AMC12 2005 12A, Problem 15

Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$. Let $D$ and $E$ be points on the circle such that $\overline{DC} \perp \overline{AB}$ and $\overline{DE}$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$? [asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,E); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); draw(rightanglemark(D,C,B,2));[/asy]

Solution: [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -4.24313994860289, xmax = 6.360350402147026, ymin = -8.17642986522568, ymax = 4.1323989018072735;  /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);   /* draw figures */ draw(circle((0.9223776980185863,-1.084225871478444), 3.171161249925393), linewidth(2) + wrwrwr);  draw((-0.5623104956355617,1.717909856970905)--(-0.39884850438732933,-3.9670413347199647), linewidth(2) + wrwrwr);  draw((-2.2487343499798245,-1.1018908670262892)--(4.0935388680341624,-1.0849377800575102), linewidth(2) + wrwrwr);  draw((-0.4813673187299407,-1.0971666418223938)--(2.1729847859273126,-3.904641555130568), linewidth(2) + wrwrwr);  draw((-0.5623104956355617,1.717909856970905)--(2.1561002471522333,-3.887757016355488), linewidth(2) + wrwrwr);  draw((-2.2487343499798245,-1.1018908670262892)--(-0.5623104956355617,1.717909856970905), linewidth(2) + wrwrwr);  draw((-0.5623104956355617,1.717909856970905)--(4.0935388680341624,-1.0849377800575102), linewidth(2) + wrwrwr);  draw(circle((-2.858607769046372,-1.1524617347926092), 0.45463011998128017), linewidth(2) + wrwrwr);  draw(circle((4.790088296064635,-1.144019465405069), 0.45463011998128006), linewidth(2) + wrwrwr);  draw(circle((-0.9168858099122313,-1.6336710898823752), 0.45463011998127983), linewidth(2) + wrwrwr);  draw(circle((1.4976032349241348,-1.3128648531558642), 0.4546301199812797), linewidth(2) + wrwrwr);  draw(circle((-0.815578577261755,2.334195522261307), 0.4546301199812809), linewidth(2) + wrwrwr);  draw(circle((2.6119827940793803,-4.5546962979711285), 0.45463011998128033), linewidth(2) + wrwrwr);   /* dots and labels */ dot((0.9223776980185863,-1.084225871478444),dotstyle);  dot((-0.5623104956355617,1.717909856970905),dotstyle);  label("$D$", (-0.49477234053524466,1.8867552447217006), NE * labelscalefactor);  dot((-0.39884850438732933,-3.9670413347199647),dotstyle);  dot((-2.2487343499798245,-1.1018908670262892),dotstyle);  label("A", (-2.183226218043193,-0.9329627307165757), NE * labelscalefactor);  dot((4.0935388680341624,-1.0849377800575102),dotstyle);  label("B", (4.165360361386693,-0.9160781919414961), NE * labelscalefactor);  dot((-0.4813673187299407,-1.0971666418223938),linewidth(4pt) + dotstyle);  label("C", (-0.41034964665984724,-0.9667318082667347), NE * labelscalefactor);  dot((2.1729847859273126,-3.904641555130568),dotstyle);  dot((2.1561002471522333,-3.887757016355488),dotstyle);  label("E", (2.2236384022525524,-3.7189116286046926), NE * labelscalefactor);  label("2", (-2.9261459241466903,-1.2031153511178476), NE * labelscalefactor,wrwrwr);  label("1", (4.722550140964316,-1.186230812342768), NE * labelscalefactor,wrwrwr);  label("3", (-0.9844239650125497,-1.6758824368200735), NE * labelscalefactor,wrwrwr);  label("4", (1.4300650798238166,-1.355076200093563), NE * labelscalefactor,wrwrwr);  label("2", (-0.8831167323620728,2.2919841753236083), NE * labelscalefactor,wrwrwr);  label("2", (2.5444446389790625,-4.5969076449088275), NE * labelscalefactor,wrwrwr);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy] We set point $A$ as a mass of 2. This means that point $B$ has a mass of $1$ since $2\times{AC} = 1\times{BC}$. This implies that point $C$ has a mass of $2+1 = 3$ and the center of the circle has a mass of $3+1 = 4$. After this, we notice that points $D$ and $E$ both must have a mass of $2$ since $2+2 = 4$ and they are both radii of the circle. To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply $\frac{3\times2\times2}{2\times2\times1}$ which is $\fbox{\frac {1}{3}}$ (the reciprocal of 3)

Answer: \frac {1}{3}