2006 SMT/Geometry Problems/Problem 4
Problem
The distance is
. Find the area of the locus of points
such that
and
is on the same side of line
as a given point
.
Solution
Consider the locus of points such that
. As per the above diagram, let circle
be a circle and
a point on that circle such that
. Any point on that circle will intercept the same arc, and clearly any point not on that circle will have
either less than or greater than
. Therefore, this circle
is the desired locus. We define circle
the same way, except such that
.
Since is an inscribed angle, we have
. Therefore,
must lie on circle
. We now want the area between circles
and
. Let
be the midpoint of
, so that
.
First, let's try to find the radii of the circles. . From the half-angle identities, we have
, so
, and
. Therefore, from the Pythagorean Theorem on
, we have
.
We do the same thing on circle . However, notice that
is a
right triangle, so we quickly find that
.
Therefore, we have the radii of the circles. We now just need to find the area between them. We can do this by finding the area of sector , subtract the area of the portion of circle
cut off by chord
, and multiply the resulting difference by
. Since
, we have
, so the area of sector
is
.
We now need to find the area of circle cut off by chord
. We have
and
. Therefore, the area of sector
is
, and the area of
is
. Thus, the area of circle
cut off by chord
is
.
Now, we subtract this from the area of sector to get
, and we multiply this by
to get the total area, which is
.