Difference between revisions of "1985 AHSME Problems/Problem 22"
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==Problem== | ==Problem== | ||
− | In a circle with center <math> O </math>, <math> AD </math> is a | + | 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== | ||
− | + | 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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− | Since <math> \angle CAD </math> is inscribed | ||
==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 , is a diameter, is a chord, and . Then the length of is
Solution
Since is an angle inscribed in a arc, we obtain , so is a -- right triangle. This gives and , and now since is a diameter, . The fact that is a diameter also means that , so is again a -- right triangle, yielding and , which finally gives .
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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