Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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+ | ==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 still have the same x-coordinate. 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{\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] | ||
==See Also== | ==See Also== |
Revision as of 19:54, 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 still have the same x-coordinate. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.