Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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<math>CO = AO - AC = 1/2</math>. | <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)}3</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>\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) } 3 } {\sqrt2} = \boxed{1/ | + | <math>\frac {\frac{ (1/2)(\sqrt2) } {\frac 3 2}</math>} {\sqrt2} = \boxed{1/3}$. |
~oinava | ~oinava |
Revision as of 19:57, 9 July 2024
Contents
[hide]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
Let the point G be the reflection of point across . (Point G is on the circle).
Let , then . The diameter is . To find , there are two ways (presented here):
1. Since is the diameter, . Using power of points, 2. Use the geometric mean theorem, (These are the same equations but obtained through different formulae)
Therefore . Since is a diameter, is right. By the Pythagorean theorem,
As established before, is right (if you are unsure, look up "inscribed angle theorem", this is a special case of the theorem where the central angle measures ) so is the altitude of , and is the base. Therefore
is the base of and is the height.
The required ratio is
The answer is .
~JH. L
Solution 5
Assume the diameter is .
Get the height via power of a point.
.
By altitude of right triangle : Altitude from to is same as altitude from to is .
and have the same (diameter) hypotenuse length, so their area ratio is their altitude ratio is $\frac {\frac{ (1/2)(\sqrt2) } {\frac 3 2}$ (Error compiling LaTeX. Unknown error_msg)} {\sqrt2} = \boxed{1/3}$.
~oinava
Video solution
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.