Difference between revisions of "2013 AMC 10A Problems/Problem 18"
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First, we shall find the area of quadrilateral <math>ABCD</math>. This can be done in any of three ways: | First, we shall find the area of quadrilateral <math>ABCD</math>. This can be done in any of three ways: | ||
− | Pick's Theorem: <math>[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{ | + | Pick's Theorem: <math>[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.</math> |
Splitting: Drop perpendiculars from <math>B</math> and <math>C</math> to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is <math>1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.</math> | Splitting: Drop perpendiculars from <math>B</math> and <math>C</math> to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is <math>1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.</math> |
Revision as of 11:40, 16 February 2015
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, we shall find the area of quadrilateral . This can be done in any of three ways:
Pick's Theorem:
Splitting: Drop perpendiculars from and to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is
Shoelace Theorem: The area is half of , or .
. 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 |
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