2003 AMC 12A Problems/Problem 17
Contents
Problem
Square has sides of length , and is the midpoint of . A circle with radius and center intersects a circle with radius and center at points and . What is the distance from to ?
Solution 2
obviously forms a kite. Let the intersection of the diagonals be . Let . Then, .
By Pythagorean Theorem, . Thus, . Simplifying, . By Pythagoras again, . Then, the area of is .
Using instead as the base, we can drop a altitude from P. . . Thus, the horizontal distance is
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Solution 3
Note that is merely a reflection of over . Call the intersection of and . Drop perpendiculars from and to , and denote their respective points of intersection by and . We then have , with a scale factor of 2. Thus, we can find and double it to get our answer. With some analytical geometry, we find that , implying that .
Solution 4
As in Solution 2, draw in and and denote their intersection point . Next, drop a perpendicular from to and denote the foot as . as they are both radii and similarly so is a kite and by a well-known theorem.
Pythagorean theorem gives us . Clearly by angle-angle and by Hypotenuse Leg. Manipulating similar triangles gives us
Solution 5
Using the double-angle formula for sine, what we need to find is .
Solution 6(LoC)
We use the Law of Cosines:
Solution 7
Let be the foot of the perpendicular from to , and let . Then we have , and . Since , we have , or . Solving gives .
Solution 8
Draw , , and . is parallel with
,
by ,
,
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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All AMC 12 Problems and Solutions |
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