HMMT November, 2022

Algebra Difficulty 4.7 AIME Prove it United States

Problem:
Let $P(x)$ be a quadratic polynomial with real coefficients. Suppose that $P(1)=20$ , $P(-1)=22$ , and $P(P(0))=400$ . Compute the largest possible value of $P(10)$ .

Solution

Solution:
Let $P(x) = a x^{2} + b x + c$ . The given equations give us:
$\begin{aligned} & a + b + c = 20 \\ & a - b + c = 22 \end{aligned}$
Hence $b = -1$ , $a + c = 21$ , and so the final equation gives us $a c^{2} = 400$ . Substituting $a = 21 - c$ and solving the cubic in $c$ , we get $c = -4, 5, 20$ . Of these, the smallest value $c = -4$ (and hence $P(x) = 25 x^{2} - x - 4$ ) ends up giving the largest value of $P(10)$ .

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