2005 AMC 12A Problems/Problem 15

Problem

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$",C,SW); draw(rightanglemark(D,C,B,2));[/asy]$

$(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}$

Solution

Solution 1

Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or $\frac{CF}{CD}$ ( $F$ is the foot of the perpendicular from $C$ to $DE$ ).

Call the radius $r$ . Then $AC = \frac 13(2r) = \frac 23r$ , $CO = \frac 13r$ . Using the Pythagorean Theorem in $\triangle OCD$ , we get $\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r$ .

Now we have to find $CF$ . Notice $\triangle OCD \sim \triangle OFC$ , so we can write the proportion:

$\frac{OF}{OC} = \frac{OC}{OD}$
$\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}$
$OF = \frac 19r$

By the Pythagorean Theorem in $\triangle OFC$ , we have $\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r$ .

Our answer is $\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}$ .

Solution 2

Let the center of the circle be $O$ .

Note that $2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB$ .

$O$ is midpoint of $AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB$ .

$O$ is midpoint of $DE \Rightarrow$ Area of $\triangle DCE = 2 \cdot$ Area of $\triangle DCO = 2 \cdot (\frac{1}{6} \cdot$ Area of $\triangle ABD) = \frac{1}{3} \cdot$ Area of $\triangle ABD \Longrightarrow \mathrm{(C)}$ .

Solution 3

Let $r$ be the radius of the circle. Note that $AC+BC = 2r$ so $AC = \frac{2}{3}r$ .

By Power of a Point Theorem, $CD^2= AC \cdot BC = 2\cdot AC^2$ , and thus $CD = \sqrt{2} \cdot AC = \frac{2\sqrt{2}}{3}r$

Then the area of $\triangle ABD$ is $\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2$ . Similarly, the area of $\triangle DCE$ is $\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2$ , so the desired ratio is $\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}$

Solution 4

$[asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), e=(-D.x,-D.y); pair H=(e.x,0); draw(A--B--D--cycle); draw(D--e--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",e,SSE); label("$B$",B,NE); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); label("$H$",H,SE); draw(e--H,dashed); label("O",(0,0),NE); label("1",(C--O),N); label("1",(H--O),N); label("2",(A--C),N); label("2",(H--B),N); label("3",(O--D),NE); label("3",(O--e),NE); label("$2\sqrt{2}$",(D--C),W); label("$2\sqrt{2}$",(H--e),E); draw(rightanglemark(e,(e.x,0),A,2)); draw(rightanglemark(D,C,B,2));[/asy]$

Let the center of the circle be $O$ . Without loss of generality, let the radius of the circle be equal to $3$ . Thus, $AO=3$ and $OB=3$ . As a consequence of $2(AC)=BC$ , $AC=2$ and $CO=1$ . Also, we know that $DO$ and $OE$ are both equal to $3$ due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to $\sqrt{3^2-1^2}$ or $2\sqrt{2}$ . Now we know that the area of $[ABD]$ is equal to $\frac{(3+2+1)(2\sqrt{2})}{2}$ or $6\sqrt{2}$ . Know we need to find the area of $[DCE]$ . By simple inspection $[COD]$ $\cong$ $[HOE]$ due to angles being equal and CPCTC. Thus $HE=2\sqrt{2}$ and $OH=1$ . Know we know the area of $[CHE]=\frac{(1+1)(2\sqrt{2})}{2}$ or $2\sqrt{2}$ . We also know that the area of $[OHE]=\frac{(1)(2\sqrt{2})}{2}$ or $\sqrt{2}$ . Thus the area of $[COE]=2\sqrt{2}-\sqrt{2}$ or $\sqrt{2}$ . We also can calculate the area of $[DOC]$ to be $\frac{(1)(2\sqrt{2})}{2}$ or $\sqrt{2}$ . Thus $[DCE]$ is equal to $[COE]$ + $[DOC]$ or $\sqrt{2}+\sqrt{2}$ or $2\sqrt{2}$ . The ratio between $[DCE]$ and $[ABD]$ is equal to $\frac{2\sqrt{2}}{6\sqrt{2}}$ or $\frac{1}{3}$ $\Longrightarrow \mathrm{(C)}$ .

Solution 5

We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for $D (x,y)$ , and notice how $E$ is a 180 degree rotation of $D$ , using the rotation matrix formula we get $E = (-x,-y)$ . WLOG say that this circle has radius $3$ . We can now find points $A$ , $B$ , and $C$ which are $(-3,0)$ , $(3,0)$ , and $(-1,0)$ respectively. By shoelace the area of $CED$ is $Y$ , and the area of $ADB$ is $3Y$ . Using division we get that the answer is $\mathrm{(C)}$ .

Solution 6 (Mass Points)

$[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 $\boxed{\frac{1}{3}}$ (the reciprocal of 3)

-Brudder

Solution 7 (Slight Trigonometry)

$[asy] unitsize(3.5cm); defaultpen(fontsize(12pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,NE); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); label("O",(0,0),NE); label("x",(C--O),N); label("3x",(B--O),N); label("2x",(A--C),N); label("3x",(O--D),NE); label("3x",(O--E),NE); label("$2x\sqrt{2}$",(D--C),W); draw(rightanglemark(D,C,B,2));[/asy]$

Let the center of the circle be $O$ . The area of $ADB$ is $\frac{1}{2} \cdot 6x \cdot 2x\sqrt{2}$ = $6x^2\sqrt{2}$ . The area of $DCE$ is $\frac{1}{2} \cdot DC \cdot DE \cdot$ sin $CDE$ . We find sin $CDE$ is $\frac{OC}{OD}$ = $\frac{1}{3}$ . Substituting $DC = 2x\sqrt{2}$ and $DE = 6x$ , we get $[DCE]$ = $\frac{1}{2} \cdot 2x\sqrt{2} \cdot 6x \cdot \frac{1}{3}$ = $2x^2\sqrt{2}$ . Hence, the ratio between the areas of $DCE$ and $ADB$ is equal to $\frac{2x^2\sqrt{2}}{6x^2\sqrt{2}}$ or $\frac{1}{3}$ = $\boxed{(C)}$ .

~Math_Genius_164

Solution 8

In my opinion, the solution below is the easiest and quickest.

Since both $DE$ and $AB$ are diameters, they intersect at the center of the circle. Call this center $O$ . WLOG, let $AC=2, CO=1,OB=3$ . Call the point where the extension of $DC$ hits the circle $F$ . Notice that $\angle DCB = 90^\circ$ . This implies that $DC=CF$ . WOLG, let $DC=CF=1$ . Then, $[DCE]=\frac 12 bh = \frac 12 * DF*CO = \frac 12 *2*1=1$ and $[ABD]=\frac 12 bh = \frac 12 *AB*CD = \frac 12 *6*1 = 3$ . Thus, the answer is $\frac{1}{3}$ = $\boxed{(C)}$ .

Solution by franzliszt

Solution 9 (Slick construction)

$[asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0), X = (1/3.0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(D--X); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,E); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); label("$X$",X,NE); draw(rightanglemark(D,C,B,2));[/asy]$ Let $X$ be the reflection of $C$ over the center $O.$ Since $\triangle DXO\cong\triangle ECO$ by SAS, it follows that the area of $\triangle DEC$ is equal to the area of $\triangle DCX.$ However, we know that $BC = 2AC\implies CO=\frac{1}{2}AC\implies CX=AC,$ so the ratio of the area of $\triangle DEC$ to the area of $\triangle DCX$ is $\frac{\frac{AC\cdot DC}{2}}{\frac{3AC\cdot DC}{2}}=\frac{1}{3}\implies \boxed{(C)}$

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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