2005 AIME I Problems/Problem 7

Problem

In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers, find $p+q.$

Solution

Solution 1

Draw the perpendiculars from $C$ and $D$ to $AB$ , labeling the intersection points as $E$ and $F$ . This forms 2 $30-60-90$ right triangles, so $AE = 5$ and $BF = 4$ . Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$ , we find another right triangle $\triangle DGC$ . $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$ . The Pythagorean Theorem yields that $GC^2 = 12^2 - \sqrt{3}^2 = 141$ , so $EF = GC = \sqrt{141}$ . Therefore, $AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}$ , and $p + q = \boxed{150}$ .

Solution 2

Extend $AD$ and $BC$ to an intersection at point $E$ . We get an equilateral triangle $ABE$ . We denote the length of a side of $\triangle ABE$ as $s$ and solve for it using the Law of Cosines: $12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\cos{60}$ $144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)$ This simplifies to $s^2 - 18s - 60=0$ ; the quadratic formula yields the (discard the negative result) same result of $9 + \sqrt{141}$ .

Solution 3

Extend $BC$ and $AD$ to meet at point $E$ , forming an equilateral triangle $\triangle ABE$ . Draw a line from $C$ parallel to $AB$ so that it intersects $AD$ at point $F$ . Then, apply Stewart's Theorem on $\triangle CFE$ . Let $CE=x$ . $2x(x-2) + 12^2x = 2x^2 + x^2(x-2)$ $x^3 - 2x^2 - 140x = 0$ $x^2 - 2x - 140 = 0$ By the quadratic formula (discarding the negative result), $x = 1 + \sqrt{141}$ , giving $AB = 9 + \sqrt{141}$ for a final answer of $p+q=150$ .

2005 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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