Difference between revisions of "2017 AMC 12B Problems/Problem 18"
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Latest revision as of 04:01, 23 January 2023
Contents
Problem
The diameter of a circle of radius
is extended to a point
outside the circle so that
. Point
is chosen so that
and line
is perpendicular to line
. Segment
intersects the circle at a point
between
and
. What is the area of
?
Solution 1
Let be the center of the circle. Note that
. However, by Power of a Point,
, so
. Now
. Since
.
Solution 2: Similar triangles with Pythagorean
is the diameter of the circle, so
is a right angle, and therefore by AA similarity,
.
Because of this, , so
.
Likewise, , so
.
Thus the area of .
Solution 2b: Area shortcut
Because is
and
is
, the ratio of the sides is
, meaning the ratio of the areas is thus
. We then have the proportion
Solution 3: Similar triangles without Pythagorean
Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:
Draw with
on
.
.
.
. (
ratio applied twice)
.
Solution 4 (Coordinate Geometry)
Let be at the origin
of a coordinate plane, with
being located at
, etc.
We can find the area of by finding the the altitude from line
to point
. Realize that this altitude is the
coordinate of point
on the coordinate plane, since the respective base of
is on the
-axis.
Using the diagram in solution one, the equation for circle is
.
The equation for line is then
, therefore
.
Substituting for
in the equation for circle
, we get:
We can solve for to yield the
coordinate of point
in the coordinate plane, since this is the point of intersection of the circle and line
. Note that one root will yield the intersection of the circle and line
at the origin, so we will ignore this root.
Expanding the expression and factoring, we get:
Our non-zero root is thus . Calculating the area of
with
as the length of
and
as the altitude, we get:
.
-Solution by Joeya
Video Solution by OmegaLearn (Similar Triangles)
https://youtu.be/NsQbhYfGh1Q?t=512
~ pi_is_3.14
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.