Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math> | <math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math> | ||
− | == Solution == | + | == Solution 1== |
Let <math>D</math> be the origin. <math>A</math> is the point <math>(0,4)</math> and <math>M</math> is the point <math>(2,0)</math>. We are given the radius of the quarter circle and semicircle as <math>4</math> and <math>2</math>, respectively, so their equations, respectively, are: | Let <math>D</math> be the origin. <math>A</math> is the point <math>(0,4)</math> and <math>M</math> is the point <math>(2,0)</math>. We are given the radius of the quarter circle and semicircle as <math>4</math> and <math>2</math>, respectively, so their equations, respectively, are: |
Revision as of 22:35, 6 December 2012
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:
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
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
.