Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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− | Solving each factor for 0 yields <math>x = 0 , \frac{16}{5}</math>. The first value of <math>0</math> is obviously referring to the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x | + | Solving each factor for 0 yields <math>x = 0 , \frac{16}{5}</math>. The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math> |
== See Also == | == See Also == |
Revision as of 21:56, 31 May 2008
Problem
Square has sides of length , and is the midpoint of . A circle with radius and center intersects a circle with raidus and center at points and . What is the distance from to ?
Solution
Let be the origin. is the point and is the point . We are given the radius of the quarter circle and semicircle as and , respectively, so their equations, respectively, are:
Algebraically manipulating the second equation gives:
Substituting this back into the first equation:
Solving each factor for 0 yields . The first value of is obviously referring to the x-coordinate of the point where the circles intersect at the origin, , so the second value must be referring to the x coordinate of . Since is the y-axis, the distance to it from is the same as the x-value of the coordinate of , so the distance from to is