Difference between revisions of "2013 AMC 10A Problems/Problem 18"

Revision as of 16:09, 27 December 2017

Problem

Let points $A = (0, 0)$ , $B = (1, 2)$ , $C=(3, 3)$ , and $D = (4, 0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $(\frac{p}{q}, \frac{r}{s})$ , where these fractions are in lowest terms. What is $p+q+r+s$ ?

$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$

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]$

First, we shall find the area of quadrilateral $ABCD$ . This can be done in any of three ways:

Pick's Theorem: $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$

Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is $1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.$

Shoelace Method: The area is half of $|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15$ , or $\dfrac{15}{2}$ .

$[ABCD] = \frac{15}{2}$ . Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$ .

Call the point where the line through $A$ intersects $\overline{CD}$ $E$ . We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$ . Furthermore, we know that $b = 4$ , as $AD = 4$ . Thus, solving for $h$ , we find that $2h = \frac{15}{4}$ , so $h = \frac{15}{8}$ . This gives that the y coordinate of E is $\frac{15}{8}$ .

Line CD can be expressed as $y = -3x+12$ , so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$ . Solving for $x$ , we find that $x = \frac{27}{8}$ .

From this, we know that $E = \left(\frac{27}{8}, \frac{15}{8}\right)$ . $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$

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.

Page

Toolbox

Search

Difference between revisions of "2013 AMC 10A Problems/Problem 18"

Revision as of 16:09, 27 December 2017

Problem

Solution

See Also

@@ Line 8: / Line 8: @@
 ==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:
@@ 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>.
-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==