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

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Let us use mass points:
 
Let us use mass points:
 
Assign <math>B</math> mass <math>1</math>.  Thus, because <math>E</math> is the midpoint of <math>AB</math>, <math>A</math> also has a mass of <math>1</math>.  Similarly, <math>C</math> has a mass of <math>1</math>.  <math>D</math> and <math>E</math> each have a mass of <math>2</math> because they are between <math>B</math> and <math>C</math> and <math>A</math> and <math>B</math> respectively.  Note that the mass of <math>D</math> is twice the mass of <math>A</math>, so AP must be twice as long as <math>PD</math>.  PD has length <math>2</math>, so <math>AP</math> has length <math>4</math> and <math>AD</math> has length <math>6</math>.  Similarly, <math>CP</math> is twice <math>PE</math> and <math>PE=1.5</math>, so <math>CP=3</math> and <math>CE=4.5</math>.  Now note that triangle <math>PED</math> is a <math>3-4-5</math> right triangle with the right angle <math>DPE</math>.  This means that the quadrilateral <math>AEDC</math> is a kite.  The area of a kite is half the product of the diagonals, <math>AD</math> and <math>CE</math>.  Recall that they are <math>6</math> and <math>4.5</math> respectively, so the area of <math>AEDC</math> is <math>6*4.5/2=\boxed{\textbf{(B)} 13.5}</math>
 
Assign <math>B</math> mass <math>1</math>.  Thus, because <math>E</math> is the midpoint of <math>AB</math>, <math>A</math> also has a mass of <math>1</math>.  Similarly, <math>C</math> has a mass of <math>1</math>.  <math>D</math> and <math>E</math> each have a mass of <math>2</math> because they are between <math>B</math> and <math>C</math> and <math>A</math> and <math>B</math> respectively.  Note that the mass of <math>D</math> is twice the mass of <math>A</math>, so AP must be twice as long as <math>PD</math>.  PD has length <math>2</math>, so <math>AP</math> has length <math>4</math> and <math>AD</math> has length <math>6</math>.  Similarly, <math>CP</math> is twice <math>PE</math> and <math>PE=1.5</math>, so <math>CP=3</math> and <math>CE=4.5</math>.  Now note that triangle <math>PED</math> is a <math>3-4-5</math> right triangle with the right angle <math>DPE</math>.  This means that the quadrilateral <math>AEDC</math> is a kite.  The area of a kite is half the product of the diagonals, <math>AD</math> and <math>CE</math>.  Recall that they are <math>6</math> and <math>4.5</math> respectively, so the area of <math>AEDC</math> is <math>6*4.5/2=\boxed{\textbf{(B)} 13.5}</math>
 +
<asy>
 +
pair A,B,C,D,E,P;
 +
A=(0,0);
 +
B=(80,0);
 +
C=(20,40);
 +
D=(50,20);
 +
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, mass 1",A,NNW);
 +
label("B, mass 1",B,NNE);
 +
label("C, mass 1",C,ENE);
 +
label("D, mass 2",D,ESE);
 +
label("E, mass 2",E,SSE);
 +
label("P, mass 3",P,SSE);
 +
</asy>
  
 
==Solution 2==
 
==Solution 2==
 
Note that triangle <math>DPE</math> is a right triangle, and that the four angles that have point <math>P</math> are all right angles. Using the fact that the centroid (<math>P</math>) divides each median in a <math>2:1</math> ratio, <math>AP=4</math> and <math>CP=3</math>. Quadrilateral <math>AEDC</math> is now just four right triangles. The area is <math>\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}</math>
 
Note that triangle <math>DPE</math> is a right triangle, and that the four angles that have point <math>P</math> are all right angles. Using the fact that the centroid (<math>P</math>) divides each median in a <math>2:1</math> ratio, <math>AP=4</math> and <math>CP=3</math>. Quadrilateral <math>AEDC</math> is now just four right triangles. The area is <math>\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}</math>

Revision as of 19:34, 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 $AEDC$?

$\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}15$ [asy] pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); 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 $6*4.5/2=\boxed{\textbf{(B)} 13.5}$ [asy] pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); 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, mass 1",A,NNW); label("B, mass 1",B,NNE); label("C, mass 1",C,ENE); label("D, mass 2",D,ESE); label("E, mass 2",E,SSE); label("P, mass 3",P,SSE); [/asy]

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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