2019 AMC 12B Problems/Problem 20
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 1
First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is , the Pythagorean Theorem gives
. This simplifies to
.
Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) defined by the circle's center, ,
, and
is cyclic. Therefore, we can apply Ptolemy's Theorem to give
, where
is the radius of the circle and
is the distance between the circle's center and
. Therefore,
. Using the Pythagorean Theorem on the triangle formed by the point
, either one of
or
, and the circle's center, we find that
, so
, and thus the answer is
.
Solution 2 (coordinate bash)
We firstly obtain as in Solution 1. Label the point
as
. The midpoint
of segment
is
. Notice that the center of the circle must lie on the line passing through the points
and
. Thus, the center of the circle lies on the line
.
Line is
. Therefore, the slope of the line perpendicular to
is
, so its equation is
.
But notice that this line must pass through and
. Hence
. So the center of the circle is
.
Finally, the distance between the center, , and point
is
. Thus the area of the circle is
.
Solution 3
The midpoint of is
. Let the tangent lines at
and
intersect at
on the
-axis. Then
is the perpendicular bisector of
. Let the center of the circle be
. Then
is similar to
, so
.
The slope of
is
, so the slope of
is
. Hence, the equation of
is
. Letting
, we have
, so
.
Now, we compute ,
, and
.
Therefore ,
and consequently, the area of the circle is
.
Solution 4 (how fast can you multiply two-digit numbers?)
Let be the intersection on the x-axis. By Power of a Point Theorem,
. Then the equations are
and
for the tangent lines passing
and
respectively. Then the lines normal to them are
and
. Thus,
After condensing, . Then, the center of
is
. Apply distance formula. WLOG, assume you use
. Then, the area of
is
Solution 5 (power of a point)
Firstly, the point of intersection of the two tangent lines has an equal distance to points and
due to power of a point theorem. This means we can easily find the point, which is
. Label this point
.
is an isosceles triangle with lengths,
,
, and
. Label the midpoint of segment
as
. The height of this triangle, or
, is
. Since
bisects
,
contains the diameter of circle
. Let the two points on circle
where
intersects be
and
with
being the shorter of the two. Now let
be
and
be
. By Power of a Point on
and
,
. Applying Power of a Point again on
and
,
. Expanding while using the fact that
,
. Plugging this into
,
. Using the quadratic formula,
, and since
,
. Since this is the diameter, the radius of circle
is
, and so the area of circle
is
.
Video Solution
For those who want a video solution: (Is similar to Solution 1) https://youtu.be/WI2NVuIp1Ik
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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