2017 AMC 12B Problems/Problem 23

Problem 23

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 $f(0)$ ?

$\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8$

Solution

First, we can define $f(x) = a(x-2)(x-3)(x-4) +x^2$ , which contains points $A$ , $B$ , and $C$ . Now we find that lines $AB$ , $AC$ , and $BC$ are defined by the equations $y = 5x - 6$ , $y= 6x-8$ , and $y=7x-12$ respectively. Since we want to find the $x$ -coordinates of the intersections of these lines and $f(x)$ , we set each of them to $f(x)$ , and synthetically divide by the solutions we already know exist (eg. if we were looking at line $AB$ , we would synthetically divide by the solutions $x=2$ and $x=3$ , because we already know $AB$ intersects the graph at $A$ and $B$ , which have $x$ -coordinates of $2$ and $3$ ). After completing this process on all three lines, we get that the $x$ -coordinates of $D$ , $E$ , and $F$ are $\frac{4a-1}{a}$ , $\frac{3a-1}{a}$ , and $\frac{2a-1}{a}$ respectively. Adding these together, we get $\frac{9a-3}{a} = 24$ which gives us $a = -\frac{1}{5}$ . Substituting this back into the original equation, we get $f(x) = -\frac{1}{5}(x-2)(x-3)(x-4) + x^2$ , and $f(0) = -\frac{1}{5}(-2)(-3)(-4) + 0 = \boxed{\textbf{(D)}\frac{24}{5}}$

Solution by: vedadehhc and tdeng

Page

Toolbox

Search

2017 AMC 12B Problems/Problem 23

Problem 23

Solution