Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 10B Problems/Problem 16"
(→‎Solution 1)
 
(One intermediate revision by one other user not shown)
Line 33: Line 33:
 
<math>\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15</math>
 
<math>\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15</math>
  
−
==Solution 1==
+
==Solution 1 ( mass points) ==
 
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>. Since the diagonals of quadrilaterals <math>AEDC</math>, <math>AD</math> and <math>CE</math>, are perpendicular, the area of <math>AEDC</math> is <math>\frac{6 \times 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 <math>AP</math> 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>. Since the diagonals of quadrilaterals <math>AEDC</math>, <math>AD</math> and <math>CE</math>, are perpendicular, the area of <math>AEDC</math> is <math>\frac{6 \times 4.5}{2}=\boxed{\textbf{(B)} 13.5}</math>
  
 
==Solution 2==
 
==Solution 2==

Latest revision as of 15:28, 30 October 2024

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 $AEDC$?

[asy] unitsize(0.2cm); 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,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,SSE); label("P",P,SSW); [/asy]

$\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15$

Solution 1 ( mass points)

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 $DPE$. Since the diagonals of quadrilaterals $AEDC$, $AD$ and $CE$, are perpendicular, the area of $AEDC$ is $\frac{6 \times 4.5}{2}=\boxed{\textbf{(B)} 13.5}$

Solution 2

Note that triangle $DPE$ is a right triangle, and that the four angles (angles $APC, CPD, DPE,$ and $EPA$) 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 $\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}$

Solution 3

From the solution above, we can find that the lengths of the diagonals are $6$ and $4.5$. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is $\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{\textbf{(B)} 13.5}$

Solution 4

From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.

Solution 5

We know that $[AEDC]=\frac{3}{4}[ABC]$, and $[ABC]=3[PAC]$ using median properties. So Now we try to find $[PAC]$. Since $\triangle PAC\sim \triangle PDE$, then the side lengths of $\triangle PAC$ are twice as long as $\triangle PDE$ since $D$ and $E$ are midpoints. Therefore, $\frac{[PAC]}{[PDE]}=2^2=4$. It suffices to compute $[PDE]$. Notice that $(1.5, 2, 2.5)$ is a Pythagorean Triple, so $[PDE]=\frac{1.5\times 2}{2}=1.5$. This implies $[PAC]=1.5\cdot 4=6$, and then $[ABC]=3\cdot 6=18$. Finally, $[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}$.

~CoolJupiter

Solution 6

As from Solution 4, we find the area of $\triangle DPE$ to be $\frac{3}{2}$. Because $DE = \frac{5}{2}$, the altitude perpendicular to $DE = \frac{6}{5}$. Also, because $DE || AC$, $\triangle ABC$ is similar to $\triangle{DBE}$ with side length ratio $2:1$, so $AC=5$ and the altitude perpendicular to $AC = \frac{12}{5}$. The altitude of trapezoid $ACDE$ is then $\frac{18}{5}$ and the bases are $\frac{5}{2}$ and $5$. So, we use the formula for the area of a trapezoid to find the area of $ACDE = \boxed{13.5}$

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_16&oldid=231022"