Difference between revisions of "2013 AMC 10A Problems/Problem 18"
(→Solution) |
m (→Problem) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Let points <math>A = (0, 0)</math>, <math>B = (1, 2)</math>, <math>C=(3, 3)</math>, and <math>D = (4, 0)</math>. Quadrilateral <math>ABCD</math> is cut into equal area pieces by a line passing through <math>A</math>. This line intersects <math>\overline{CD}</math> at point <math>(\frac{p}{q}, \frac{r}{s})</math>, where these fractions are in lowest terms. What is <math>p+q+r+s</math>? | + | Let points <math>A = (0, 0)</math>, <math>B = (1, 2)</math>, <math>C=(3, 3)</math>, and <math>D = (4, 0)</math>. Quadrilateral <math>ABCD</math> is cut into equal area pieces by a line passing through <math>A</math>. This line intersects <math>\overline{CD}</math> at point <math>\left(\frac{p}{q}, \frac{r}{s}\right)</math>, where these fractions are in lowest terms. What is <math>p+q+r+s</math>? |
Revision as of 16:10, 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
![[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]](http://latex.artofproblemsolving.com/0/0/d/00dd16fce9ef7f4092eb4e5ec858cda56c18b9f9.png)
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.