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1964 AHSME Problems/Problem 22: Difference between revisions

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Created page with "== Problem == Given parallelogram <math>ABCD</math> with <math>E</math> the midpoint of diagonal <math>BD</math>. Point <math>E</math> is connected to a point <math>F</math>..."
 
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== Solution==
== Solution==


If it works for a parallelogram <math>ABCD</math>, it should also work for a unit square, with <math>A(0, 0), B(0, 1), C(1, 1), D(1, 0)</math>.  We are given that <math>E</math> is the midpoint of <math>BD</math>, so <math>E(0.5, 0.5)</math>.  If <math>F</math> is on <math>DA</math>, then <math>F(x, 0)</math>.  We note that <math>DF = 1-x</math> and <math>DA = 1</math>, so <math>\frac{DF} = \frac{1}{3}DA</math> means <math>1-x = \frac{1}{3}</math>, or <math>x = \frac{2}{3}</math>, and hence <math>F(\frac{2}{3}, 0)</math>.
If it works for a parallelogram <math>ABCD</math>, it should also work for a unit square, with <math>A(0, 0), B(0, 1), C(1, 1), D(1, 0)</math>.  We are given that <math>E</math> is the midpoint of <math>BD</math>, so <math>E(0.5, 0.5)</math>.  If <math>F</math> is on <math>DA</math>, then <math>F(x, 0)</math>.  We note that <math>DF = 1-x</math> and <math>DA = 1</math>, so <math>DF = \frac{1}{3}DA</math> means <math>1-x = \frac{1}{3}</math>, or <math>x = \frac{2}{3}</math>, and hence <math>F(\frac{2}{3}, 0)</math>.


We note that <math>\triangle DFE</math> has a base <math>DF</math> that is <math>\frac{1}{3}</math> and an altitude from <math>E</math> to <math>DF</math> that is <math>\frac{1}{2}</math>.  Therefore, <math>[DEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{12}</math>.
We note that <math>\triangle DFE</math> has a base <math>DF</math> that is <math>\frac{1}{3}</math> and an altitude from <math>E</math> to <math>DF</math> that is <math>\frac{1}{2}</math>.  Therefore, <math>[DEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{12}</math>.
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Quadrilateral <math>ABEF</math> can be split into <math>\triangle ABE</math> and <math>\triangle AEF</math>.  The first triangle is <math>\frac{1}{4}</math> of the unit square cut diagonally, so <math>[ABE] = \frac{1}{4}</math>.  The second triangle has base <math>AF</math> that is <math>\frac{2}{3}</math> and height <math>E</math> to <math>AF</math> that is <math>\frac{1}{2}</math>.  Therefore, <math>[AEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{6}</math>.
Quadrilateral <math>ABEF</math> can be split into <math>\triangle ABE</math> and <math>\triangle AEF</math>.  The first triangle is <math>\frac{1}{4}</math> of the unit square cut diagonally, so <math>[ABE] = \frac{1}{4}</math>.  The second triangle has base <math>AF</math> that is <math>\frac{2}{3}</math> and height <math>E</math> to <math>AF</math> that is <math>\frac{1}{2}</math>.  Therefore, <math>[AEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{6}</math>.


The entire quadrilateral <math>ABEF</math> has area <math>\frac{1}{4} + \frac{1}{6} = \frac{5}{12}</math>.  This is <math>5</math> times larger than the area if <math>\triangle DFE</math>, so the ratio is <math>1:5</math>, or <math>\boxed{\textbf{(C)}}</math>.
The entire quadrilateral <math>ABEF</math> has area <math>\frac{1}{4} + \frac{1}{6} = \frac{5}{12}</math>.  This is <math>5</math> times larger than the area of <math>\triangle DFE</math>, so the ratio is <math>1:5</math>, or <math>\boxed{\textbf{(C)}}</math>.
 
