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

m (Solution)
(10 intermediate revisions by 8 users not shown)
Line 1: Line 1:
==Problem 23==
+
==Problem==
 
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?
 
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?
  
Line 8: Line 8:
  
 
Solution by vedadehhc
 
Solution by vedadehhc
 +
 +
==Solution 2==
 +
<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>, <math>x_1</math> are the three roots of <math>f(x) - bx-c</math>. Apparently the value 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>
 +
Add them up we have
 +
<cmath> 3\frac{9a-1}{a} = 18 + x_1+x_2+x_3 = 18 +24</cmath>
 +
Solve it, we get <math>a = -\frac{1}{5}</math>.
 +
<math>\boxed{\textbf{(D)}\frac{24}{5}}</math>.
 +
 +
- Mathdummy
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2017|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Intermediate Algebra Problems]]

Revision as of 13:56, 15 February 2021

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

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

Solution 2

$\boxed{\textbf{No need to find the equations for the lines, really.}}$ 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 $2$, $3$, $x_1$ are the three roots of $f(x) - bx-c$. Apparently the value 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, \[\frac{9a-1}{a} = 2+3 + x_1=3+4+x_2 = 2+4+x_3\] Add them up we have \[3\frac{9a-1}{a} = 18 + x_1+x_2+x_3 = 18 +24\] Solve it, we get $a = -\frac{1}{5}$. $\boxed{\textbf{(D)}\frac{24}{5}}$.

- Mathdummy

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png