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

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==Solution 2==
 
==Solution 2==
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<math>\boxed{\textbf{No need to find the equations for the lines, really.}}</math> First of all, <math>f(x) = a(x-2)(x-3)(x-4) +x^2</math>. Let's say the line <math>AB</math> is <math>y=bx+c</math>, and <math>x_1</math> is the <math>x</math> coordinate of the third intersection, then <math>2</math>, <math>3</math>, and <math>x_1</math> are the three roots of <math>f(x) - bx-c</math>. The values of <math>b</math> and <math>c</math> have no effect on the sum of the 3 roots, because the coefficient of the <math>x^2</math> term is always <math>-9a+1</math>. So we have  
+
No need to find the equations for the lines, really. First of all, <math>f(x) = a(x-2)(x-3)(x-4) +x^2</math>. Let's say the line <math>AB</math> is <math>y=bx+c</math>, and <math>x_1</math> is the <math>x</math> coordinate of the third intersection, then <math>2</math>, <math>3</math>, and <math>x_1</math> are the three roots of <math>f(x) - bx-c</math>. The values of <math>b</math> and <math>c</math> have no effect on the sum of the 3 roots, because the coefficient of the <math>x^2</math> term is always <math>-9a+1</math>. So we have  
 
<cmath> \frac{9a-1}{a} = 2 + 3 + x_1= 3 + 4 + x_2 = 2 + 4 + x_3</cmath>
 
<cmath> \frac{9a-1}{a} = 2 + 3 + x_1= 3 + 4 + x_2 = 2 + 4 + x_3</cmath>
 
Adding all three equations up, we get  
 
Adding all three equations up, we get  
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Cleaned up by SSding
 
Cleaned up by SSding
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==Solution 3==
 +
Map every point <math>(x,y)</math> to <math>(x, y - x^2)</math>. Note that the x-coordinates do not change. Under this map, <math>A</math> goes to <math>(2,0)</math>, <math>B</math> goes to <math>(3, 0)</math> and <math>C</math> goes to <math>(4,0)</math>. The cubic through <math>A</math>, <math>B</math>, and <math>C</math> 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 <math>k(x-2)(x-3)(x-4)</math>. The quadratic through <math>A</math> and <math>B</math> is <math>c(x-2)(x-3)</math>. Note that <math>c(x-2)(x-3) + x^2</math> must be a line, so <math>c = -1</math> to cancel out the squared terms. The intersection of the quadratic and cubic is solved by
 +
<cmath>-(x-2)(x-3) = k(x-2)(x-3)(x-4) \implies x = 4 - \frac{1}{k}</cmath>
 +
Similarly, the other x-coordinates are <math>3 - \frac{1}{k}</math> and <math>2 - \frac{1}{k}</math>. Summing, we have
 +
<cmath>9 - \frac{3}{k} = 24 \implies k = -\frac{1}{5}</cmath>
 +
We have <math>f(x) = -\frac{1}{5} (x-2)(x-3)(x-4) + x^2</math> so <math>f(0) = 2 \cdot 3 \cdot 4 / 5 = \boxed{\textbf{(D)}\frac{24}{5}}</math>.
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 +
If the mapping is too complicated, this solution is equivalent to realizing that the line <math>AB</math> has the equation <math>y = x^2 - (x-2)(x-3)</math> and solving for the intersection points.
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez CrazyVideoGamez]
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 +
==Solution 4 (Mindless Vieta's Theorem)==
 +
 +
Since <math>f(x)</math> is a third degree polynomial, let <math>f(x)=ax^3+bx^2+cx+d</math>. We want to solve for <math>d</math>.
 +
 +
Notice that the 3 solutions to <math>f(x)=x^2</math> are <math>2, 3, 4</math>. Hence the polynomial <math>ax^3+(b-1)x^2+cx+d</math> has roots 2, 3, 4. By Vieta's theorem we get <math>-\frac{d}{a}=24</math>. It's not hard to get that <math>AB</math>, <math>AC</math>, and <math>BC</math> are given by the equations <math>y = 5x - 6</math>, <math>y= 6x-8</math>, and <math>y=7x-12</math> respectively. The 3 solutions to <math>f(x)=5x-6</math> are <math>2, 3, x_D</math>. Like before, using Vieta's theorem we get <math>-\frac{d+6}{a}=6x_D</math>. Similarly we get <math>-\frac{d+8}{a}=8x_E</math> and <math>-\frac{d+12}{a}=12x_F</math>.
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At this point we have 5 unknowns: <math>a, d, x_D, x_E, x_F</math>, and 5 equations:
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\begin{align}
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& -\frac{d}{a}=24\
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\
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& -\frac{d+6}{a}=6x_D\
 +
\
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& -\frac{d+8}{a}=8x_E\
 +
\
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&-\frac{d+12}{a}=12x_F\
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\
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&x_D+x_E+x_F=24
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\end{align}
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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>
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 +
~tsun26
  
 
==See Also==
 
==See Also==

Latest revision as of 09:10, 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 $-\frac{d+8}{a}=8x_E$ and $-\frac{d+12}{a}=12x_F$.

At this point we have 5 unknowns: $a, d, x_D, x_E, x_F$, and 5 equations: −da=24−d+6a=6xD−d+8a=8xE−d+12a=12xFxD+xE+xF=24

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

~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
All AMC 12 Problems and Solutions

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