Difference between revisions of "2013 AMC 10A Problems/Problem 18"
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==Solution== | ==Solution== | ||
+ | |||
+ | <center><asy> | ||
+ | size(8cm); | ||
+ | pair A, B, C, D, E, EE; | ||
+ | A = (0,0); | ||
+ | B = (1,2); | ||
+ | C = (3,3); | ||
+ | D = (4,0); | ||
+ | E = (27/8,15/8); | ||
+ | EE = (27/8,0); | ||
+ | draw(A--B--C--D--A--E); | ||
+ | draw(E--EE,linetype("8 8")); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | draw(rightanglemark(E,EE,D,4)); | ||
+ | label("A",A,SW); | ||
+ | label("B",B,NW); | ||
+ | label("C",C,NE); | ||
+ | label("D",D,SE); | ||
+ | label("E",E,NE); | ||
+ | label("$4$",(A+D)/2,S); | ||
+ | label("$\frac{27}{8}$",(D+EE)/2,S); | ||
+ | label("$\frac{15}{8}$",(E+EE)/2,W); | ||
+ | </asy></center> | ||
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: | ||
Line 23: | Line 50: | ||
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{(B) }58}</math> | + | From this, we know that <math>E = \left(\frac{27}{8}, \frac{15}{8}\right)</math>. <math>27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}</math> |
==See Also== | ==See Also== |
Revision as of 17:09, 27 December 2017
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 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 .
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.