2006 AIME II Problems/Problem 12

Revision as of 12:18, 7 January 2019 by Mudhaniu (talk | contribs) (Solution 3: Trig)


Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r.$


[asy] size(250); pointpen = black; pathpen = black + linewidth(0.65); pen s = fontsize(8); pair A=(0,0),B=(-3^.5,-3),C=(3^.5,-3),D=13*expi(-2*pi/3),E1=11*expi(-pi/3),F=E1+D; path O = CP((0,-2),A); pair G = OP(A--F,O); D(MP("A",A,N,s)--MP("B",B,W,s)--MP("C",C,E,s)--cycle);D(O); D(B--MP("D",D,W,s)--MP("F",F,s)--MP("E",E1,E,s)--C); D(A--F);D(B--MP("G",G,SW,s)--C); MP("11",(A+E1)/2,NE);MP("13",(A+D)/2,NW);MP("l_1",(D+F)/2,SW);MP("l_2",(E1+F)/2,SE); [/asy]

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

Solution by e_power_pi_times_i

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 $\boxed{865}$

Solution 3: Trig

Lines $l1$ and $l2$ 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 = $\boxed{865}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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