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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 22"

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== Problem ==
 
== Problem ==
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<div style="float:right">
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[[Image:2006 CyMO-22.PNG|250px|right]]
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[[Image:2006 CyMO-22.PNG|250px]]
 
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</div>
 
  
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<math>ABCD</math> is rectangular and the points <math>K,L,M,N</math> lie on the sides <math>AB, BC, CD, DA</math> respectively so that <math>\frac{AK}{KB}=\frac{BL}{LC}=\frac{CM}{MD}=\frac{DN}{NA}=2</math>. If <math>E_1</math> is the area of <math>KLMN</math> and <math>E_2</math> is the area of the rectangle <math>ABCD</math>, the ratio <math>\frac{E_1}{E_2}</math> equals
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<math>AB\Gamma \Delta</math> is rectangular and the points <math>K,\Lambda ,M,N</math> lie on the sides <math>AB, B\Gamma , \Gamma \Delta, \Delta A</math> respectively so that <math>\frac{AK}{KB}=\frac{BL}{L\Gamma}=\frac{\Gamma M}{M\Delta}=\frac{\Delta N}{NA}=2</math>. If <math>E_1</math> is the area of <math>K\Lambda MN</math> and <math>E_2</math> is the area of the rectangle <math>AB\Gamma \Delta</math>, the ratio <math>\frac{E_1}{E_2}</math> equals
  
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A. <math>\frac{5}{9}</math>
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<math>\mathrm{(A)}\ \frac{5}{9}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{9}{5}\qquad\mathrm{(D)}\ \frac{3}{5}\qquad\mathrm{(E)}\ \text{None of these}</math>
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B. <math>\frac{1}{3}</math>
 
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C. <math>\frac{9}{5}</math>
 
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D. <math>\frac{3}{5}</math>
 
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E. None of these
 
  
 
==Solution==
 
==Solution==
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Let <math>AB = CD = x</math>, <math>BC = AD = y</math>. Using the [[Pythagorean Theorem]], <math>KM = \sqrt{\frac{x^2}{9} + y^2}</math>, <math>LN = \sqrt{x^2 + \frac{y^2}{9}}</math>. Using the formula <math>A = \frac{1}{2}d_1d_2</math> for a [[rhombus]], we get <math>\frac{1}{2}\sqrt{\left(x^2 + \frac{y^2}{9}\right)\left(x^2 + \frac{y^2}{9}\right)} = \frac{1}{2}\sqrt{\frac{x^4}{9} + \frac{y^4}{9} + \frac{82}{81}x^2y^2} = \frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18}</math>. Thus the ratio is <math>\frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18xy}</math>. There is no way we can simplify this further, and in fact we can plug in different values of <math>x,y</math> to see that the answer is <math>\mathrm{E}</math>.  
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Let <math>AB = \Gamma \Delta = x</math>, <math>B\Gamma = A\Delta = y</math>. Using the [[Pythagorean Theorem]], <math>KM = \sqrt{\frac{x^2}{9} + y^2}</math>, <math>\Lambda N = \sqrt{x^2 + \frac{y^2}{9}}</math>. Using the formula <math>A = \frac{1}{2}d_1d_2</math> for a [[rhombus]], we get <math>\frac{1}{2}\sqrt{\left(x^2 + \frac{y^2}{9}\right)\left(x^2 + \frac{y^2}{9}\right)} = \frac{1}{2}\sqrt{\frac{x^4}{9} + \frac{y^4}{9} + \frac{82}{81}x^2y^2} = \frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18}</math>. Thus the ratio is <math>\frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18xy}</math>. There is no way we can simplify this further, and in fact we can plug in different values of <math>x,y</math> to see that the answer is <math>\mathrm{E}</math>.  
  
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Be careful not to just try a couple of simple examples like <math>ABCD</math> being a square, where we will get the answer <math>5/9</math>, which is incorrect in general.  
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Be careful not to just try a couple of simple examples like <math>AB\Gamma \Delta</math> being a square, where we will get the answer <math>5/9</math>, which is incorrect in general.
  
 
==See also==
 
==See also==

Latest revision as of 12:19, 26 April 2008

Problem

2006 CyMO-22.PNG

$AB\Gamma \Delta$ is rectangular and the points $K,\Lambda ,M,N$ lie on the sides $AB, B\Gamma , \Gamma \Delta, \Delta A$ respectively so that $\frac{AK}{KB}=\frac{BL}{L\Gamma}=\frac{\Gamma M}{M\Delta}=\frac{\Delta N}{NA}=2$. If $E_1$ is the area of $K\Lambda MN$ and $E_2$ is the area of the rectangle $AB\Gamma \Delta$, the ratio $\frac{E_1}{E_2}$ equals

$\mathrm{(A)}\ \frac{5}{9}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{9}{5}\qquad\mathrm{(D)}\ \frac{3}{5}\qquad\mathrm{(E)}\ \text{None of these}$

Solution

Let $AB = \Gamma \Delta = x$, $B\Gamma = A\Delta = y$. Using the Pythagorean Theorem, $KM = \sqrt{\frac{x^2}{9} + y^2}$, $\Lambda N = \sqrt{x^2 + \frac{y^2}{9}}$. Using the formula $A = \frac{1}{2}d_1d_2$ for a rhombus, we get $\frac{1}{2}\sqrt{\left(x^2 + \frac{y^2}{9}\right)\left(x^2 + \frac{y^2}{9}\right)} = \frac{1}{2}\sqrt{\frac{x^4}{9} + \frac{y^4}{9} + \frac{82}{81}x^2y^2} = \frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18}$. Thus the ratio is $\frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18xy}$. There is no way we can simplify this further, and in fact we can plug in different values of $x,y$ to see that the answer is $\mathrm{E}$.

Be careful not to just try a couple of simple examples like $AB\Gamma \Delta$ being a square, where we will get the answer $5/9$, which is incorrect in general.

See also

2006 Cyprus MO, Lyceum (Problems)
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
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