Difference between revisions of "2005 AMC 10A Problems/Problem 23"

Latest revision as of 03:05, 11 November 2024

Problem

Let $AB$ be a diameter of a circle and let $C$ be a point on $AB$ with $2\cdot AC=BC$ . Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ 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$",C,SW); draw(rightanglemark(D,C,B,2));[/asy]$

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

Solution 1

WLOG, let us assume that the diameter is of length $1$ .

The length of $AC$ is $\frac{1}{3}$ and $CO$ is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$ .

$OD$ is the radius of the circle, which is $\frac{1}{2}$ , so using the Pythagorean Theorem the height $CD$ of $\triangle DCO$ is $\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{6}\right)^2} = \frac{\sqrt{2}}{3}$ . This is also the height of the $\triangle ABD$ .

The area of $\triangle DCO$ is $\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}$ = $\frac{\sqrt{2}}{36}$ .

The height of $\triangle DCE$ can be found using the area of $\triangle DCO$ and $DO$ as base.

Hence, the height of $\triangle DCE$ is $\dfrac{\dfrac{\sqrt{2}}{36}}{\dfrac{1}{2}\cdot\dfrac{1}{2}}$ = $\dfrac{\sqrt{2}}{9}$ .

The diameter is the base for both the triangles $\triangle DCE$ and $\triangle ABD$ ,

Hence, the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$ is $\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}$ = $\boxed{\textbf{(C) }\frac{1}{3}}$

Solution 2

Since $\triangle DCE$ and $\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$ .

$[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\frac{1}{3}r$ .

Since $m\angle DCO=m\angle DFC=90^\circ$ , then $\triangle DCO\sim \triangle DFC$ . So the ratio of the two altitudes is $\frac{CF}{DC}=\frac{OC}{DO}=\boxed{\textbf{(C) }\frac{1}{3}}$

Solution 3

Let the center of the circle be point $O$ ; Without loss of generality, assume $AC=2$ , so $CB=4$ and the diameter and radius are $6$ and $3$ , respectively. Therefore, $CO=1$ , and $DO=3$ . The area of $\triangle DCE$ can be expressed as $\frac{1}{2}(CD)(6)\text{sin }(CDE).$ $\frac{1}{2}(CD)(6)$ happens to be the area of $\triangle ABD$ . Furthermore, $\text{sin } CDE = \frac{CO}{DO},$ or $\frac{1}{3}.$ Therefore, the ratio is $\boxed{\textbf{(C) }\frac{1}{3}}.$

Solution 4

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

Let the point G be the reflection of point $D$ across $\overline{AB}$ . (Point G is on the circle).

Let $AC=x$ , then $BC=2x$ . The diameter is $3x$ . To find $DC$ , there are two ways (presented here):

1. Since $\overline{AB}$ is the diameter, $CD=CG$ . Using power of points, $AC\cdot BC=x\cdot2x=2x^{2}=CD^{2} \longrightarrow CD=x\sqrt{2}$ 2. Use the geometric mean theorem, $AC\cdot BC=x\cdot2x=2x^{2}=CD^{2} \longrightarrow CD=x\sqrt{2}$ (These are the same equations but obtained through different formulae)

Therefore $DG=2x\sqrt{2}$ . Since $\overline{DE}$ is a diameter, $\triangle DGE$ is right. By the Pythagorean theorem, $DE^{2}=GD^{2}+GE^{2} \longrightarrow \left(3x\right)^{2}=\left(2x\sqrt{2}\right)^{2}+GE^{2}$ $9x^{2}=8x^{2}+GE^{2} \longrightarrow GE^{2}=x^{2} \longrightarrow GE=x$

As established before, $\angle DGE$ is right (if you are unsure, look up "inscribed angle theorem", this is a special case of the theorem where the central angle measures $180^{\circ}$ ) so $GE=x$ is the altitude of $\triangle DCE$ , and $DC=x\sqrt{2}$ is the base. Therefore $\left[DCE\right]=\frac{1}{2}\cdot DC\cdot GE=\frac{1}{2}\cdot x\sqrt{2}\cdot x=\frac{x^{2}\sqrt{2}}{2}$

$AB=3x$ is the base of $\triangle ABD$ and $CD=x\sqrt{2}$ is the height. $\left[ABD\right]=\frac{1}{2}\cdot3x\cdot x\sqrt{2}=\frac{3x^{2}\sqrt{2}}{2}$

The required ratio is $\frac{\left[DCE\right]}{\left[ABD\right]}=\frac{\frac{x^{2}\sqrt{2}}{2}}{\frac{3x^{2}\sqrt{2}}{2}}=\frac{x^{2}\sqrt{2}}{2}\cdot\frac{2}{3x^{2}\sqrt{2}}=\frac{x^{2}\sqrt{2}}{3x^{2}\sqrt{2}}=\frac{1}{3}$ The answer is $\boxed{\textbf{(C) } \frac{1}{3}}$ .

