Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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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) | ||
− | AC=sqrt((6-5)^2+(13-0)^2)=sqrt(170) | + | AC=<math>sqrt((6-5)^2+(13-0)^2)=sqrt(170)</math> |
AD=sqrt((6-9)^2)+(13-12)^2)=sqrt(10) | AD=sqrt((6-9)^2)+(13-12)^2)=sqrt(10) | ||
DC=sqrt((9-5)^2+(12-0)^2)=aqrt(160) | DC=sqrt((9-5)^2+(12-0)^2)=aqrt(160) |
Revision as of 20:56, 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)
AC= AD=sqrt((6-9)^2)+(13-12)^2)=sqrt(10) DC=sqrt((9-5)^2+(12-0)^2)=aqrt(160) Therefore OA=AC*AD/DC=sqrt(85/5) Consequently, the area of the circle is pi*OA^2=pi*85/5 (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.
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