Difference between revisions of "1993 AHSME Problems/Problem 17"
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== Solution == | == Solution == | ||
+ | Assume the length of the side of the square is 2, WLOG. This means the side of one t section is 1. As the lines are at clock face positions, each section has a <math>\tfrac{360}{12} = 30</math> degree angle from the center. So each section t is a <math>30-60-90</math> triangle with a long leg of 1. Therefore, the short leg is <math>\tfrac{1}{\sqrt3}</math>. | ||
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+ | This makes the area of each <math>t = \tfrac{1}{2}\cdot b \cdot h = \tfrac{1}{2}\cdot 1 \cdot \tfrac{1}{\sqrt3} = \tfrac{1}{2\sqrt3}</math> | ||
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+ | The total area comprises <math>4q+8t</math>, so <cmath>4q+(8\cdot \tfrac{1}{2\sqrt3}) = 2^2=4</cmath> | ||
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+ | <cmath>4q + \tfrac{4}{\sqrt3} = 4</cmath> | ||
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+ | <cmath>4q = 4 - \tfrac{4}{\sqrt3}</cmath> | ||
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+ | <cmath>q = 1 - \tfrac{1}{\sqrt3} = \tfrac{\sqrt3 - 1 }{\sqrt3}</cmath> | ||
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+ | <cmath>\frac{q}{t} = \frac{\tfrac{\sqrt3 - 1 }{\sqrt3}}{\tfrac{1}{2\sqrt3}} = 2\cdot (\sqrt3 - 1) = \boxed{2\sqrt3-2}</cmath> | ||
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<math>\fbox{A}</math> | <math>\fbox{A}</math> | ||
Latest revision as of 02:42, 12 December 2018
Problem
Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then
Solution
Assume the length of the side of the square is 2, WLOG. This means the side of one t section is 1. As the lines are at clock face positions, each section has a degree angle from the center. So each section t is a triangle with a long leg of 1. Therefore, the short leg is .
This makes the area of each
The total area comprises , so
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 19 | |
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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