Difference between revisions of "2013 AMC 10A Problems/Problem 18"
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label("E",E,NE); | label("E",E,NE); | ||
label("$4$",(A+D)/2,S); | label("$4$",(A+D)/2,S); | ||
− | label("$\frac{27}{8}$",( | + | label("$\frac{27}{8}$",(A+EE)/2,S); |
label("$\frac{15}{8}$",(E+EE)/2,W); | label("$\frac{15}{8}$",(E+EE)/2,W); | ||
</asy></center> | </asy></center> | ||
Line 51: | Line 51: | ||
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> | 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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <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("F",EE,S); | ||
+ | label("$4$",(A+D)/2,S); | ||
+ | label("$x$",(A+EE)/2,S; | ||
+ | label("$4-x$",(D+EE)/2,S); | ||
+ | label("$\frac{15}{8}$",(E+EE)/2,W); | ||
+ | </asy></center> | ||
+ | |||
+ | Following the steps above, you can find that the height of triangle <math>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>. 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 19:44, 1 January 2019
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
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
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("F",EE,S); label("$4$",(A+D)/2,S); label("$x$",(A+EE)/2,S; label("$4-x$",(D+EE)/2,S); label("$\frac{15}{8}$",(E+EE)/2,W); (Error compiling LaTeX. label("$x$",(A+EE)/2,S; ^ de021056a2cc2ef4e3e5c00687df10ec92699974.asy: 26.23: syntax error error: could not load module 'de021056a2cc2ef4e3e5c00687df10ec92699974.asy')
Following the steps above, you can find that the height of triangle is , and from there split the base into two parts, , and , such that is the segment from the origin to the 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.