AIME I

Algebra Level pending

Let P ( x ) P(x) be a quadratic polynomial with real coefficients satisfying:

x 2 − 2 x + 2 ≤ P ( x ) ≤ 2 x 2 − 4 x + 3 x^2-2x+2 \le P(x) \le 2x^2-4x+3

for all real numbers x x and suppose P ( 11 ) = 181 P(11)= 181 .

Find P ( 16 ) = ? P(16)=?

406 408 410 405 409

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Chew-Seong Cheong
Apr 26, 2017

x 2 − 2 x + 2 ≤ P ( x ) ≤ 2 x 2 − 4 x + 3 ( x − 1 ) 2 + 1 ≤ P ( x ) ≤ 2 ( x − 1 ) 2 + 1 ⟹ ( x − 1 ) 2 + 1 ≤ a ( x − 1 ) 2 + 1 ≤ 2 ( x − 1 ) 2 + 1 where 1 < a < 2 \begin{aligned} x^2-2x + 2 \le P(x&) \le 2x^2-4x+3 \\ (x-1)^2 + 1 \le P(x&) \le 2(x-1)^2 + 1 \\ \implies (x-1)^2 + 1 \le {\color{#3D99F6}a}(x-1&)^2 + 1 \le 2(x-1)^2 + 1 & \small \color{#3D99F6} \text{where }1 < a < 2 \end{aligned}

Now, we have:

P ( x ) = a ( x − 1 ) 2 + 1 P ( 11 ) = 181 ⟹ 100 a + 1 = 181 ⟹ a = 9 5 ⟹ P ( x ) = 9 5 ( x − 1 ) 2 + 1 P ( 16 ) = 9 5 ( 16 − 1 ) 2 + 1 = 406 \begin{aligned} P(x) & = a(x-1)^2 + 1 \\ P(11) & = 181 \\ \implies 100a+1 & = 181 \\ \implies a & = \frac 95 \\ \implies P(x) & = \frac 95 (x-1)^2 + 1 \\ P(16) & = \frac 95 (16-1)^2 + 1 = \boxed{406} \end{aligned}

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Thank you so much.

Hana Wehbi - 4 years, 1 month ago

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