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Difference between revisions of "2008 AMC 10B Problems/Problem 10"
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==Problem==
 
==Problem==
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Suppose that <math>(u_n)</math> is a serquence of real numbers satifying <math>u_{n+2}=2u_{n+1}+u_n</math>,
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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>?
  
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and that <math>u_3=9</math> and <math>u_6=128</math>. What is <math>u_5</math>?
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<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>
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(A) 40 (B) 53 (C) 68 (D) 88 (E) 104
 
  
 
==Solution==
 
==Solution==
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We know that <math>u_6=128</math>, so we plug in <math>n=4</math> to get <math>128=2u_5+u_4</math>. We plug in <math>n=3</math> to get
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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 of the circle.
  
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<math>u_5=2u_4+9</math>. Substituting gives
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By the [[Pythagorean Theorem]], <math>OD = \sqrt{OA^2 - DA^2} = 4</math>, and by subtraction, <math>CD=OC - OD = 1</math>.
  
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<math>128=5u_4+18 \rightarrow u_4=22</math>
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Using the Pythagorean Theorem again, <math>AC= \sqrt{AD^2 + CD^2} = \sqrt{3^2+1^2}=\sqrt{10} \Longrightarrow \textbf{(A)}</math>.
  
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This gives <math>u_5=\frac{128-22}{2}=53</math>.
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<center><asy>
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pen d = linewidth(0.7); pathpen = d; pointpen = black; pen f = fontsize(9);
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Answer B is the correct answer
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path p = CR((0,0),5);
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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);
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NOTE: This is my (BOGTRO) solution, not the official one,
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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);
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</asy></center>
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and should be ignored in view of a better solution.
 
  
 
==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}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 11:36, 7 June 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 $AC$?

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

Solution

Let the center of the circle be $O$, and let $D$ be the intersection of $\overline{AB}$ and $\overline{OC}$ (then $D$ is the midpoint of $\overline{AB}$). $OA=OB=5$, since they are both radii of the circle.

By the Pythagorean Theorem, $OD = \sqrt{OA^2 - DA^2} = 4$, and by subtraction, $CD=OC - OD = 1$.

Using the Pythagorean Theorem again, $AC= \sqrt{AD^2 + CD^2} = \sqrt{3^2+1^2}=\sqrt{10} \Longrightarrow \textbf{(A)}$.

[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 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

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