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

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== Problem ==
 
== Problem ==
In trapezoid we have parallel to , as the midpoint of , and as the midpoint of . The area of is twice the area of . What is ?
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In trapezoid <math>ABCD</math> we have <math>\overline{AB}</math> parallel to <math>\overline{DC}</math>, <math>E</math> as the midpoint of <math>\overline{BC}</math>, and <math>F</math> as the midpoint of <math>\overline{DA}</math>. The area of <math>ABEF</math> is twice the area of <math>FECD</math>. What is <math>AB/DC</math>?
  
\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8
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<math>\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8 </math>
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== Solution 1==
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Since the height of both trapezoids are equal, and the area of <math>ABEF</math> is twice the area of <math>FECD</math>,
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<math>\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)</math>.
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<math>\frac{AB+EF}{2}=DC+EF</math>, so
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<math>AB+EF=2DC+2EF</math>.
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<math>EF</math> is exactly halfway between <math>AB</math> and <math>DC</math>, so <math>EF=\frac{AB+DC}{2}</math>.
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<math>AB+\frac{AB+DC}{2}=2DC+AB+DC</math>, so
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<math>\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB</math>, and
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<math>\frac{1}{2}AB=\frac{5}{2}DC</math>.
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<math>\frac{AB}{DC} = \boxed{5}</math>.
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==Solution 2==
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Mark <math>DC=z</math>, <math>AB=x</math>, and <math>FE=y.</math>
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Note that the heights of trapezoids <math>ABEF</math> & <math>FECD</math> are the same. Mark the height to be <math>h</math>.
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Then, we have that <math>\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)</math>.
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From this, we get that <math>x=2z+y</math>.
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We also get that <math>\tfrac{x+z}{2} \cdot h= 3(\tfrac{y+z}{2} \cdot h)</math>.
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Simplifying, we get that <math>2x=z+3y</math>
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Notice that we want <math>\tfrac{AB}{DC}=\tfrac{x}{z}</math>.
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Dividing the first equation by <math>z</math>, we get that <math>\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})</math>.
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Dividing the second equation by <math>z</math>, we get that <math>2(\tfrac{x}{z})=1+3(\tfrac{y}{z})</math>.
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Now, when we subtract the top equation from the bottom, we get that <math>\tfrac{x}{z}=5</math>
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Hence, the answer is <math>\boxed{5}</math>
  
== Solution ==
 
 
== See Also ==
 
== See Also ==
*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}}
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{{MAA Notice}}

Latest revision as of 20:22, 26 December 2020

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

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

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

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 $\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)$.

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

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

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

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

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

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

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

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

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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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