Difference between revisions of "2012 AMC 12B Problems/Problem 20"
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In the first case, <math>\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14</math>, so <math>\sin (\angle BCD) = \sqrt{1-121/196} = 5\sqrt{3}/14</math>. Therefore the area of this trapezoid is <math>\frac{1}{2} (3+11) \cdot 7 \cdot 5\sqrt{3}/14 = \frac{35}{2}\sqrt{3}</math>. | In the first case, <math>\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14</math>, so <math>\sin (\angle BCD) = \sqrt{1-121/196} = 5\sqrt{3}/14</math>. Therefore the area of this trapezoid is <math>\frac{1}{2} (3+11) \cdot 7 \cdot 5\sqrt{3}/14 = \frac{35}{2}\sqrt{3}</math>. | ||
− | In the | + | In the second case, <math>\cos(\angle BCD) = (6^2+7^2-3^2)/(2\cdot 6\cdot 7) = 19/21</math>, so <math>\sin (\angle BCD) = \sqrt{1-361/441} = 4\sqrt{5}/21</math>. Therefore the area of this trapezoid is <math>\frac{1}{2} (5+11) \cdot 7 \cdot 4\sqrt{5}/21 =\frac{32}{3}\sqrt{5}</math>. |
− | In the | + | In the third case, <math>\angle BCD = 90^{\circ}</math>, therefore the area of this trapezoid is <math>\frac{1}{2} (7+11) \cdot 3 = 27</math>. |
So <math>r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5</math>, which is rounded down to <math>63... \framebox{D}</math>. | So <math>r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5</math>, which is rounded down to <math>63... \framebox{D}</math>. |
Revision as of 17:41, 17 February 2014
Problem 20
A trapezoid has side lengths 3, 5, 7, and 11. The sums of all the possible areas of the trapezoid can be written in the form of , where , , and are rational numbers and and are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to ?
Solution
Name the trapezoid , where is parallel to , , and . Draw a line through parallel to , crossing the side at . Then , . One needs to guarantee that , so there are only three possible trapezoids:
In the first case, , so . Therefore the area of this trapezoid is .
In the second case, , so . Therefore the area of this trapezoid is .
In the third case, , therefore the area of this trapezoid is .
So , which is rounded down to .
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
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