2012 AMC 12B Problems/Problem 20

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 first 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 first 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 is rounded down to 63.

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2012 AMC 12B Problems/Problem 20

Problem 20

Solution