2006 AMC 12A Problems/Problem 17
Contents
Problem
Square has side length , a circle centered at has radius , and and are both rational. The circle passes through , and lies on . Point lies on the circle, on the same side of as . Segment is tangent to the circle, and . What is ?
Solutions
Solution 1
One possibility is to use the coordinate plane, setting at the origin. Point will be and will be since , and are collinear and contain a diagonal of . The Pythagorean theorem results in
This implies that and ; dividing gives us .
Solution 2
First note that angle is right since is tangent to the circle. Using the Pythagorean Theorem on , then, we see
But it can also be seen that . Therefore, since lies on , . Using the Law of Cosines on , we see
Thus, since and are rational, and . So , , and .
Solution 3
(Similar to Solution 1) First, draw line AE and mark a point Z that is equidistant from E and D so that and that line includes point D. Since DE is equal to the radius ,
Note that triangles and share the same hypotenuse , meaning that Plugging in our values we have: By logic and
Therefore,
Solution 4 - Alcumus
Let , , , , and . Apply the Pythagorean Theorem to to obtain from which . Because and are rational, it follows that and , so .
OR
Extend past to meet the circle at . Because is collinear with and , Also, which implies , so is an isosceles right triangle. Thus . By the Power of a Point Theorem, As in the first solution, we conclude that .
Solution 5 - Answer Choices
(Last minute solution) We roughly measure the distance of and the distance of . Since is clearly less than , we can eliminate answer choices (D) and (E). Next, if we compare the distances, seems to be just a little more than half of , thus eliminating answer choice (A). is only a little bit bigger than half of , so we can reasonably assume that their ratio is less than . That leaves us with answer choice , or .
See Also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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