Difference between revisions of "2005 AMC 12A Problems/Problem 15"
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− | Let the | + | Let the center of the circle be <math>O</math>. |
Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>. | Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>. |
Revision as of 18:37, 5 October 2016
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 center of the circle be .
Note that .
is midpoint of .
is midpoint of Area of Area of Area of Area of .
Solution 3
Let be the radius of the circle. Note that so .
By Power of a Point Theorem, , and thus
Then the area of is . Similarly, the area of is , so the desired ratio is
See also
2005 AMC 12A (Problems • Answer Key • 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 | |
All AMC 12 Problems and Solutions |
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