Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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+ | First, we can define <math>f(x) = a(x-2)(x-3)(x-4) +x^2</math>, which contains points <math>A</math>, <math>B</math>, and <math>C</math>. Now we find that lines <math>AB</math>, <math>AC</math>, and <math>BC</math> are defined by the equations <math>y = 5x - 6</math>, <math>y= 6x-8</math>, and <math>y=7x-12</math> respectively. 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 (eg. if we were looking at line <math>AB</math>, we would synthetically divide by the solutions <math>x=2</math> and <math>x=3</math>, because we already know <math>AB</math> intersects the graph at <math>A</math> and <math>B</math>, which have <math>x</math>-coordinates of <math>2</math> and <math>3</math>). After completing this process on all three lines, we get that the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> are <math>\frac{4a-1}{a}</math>, <math>\frac{3a-1}{a}</math>, and <math>\frac{2a-1}{a}</math> respectively. Adding these together, we get <math>\frac{9a-3}{a} = 24</math> which gives us <math>a = -\frac{1}{5}</math>. Substituting this back into the original equation, we get <math>f(x) = -\frac{1}{5}(x-2)(x-3)(x-4) + x^2</math>, and <math>f(0) = -\frac{1}{5}(-2)(-3)(-4) + 0 = \boxed{\textbf{(D)}\frac{24}{5}}</math> | ||
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+ | Solution by: vedadehhc and tdeng |
Revision as of 22:51, 16 February 2017
Problem 23
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
First, we can define , which contains points
,
, and
. 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 (eg. if we were looking at line
, we would synthetically divide by the solutions
and
, because we already know
intersects the graph at
and
, which have
-coordinates of
and
). After completing this process on all three lines, we get that the
-coordinates of
,
, and
are
,
, and
respectively. Adding these together, we get
which gives us
. Substituting this back into the original equation, we get
, and
Solution by: vedadehhc and tdeng