Difference between revisions of "2008 AMC 12B Problems/Problem 9"
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− | ==Problem | + | ==Problem== |
Points <math>A</math> and <math>B</math> are on a circle of radius <math>5</math> and <math>AB = 6</math>. Point <math>C</math> is the midpoint of the minor arc <math>AB</math>. What is the length of the line segment <math>AC</math>? | Points <math>A</math> and <math>B</math> are on a circle of radius <math>5</math> and <math>AB = 6</math>. Point <math>C</math> is the midpoint of the minor arc <math>AB</math>. What is the length of the line segment <math>AC</math>? | ||
<math>\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4</math> | <math>\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4</math> | ||
− | == | + | ==Solutions== |
− | === | + | ===Solution 1=== |
− | 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 <math>AB</math>. By the law of cosines, |
− | <math>6^2=5^2+5^2-2 | + | <math>6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)</math> implies <math>\cos(\alpha) = 7/25</math>. |
− | <math>\alpha = cos^{- | + | The [[Trigonometric_identities#Half_Angle_Identities | half-angle formula]] says that |
+ | <math>\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}</math>. The law of cosines tells us <math>AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}</math>, which is answer choice <math>\boxed{\text{A}}</math>. | ||
− | + | ===Solution 2=== | |
− | <math> | + | Define <math>D</math> as the midpoint of line segment <math>\overline{AB}</math>, and <math>O</math> the center of the circle. Then <math>O</math>, <math>C</math>, and <math>D</math> are collinear, and since <math>D</math> is the midpoint of <math>AB</math>, <math>m\angle ODA=90\deg</math> and so <math>OD=\sqrt{5^2-3^2}=4</math>. Since <math>OD=4</math>, <math>CD=5-4=1</math>, and so <math>AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow \boxed{\text{A}}</math>. |
− | <math>AC = \sqrt{ | ||
− | < | + | <center><asy> |
− | + | pen d = linewidth(0.7); pathpen = d; pointpen = black; pen f = fontsize(9); | |
− | + | path p = CR((0,0),5); | |
− | + | pair O = (0,0), A=(5,0), B = IP(p,CR(A,6)), C = IP(p,CR(A,3)), D=IP(A--B,O--C); | |
− | = | + | D(p); D(MP("A",A,E)--D(MP("O",O))--MP("B",B,NE)--cycle); D(A--MP("C",C,ENE),dashed+d); D(O--C,dashed+d); D(rightanglemark(O,D(MP("D",D,W)),A)); MP("5",(A+O)/2); MP("3",(A+D)/2,SW); |
− | + | </asy></center> | |
==See Also== | ==See Also== | ||
− | {{AMC12 box|year=2008|ab=B|num-b= | + | {{AMC12 box|year=2008|ab=B|num-b=8|num-a=10}} |
+ | {{MAA Notice}} |
Latest revision as of 13:48, 15 February 2021
Problem
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 ?
Solutions
Solution 1
Let be the angle that subtends the arc . By the law of cosines, implies .
The half-angle formula says that . The law of cosines tells us , which is answer choice .
Solution 2
Define as the midpoint of line segment , and the center of the circle. Then , , and are collinear, and since is the midpoint of , 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 |
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