Difference between revisions of "2013 AMC 10A Problems/Problem 18"
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Line CD can be expressed as <math>y = -3x+12</math>, so the <math>x</math> coordinate of E satisfies <math>\frac{15}{8} = -3x + 12</math>. Solving for <math>x</math>, we find that <math>x = \frac{27}{8}</math>. | Line CD can be expressed as <math>y = -3x+12</math>, so the <math>x</math> coordinate of E satisfies <math>\frac{15}{8} = -3x + 12</math>. Solving for <math>x</math>, we find that <math>x = \frac{27}{8}</math>. | ||
− | From this, we know that <math>E = (\frac{27}{8}, \frac{15}{8})</math>. <math>27 + 15 + 8 + 8 = \boxed{\textbf{( | + | From this, we know that <math>E = (\frac{27}{8}, \frac{15}{8})</math>. <math>27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}</math> |
==See Also== | ==See Also== |
Revision as of 14:29, 8 February 2013
Problem
Let points ,
,
, and
. Quadrilateral
is cut into equal area pieces by a line passing through
. This line intersects
at point
, where these fractions are in lowest terms. What is
?
Solution
First, various area formulas (shoelace, splitting, etc) allow us to find that . Therefore, each equal piece that the line separates
into must have an area of
.
Call the point where the line through intersects
. We know that
. Furthermore, we know that
, as
. Thus, solving for
, we find that
, so
. This gives that the y coordinate of E is
.
Line CD can be expressed as , so the
coordinate of E satisfies
. Solving for
, we find that
.
From this, we know that .
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |