Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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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> | ||
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+ | ==Solution 3== | ||
+ | |||
+ | Say the center of the circle is point F; | ||
+ | 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>CF=1</math> and <math>DF=3</math>. | ||
+ | The area of <math>\triangle CDE</math> can be expressed as <math>\frac{1}{2}(CD)(2r)sin(CDE)</math>, where <math>r</math> is the radius of circle <math>F</math>. <math>\frac{1}{2}(CD)(2r)</math> happens to be the area of <math>\triangle ADB</math>. Furthermore, <math>sin CDE = \frac{CF}{DF},</math> or <math>\frac{1}{3}.</math> Therefore, the ratio is <math>\frac{1}{3}.</math> | ||
==See also== | ==See also== |
Revision as of 13:46, 14 September 2015
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
Let us assume that the diameter is of length .
is of diameter and is .
is the radius of the circle, so using the Pythagorean theorem height of is . 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
Solution 3
Say the center of the circle is point F; Without loss of generality, assume , so and the diameter and radius are and , respectively. Therefore, and . The area of can be expressed as , where is the radius of circle . happens to be the area of . Furthermore, or Therefore, the ratio 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 |
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