Difference between revisions of "2017 AMC 12B Problems/Problem 23"

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Solution by gorefeebuddie
 
Solution by gorefeebuddie
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Note: This is a really good AMC 12 problem. It is one of those problems that have it every year.
  
 
==See Also==
 
==See Also==

Revision as of 14:48, 1 December 2018

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 have it every year.

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
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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