Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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− | The mid point of AB is D(9,12) | + | The mid point of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>X</math> axis. Then <math>CD</math> would be the perpendicular bisector of <math>AB</math>. Let the center of circle be O. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, that is <math>\frac{OA}{AC} = \frac{AD}{DC}.</math> |
− | The slope of AB is | + | The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, therefore the slope of CD will be 3. The equation of <math>CD</math> is <math>y-12=3*(x-9)</math>, that is <math>y=3x-15</math>. Let <math>y=0</math>. Then we have <math>x=5</math>, which is the <math>x</math> coordinate of <math>C(5,0)</math>. |
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math> | <math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math> |
Revision as of 21:12, 15 February 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, 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
.
Solution 3
The mid point of is
. Let the tangent lines at
and
intersect at
on the
axis. Then
would be the perpendicular bisector of
. Let the center of circle be O. Then
is similar to
, that is
The slope of
is
, therefore the slope of CD will be 3. The equation of
is
, that is
. Let
. Then we have
, which is the
coordinate of
.
Therefore
Consequently, the area of the circle is
(by Zhen Qin)
(P.S. Will someone please Latex this?)
(ed by a pewdiepie subscriber)
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.
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