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>. | 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> | 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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| − | <math>4q + \tfrac{4}{\sqrt3} = 4</ | + | 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> | ||
| − | < | + | <cmath>4q = 4 - \tfrac{4}{\sqrt3}</cmath> |
| − | < | + | <cmath>q = 1 - \tfrac{1}{\sqrt3} = \tfrac{\sqrt3 - 1 }{\sqrt3}</cmath> |
| − | < | + | <cmath>\frac{q}{t} = \frac{\tfrac{\sqrt3 - 1 }{\sqrt3}}{\tfrac{1}{2\sqrt3}} = 2\cdot (\sqrt3 - 1) = \boxed{2\sqrt3-2}</cmath> |
Latest revision as of 02:42, 12 December 2018
Problem
![[asy] draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle, black+linewidth(.75)); draw((0,-1)--(0,1), black+linewidth(.75)); draw((-1,0)--(1,0), black+linewidth(.75)); draw((-1,-1/sqrt(3))--(1,1/sqrt(3)), black+linewidth(.75)); draw((-1,1/sqrt(3))--(1,-1/sqrt(3)), black+linewidth(.75)); draw((-1/sqrt(3),-1)--(1/sqrt(3),1), black+linewidth(.75)); draw((1/sqrt(3),-1)--(-1/sqrt(3),1), black+linewidth(.75)); [/asy]](http://latex.artofproblemsolving.com/b/a/6/ba646966f2f2a41a55ba82a65b59d08dcf4e8076.png)
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 ![\[4q+(8\cdot \tfrac{1}{2\sqrt3}) = 2^2=4\]](http://latex.artofproblemsolving.com/e/6/7/e67682e0f26d2c39a309887ebeec6f6418e256bc.png)
![\[4q + \tfrac{4}{\sqrt3} = 4\]](http://latex.artofproblemsolving.com/c/c/1/cc1abdc155286a805afc51f6d85dd6967ba4dfb3.png)
![\[4q = 4 - \tfrac{4}{\sqrt3}\]](http://latex.artofproblemsolving.com/0/a/d/0adffec9dd6f644aa60fb8ddb659373c07339e8f.png)
![\[q = 1 - \tfrac{1}{\sqrt3} = \tfrac{\sqrt3 - 1 }{\sqrt3}\]](http://latex.artofproblemsolving.com/7/6/a/76abddd899465005f603f00da07591eb97865cb8.png)
![\[\frac{q}{t} = \frac{\tfrac{\sqrt3 - 1 }{\sqrt3}}{\tfrac{1}{2\sqrt3}} = 2\cdot (\sqrt3 - 1) = \boxed{2\sqrt3-2}\]](http://latex.artofproblemsolving.com/d/a/8/da8c428762dd57b48da718704c9dec8a8719c099.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
