2005 AMC 12A Problems/Problem 15
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Problem
Let 
be a diameter of a circle and 
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
Solution 1
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or 
(
is the foot of the perpendicular from to 
).
Call the radius 
. Then 
, 
. Using the Pythagorean Theorem in 
, we get 
.
Now we have to find 
. Notice 
, so we can write the proportion:
By the Pythagorean Theorem in 
, we have 
.
Our answer is 
.
Solution 2
Let the centre of the circle be 
.
Note that 
.
is midpoint of 
.
is midpoint of 
Area of 
Area of 
Area of 
Area of 
.
See also
| 2005 AMC 12A (Problems • Resources) | ||
| Preceded by Problem 14 | Followed by Problem 16 | |
| 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 | ||
