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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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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==