Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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==Solution== | ==Solution== | ||
+ | First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>. | ||
+ | |||
+ | Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem: | ||
+ | |||
+ | <math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>. | ||
==See Also== | ==See Also== |
Revision as of 17:07, 14 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
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
.
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.