Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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<math>x^2 + (y - 4)^2 - (x - 2)^2 - y^2 = 12</math> | <math>x^2 + (y - 4)^2 - (x - 2)^2 - y^2 = 12</math> | ||
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<math>(2y)^2 + (y - 4)^2 = 16</math> | <math>(2y)^2 + (y - 4)^2 = 16</math> | ||
Revision as of 12:03, 14 February 2014
Contents
[hide]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 1
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:
Subtract the second equation from the first:
Then substitute:
Thus and
making
and
.
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
Solution 2
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 3
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
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.