~JH. L

Solution 5

Assume the diameter is $3$ .

$AC = 1$

Get the height $CD = \sqrt{(AC)(BC)} = \sqrt2$ via power of a point.

$CO = AO - AC = 1/2$ .

By altitude of right triangle $\triangle CDO$ : Altitude from $C$ to $DE$ is same as altitude from $C$ to $DO$ is $\frac{(CO)(CD)}{DO} = \frac{(1/2)(\sqrt2)}{\frac 3 2}$ .

$\triangle DCE$ and $\triangle ABD$ have the same (diameter) hypotenuse length, so their area ratio is their altitude ratio is $\frac {\frac{ (1/2)(\sqrt2) } {\frac 3 2}} {\sqrt2} = \boxed{1/3}$ .

~oinava

Video solution

https://youtu.be/i6eooSSJF64

@@ Line 6: / Line 6: @@
 defaultpen(fontsize(10pt)+linewidth(.8pt));
 dotfactor=3;
-pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);
+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);
@@ Line 22: / Line 22: @@
 draw(rightanglemark(D,C,B,2));</asy>
-<math> \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math>
+<math> \textbf{(A) } \frac{1}{6}\qquad \textbf{(B) } \frac{1}{4}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \frac{2}{3} </math>
 ==Solution 1==
 [[File:Circlenc1.png]]
-WLOG, Let us assume that the diameter is of length <math>1</math>.
+WLOG, let us assume that the diameter is of length <math>1</math>.
 The length of <math>AC</math> is <math>\frac{1}{3}</math> and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>.
-<math>OD</math> is the radius of the circle, which is <math>\frac{1}{2}</math>, so using the Pythagorean Theorem the height <math>CD</math> of <math>\triangle DCO</math> is <math>\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{6}\right)^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>.
+<math>OD</math> is the radius of the circle, which is <math>\frac{1}{2}</math>, so using the [[Pythagorean Theorem]] the height <math>CD</math> of <math>\triangle DCO</math> is <math>\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{6}\right)^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>.
 The area of <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>.
 The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base.
-Hence the height of <math>\triangle DCE</math> is <math>\dfrac{\dfrac{\sqrt{2}}{36}}{\dfrac{1}{2}\cdot\dfrac{1}{2}}</math> = <math>\dfrac{\sqrt{2}}{9}</math>.
+Hence, the height of <math>\triangle DCE</math> is <math>\dfrac{\dfrac{\sqrt{2}}{36}}{\dfrac{1}{2}\cdot\dfrac{1}{2}}</math> = <math>\dfrac{\sqrt{2}}{9}</math>.
-The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math>.
+The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math>,
 Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is
-<math>\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}</math> = <math>\dfrac{1}{3} \Rightarrow C</math>
+<math>\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}</math> = <math>\boxed{\textbf{(C) }\frac{1}{3}}</math>
 ==Solution 2==
@@ Line 61: / Line 61: @@
 <math>OD=r, OC=\frac{1}{3}r</math>.
-Since <math>m\angle DCO=m\angle DFC=90^\circ</math>, then <math>\triangle DCO\cong \triangle DFC</math>. So the ratio of the two altitudes is <math>\frac{CF}{DC}=\frac{OC}{DO}=\frac{1}{3}\Rightarrow \text{(C)}</math>
+Since <math>m\angle DCO=m\angle DFC=90^\circ</math>, then <math>\triangle DCO\sim \triangle DFC</math>. So the ratio of the two altitudes is <math>\frac{CF}{DC}=\frac{OC}{DO}=\boxed{\textbf{(C) }\frac{1}{3}}</math>
 ==Solution 3==
-Say the center of the circle is point <math>O</math>;
+Let the center of the circle be point <math>O</math>;
 Without loss of generality, assume <math>AC=2</math>, so <math>CB=4</math> and the diameter and radius are <math>6</math> and <math>3</math>, respectively. Therefore, <math>CO=1</math>, and <math>DO=3</math>.
-The area of <math>\triangle DCE</math> can be expressed as <math>\frac{1}{2}(CD)(6)\text{sin }(CDE).</math> <math>\frac{1}{2}(CD)(6)</math> happens to be the area of <math>\triangle ABD</math>. Furthermore, <math>\text{sin } CDE = \frac{CO}{DO},</math> or <math>\frac{1}{3}.</math> Therefore, the ratio is <math>\frac{1}{3}.</math>
