Difference between revisions of "2005 AMC 12B Problems/Problem 14"
m (→See also) |
Arthy00009 (talk | contribs) m (→Solution) |
||
(One intermediate revision by one other user not shown) | |||
Line 11: | Line 11: | ||
== Solution == | == Solution == | ||
− | Let <math>R</math> be the radius of the circle. Draw the two radii that meet the points of tangency to the lines <math>y = \pm x</math>. We can see that a square is formed by the origin, two tangency points, and the center of the circle. The side lengths of this square are <math>R</math> and the diagonal is <math>k = R+6</math>. The diagonal of a square is <math>\sqrt{2}</math> times the side length. Therefore, <math>R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow E</math>. | + | Let <math>R</math> be the radius of the circle. Draw the two radii that meet the points of tangency to the lines <math>y = \pm x</math>. We can see that a square is formed by the origin, two tangency points, and the center of the circle. The side lengths of this square are <math>R</math> and the diagonal is <math>k = R+6</math>. The diagonal of a square is <math>\sqrt{2}</math> times the side length. Therefore, <math>R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow \boxed{\mathrm{E}}</math>. |
<center> | <center> | ||
Line 39: | Line 39: | ||
== See also == | == See also == | ||
{{AMC12 box|year=2005|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2005|ab=B|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:12, 24 December 2020
Problem
A circle having center , with
, is tangent to the lines
,
and
. What is the radius of this circle?
Solution
Let be the radius of the circle. Draw the two radii that meet the points of tangency to the lines
. We can see that a square is formed by the origin, two tangency points, and the center of the circle. The side lengths of this square are
and the diagonal is
. The diagonal of a square is
times the side length. Therefore,
.
See also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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.