2005 AMC 10A Problems/Problem 23
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 5
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 6 (Video)
Video solution: https://youtu.be/i6eooSSJF64
See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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