AIME 2005 I · 第 7 题

Solution

Solution 1

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 $p+q=150$.

Solution 5 (EASY EASY VERY TREMENDOUSLY EASY)

Draw a line from point $C$ to a new point $E$ on $AD$ parallel to $AB$. Draw a line from point $E$ to a new point $F$ on $AB$ parallel to $CD$. This creates parallelogram $CEFB$.

The reasoning for this is to connect the two angles that are congruent.

$\angle CEF = 60^{\circ}$ and triangle $AEF$ is an equilateral triangle with side length 8. Thus, $AF = 8$.

$ED = 10 - 8 = 2$. By the Law of Cosines, $CD^2 = ED^2 + EC^2 - 2 \cdot {ED} \cdot {EC} \cdot \cos{\angle DEC}$. Thus, $144 = 4 + c^2 - 2c$ and $c = 1 + \sqrt{141}$. $EC = FB$.

$AB = AF + FB = 8 + 1 + \sqrt{141} = 9 + \sqrt{141}$ so the answer is $p+q=150$.

-unhappyfarmer