2013 AMC 10A Problems/Problem 18
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
![[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("$\frac{27}{8}$",(A+EE)/2,S); label("$\frac{15}{8}$",(E+EE)/2,W); [/asy]](http://latex.artofproblemsolving.com/8/b/3/8b3b8e056ad8b3c97066b76d8a5d0c23c0cf43be.png)
First, we shall find the area of quadrilateral . This can be done in any of three ways:
Pick's Theorem: ![$[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$](http://latex.artofproblemsolving.com/f/a/0/fa08a023a9b7850954a34f84a178ae7b0f0f8957.png)
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 
.
![$[ABCD] = \frac{15}{2}$](http://latex.artofproblemsolving.com/0/1/d/01d425fd257b26e8f604a40eeb6bcb56995bdef9.png)
. 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 ![$[ADE] = \frac{15}{4} = \frac{bh}{2}$](http://latex.artofproblemsolving.com/7/5/b/75b6dc7ed815aeb31c85e08c1ca5b86e3b0ebc67.png)
. 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, F;
A = (0,0);
B = (1,2);
C = (3,3);
D = (4,0);
E = (27/8,15/8);
F = (27/8,0);
draw(A--B--C--D--A--E);
draw(E--F,linetype("8 8"));
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
draw(rightanglemark(E,F,D,4));
label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);
label("E",E,NE);
label("F",F,S);
label("$4$",(A+D)/2,S);
label("$x$",(A+F)/2,S;
label("$4-x$",(D+F)/2,S);
label("$\frac{15}{8}$",(E+F)/2,W);
(Error making remote request. Unknown error_msg)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. 
