Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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Pythagorean theorem gives us <math>AM=2 \sqrt{5}</math>. Clearly <math>\triangle XMD \sim \triangle XDA \sim \triangle DMA \sim \triangle ZDP</math> by angle-angle and <math>\triangle XMD \cong \triangle XMP</math> by Hypotenuse Leg. | Pythagorean theorem gives us <math>AM=2 \sqrt{5}</math>. Clearly <math>\triangle XMD \sim \triangle XDA \sim \triangle DMA \sim \triangle ZDP</math> by angle-angle and <math>\triangle XMD \cong \triangle XMP</math> by Hypotenuse Leg. | ||
Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math> | Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | Using the double-angle formula for sine, what we need to find is <math>AD\cdot \sin(DAP) = AD\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}</math>. | ||
== See Also == | == See Also == |
Revision as of 20:39, 10 February 2015
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
Solution 4
Using the double-angle formula for sine, what we need to find is .
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.