+The area of <math>\triangle DCE</math> can be expressed as <math>\frac{1}{2}(CD)(6)\text{sin }(CDE).</math> <math>\frac{1}{2}(CD)(6)</math> happens to be the area of <math>\triangle ABD</math>. Furthermore, <math>\text{sin } CDE = \frac{CO}{DO},</math> or <math>\frac{1}{3}.</math> Therefore, the ratio is <math>\boxed{\textbf{(C) }\frac{1}{3}}.</math>
+==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);
+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>
+Let the point G be the reflection of point <math>D</math> across <math>\overline{AB}</math>. (Point G is on the circle).
+Let <math>AC=x</math>, then <math>BC=2x</math>. The diameter is <math>3x</math>. To find <math>DC</math>, there are two ways (presented here):
+. Since <math>\overline{AB}</math> is the diameter, <math>CD=CG</math>. Using power of points,
+<cmath>AC\cdot BC=x\cdot2x=2x^{2}=CD^{2} \longrightarrow CD=x\sqrt{2}</cmath>
+. Use the geometric mean theorem,
+<cmath>AC\cdot BC=x\cdot2x=2x^{2}=CD^{2} \longrightarrow CD=x\sqrt{2}</cmath>
+(These are the same equations but obtained through different formulae)
+Therefore <math>DG=2x\sqrt{2}</math>. Since <math>\overline{DE}</math> is a diameter, <math>\triangle DGE</math> is right. By the Pythagorean theorem,
+<cmath>DE^{2}=GD^{2}+GE^{2} \longrightarrow \left(3x\right)^{2}=\left(2x\sqrt{2}\right)^{2}+GE^{2}</cmath>
+<cmath>9x^{2}=8x^{2}+GE^{2} \longrightarrow GE^{2}=x^{2} \longrightarrow GE=x</cmath>
+As established before, <math>\angle DGE</math> is right (if you are unsure, look up "inscribed angle theorem", this is a special case of the theorem where the central angle measures <math>180^{\circ}</math>) so <math>GE=x</math> is the altitude of <math>\triangle DCE</math>, and <math>DC=x\sqrt{2}</math> is the base. Therefore
+<cmath>\left[DCE\right]=\frac{1}{2}\cdot DC\cdot GE=\frac{1}{2}\cdot x\sqrt{2}\cdot x=\frac{x^{2}\sqrt{2}}{2}</cmath>
+<math>AB=3x</math> is the base of <math>\triangle ABD</math> and <math>CD=x\sqrt{2}</math> is the height.
+<cmath>\left[ABD\right]=\frac{1}{2}\cdot3x\cdot x\sqrt{2}=\frac{3x^{2}\sqrt{2}}{2}</cmath>
+The required ratio is
+<cmath>\frac{\left[DCE\right]}{\left[ABD\right]}=\frac{\frac{x^{2}\sqrt{2}}{2}}{\frac{3x^{2}\sqrt{2}}{2}}=\frac{x^{2}\sqrt{2}}{2}\cdot\frac{2}{3x^{2}\sqrt{2}}=\frac{x^{2}\sqrt{2}}{3x^{2}\sqrt{2}}=\frac{1}{3}</cmath>
+The answer is <math>\boxed{\textbf{(C) } \frac{1}{3}}</math>.
+~JH. L
+== Solution 5 ==
+Assume the diameter is <math>3</math>.
+<math>AC = 1</math>
+Get the height <math>CD = \sqrt{(AC)(BC)} = \sqrt2</math> via power of a point.
+<math>CO = AO - AC = 1/2</math>.
+By altitude of right triangle <math>\triangle CDO</math>: Altitude from <math>C</math> to <math>DE</math> is same as altitude from <math>C</math> to <math>DO</math> is <math>\frac{(CO)(CD)}{DO} = \frac{(1/2)(\sqrt2)}{\frac 3 2}</math>.
+<math>\triangle DCE</math> and <math>\triangle ABD</math> have the same (diameter) hypotenuse length, so their area ratio is their altitude ratio is
+<math>\frac {\frac{ (1/2)(\sqrt2) } {\frac 3 2}} {\sqrt2} = \boxed{1/3}</math>.
-== Solution 4 ==
+~oinava
-WLOG, let <math>AC=1</math>, <math>BC=2</math>, so radius of the circle is <math>\frac{3}{2}</math> and <math>OC=\frac{1}{2}</math>. As in solution 1, By same altitude, the ratio <math>[DCE]/[ABD]=PE/AB</math>, where <math>P</math> is the point where <math>DC</math> extended meets circle <math>O</math>. Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is <math>\frac{1}{3}</math>.
+== Video solution ==
+https://youtu.be/i6eooSSJF64
 ==See Also==
@@ Line 77: / Line 150: @@
 [[Category:Introductory Geometry Problems]]
-[[Category:Area Ratio Problems]]
+[[Category:Triangle Area Ratio Problems]]
 {{MAA Notice}}

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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