AIME 2006 II · 第 12 题

Solution 1

Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$. Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$. So $\Delta{GBC} \sim \Delta{EAF}$.

$$[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.$$ Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$. Therefore, the answer is $429+433+3=\boxed{865}$.

Solution 2: Analytic Geometry/Coord Bash

Solution by e_power_pi_times_i/edited by srisainandan6

Let the center of the circle be $O$ and the origin. Then, $A (0,2)$, $B (-\sqrt{3}, -1)$, $C (\sqrt{3}, -1)$. $D$ and $E$ can be calculated easily knowing $AD$ and $AE$, $D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})$, $E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})$. As $DF$ and $EF$ are parallel to $AE$ and $AD$, $F (-1, -12\sqrt{3}+2)$. $G$ and $A$ is the intersection between $AF$ and circle $O$. Therefore $G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})$. Using the Shoelace Theorem, $[CBG] = \dfrac{429\sqrt{3}}{433}$, so the answer is $\boxed{865}$. Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation.

Note by chrisdiamond10: We can save time calculating the area of the triangle once we have the coordinates of $B,C,G$ by using $\frac{b\cdot h}{2}$. Use $BC$ as the base, then the base is $2\sqrt{3}$. The height is easily calculated as $-1-\left(-\frac{862}{433}\right)=-1+\frac{862}{433}=\frac{429}{433}$, so multiplying base by height and dividing by two we find that the total area is $\frac{429\sqrt{3}}{433}$, and our answer is $\boxed{865}$.

Solution 3: Trig

Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$. Since $BAC$ is equilateral with angle of $60^{\circ}$, angle $D$ is $120^{\circ}$. Use law of cosines to find $AF = \sqrt{433}$. Then use law of sines to find angle $BAG$ and $GAC$. Next we use Ptolemy's Theorem on $ABGC$ to find that $CG + BG = AG$. Next we use law of cosine on triangles $BAG$ and $GAC$, solving for BG and CG respectively. Subtract the two equations and divide out a $BG + CG$ to find the value of $CG - BG$. Next, $AG = 2\cdot R \cos{\theta}$, where R is radius of circle $= 2$ and $\theta =$ angle $BAG$. We already know sine of the angle so find cosine, hence we have found $AG$. At this point it is system of equation yielding $CG = \frac{26\sqrt{3}}{\sqrt{433}}$ and $BG = \frac{22\sqrt{3}}{\sqrt{433}}$. Given $[CBG] = \frac{BC \cdot CG \cdot BG}{4R}$, and $BC = 2\sqrt{3}$ by $30-60-90$ triangle, we can evaluate to find $[CBG] = \frac{429\sqrt{3}}{433}$, to give answer = $\boxed{865}$.

Solution 4

Note that $AB=2\sqrt3$, $DF=11$, and $EF=13$. If we take a homothety of the parallelogram with respect to $A$, such that $F$ maps to $G$, we see that $\frac{[ABG]}{[ACG]}=\frac{11}{13}$. Since $\angle AGB=\angle AGC=60^{\circ}$, from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$. Let $BG=11k$ and $CG=13k$.

By Law of Cosines on $\triangle BGC$, we have

$$12=k^2(11^2+11\cdot13+13^2)=433k^2\implies k^2=\frac{12}{433}$$ Thus, $[CBG]=\frac12 (11k)(13k)\sin 120^{\circ} = \frac{\sqrt3}{4}\cdot 143\cdot \frac{12}{433}=\frac{429\sqrt3}{433}\implies\boxed{865}$.

~rayfish

Solution 5: Easy Similar triangles

First, by properties of cyclic quadrilterals, $. Next, we notice quadrilateral AEFD is a parallelogram, and opposites angles are supplementary, and therefore$. Finally, by inscribed angles, $. Thus we can conclude$\triangle CBG \sim \triangle AFD$ by AA similarity.

Because $\triangle FAE \cong \triangle AFD$, $\triangle FAE \sim \triangle CBG$ aswell. So, we can write $\frac{BC}{AF} = \frac{BG}{DF} \to \frac{2\sqrt{3}}{\sqrt{433}} = \frac{BG}{11}$ since $AF = \sqrt{433}$ by the Law of Cosines. This gives $BG = \frac{22\sqrt{3}}{\sqrt{433}}$ and doing an identical process yields $CG = \frac{26\sqrt{3}}{\sqrt{433}}$. Thus, $[CBG] = \frac{26\sqrt{3}}{\sqrt{433}} \cdot \frac{22\sqrt{3}}{\sqrt{433}} \cdot \sin 120 = \frac{429\sqrt{3}}{433} \to p+q+r = \boxed{865}$.

~BossLu99