Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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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>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> | + | 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>. | 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>. |
Revision as of 11:48, 3 July 2024
Contents
[hide]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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.