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>? | ||
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==Solution 2== | ==Solution 2== | ||
− | 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>, <math>x_1</math> are the three roots of <math>f(x) - bx-c</math>. | + | <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> | + | <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\frac{9a-1}{a} = 18 + x_1+x_2+x_3 = 18 +24</cmath> | + | <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>. | <math>\boxed{\textbf{(D)}\frac{24}{5}}</math>. | ||
- Mathdummy | - Mathdummy | ||
+ | |||
+ | Cleaned up by SSding | ||
==See Also== | ==See Also== |
Latest revision as of 23:25, 4 November 2021
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
First, we can define , which contains points , , and . 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 (eg. if we were looking at line , we would synthetically divide by the solutions and , because we already know intersects the graph at and , which have -coordinates of and ). After completing this process on all three lines, we get that the -coordinates of , , and are , , and respectively. Adding these together, we get which gives us . 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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