Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 12B Problems/Problem 9"
(→‎Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
 +
=Trig Solution:=
 
Let <math>\alpha</math> be the angle that subtends the arc AB. By the law of cosines,
 
Let <math>\alpha</math> be the angle that subtends the arc AB. By the law of cosines,
 
<math>6^2=5^2+5^2-2*5*5cos(\alpha)</math>
 
<math>6^2=5^2+5^2-2*5*5cos(\alpha)</math>
Line 18: Line 19:
 
<math>AC = \sqrt{10}</math>, which is answer choice A.
 
<math>AC = \sqrt{10}</math>, which is answer choice A.
  
 +
=Other 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>, 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=7|num-a=9}}
 
{{AMC12 box|year=2008|ab=B|num-b=7|num-a=9}}

Revision as of 19:57, 30 November 2008

Problem 9

Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6$. Point $C$ is the midpoint of the minor arc $AB$. What is the length of the line segment $AC$?

$\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4$

Solution

Trig Solution:

Let $\alpha$ be the angle that subtends the arc AB. By the law of cosines, $6^2=5^2+5^2-2*5*5cos(\alpha)$

$\alpha = cos^{-1}(7/25)$

The half-angle formula says that $cos(\alpha/2) = \frac{\sqrt{1+cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ $AC = \sqrt{5^2+5^2-2*5*5*\frac{4}{5}}$

$AC = \sqrt{50-50\frac{4}{5}}$

$AC = \sqrt{10}$, which is answer choice A.

Other 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, $m\angle RDA=90\deg$, and so $RD=\sqrt{5^2-3^2}=4$. Since $RD=4$, $CD=5-4=1$, and so $AC=sqrt{3^2+1^2}=\sqrt{10} \rightarrow A$

See Also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_9&oldid=28442"