# 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 $[asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("A",(0,0),SW); label("B",(20.87,0),SE); label("E",(15.87,8.66),NE); label("D",(5,8.66),NW); label("P",(5,0),S); label("Q",(15.87,0),S); label("C",(16.87,7),E); label("12",(10.935,7.794),S); label("10",(2.5,4.5),W); label("10",(18.37,4.5),E); [/asy]$

Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$. Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$, and $BC=8$ and $EC=2$. We are given that $DC=12$. Since $\angle CED = 120^{\circ}$, using Law of Cosines on $\bigtriangleup CED$ gives $$12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})$$ which gives $$144-4=DE^2+2DE$$. Adding $1$ to both sides gives $141=(DE+1)^2$, so $DE=\sqrt{141}-1$. $\bigtriangleup DAP$ and $\bigtriangleup EBQ$ are both $30-60-90$, so $AP=5$ and $BQ=5$. $PQ=DE$, and therefore $AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}$.

### Solution 2 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 3 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 4

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 $CE=x$. $$2x(x-2) + 12^2x = 2x^2 + x^2(x-2)$$ $$x^3 - 2x^2 - 140x = 0$$ By the quadratic formula (discarding the negative result), $x = 1 + \sqrt{141}$, giving $AB = 9 + \sqrt{141}$ for a final answer of $p+q=150$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 