Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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Since we want to find the <math>x</math>-coordinates of the intersections of these lines and <math>f(x)</math>, we set each of them to <math>f(x)</math> and synthetically divide by the solutions we already know exist. | Since we want to find the <math>x</math>-coordinates of the intersections of these lines and <math>f(x)</math>, we set each of them to <math>f(x)</math> and synthetically divide by the solutions we already know exist. | ||
− | In the case of line <math>AB</math> | + | In the case of line <math>AB</math>, we may write <math>a(x-2)(x-3)(x-4)+x^2-5x+6 = a(x-2)(x-3)(x-r_1)</math> for some real number <math>r_1</math>. Dividing both sides by <math>(x-2)(x-3)</math> gives <math>a(x-4)+1 = a(x-r_1)</math> or <math>r_1 = \frac {4a-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>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>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>. | ||
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==Solution 2== | ==Solution 2== | ||
− | + | No need to find the equations for the lines, really. 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 | ||
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Cleaned up by SSding | Cleaned up by SSding | ||
+ | |||
+ | ==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 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 <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> | ||
+ | Similarly, the other x-coordinates are <math>3 - \frac{1}{k}</math> and <math>2 - \frac{1}{k}</math>. Summing, we have | ||
+ | <cmath>9 - \frac{3}{k} = 24 \implies k = -\frac{1}{5}</cmath> | ||
+ | We have <math>f(x) = -\frac{1}{5} (x-2)(x-3)(x-4) + x^2</math> so <math>f(0) = 2 \cdot 3 \cdot 4 / 5 = \boxed{\textbf{(D)}\frac{24}{5}}</math>. | ||
+ | |||
+ | If the mapping is too complicated, this solution is equivalent to realizing that the line <math>AB</math> has the equation <math>y = x^2 - (x-2)(x-3)</math> and solving for the intersection points. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez CrazyVideoGamez] | ||
+ | |||
+ | ==Solution 4 (Mindless Vieta's Theorem)== | ||
+ | |||
+ | 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> 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: | ||
+ | \begin{align} | ||
+ | & -\frac{d}{a}=24\ | ||
+ | \ | ||
+ | & -\frac{d+6}{a}=6x_D\ | ||
+ | \ | ||
+ | & -\frac{d+8}{a}=8x_E\ | ||
+ | \ | ||
+ | &-\frac{d+12}{a}=12x_F\ | ||
+ | \ | ||
+ | &x_D+x_E+x_F=24 | ||
+ | \end{align} | ||
+ | |||
+ | The specific structure of this system of equations allows it to be solved with relatively ease. Solving, we get <math>d=\frac{24}{5}</math> | ||
+ | |||
+ | ~tsun26 | ||
==See Also== | ==See Also== |
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 |
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