Difference between revisions of "2013 AMC 10A Problems/Problem 18"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
<center><asy> | <center><asy> | ||
Line 78: | Line 78: | ||
label("F",F,S); | label("F",F,S); | ||
label("$4$",(A+D)/2,S); | label("$4$",(A+D)/2,S); | ||
− | label("$x$",(A | + | label("$x$",(A+F)/2,S); |
− | |||
label("$\frac{15}{8}$",(E+F)/2,W); | label("$\frac{15}{8}$",(E+F)/2,W); | ||
</asy></center> | </asy></center> | ||
− | + | Let the point where the altitude from <math>E</math> to <math>\overline{AD}</math> be labeled <math>F</math>. | |
− | Following the steps above, you can find that the height of | + | Following the steps above, you can find that the height of <math>\triangle ADE</math> is <math>\frac{15}{8}</math>, and from there split the base into two parts, <math>x</math>, and <math>4-x</math>, such that <math>x</math> is the segment from the origin to the point <math>F</math>, and <math>4-x</math> is the segment from point <math>F</math> to point <math>D</math>. Then, by the Pythagorean Theorem, <math>x=\frac{27}{8}</math>, and the answer is <math>\boxed{\textbf{(B) }58}</math> |
==See Also== | ==See Also== |
Revision as of 11:18, 18 April 2020
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 1
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 Method: 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 .
Solution 2
Let the point where the altitude from to be labeled . Following the steps above, you can find that the height of is , and from there split the base into two parts, , and , such that is the segment from the origin to the point , and is the segment from point to point . Then, by the Pythagorean Theorem, , and the answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.