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Difference between revisions of "2005 AMC 10B Problems/Problem 23"

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<math>\frac{AB}{DC} = \boxed{5}</math>.
 
<math>\frac{AB}{DC} = \boxed{5}</math>.
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==~@rrusczyk==
  
 
==Solution 2==
 
==Solution 2==

Revision as of 12:54, 23 December 2019

Problem

In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$, $E$ as the midpoint of $\overline{BC}$, and $F$ as the midpoint of $\overline{DA}$. The area of $ABEF$ is twice the area of $FECD$. What is $AB/DC$?

$\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8$

Solution 1

Since the height of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$,

$\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)$.

$\frac{AB+EF}{2}=DC+EF$, so

$AB+EF=2DC+2EF$.

$EF$ is exactly halfway between $AB$ and $DC$, so $EF=\frac{AB+DC}{2}$.

$AB+\frac{AB+DC}{2}=2DC+AB+DC$, so

$\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB$, and

$\frac{1}{2}AB=\frac{5}{2}DC$.

$\frac{AB}{DC} = \boxed{5}$.

~@rrusczyk

Solution 2

Mark $DC=z$, $AB=x$, and $FE=y.$ Note that the heights of trapezoids $ABEF$ & $FECD$ are the same. Mark the height to be $h$.

Then, we have that $\frac{x+y}{2}\cdot h=2(\frac{y+z}{2} \cdot h)$.

From this, we get that $x=2z+y$.

We also get that $\frac{x+z}{2} \cdot h= 3(\frac{y+z}{2} \cdot h)$.

Simplifying, we get that $2x=z+3y$

Notice that we want $\frac{AB}{DC}=\frac{x}{z}$.

Dividing the first equation by $z$, we get that $\frac{x}{z}=2+\frac{y}{z}\implies 3(\frac{x}{z})=6+3(\frac{y}{z})$.

Dividing the second equation by $z$, we get that $2(\frac{x}{z})=1+3(\frac{y}{z})$.

Now, when we subtract the top equation from the bottom, we get that $\frac{x}{z}=5$

Hence, the answer is $\boxed{5}$

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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

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