Difference between revisions of "1964 AHSME Problems/Problem 22"
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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> | + | 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 | + | 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
Contents
[hide]Problem
Given parallelogram with
the midpoint of diagonal
. Point
is connected to a point
in
so that
. What is the ratio of the area of
to the area of quadrilateral
?
Solution
If it works for a parallelogram , it should also work for a unit square, with
. We are given that
is the midpoint of
, so
. If
is on
, then
. We note that
and
, so
means
, or
, and hence
.
We note that has a base
that is
and an altitude from
to
that is
. Therefore,
.
Quadrilateral can be split into
and
. The first triangle is
of the unit square cut diagonally, so
. The second triangle has base
that is
and height
to
that is
. Therefore,
.
The entire quadrilateral has area
. This is
times larger than the area of
, so the ratio is
, or
.
Solution 2
Therefore,
, giving us the answer
. -nullptr07
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AHSME Problems and Solutions |
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