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Difference between revisions of "2013 AMC 10B Problems/Problem 16"

(→‎Solution 2)
(→‎Problem)
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<math>\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}</math>
 
<math>\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}</math>
 +
<asy>
 +
pair A,B,C,D,E,P;
 +
A=(0,0);
 +
B=(80,0);
 +
C=(20,40);
 +
D=(50,30);
 +
E=(40,0);
 +
P=(33.3,13.3);
 +
draw(A--B);
 +
draw(B--C);
 +
draw(A--C);
 +
draw(C--E);
 +
draw(A--D);
 +
draw(D--E);
 +
dot(A);
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dot(B);
 +
dot(C);
 +
dot(D);
 +
dot(E);
 +
dot(P);
 +
label("A",A,NNW);
 +
label("B",B,NNE);
 +
label("C",C,ENE);
 +
label("D",D,ESE);
 +
label("E",E,SSE);
 +
label("P",P,SSE);
 +
</asy>
  
 
==Solution==
 
==Solution==

Revision as of 20:31, 22 February 2013

Problem

In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$?

$\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}$ [asy] pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,30); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("P",P,SSE); [/asy]

Solution

Let us use mass points: Assign $B$ mass $1$. Thus, because $E$ is the midpoint of $AB$, $A$ also has a mass of $1$. Similarly, $C$ has a mass of $1$. $D$ and $E$ each have a mass of $2$ because they are between $B$ and $C$ and $A$ and $B$ respectively. Note that the mass of $D$ is twice the mass of $A$, so AP must be twice as long as $PD$. PD has length $2$, so $AP$ has length $4$ and $AD$ has length $6$. Similarly, $CP$ is twice $PE$ and $PE=1.5$, so $CP=3$ and $CE=4.5$. Now note that triangle $PED$ is a $3-4-5$ right triangle with the right angle $DPE$. This means that the quadrilateral $AEDC$ is a kite. The area of a kite is half the product of the diagonals, $AD$ and $CE$. Recall that they are $6$ and $4.5$ respectively, so the area of $AEDC$ is $6*4.5/2=\boxed{\textbf{(B)} 13.5}$

Solution 2

Note that triangle $DPE$ is a right triangle, and that the four angles that have point $P$ are all right angles. Using the fact that the centroid ($P$) divides each median in a $2:1$ ratio, $AP=4$ and $CP=3$. Quadrilateral $AEDC$ is now just four right triangles. The area is $\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}$

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