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> |
Revision as of 19:56, 10 August 2020
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
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
.
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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