Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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<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>. The area of | <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>. The area of | ||
− | ==Solution== | + | ==Solution 1== |
http://img443.imageshack.us/img443/8034/circlenc1.png | http://img443.imageshack.us/img443/8034/circlenc1.png | ||
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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>\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}</math> = <math>\frac{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); | ||
+ | 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{DC}{CF}=\frac{DO}{OC}=\frac{1}{3}\Rightarrow \text{(C)}</math> | ||
==See also== | ==See also== |
Revision as of 19:58, 5 February 2010
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 ?
. The area of
Solution 1
http://img443.imageshack.us/img443/8034/circlenc1.png
is of diameter and is - = .
is the radius of the circle, so using the Pythagorean theorem height of is $\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2$ (Error compiling LaTeX. ! Missing } inserted.) = . This is also the height of the .
Area of the 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
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 |