Difference between revisions of "2008 AMC 10B Problems/Problem 10"
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− | Let the center of the circle be <math>O</math>, and let <math>D</math> be the intersection of <math>\overline{AB}</math> and <math>\overline{OC}</math> (then <math>D</math> is the midpoint of <math>\overline{AB}</math>). <math>OA=OB=5</math>, since they are both radii. | + | Let the center of the circle be <math>O</math>, and let <math>D</math> be the intersection of <math>\overline{AB}</math> and <math>\overline{OC}</math> (then <math>D</math> is the midpoint of <math>\overline{AB}</math>). <math>OA=OB=5</math>, since they are both radii of the circle. |
By the [[Pythagorean Theorem]], <math>OD = \sqrt{OA^2 - DA^2} = 4</math>, and by subtraction, <math>CD=OC - OD = 1</math>. | By the [[Pythagorean Theorem]], <math>OD = \sqrt{OA^2 - DA^2} = 4</math>, and by subtraction, <math>CD=OC - OD = 1</math>. |
Latest revision as of 12:36, 7 June 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
?
Solution
Let the center of the circle be , and let
be the intersection of
and
(then
is the midpoint of
).
, since they are both radii of the circle.
By the Pythagorean Theorem, , and by subtraction,
.
Using the Pythagorean Theorem again, .
![[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]](http://latex.artofproblemsolving.com/7/2/f/72fbe8b90fcde66500984b26ba2b38b0f7765119.png)
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.