Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | <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>? |
+ | |||
+ | <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> \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> | ||
− | ==Solution== | + | ==Solution 1== |
− | {{solution}} | + | [[File:Circlenc1.png]] |
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Since <math>\triangle DCE</math> and <math>\triangle ABD</math> share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from <math>C</math> to <math>DE</math>. | ||
+ | |||
+ | <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> | ||
+ | <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> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Say the center of the circle is 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> | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | 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>. | ||
− | == | + | == Solution 5 (Video) == |
+ | Video solution: https://youtu.be/i6eooSSJF64 | ||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Ratio Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:16, 26 November 2020
Contents
Problem
Let be a diameter of a circle and let be a point on with . Let and be points on the circle such that and is a second diameter. What is the ratio of the area of to the area of ?
Solution 1
WLOG, Let us assume that the diameter is of length .
The length of is and is .
is the radius of the circle, which is , so using the Pythagorean Theorem the height of is . This is also the height of the .
The area of is = .
The height of can be found using the area of and as base.
Hence, the height of is = .
The diameter is the base for both the triangles and ,
Hence, the ratio of the area of to the area of is =
Solution 2
Since and share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from to .
.
Since , then . So the ratio of the two altitudes is
Solution 3
Say the center of the circle is point ; Without loss of generality, assume , so and the diameter and radius are and , respectively. Therefore, , and . The area of can be expressed as happens to be the area of . Furthermore, or Therefore, the ratio is
Solution 4
WLOG, let , , so radius of the circle is and . As in solution 1, By same altitude, the ratio , where is the point where extended meets circle . Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is .
Solution 5 (Video)
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 |
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