Difference between revisions of "2009 AMC 10B Problems/Problem 16"
(→Solution 3) |
|||
(2 intermediate revisions by one other user not shown) | |||
Line 69: | Line 69: | ||
== Solution 3== | == Solution 3== | ||
− | <math>\triangle BAO\cong\triangle BCO</math> by SSS congruence, so <math>\angle ABO = \angle CBO = \frac{60}{2} = 30 \ | + | <math>\triangle BAO\cong\triangle BCO</math> by SSS congruence, so <math>\angle ABO = \angle CBO = \frac{60}{2} = 30 ^\circ</math>. Since <math>BA</math> is tangent to the circle, it is perpendicular to <math>AO</math>. This means that <math>\triangle BAO</math> is a 30-60-90 triangle. The ratio of the side-lengths of a 30-60-90 triangle is <math>1:\sqrt3:2</math>, so <math>BO=2</math>. <math>BD = BO-DO</math>. Since <math>DO</math> is the radius of the circle, <math>DO=1</math> and <math>BD = 2-1=1</math>. Hence, <math>\frac{BD}{BO} = \frac{1}{2}</math>, and the answer is <math>\boxed{\textbf{(B) } \frac{1}{2}}</math> ~azc1027 |
== See Also == | == See Also == |
Latest revision as of 19:59, 25 November 2023
Problem
Points and
lie on a circle centered at
, each of
and
are tangent to the circle, and
is equilateral. The circle intersects
at
. What is
?
Solution
Solution 1
As is equilateral, we have
, hence
. Then
, and from symmetry we have
. Thus, this gives us
.
We know that , as
lies on the circle. From
we also have
, Hence
, therefore
, and
.
Solution 2
As in the previous solution, we find out that . Hence
and
are both equilateral.
We then have , hence
is the incenter of
, and as
is equilateral,
is also its centroid. Hence
, and as
, we have
, therefore
, and as before we conclude that
.
Solution 3
by SSS congruence, so
. Since
is tangent to the circle, it is perpendicular to
. This means that
is a 30-60-90 triangle. The ratio of the side-lengths of a 30-60-90 triangle is
, so
.
. Since
is the radius of the circle,
and
. Hence,
, and the answer is
~azc1027
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.