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Difference between revisions of "1985 AHSME Problems/Problem 22"
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==Problem==
 
==Problem==
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In a circle with center <math> O </math>, <math> AD </math> is a [[diameter]], <math> ABC </math> is a [[chord]], <math> BO=5 </math>, and <math> \angle ABO=\stackrel{\frown}{CD}=60^\circ </math>. Then the length of <math> BC </math> is:
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In a circle with center <math> O </math>, <math> AD </math> is a [[diameter]], <math> ABC </math> is a [[chord]], <math> BO=5 </math> and <math> \angle ABO= \ \stackrel{\frown}{CD} \ =60^\circ </math>. Then the length of <math> BC </math> is
  
 
<asy>
 
<asy>

Revision as of 00:08, 3 April 2018

Problem

In a circle with center $O$, $AD$ is a diameter, $ABC$ is a chord, $BO=5$ and $\angle ABO= \ \stackrel{\frown}{CD} \ =60^\circ$. Then the length of $BC$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^Circle(O,1)); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(305)); label("$5$", B--O, dir(O--B)*dir(90)); label("$60^\circ$", dir(185), dir(185)); label("$60^\circ$", B+0.05*dir(-25), dir(-25));[/asy]

$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }3+\sqrt{3} \qquad \mathrm{(C) \ } 5-\frac{\sqrt{3}}{2} \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of the above}$

Solution

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^Circle(O,1)); pair point=O; draw(C--D); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(305)); label("$5$", B--O, dir(O--B)*dir(90)); label("$60^\circ$", dir(185), dir(185)); label("$60^\circ$", B+0.05*dir(-25), dir(-25));[/asy]

Since $\angle CAD$ is inscribed and intersects an arc of length $60^\circ$, $\angle CAD=30^\circ$. Thus, $\triangle ABO$ is a $30-60-90$ right triangle. Thus, $AO=BO\sqrt{3}=5\sqrt{3}$ and $AB=2BO=10$. Since $AO$ and $DO$ are both radii, $DO=AO=5\sqrt{3}$ and $AD=10\sqrt{3}$. Since $\angle ACD$ is inscribed in a semicircle, it's a right angle, and $\triangle ACD$ is also a $30-60-90$ right triangle. Thus, $CD=\frac{AD}{2}=5\sqrt{3}$ and $AC=CD\sqrt{3}=15$. Finally, $BC=AC-AB=15-10=5, \boxed{\text{D}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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