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>. |
==See Also== | ==See Also== |
Latest revision as of 02:02, 24 July 2019
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 .
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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