2019 AMC 10B Problems/Problem 23
- The following problem is from both the 2019 AMC 10B #23 and 2019 AMC 12B #20, so both problems redirect to this page.
Contents
Problem
Points and
lie on circle
in the plane. Suppose that the tangent lines to
at
and
intersect at a point on the
-axis. What is the area of
?
Solution
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was . Using Pythagorean Theorem gives
.
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, ,
, and
form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:
, where
represents the distance between circle center and
. Therefore,
. Using Pythagorean Theorem on
, either one of
or
, and the circle center, we realize that
, at which point
, so the answer is
.
Solution 2
First, follow solution 1 and obtain . Label the point
as point
. The midpoint
of segment
is
. Notice that the center of the circle must lie on the line that goes through the points
and
. Thus, the center of the circle lies on the line
.
Line is
. The perpendicular line must pass through
and
. The slope of the perpendicular line is
. The line is hence
. The point
lies on this line. Therefore,
. Solving this equation tells us that
. So the center of the circle is
. The distance between the center,
, and point A is
. Hence, the area is
. The answer is
. \\
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See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
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