Difference between revisions of "2019 AMC 10B Problems/Problem 23"
m (→Solution 3) |
(→Solution 3) |
||
Line 24: | Line 24: | ||
The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0) | The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0) | ||
− | + | <math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math> | |
− | AD=sqrt | + | <math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math> |
− | DC=sqrt | + | <math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math> |
− | Therefore OA=AC*AD | + | Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math> |
− | Consequently, the area of the circle is pi*OA^2=pi*85/ | + | Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}</math> |
(by Zhen Qin) | (by Zhen Qin) | ||
(P.S. Will someone please Latex this?) | (P.S. Will someone please Latex this?) | ||
+ | |||
+ | (<math>\LaTeX</math>) | ||
==See Also== | ==See Also== |
Revision as of 22:04, 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.
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 AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC is similiar to DAC, that is OA/AC=AD/DC. The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0)
Therefore
Consequently, the area of the circle is
(by Zhen Qin)
(P.S. Will someone please Latex this?)
()
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.
SUB2PEWDS