2017 AMC 12B Problems/Problem 23
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
No need to find the equations for the lines, really. 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.
Solution 4 (Mindless Vieta's Theorem)
Since is a third degree polynomial, let . We want to solve for .
Notice that the 3 solutions to are . Hence the polynomial has roots 2, 3, 4. By Vieta's theorem we get . It's not hard to get that , , and are given by the equations , , and respectively. The 3 solutions to are . Like before, using Vieta's theorem we get . Similarly we get and .
At this point we have 5 unknowns: , and 5 equations:
The specific structure of this system of equations allows it to be solved with relatively ease. Solving, we get
~tsun26
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.