Difference between revisions of "2008 AMC 12B Problems/Problem 9"
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===Solution 2=== | ===Solution 2=== | ||
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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>. | 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>. | ||
+ | |||
+ | <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=8|num-a=10}} | {{AMC12 box|year=2008|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:58, 21 July 2020
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 ?
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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