Difference between revisions of "1973 AHSME Problems/Problem 15"
Rockmanex3 (talk | contribs) (Solution to Problem 15 — drawing diagram is hard) |
Rockmanex3 (talk | contribs) (Solution to Problem 15 (credit to official solution)) |
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Line 18: | Line 18: | ||
draw((0,0)--(0,-125)); | draw((0,0)--(0,-125)); | ||
draw(arc((0,0),200,233.13,306.87)); | draw(arc((0,0),200,233.13,306.87)); | ||
+ | dot((0,0)); | ||
+ | label("O",(0,0),N); | ||
+ | dot((-120,-160)); | ||
+ | label("A",(-120,-160),SW); | ||
+ | dot((120,-160)); | ||
+ | label("B",(120,-160),SE); | ||
</asy> | </asy> | ||
Revision as of 13:37, 3 August 2018
Problem
A sector with acute central angle is cut from a circle of radius 6. The radius of the circle circumscribed about the sector is
Solution
Let be the center of the circle and be two points on the circle such that . If the circle circumscribes the sector, then the circle must circumscribe .
Draw the perpendicular bisectors of and and mark the intersection as point , and draw a line from to . By HL Congruency and CPCTC, .
Let be the circumradius of the triangle. Using the definition of cosine for right triangles, Answer choices A, C, and E are smaller, so they are eliminated. However, as aproaches , the value would approach infinity while would approach . A super large circle would definitely not be a circumcircle if is close to , so we can confirm that the answer is .
See Also
1973 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |