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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>
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==Solution==
 
==Solution==
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<asy>
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Since <math>\angle CAD</math> is an angle inscribed in a <math> 60{^\circ}</math> arc, we obtain <math>\angle CAD =\frac{60^{\circ}}{2} = 30^{\circ}</math>, so <math>\triangle ABO</math> is a <math>30^{\circ}</math>-<math>60^{\circ}</math>-<math>90^{\circ}</math> right triangle. This gives <math>AO = BO\sqrt{3} = 5\sqrt{3}</math> and <math>AB = 2BO = 10</math>, and now since <math>AD</math> is a diameter, <math>AD = 2AO = 10\sqrt{3}</math>. The fact that <math>AD</math> is a diameter also means that <math>\angle ACD = 90^{\circ}</math>, so <math>\triangle ACD</math> is again a <math>30^{\circ}</math>-<math>60^{\circ}</math>-<math>90^{\circ}</math> right triangle, yielding <math>CD = \frac{1}{2}AD = 5\sqrt{3}</math> and <math>AC = \frac{\sqrt{3}}{2}AD = 15</math>, which finally gives <math>BC = AC-AB = 15-10 = \boxed{\text{(D)} \ 5}</math>.
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defaultpen(linewidth(0.7)+fontsize(10));
 
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pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C);
 
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draw(C--A--D^^B--O^^Circle(O,1));
 
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pair point=O;
 
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draw(C--D);
 
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label("$A$", A, dir(point--A));
 
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label("$B$", B, dir(point--B));
 
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label("$C$", C, dir(point--C));
 
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label("$D$", D, dir(point--D));
 
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label("$O$", O, dir(305));
 
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label("$5$", B--O, dir(O--B)*dir(90));
 
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label("$60^\circ$", dir(185), dir(185));
 
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label("$60^\circ$", B+0.05*dir(-25), dir(-25));</asy>
 
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Since <math> \angle CAD </math> is inscribed and intersects an arc of length <math> 60^\circ </math>, <math> \angle CAD=30^\circ </math>. Thus, <math> \triangle ABO </math> is a <math> 30-60-90 </math> right triangle. Thus, <math> AO=BO\sqrt{3}=5\sqrt{3} </math> and <math> AB=2BO=10 </math>. Since <math> AO </math> and <math> DO </math> are both radii, <math> DO=AO=5\sqrt{3} </math> and <math> AD=10\sqrt{3} </math>. Since <math> \angle ACD </math> is inscribed in a semicircle, it's a right angle, and <math> \triangle ACD </math> is also a <math> 30-60-90 </math> right triangle. Thus, <math> CD=\frac{AD}{2}=5\sqrt{3} </math> and <math> AC=CD\sqrt{3}=15 </math>. Finally, <math> BC=AC-AB=15-10=5, \boxed{\text{D}} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=21|num-a=23}}
 
{{AHSME box|year=1985|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:19, 19 March 2024

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

Since $\angle CAD$ is an angle inscribed in a $60{^\circ}$ arc, we obtain $\angle CAD =\frac{60^{\circ}}{2} = 30^{\circ}$, so $\triangle ABO$ is a $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ right triangle. This gives $AO = BO\sqrt{3} = 5\sqrt{3}$ and $AB = 2BO = 10$, and now since $AD$ is a diameter, $AD = 2AO = 10\sqrt{3}$. The fact that $AD$ is a diameter also means that $\angle ACD = 90^{\circ}$, so $\triangle ACD$ is again a $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ right triangle, yielding $CD = \frac{1}{2}AD = 5\sqrt{3}$ and $AC = \frac{\sqrt{3}}{2}AD = 15$, which finally gives $BC = AC-AB = 15-10 = \boxed{\text{(D)} \ 5}$.

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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