Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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− | ==Problem | + | ==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>? | ||
<math>\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</math> | <math>\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</math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | Note that <math>f(x) - x^2</math> has roots <math>2, 3</math>, and <math>4</math>. Therefore, we may write <math>f(x) = a(x-2)(x-3)(x-4) +x^2</math>. Now we find that lines <math>AB</math>, <math>AC</math>, and <math>BC</math> are defined by the equations <math>y = 5x - 6</math>, <math>y= 6x-8</math>, and <math>y=7x-12</math> respectively. | ||
+ | |||
+ | Since we want to find the <math>x</math>-coordinates of the intersections of these lines and <math>f(x)</math>, we set each of them to <math>f(x)</math> and synthetically divide by the solutions we already know exist. | ||
+ | |||
+ | In the case of line <math>AB</math>, we may write <math>a(x-2)(x-3)(x-4)+x^2-5x+6 = a(x-2)(x-3)(x-r_1)</math> for some real number <math>r_1</math>. Dividing both sides by <math>(x-2)(x-3)</math> gives <math>a(x-4)+1 = a(x-r_1)</math> or <math>r_1 = \frac {4a-1}{a}</math>. | ||
+ | |||
+ | For line <math>AC</math>, we have <math>a(x-2)(x-3)(x-4)+x^2-6x+8 = a(x-2)(x-4)(x-r_2)</math> for some real number <math>r_2</math>, which gives <math>a(x-3)+1 = a(x-r_2)</math> or <math>r_2 = \frac {3a-1}{a}</math>. | ||
+ | |||
+ | For line <math>BC</math>, we have <math>a(x-2)(x-3)(x-4)+x^2-7x+12 = a(x-3)(x-4)(x-r_3)</math> for some real number <math>r_3</math>, which gives <math>a(x-2)+1 = a(x-r_3)</math> or <math>r_3 = \frac {2a-1}{a}</math>. | ||
+ | |||
+ | Since <math>r_1 + r_2 + r_3 = 24</math>, we have <math> \frac {4a-1}{a} + \frac {3a-1}{a} + \frac {2a-1}{a} = 24</math> or <math> \frac {9a-3}{a} = 24</math>. Solving for <math>a</math> gives <math>a = - \frac{1}{5}</math>. | ||
+ | |||
+ | Substituting this back into the original equation, we get <math>f(x) = -\frac{1}{5}(x-2)(x-3)(x-4) + x^2</math>, and <math>f(0) = -\frac{1}{5}(-2)(-3)(-4) + 0 = \boxed{\textbf{(D)}\frac{24}{5}}</math> | ||
+ | |||
+ | 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>, 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> | ||
+ | Adding all three equations up, we get | ||
+ | <cmath> 3\left(\frac{9a-1}{a}\right) = 18 + x_1 + x_2 + x_3 = 18 + 24</cmath> | ||
+ | Solving this equation, we get <math>a = -\frac{1}{5}</math>. We finish as Solution 1 does. | ||
+ | <math>\boxed{\textbf{(D)}\frac{24}{5}}</math>. | ||
+ | |||
+ | - Mathdummy | ||
+ | |||
+ | Cleaned up by SSding | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2017|ab=B|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 12:48, 3 July 2024
Contents
Problem
The graph of , where
is a polynomial of degree
, contains points
,
, and
. Lines
,
, and
intersect the graph again at points
,
, and
, respectively, and the sum of the
-coordinates of
,
, and
is 24. What is
?
Solution 1
Note that has roots
, and
. Therefore, we may write
. Now we find that lines
,
, and
are defined by the equations
,
, and
respectively.
Since we want to find the -coordinates of the intersections of these lines and
, we set each of them to
and synthetically divide by the solutions we already know exist.
In the case of line , we may write
for some real number
. Dividing both sides by
gives
or
.
For line , we have
for some real number
, which gives
or
.
For line , we have
for some real number
, which gives
or
.
Since , we have
or
. Solving for
gives
.
Substituting this back into the original equation, we get , and
Solution by vedadehhc
Solution 2
First of all,
. Let's say the line
is
, and
is the
coordinate of the third intersection, then
,
, and
are the three roots of
. The values of
and
have no effect on the sum of the 3 roots, because the coefficient of the
term is always
. So we have
Adding all three equations up, we get
Solving this equation, we get
. We finish as Solution 1 does.
.
- Mathdummy
Cleaned up by SSding
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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