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2012 AMC 12B Problems/Problem 20

Revision as of 09:48, 5 February 2017 by Bananapnaplsxie (talk | contribs) (→‎Solution)

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 $r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3$, where $r_1$, $r_2$, and $r_3$ are rational numbers and $n_1$ and $n_2$ are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to $r_1+r_2+r_3+n_1+n_2$?

$\textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65$

Solution

Name the trapezoid $ABCD$, where $AB$ is parallel to $CD$, $AB<CD$, and $AD<BC$. Draw a line through $B$ parallel to $AD$, crossing the side $CD$ at $E$. Then $BE=AD$, $EC=DC-DE=DC-AB$. One needs to guarantee that $BE+EC>BC$, so there are only three possible trapezoids:

\[AB=3, BC=7, CD=11, DA=5, CE=8\] \[AB=5, BC=7, CD=11, DA=3, CE=6\] \[AB=7, BC=5, CD=11, DA=3, CE=4\]


In the first case, $\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14$, so $\sin (\angle BCD) = \sqrt{1-121/196} = 5\sqrt{3}/14$. Therefore the area of this trapezoid is $\frac{1}{2} (3+11) \cdot 7 \cdot 5\sqrt{3}/14 = \frac{35}{2}\sqrt{3}$.

In the second case, $\cos(\angle BCD) = (6^2+7^2-3^2)/(2\cdot 6\cdot 7) = 19/21$, so $\sin (\angle BCD) = \sqrt{1-361/441} = 4\sqrt{5}/21$. Therefore the area of this trapezoid is $\frac{1}{2} (5+11) \cdot 7 \cdot 4\sqrt{5}/21 =\frac{32}{3}\sqrt{5}$.

In the third case, $\angle BCD = 90^{\circ}$, therefore the area of this trapezoid is $\frac{1}{2} (7+11) \cdot 3 = 27$.

So $r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5$, which rounds down to $\boxed{\textbf{(D)}\ 63}$.

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AMC 12 Problems and Solutions

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