2005 AIME I Problems/Problem 7

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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


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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. $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$. 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 $p + q = 150$.

Solution 2


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

See also

2005 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions