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Difference between revisions of "1993 AHSME Problems/Problem 17"

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== Solution ==
 
== Solution ==
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Assume the length of the side of the square is 4, WLOG.  This means the side of one t section is 2.  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 2.  Therefore, the short leg is <math>\tfrac{2}{\sqrt3}</math>.
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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 2 \cdot \tfrac{2}{\sqrt3} =  \tfrac{2}{\sqrt3}</math>
 
  
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The total area comprises <math>4q+8t</math>, so <math>4q+(8\cdot \tfrac{2}{\sqrt3}) = 4^2=16</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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<math>4q + \tfrac{16}{\sqrt3} = 16</math>
 
  
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<math>4q = 16 - \tfrac{16}{\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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<math>q = 4 - \tfrac{4}{\sqrt3} = \tfrac{4\sqrt3 - 4 }{\sqrt3}</math>
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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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<math>\frac{q}{t} = \frac{\tfrac{4\sqrt3 - 4 }{\sqrt3}}{\tfrac{2}{\sqrt3}} = \frac{4\sqrt3 - 4}{2} = \boxed{2\sqrt3-2}</math>
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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>
  
  

Latest revision as of 03:42, 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]

Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If $t$ is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and $q$ is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then $\frac{q}{t}=$

$\text{(A) } 2\sqrt{3}-2\quad \text{(B) } \frac{3}{2}\quad \text{(C) } \frac{\sqrt{5}+1}{2}\quad \text{(D) } \sqrt{3}\quad \text{(E) } 2$

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 $\tfrac{360}{12} = 30$ degree angle from the center. So each section t is a $30-60-90$ triangle with a long leg of 1. Therefore, the short leg is $\tfrac{1}{\sqrt3}$.


This makes the area of each $t = \tfrac{1}{2}\cdot b \cdot h = \tfrac{1}{2}\cdot 1 \cdot \tfrac{1}{\sqrt3} = \tfrac{1}{2\sqrt3}$


The total area comprises $4q+8t$, so \[4q+(8\cdot \tfrac{1}{2\sqrt3}) = 2^2=4\]

\[4q + \tfrac{4}{\sqrt3} = 4\]

\[4q = 4 - \tfrac{4}{\sqrt3}\]

\[q = 1 - \tfrac{1}{\sqrt3} = \tfrac{\sqrt3 - 1 }{\sqrt3}\]


\[\frac{q}{t} = \frac{\tfrac{\sqrt3 - 1 }{\sqrt3}}{\tfrac{1}{2\sqrt3}} = 2\cdot (\sqrt3 - 1) = \boxed{2\sqrt3-2}\]


$\fbox{A}$

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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