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2017 AMC 12B Problems/Problem 23

Revision as of 22:05, 23 December 2018 by Skonar (talk | contribs) (→‎Solution)

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 gorefeebuddie

Note: This is a really good AMC 12 problem. It is one of those problems that they have every year.

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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