Difference between revisions of "2008 AMC 12B Problems/Problem 9"
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<math>AC = \sqrt{10}</math>, which is answer choice A. | <math>AC = \sqrt{10}</math>, which is answer choice A. | ||
| − | === | + | ===More Elegant Solution=== |
| − | Define D as the midpoint of AB, and R the center of the circle. R, C, and D are collinear, and since D is the midpoint of AB, <math>m\angle RDA=90\deg</math> | + | Define D as the midpoint of AB, and R the center of the circle. R, C, and D are collinear, and since D is the midpoint of AB, |
| + | |||
| + | <math>m\angle RDA=90\deg</math> | ||
| + | |||
| + | and so | ||
| + | |||
| + | <math>RD=\sqrt{5^2-3^2}=4</math>. | ||
| + | |||
| + | Since | ||
| + | |||
| + | <math>RD=4</math>, | ||
| + | |||
| + | <math>CD=5-4=1</math>, | ||
| + | |||
| + | and so | ||
| + | |||
| + | <math>AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow A</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2008|ab=B|num-b=8|num-a=10}} | ||
Revision as of 16:23, 14 August 2011
Problem 9
Points 
and 
are on a circle of radius 
and 
. Point 
is the midpoint of the minor arc 
. What is the length of the line segment 
?
Solution
Trig Solution:
Let 
be the angle that subtends the arc AB. By the law of cosines,
The half-angle formula says that
, which is answer choice A.
More Elegant Solution
Define D as the midpoint of AB, and R the center of the circle. R, C, and D are collinear, and since D is the midpoint of AB,
and so
.
Since
,
,
,
and so
See Also
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 | |
