2012 AMC 12B Problems/Problem 20

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 $r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3$ , 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 1

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, by Law of Cosines, $\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14$ , 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 $\boxed{\textbf{(D)}\ 63}$ .

Solution 2

Let the area of the trapezoid be $S$ , the area of the triangle be $S_1$ , the area of the parallelogram be $S_2$ .

By Heron's Formula $S_1 = \sqrt{\frac{b+c+d-a}{2} \cdot \frac{c+d-a-b}{2} \cdot \frac{a+b+d-c}{2} \cdot \frac{b+c-a-d}{2}}$

$S_2 = \frac{S_1 \cdot 2}{c-a} \cdot a = \frac{2aS_1}{c-a}$

$S = S_1 + S_2 = S_1(1+\frac{2a}{c-a}) = S_1 \cdot \frac{c+a}{c-a} = \frac14 \cdot \frac{c+a}{c-a} \cdot \sqrt{(b+c+d-a)(c+d-a-b)(a+b+d-c)(b+c-a-d)}$

If $a = 3$ , $b = 7$ , $c = 11$ , $d = 5$

$S = \frac14 \cdot \frac{14}{8} \cdot \sqrt{(7+11+5-3)(11+5-3-7)(3+7+5-11)(7+11-3-5)} = \frac{35\sqrt{3}}{2}$

If $a = 3$ , $b = 11$ , $c = 5$ , $d = 7$

$S = \frac14 \cdot \frac{8}{2} \cdot \sqrt{(11+5+7-3)(5+7-3-11)(3+5+7-11)(11+5-3-7)}$ , which is impossible as $5+7-3-11 = -2$

If $a = 3$ , $b = 5$ , $c = 7$ , $d = 11$

$S = \frac14 \cdot \frac{10}{4} \cdot \sqrt{(5+7+11-3)(7+11-3-5)(3+5+11-7)(5+7-11-3)}$ , which is impossible as $5+7-11-3 = -2$

If $a = 5$ , $b = 11$ , $c = 7$ , $d = 3$

$S = \frac14 \cdot \frac{12}{2} \cdot \sqrt{(11+7+3-5)(7+3-5-11)(5+11+3-7)(11+7-5-3)}$ , which is impossible as $7+3-5-11 = -6$

If $a = 5$ , $b = 3$ , $c = 11$ , $d = 7$

$S = \frac14 \cdot \frac{16}{6} \cdot \sqrt{(3+11+7-5)(11+7-5-3)(5+3+7-11)(3+11-5-7)} = \frac{32\sqrt{5}}{3}$

If $a = 7$ , $b = 3$ , $c = 11$ , $d = 5$

$S = \frac14 \cdot \frac{18}{4} \cdot \sqrt{(3+11+5-7)(11+5-7-)(7+5+5-11)(3+11-7-5)} = 27$

Thus the answer is $\frac{35}{2} + \frac{32}{3} + 27 + 3 + 5$ , which rounds down to $\boxed{\textbf{(D) } 63}$

~isabelchen

Note: the three invalid cases can also be determined by the triangle inequality.

Video Solution

https://youtu.be/8w1vrsD2urs ~Math Problem Solving Skills

2012 AMC 12B (Problems • Answer Key • Resources)
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2012 AMC 12B Problems/Problem 20

Contents

Problem 20

Solution 1

Solution 2

Video Solution

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