Difference between revisions of "2013 AMC 12A Problems/Problem 13"
Epicwisdom (talk | contribs) m (moved 2013 AMC 12A Problems/Problems 13 to 2013 AMC 12A Problems/Problem 13: Wrong title ("Problems 13")) |
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+ | == Problem== | ||
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+ | Let points <math> A = (0,0) , \ B = (1,2), \ C = (3,3), </math> and <math> D = (4,0) </math>. Quadrilateral <math> ABCD </math> is cut into equal area pieces by a line passing through <math> A </math>. This line intersects <math> \overline{CD} </math> at point <math> \left (\frac{p}{q}, \frac{r}{s} \right ) </math>, where these fractions are in lowest terms. What is <math> p + q + r + s </math>? | ||
+ | |||
+ | <math> \textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | |||
If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods. | If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods. | ||
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Therefore the point of intersection is (<math>\frac{27}{8}</math>, <math>\frac{15}{8}</math>), and our desired result is <math>27+8+15+8=58</math>, which is <math>B</math>. | Therefore the point of intersection is (<math>\frac{27}{8}</math>, <math>\frac{15}{8}</math>), and our desired result is <math>27+8+15+8=58</math>, which is <math>B</math>. | ||
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+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=12|num-a=14}} |
Revision as of 18:40, 22 February 2013
Contents
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
Solution 1
If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods.
Pick's Theorem states that
= - , where is the number of lattice points in the interior of the polygon, and is the number of lattice points on the boundary of the polygon.
In this case,
= - =
so
=
The bottom half of the quadrilateral makes a triangle with base and half the total area, so we can deduce that the height of the triangle must be in order for its area to be . This height is the y coordinate of our desired intersection point.
Note that segment CD lies on the line . Substituting in for y, we can find that the x coordinate of our intersection point is .
Therefore the point of intersection is (, ), and our desired result is , which is .
See also
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 |