Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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==Solution 3== | ==Solution 3== | ||
− | Map every point <math>(x,y)</math> to <math>(x, y - x^2)</math>. Note that the x-coordinates do not change. Under this map, <math>A</math> goes to <math>(2,0)</math>, <math>B</math> goes to <math>(3, 0)</math> and <math>C</math> goes to <math>(4,0)</math>. The cubic through <math>A</math>, <math>B</math>, and <math>C</math> remains a cubic, while the lines between two points turn into quadratics. Finally, note that the intersection points of the lines and the cubic | + | Map every point <math>(x,y)</math> to <math>(x, y - x^2)</math>. Note that the x-coordinates do not change. Under this map, <math>A</math> goes to <math>(2,0)</math>, <math>B</math> goes to <math>(3, 0)</math> and <math>C</math> goes to <math>(4,0)</math>. The cubic through <math>A</math>, <math>B</math>, and <math>C</math> 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. The cubic under this new coordinate plane has equation <math>k(x-2)(x-3)(x-4)</math>. The quadratic through <math>A</math> and <math>B</math> is <math>c(x-2)(x-3)</math>. Note that <math>c(x-2)(x-3) + x^2</math> must be a line, so <math>c = -1</math> to cancel out the squared terms. The intersection of the quadratic and cubic is solved by |
<cmath>-(x-2)(x-3) = k(x-2)(x-3)(x-4) \implies x = 4 - \frac{1}{k}</cmath> | <cmath>-(x-2)(x-3) = k(x-2)(x-3)(x-4) \implies x = 4 - \frac{1}{k}</cmath> | ||
Similarly, the other x-coordinates are <math>3 - \frac{1}{k}</math> and <math>2 - \frac{1}{k}</math>. Summing, we have | Similarly, the other x-coordinates are <math>3 - \frac{1}{k}</math> and <math>2 - \frac{1}{k}</math>. Summing, we have |
Revision as of 19:55, 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. 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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