Difference between revisions of "2008 AMC 10B Problems/Problem 10"
(→Solution) |
|||
(5 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
==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>? | |
− | + | <math>\mathrm{(A)}\ \sqrt{10}\qquad\mathrm{(B)}\ \frac{7}{2}\qquad\mathrm{(C)}\ \sqrt{14}\qquad\mathrm{(D)}\ \sqrt{15}\qquad\mathrm{(E)}\ 4</math> | |
− | |||
− | (A) | ||
==Solution== | ==Solution== | ||
− | + | 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. | |
− | <math> | + | By the [[Pythagorean Theorem]], <math>OD = \sqrt{OA^2 - DA^2} = 4</math>, and by subtraction, <math>CD=OC - OD = 1</math>. |
− | <math> | + | Using the Pythagorean Theorem again, <math>AC= \sqrt{AD^2 + CD^2} = \sqrt{3^2+1^2}=\sqrt{10} \Longrightarrow \textbf{(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== | ||
{{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
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.