Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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Therefore <math>OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}</math>, | Therefore <math>OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}</math>, | ||
and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>. | and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>. | ||
+ | |||
+ | ==Solution 4 (Power of a Point)== | ||
+ | |||
+ | Firstly, the point of intersection of the two tangent lines has an equal distance to points <math>A</math> and <math>B</math> due to power of a point theorem. This means we can easily find the point, which is <math>(5, 0)</math>. Label this point <math>X</math>. <math>\triangle{XAB}</math> is an isosceles triangle with lengths, <math>\sqrt{170}</math>, <math>\sqrt{170}</math>, and <math>2\sqrt{10}</math>. Label the midpoint of segment <math>AB</math> as <math>M</math>. The height of this triangle, or <math>\overline{XM}</math>, is <math>4\sqrt{10}</math>. Since <math>\overline{XM}</math> bisects <math>\overline{AB}</math>, <math>\overleftrightarrow{XM}</math> contains the diameter of circle <math>\omega</math>. Let the two points on circle <math>\omega</math> where <math>\overleftrightarrow{XM}</math> intersects be <math>P</math> and <math>Q</math> with <math>\overline{XP}</math> being the shorter of the two. Now let <math>\overline{MP}</math> be <math>x</math> and <math>\overline{MQ}</math> be <math>y</math>. By Power of a Point on <math>\overline{PQ}</math> and <math>\overline{AB}</math>, <math>xy = (\sqrt{10})^2 = 10</math>. Applying Power of a Point again on <math>\overline{XQ}</math> and <math>\overline{XA}</math>, <math>(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170</math>. Expanding while using the fact that <math>xy = 10</math>, <math>y=x+\frac{\sqrt{10}}{2}</math>. Plugging this into <math>xy=10</math>, <math>2x^2+\sqrt{10}x-20=0</math>. Using the quadratic formula, <math>x = \frac{\sqrt{170}-\sqrt{10}}{4}</math>, and since <math>x+y=2x+\frac{\sqrt{10}}{2}</math>, <math>x+y=\frac{\sqrt{170}}{2}</math>. Since this is the diameter, the radius of circle <math>\omega</math> is <math>\frac{\sqrt{170}}{4}</math>, and so the area of circle <math>\omega</math> is <math>\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>. | ||
+ | |||
+ | -bradleyguo | ||
==See Also== | ==See Also== |
Revision as of 17:28, 30 May 2019
- The following problem is from both the 2019 AMC 10B #23 and 2019 AMC 12B #20, so both problems redirect to this page.
Contents
[hide]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
.
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 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
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 (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
.
-bradleyguo
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.