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

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(Solution 5)
 
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==Problem==
 
==Problem==
 
Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?
 
Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?
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<asy>
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unitsize(2.5cm);
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defaultpen(fontsize(10pt)+linewidth(.8pt));
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dotfactor=3;
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pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);
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pair D=dir(aCos(C.x)), E=(-D.x,-D.y);
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draw(A--B--D--cycle);
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draw(D--E--C);
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draw(unitcircle,white);
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drawline(D,C);
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dot(O);
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clip(unitcircle);
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draw(unitcircle);
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label("$E$",E,SSE);
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label("$B$",B,E);
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label("$A$",A,W);
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label("$D$",D,NNW);
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label("$C$",C,SW);
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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> \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>
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[[File:Circlenc1.png]]
 
[[File:Circlenc1.png]]
  
<math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>.  
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WLOG, Let us assume that the diameter is of length <math>1</math>.
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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, so using the Pythagorean theorem height <math>CD</math> of <math>\triangle AOC</math> is <math>\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>.
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<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>.
  
Area of the <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>.
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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.  
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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>\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}</math> = <math>\frac{\sqrt{2}}{9}</math>.  
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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>.
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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
 
Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is
<math>\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}</math> = <math>\frac{1}{3} \Rightarrow C</math>
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<math>\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}</math> = <math>\dfrac{1}{3} \Rightarrow C</math>
  
 
==Solution 2==
 
==Solution 2==
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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\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>
  
==See also==
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==Solution 3==
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Say the center of the circle is point <math>O</math>;
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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>.
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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>
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== Solution 4 ==
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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>.
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== Solution 5 (Video) ==
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Video solution: https://youtu.be/i6eooSSJF64
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==See Also==
 
{{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}}
 
{{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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[[Category:Area Ratio Problems]]
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{{MAA Notice}}

Latest revision as of 14:16, 26 November 2020

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]

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

Solution 1

Circlenc1.png

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 $\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}$ = $\dfrac{1}{3} \Rightarrow C$

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 $\triangle DCO\cong \triangle DFC$. So the ratio of the two altitudes is $\frac{CF}{DC}=\frac{OC}{DO}=\frac{1}{3}\Rightarrow \text{(C)}$

Solution 3

Say the center of the circle is 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 $\frac{1}{3}.$

Solution 4

WLOG, let $AC=1$, $BC=2$, so radius of the circle is $\frac{3}{2}$ and $OC=\frac{1}{2}$. As in solution 1, By same altitude, the ratio $[DCE]/[ABD]=PE/AB$, where $P$ is the point where $DC$ extended meets circle $O$. Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is $\frac{1}{3}$.

Solution 5 (Video)

Video solution: https://youtu.be/i6eooSSJF64

See Also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions

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