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

Revision as of 08:09, 3 November 2024

Problem

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 1

Note that $f(x) - x^2$ has roots $2, 3$ , and $4$ . Therefore, we may write $f(x) = a(x-2)(x-3)(x-4) +x^2$ . 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.

In the case of line $AB$ , we may write $a(x-2)(x-3)(x-4)+x^2-5x+6 = a(x-2)(x-3)(x-r_1)$ for some real number $r_1$ . Dividing both sides by $(x-2)(x-3)$ gives $a(x-4)+1 = a(x-r_1)$ or $r_1 = \frac {4a-1}{a}$ .

For line $AC$ , we have $a(x-2)(x-3)(x-4)+x^2-6x+8 = a(x-2)(x-4)(x-r_2)$ for some real number $r_2$ , which gives $a(x-3)+1 = a(x-r_2)$ or $r_2 = \frac {3a-1}{a}$ .

For line $BC$ , we have $a(x-2)(x-3)(x-4)+x^2-7x+12 = a(x-3)(x-4)(x-r_3)$ for some real number $r_3$ , which gives $a(x-2)+1 = a(x-r_3)$ or $r_3 = \frac {2a-1}{a}$ .

Since $r_1 + r_2 + r_3 = 24$ , we have $\frac {4a-1}{a} + \frac {3a-1}{a} + \frac {2a-1}{a} = 24$ or $\frac {9a-3}{a} = 24$ . Solving for $a$ gives $a = - \frac{1}{5}$ .

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

Solution 2

No need to find the equations for the lines, really. First of all, $f(x) = a(x-2)(x-3)(x-4) +x^2$ . Let's say the line $AB$ is $y=bx+c$ , and $x_1$ is the $x$ coordinate of the third intersection, then $2$ , $3$ , and $x_1$ are the three roots of $f(x) - bx-c$ . The values of $b$ and $c$ have no effect on the sum of the 3 roots, because the coefficient of the $x^2$ term is always $-9a+1$ . So we have $\frac{9a-1}{a} = 2 + 3 + x_1= 3 + 4 + x_2 = 2 + 4 + x_3$ Adding all three equations up, we get $3\left(\frac{9a-1}{a}\right) = 18 + x_1 + x_2 + x_3 = 18 + 24$ Solving this equation, we get $a = -\frac{1}{5}$ . We finish as Solution 1 does. $\boxed{\textbf{(D)}\frac{24}{5}}$ .

- Mathdummy

Cleaned up by SSding

Solution 3

Map every point $(x,y)$ to $(x, y - x^2)$ . Note that the x-coordinates do not change. Under this map, $A$ goes to $(2,0)$ , $B$ goes to $(3, 0)$ and $C$ goes to $(4,0)$ . The cubic through $A$ , $B$ , and $C$ remains a cubic, while the lines between two points turn into quadratics. Finally, note that the intersection points of the lines and the cubic have the same x-coordinates as the intersection points of the quadratics and the cubic after applying the mapping. The cubic under this new coordinate plane has equation $k(x-2)(x-3)(x-4)$ . The quadratic through $A$ and $B$ is $c(x-2)(x-3)$ . Note that $c(x-2)(x-3) + x^2$ must be a line, so $c = -1$ to cancel out the squared terms. The intersection of the quadratic and cubic is solved by $-(x-2)(x-3) = k(x-2)(x-3)(x-4) \implies x = 4 - \frac{1}{k}$ Similarly, the other x-coordinates are $3 - \frac{1}{k}$ and $2 - \frac{1}{k}$ . Summing, we have $9 - \frac{3}{k} = 24 \implies k = -\frac{1}{5}$ We have $f(x) = -\frac{1}{5} (x-2)(x-3)(x-4) + x^2$ so $f(0) = 2 \cdot 3 \cdot 4 / 5 = \boxed{\textbf{(D)}\frac{24}{5}}$ .

If the mapping is too complicated, this solution is equivalent to realizing that the line $AB$ has the equation $y = x^2 - (x-2)(x-3)$ and solving for the intersection points.

~CrazyVideoGamez

Solution 4 (Mindless Vieta's Theorem)

Since $f(x)$ is a third degree polynomial, let $f(x)=ax^3+bx^2+cx+d$ . We want to solve for $d$ .

Notice that the 3 solutions to $f(x)=x^2$ are $2, 3, 4$ . Hence the polynomial $ax^3+(b-1)x^2+cx+d$ has roots 2, 3, 4. By Vieta's theorem we get $-\frac{d}{a}=24$ . It's not hard to get that $AB$ , $AC$ , and $BC$ are given by the equations $y = 5x - 6$ , $y= 6x-8$ , and $y=7x-12$ respectively. The 3 solutions to $f(x)=5x-6$ are $2, 3, x_D$ . Like before, using Vieta's theorem we get $-\frac{d+6}{a}=6x_D$ . Similarly we get $-\frac{d+8}{a}=8x_E$ , $-\frac{d+12}{a}=12x_F$ .

At this point we have 5 unknowns: $a, d, x_D, x_E, x_F$ , and 5 equations: \begin{align} & -\frac{d}{a}=24\\ \\ & -\frac{d+6}{a}=6x_D\\ \\ & -\frac{d+8}{a}=8x_E\\ \\ &-\frac{d+12}{a}=12x_F\\ \\ &x_D+x_E+x_F=24 \end{align}

The specific structure of this system of equations allows it to be solved with relatively ease. Solving, we get $d=\frac{24}{5}$

~tsun26

@@ Line 65: / Line 65: @@
 The specific structure of this system of equations allows it to be solved with relatively ease. Solving, we get <math>d=\frac{24}{5}</math>
+~tsun26
 ==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
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