Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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==Solution 2== | ==Solution 2== | ||
− | <math> | + | <math>\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 | Adding all three equations up, we get |
Revision as of 19:56, 1 October 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
Solution 3
Map every point to . Note that the x-coordinates do not change. Under this map, goes to , goes to and goes to . The cubic through , , and 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 . The quadratic through and is . Note that must be a line, so to cancel out the squared terms. The intersection of the quadratic and cubic is solved by Similarly, the other x-coordinates are and . Summing, we have We have so .
If the mapping is too complicated, this solution is equivalent to realizing that the line has the equation and solving for the intersection points.
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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