Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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Since <math>f(x)</math> is a third degree polynomial, let <math>f(x)=ax^3+bx^2+cx+d</math>. We want to solve for <math>d</math>. | Since <math>f(x)</math> is a third degree polynomial, let <math>f(x)=ax^3+bx^2+cx+d</math>. We want to solve for <math>d</math>. | ||
− | Notice that the 3 solutions to <math>f(x)=x^2</math> are <math>2, 3, 4</math>. Hence the polynomial <math>ax^3+(b-1)x^2+cx+d</math> has roots 2, 3, 4. By Vieta's theorem we get <math>-\frac{d}{a}=24</math>. It's not hard to get that <math>AB</math>, <math>AC</math>, and <math>BC</math> are given by the equations <math>y = 5x - 6</math>, <math>y= 6x-8</math>, and <math>y=7x-12</math> respectively. The 3 solutions to <math>f(x)=5x-6</math> are <math>2, 3, x_D</math>. Like before, using Vieta's theorem we get <math>-\frac{d+6}{a}=6x_D</math>. Similarly we get <math>-\frac{d+8}{a}=8x_E</math> | + | Notice that the 3 solutions to <math>f(x)=x^2</math> are <math>2, 3, 4</math>. Hence the polynomial <math>ax^3+(b-1)x^2+cx+d</math> has roots 2, 3, 4. By Vieta's theorem we get <math>-\frac{d}{a}=24</math>. It's not hard to get that <math>AB</math>, <math>AC</math>, and <math>BC</math> are given by the equations <math>y = 5x - 6</math>, <math>y= 6x-8</math>, and <math>y=7x-12</math> respectively. The 3 solutions to <math>f(x)=5x-6</math> are <math>2, 3, x_D</math>. Like before, using Vieta's theorem we get <math>-\frac{d+6}{a}=6x_D</math>. Similarly we get <math>-\frac{d+8}{a}=8x_E</math> and <math>-\frac{d+12}{a}=12x_F</math>. |
At this point we have 5 unknowns: <math>a, d, x_D, x_E, x_F</math>, and 5 equations: | At this point we have 5 unknowns: <math>a, d, x_D, x_E, x_F</math>, and 5 equations: |
Latest revision as of 08:10, 3 November 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
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.