== Solution 2 ==
<cmath>
\begin{align*}
\frac{A_{DFE}}{A_{ADE}} &= \frac{\frac{1}{3}DA}{DA} = \frac{1}{3} \\
A_{DFE} &= \frac{1}{3}A_{ADE} \\
A_{ADE} &= \frac{1}{2}A_{ADB} = \frac{1}{4}A_{ABCD} \\
\implies A_{DFE} &= \frac{1}{3} \cdot \frac{1}{4} A_{ABCD} = \frac{1}{12} A_{ABCD} \\
A_{ABEF} &= A_{ABD} - A_{DFE} \\
&= \frac{1}{2}A_{ABCD} - \frac{1}{12}A_{ABCD} \\
&= \frac{5}{12}A_{ABCD}
\end{align*}
</cmath>
Therefore, <math>\frac{A_{DFE}}{A_{ABEF}} = \frac{\frac{1}{12}}{\frac{5}{12}} = \frac{1}{5}</math>, giving us the answer <math>\boxed{\textbf{(C)}}</math>. -nullptr07


==See Also==
==See Also==

Latest revision as of 22:28, 29 June 2023

Problem

Given parallelogram $ABCD$ with $E$ the midpoint of diagonal $BD$. Point $E$ is connected to a point $F$ in $DA$ so that $DF=\frac{1}{3}DA$. What is the ratio of the area of $\triangle DFE$ to the area of quadrilateral $ABEF$?

$\textbf{(A)}\ 1:2 \qquad \textbf{(B)}\ 1:3 \qquad \textbf{(C)}\ 1:5 \qquad \textbf{(D)}\ 1:6 \qquad \textbf{(E)}\ 1:7$

Solution

If it works for a parallelogram $ABCD$, it should also work for a unit square, with $A(0, 0), B(0, 1), C(1, 1), D(1, 0)$. We are given that $E$ is the midpoint of $BD$, so $E(0.5, 0.5)$. If $F$ is on $DA$, then $F(x, 0)$. We note that $DF = 1-x$ and $DA = 1$, so $DF = \frac{1}{3}DA$ means $1-x = \frac{1}{3}$, or $x = \frac{2}{3}$, and hence $F(\frac{2}{3}, 0)$.

We note that $\triangle DFE$ has a base $DF$ that is $\frac{1}{3}$ and an altitude from $E$ to $DF$ that is $\frac{1}{2}$. Therefore, $[DEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{12}$.

Quadrilateral $ABEF$ can be split into $\triangle ABE$ and $\triangle AEF$. The first triangle is $\frac{1}{4}$ of the unit square cut diagonally, so $[ABE] = \frac{1}{4}$. The second triangle has base $AF$ that is $\frac{2}{3}$ and height $E$ to $AF$ that is $\frac{1}{2}$. Therefore, $[AEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{6}$.

The entire quadrilateral $ABEF$ has area $\frac{1}{4} + \frac{1}{6} = \frac{5}{12}$. This is $5$ times larger than the area of $\triangle DFE$, so the ratio is $1:5$, or $\boxed{\textbf{(C)}}$.

Solution 2

\begin{align*} \frac{A_{DFE}}{A_{ADE}} &= \frac{\frac{1}{3}DA}{DA} = \frac{1}{3} \\ A_{DFE} &= \frac{1}{3}A_{ADE} \\ A_{ADE} &= \frac{1}{2}A_{ADB} = \frac{1}{4}A_{ABCD} \\ \implies A_{DFE} &= \frac{1}{3} \cdot \frac{1}{4} A_{ABCD} = \frac{1}{12} A_{ABCD} \\ A_{ABEF} &= A_{ABD} - A_{DFE} \\ &= \frac{1}{2}A_{ABCD} - \frac{1}{12}A_{ABCD} \\ &= \frac{5}{12}A_{ABCD} \end{align*} Therefore, $\frac{A_{DFE}}{A_{ABEF}} = \frac{\frac{1}{12}}{\frac{5}{12}} = \frac{1}{5}$, giving us the answer $\boxed{\textbf{(C)}}$. -nullptr07

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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