thus, $k-l = a b c$ is maximized when $a, b$ , and $c$ are $9, 10$ , and $11$ in some order, so the answer is $9 \cdot 10 \cdot 11 = 990$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.