Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
m (→‎Solution 1)
 
(46 intermediate revisions by 21 users not shown)
Line 2: Line 2:
 
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>?
  
−
<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>
+
<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>
 +
 
 +
<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==
 
==Solution 1==
 
[[File:Circlenc1.png]]
 
[[File:Circlenc1.png]]
  
−
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>.  
  
−
<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>.  
+
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>.
+
<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>.
+
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.  
+
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>.  
+
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
 
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>
+
<math>\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}</math> = <math>\boxed{\textbf{(C) }\frac{1}{3}}</math>
  
 
==Solution 2==
 
==Solution 2==
Line 41: Line 61:
 
<math>OD=r, OC=\frac{1}{3}r</math>.
 
<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==
 
==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>.
+
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)sin(CDE)</math>, where <math>r</math> is the radius of circle <math>F</math>. <math>\frac{1}{2}(CD)(6)</math> happens to be the area of <math>\triangle ABD</math>. Furthermore, <math>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):
 +
 
 +
1. 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>
 +
2. 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>.
 +
 
 +
~oinava
 +
 
 +
== Video solution ==
 +
https://youtu.be/i6eooSSJF64
  
−
==See also==
+
==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]]
−
[[Category:Area Ratio Problems]]
+
[[Category:Triangle Area Ratio Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

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

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 $\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}$ = $\boxed{\textbf{(C) }\frac{1}{3}}$

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

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_23&oldid=233223"