Difference between revisions of "2012 AMC 12B Problems/Problem 20"
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== Problem 20 == | == Problem 20 == | ||
− | A trapezoid has side lengths 3, 5, 7, and 11. The | + | A trapezoid has side lengths 3, 5, 7, and 11. The sum of all the possible areas of the trapezoid can be written in the form of <math>r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3</math>, where <math>r_1</math>, <math>r_2</math>, and <math>r_3</math> are rational numbers and <math>n_1</math> and <math>n_2</math> are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to <math>r_1+r_2+r_3+n_1+n_2</math>? |
<math>\textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65</math> | <math>\textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65</math> | ||
− | ==Solution== | + | ==Solution 1== |
Name the trapezoid <math>ABCD</math>, where <math>AB</math> is parallel to <math>CD</math>, <math>AB<CD</math>, and <math>AD<BC</math>. Draw a line through <math>B</math> parallel to <math>AD</math>, crossing the side <math>CD</math> at <math>E</math>. Then <math>BE=AD</math>, <math>EC=DC-DE=DC-AB</math>. One needs to guarantee that <math>BE+EC>BC</math>, so there are only three possible trapezoids: | Name the trapezoid <math>ABCD</math>, where <math>AB</math> is parallel to <math>CD</math>, <math>AB<CD</math>, and <math>AD<BC</math>. Draw a line through <math>B</math> parallel to <math>AD</math>, crossing the side <math>CD</math> at <math>E</math>. Then <math>BE=AD</math>, <math>EC=DC-DE=DC-AB</math>. One needs to guarantee that <math>BE+EC>BC</math>, so there are only three possible trapezoids: | ||
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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, by Law of Cosines, <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 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 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>. | ||
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So <math>r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5</math>, which rounds down to <math>\boxed{\textbf{(D)}\ 63}</math>. | So <math>r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5</math>, which rounds down to <math>\boxed{\textbf{(D)}\ 63}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | [[File:2012AMC12BProblem20Solution2.PNG|center|500px]] | ||
+ | |||
+ | Let the area of the trapezoid be <math>S</math>, the area of the triangle be <math>S_1</math>, the area of the parallelogram be <math>S_2</math>. | ||
+ | |||
+ | By [https://en.wikipedia.org/wiki/Heron%27s_formula Heron's Formula] <math>S_1 = \sqrt{\frac{b+c+d-a}{2} \cdot \frac{c+d-a-b}{2} \cdot \frac{a+b+d-c}{2} \cdot \frac{b+c-a-d}{2}}</math> | ||
+ | |||
+ | <math>S_2 = \frac{S_1 \cdot 2}{c-a} \cdot a = \frac{2aS_1}{c-a}</math> | ||
+ | |||
+ | <math>S = S_1 + S_2 = S_1(1+\frac{2a}{c-a}) = S_1 \cdot \frac{c+a}{c-a} = \frac14 \cdot \frac{c+a}{c-a} \cdot \sqrt{(b+c+d-a)(c+d-a-b)(a+b+d-c)(b+c-a-d)}</math> | ||
+ | |||
+ | If <math>a = 3</math>, <math>b = 7</math>, <math>c = 11</math>, <math>d = 5</math> | ||
+ | |||
+ | <math>S = \frac14 \cdot \frac{14}{8} \cdot \sqrt{(7+11+5-3)(11+5-3-7)(3+7+5-11)(7+11-3-5)} = \frac{35\sqrt{3}}{2}</math> | ||
+ | |||
+ | |||
+ | If <math>a = 3</math>, <math>b = 11</math>, <math>c = 5</math>, <math>d = 7</math> | ||
+ | |||
+ | <math>S = \frac14 \cdot \frac{8}{2} \cdot \sqrt{(11+5+7-3)(5+7-3-11)(3+5+7-11)(11+5-3-7)}</math>, which is impossible as <math>5+7-3-11 = -2</math> | ||
+ | |||
+ | |||
+ | If <math>a = 3</math>, <math>b = 5</math>, <math>c = 7</math>, <math>d = 11</math> | ||
+ | |||
+ | <math>S = \frac14 \cdot \frac{10}{4} \cdot \sqrt{(5+7+11-3)(7+11-3-5)(3+5+11-7)(5+7-11-3)}</math>, which is impossible as <math>5+7-11-3 = -2</math> | ||
+ | |||
+ | |||
+ | If <math>a = 5</math>, <math>b = 11</math>, <math>c = 7</math>, <math>d = 3</math> | ||
+ | |||
+ | <math>S = \frac14 \cdot \frac{12}{2} \cdot \sqrt{(11+7+3-5)(7+3-5-11)(5+11+3-7)(11+7-5-3)}</math>, which is impossible as <math>7+3-5-11 = -6</math> | ||
+ | |||
+ | |||
+ | If <math>a = 5</math>, <math>b = 3</math>, <math>c = 11</math>, <math>d = 7</math> | ||
+ | |||
+ | <math>S = \frac14 \cdot \frac{16}{6} \cdot \sqrt{(3+11+7-5)(11+7-5-3)(5+3+7-11)(3+11-5-7)} = \frac{32\sqrt{5}}{3}</math> | ||
+ | |||
+ | |||
+ | If <math>a = 7</math>, <math>b = 3</math>, <math>c = 11</math>, <math>d = 5</math> | ||
+ | |||
+ | <math>S = \frac14 \cdot \frac{18}{4} \cdot \sqrt{(3+11+5-7)(11+5-7-)(7+5+5-11)(3+11-7-5)} = 27</math> | ||
+ | |||
+ | Thus the answer is <math>\frac{35}{2} + \frac{32}{3} + 27 + 3 + 5</math>, which rounds down to <math>\boxed{\textbf{(D) } 63}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | Note: the three invalid cases can also be determined by the triangle inequality. | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://youtu.be/8w1vrsD2urs ~Math Problem Solving Skills | ||
+ | |||
+ | |||
== See Also == | == See Also == |
Latest revision as of 07:03, 6 September 2024
Problem 20
A trapezoid has side lengths 3, 5, 7, and 11. The sum 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 1
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, by Law of Cosines, , 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 rounds down to .
Solution 2
Let the area of the trapezoid be , the area of the triangle be , the area of the parallelogram be .
If , , ,
If , , ,
, which is impossible as
If , , ,
, which is impossible as
If , , ,
, which is impossible as
If , , ,
If , , ,
Thus the answer is , which rounds down to
Note: the three invalid cases can also be determined by the triangle inequality.
Video Solution
https://youtu.be/8w1vrsD2urs ~Math Problem Solving Skills
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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