2017 AMC 10B Problems/Problem 22
Contents
Problem
The diameter of a circle of radius is extended to a point outside the circle so that . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the area of ?
Solution 1
Notice that and are right triangles. Then . , so . We also find that (You can also use power of point ~MATHWIZARD2010), and thus the area of is .
Solution 2
We note that by similarity. Also, since the area of and , , so the area of .
Solution 3
As stated before, note that is similar to . By similarity, we note that is equivalent to . We set to and to . By the Pythagorean Theorem, . Combining, . We can add and divide to get . We square root and rearrange to get . We know that the legs of the triangle are and . Multiplying by and eventually gives us . We divide this by , since is the formula for a triangle. This gives us .
Solution 4
Let's call the center of the circle that segment is the diameter of, . Note that is an isosceles right triangle. Solving for side , using the Pythagorean theorem, we find it to be . Calling the point where segment intersects circle , the point , segment would be . Also, noting that is a right triangle, we solve for side , using the Pythagorean Theorem, and get . Using Power of Point on point , we can solve for . We can subtract from to find and then solve for using Pythagorean theorem once more.
= (Diameter of circle + ) = =
= - =
Now to solve for :
- = + = =
Note that is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases and , we get the area of triangle to be .
Solution 5 (Coordinate Geo)
Drawing the picture, we realize that the equation for the line from A to E is , and the equation for the circle is plugging in for y we get so , that means
the height is and the base is , so the area is
-harsha12345
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.