Question

The graph of $y = f(x),$ where $f(x)$ is a polynomial of degree $3,$ contains points $A(2, 4),$ $B(3, 9),$ and $C(4, 16).$ Lines $AB,$ $AC,$ and $BC$ intersect the graph again at points $D,$ $E,$ and $F,$ respectively, and the sum of the $x$ -coordinates of $D,$ $E,$ and $F$ is $24.$ What is $f(0)?$

$-2$

$0$

$2$

$\dfrac{24}{5}$

$8$

Solution:

The points $A, B, C$ lie on $y = x^2,$ so $g(x) = f(x) - x^2$ has roots $2, 3, 4:$ $g(x) = a(x-2)(x-3)(x-4)$ for some $a \ne 0.$ The coefficients of $x^3$ and $x^2$ in $f$ are $a$ and $1 - 9a,$ so by Vieta the three roots of $f(x) - L(x)$ (for any linear $L$ ) sum to $9 - \tfrac1a.$ The lines $AB, AC, BC$ meet the cubic in triples $\{2, 3, x_D\},$ $\{2, 4, x_E\},$ $\{3, 4, x_F\},$ so $x_D + x_E + x_F = 3\left(9 - \tfrac1a\right) - 2(2 + 3 + 4) = 9 - \tfrac3a = 24,$ giving $a = -\tfrac15.$ Then $f(x) = x^2 - \tfrac15(x-2)(x-3)(x-4),$ so $f(0) = 0 - \tfrac15(-2)(-3)(-4) = \tfrac{24}{5}.$

Thus, the correct answer is D.

2017 AMC 12B Problem 23

Solution:

Problem 23 in Other Years