Difference between revisions of "2009 AMC 10B Problems/Problem 16"
(New page: == Problem == Points <math>A</math> and <math>C</math> lie on a circle centered at <math>O</math>, each of <math>\overline{BA}</math> and <math>\overline{BC}</math> are tangent to the cir...) |
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\text{(E) } \frac {\sqrt3}{2} | \text{(E) } \frac {\sqrt3}{2} | ||
</math> | </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == | ||
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</asy> | </asy> | ||
− | As <math>\triangle ABC</math> is equilateral, we have <math>\angle BAC = \angle BCA = 60^\circ</math>, hence <math>\angle OAC = \angle OCA = 30^\circ</math>. Then <math>\ | + | As <math>\triangle ABC</math> is equilateral, we have <math>\angle BAC = \angle BCA = 60^\circ</math>, hence <math>\angle OAC = \angle OCA = 30^\circ</math>. Then <math>\angle AOC = 120^\circ</math>, and from symmetry we have <math>\angle AOB = \angle COB = 60^\circ</math>. Thus, this gives us <math>\angle ABO = \angle CBO = 30^\circ</math>. |
− | We know that <math>DO = AO</math>, as <math>D</math> lies on the circle. From <math>\triangle ABO</math> we also have <math>AO = BO \sin 30^\circ = \frac{BO}2</math>, Hence <math>DO = \frac{BO}2</math>, therefore <math>BD = BO - DO = \frac{BO}2</math>, and <math>\frac{BD}{BO} = \boxed{\frac 12}</math>. | + | We know that <math>DO = AO</math>, as <math>D</math> lies on the circle. From <math>\triangle ABO</math> we also have <math>AO = BO \sin 30^\circ = \frac{BO}2</math>, Hence <math>DO = \frac{BO}2</math>, therefore <math>BD = BO - DO = \frac{BO}2</math>, and <math>\frac{BD}{BO} = \boxed{\frac 12 \Longrightarrow B}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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We then have <math>\angle SCD = \angle SAD = 30^\circ</math>, hence <math>D</math> is the incenter of <math>\triangle ABC</math>, and as <math>\triangle ABC</math> is equilateral, <math>D</math> is also its centroid. Hence <math>2 \cdot SD = BD</math>, and as <math>SD = SO</math>, we have <math>2\cdot SD = SD + SO = OD</math>, therefore <math>BD=OD</math>, and as before we conclude that <math>\frac{BD}{BO} = \boxed{\frac 12}</math>. | We then have <math>\angle SCD = \angle SAD = 30^\circ</math>, hence <math>D</math> is the incenter of <math>\triangle ABC</math>, and as <math>\triangle ABC</math> is equilateral, <math>D</math> is also its centroid. Hence <math>2 \cdot SD = BD</math>, and as <math>SD = SO</math>, we have <math>2\cdot SD = SD + SO = OD</math>, therefore <math>BD=OD</math>, and as before we conclude that <math>\frac{BD}{BO} = \boxed{\frac 12}</math>. | ||
+ | |||
+ | == Solution 3== | ||
+ | <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 == | ||
{{AMC10 box|year=2009|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2009|ab=B|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
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 |
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