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"

(→‎Trig Solution:)
 
(7 intermediate revisions by 6 users not shown)
Line 1: Line 1:
−
==Problem 9==
+
==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>
  
−
==Solution==
+
==Solutions==
−
===Trig Solution:===
+
===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*5*5cos(\alpha)</math>
+
<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^{-1}(7/25)</math>
+
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>.
  
−
The half-angle formula says that
+
===Solution 2===
−
<math>cos(\alpha/2) = \frac{\sqrt{1+cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}</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{5^2+5^2-2*5*5*\frac{4}{5}}</math>
 
  
−
<math>AC = \sqrt{50-50\frac{4}{5}}</math>
+
<center><asy>
−
 
+
pen d = linewidth(0.7); pathpen = d; pointpen = black; pen f = fontsize(9);
−
<math>AC = \sqrt{10}</math>, which is answer choice A.
+
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);
−
=Other Solution=
+
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);
−
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>
+
</asy></center>
  
 
==See Also==
 
==See Also==
−
{{AMC12 box|year=2008|ab=B|num-b=7|num-a=9}}
+
{{AMC12 box|year=2008|ab=B|num-b=8|num-a=10}}
 +
{{MAA Notice}}

Latest revision as of 12:48, 15 February 2021

Problem

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 $AC$?

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

Solutions

Solution 1

Let $\alpha$ be the angle that subtends the arc $AB$. By the law of cosines, $6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$ implies $\cos(\alpha) = 7/25$.

The half-angle formula says that $\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The law of cosines tells us $AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$, which is answer choice $\boxed{\text{A}}$.

Solution 2

Define $D$ as the midpoint of line segment $\overline{AB}$, and $O$ the center of the circle. Then $O$, $C$, and $D$ are collinear, and since $D$ is the midpoint of $AB$, $m\angle ODA=90\deg$ and so $OD=\sqrt{5^2-3^2}=4$. Since $OD=4$, $CD=5-4=1$, and so $AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow \boxed{\text{A}}$.

[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]

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_9&oldid=146834"