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

(Fixed LaTex and added second solution)
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In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>.  What is the area of <math>AEDC</math>?
 
In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>.  What is the area of <math>AEDC</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>
  
 
==Solution==
 
==Solution==
 
Let us use mass points:
 
Let us use mass points:
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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>
  
 
==Solution 2==
 
==Solution 2==
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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 2+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}$
+
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 2+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}</math>

Revision as of 23:16, 21 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 $AEDC$?

$\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}$

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 $\frac{4\cdot 2+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}$

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