Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 14"

 
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==Problem==
 
==Problem==
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In square <math>ABCD</math> the segment <math>KB</math> equals a side of the square. The ratio of areas <math>\frac{S_1}{S_2}</math> is
 
In square <math>ABCD</math> the segment <math>KB</math> equals a side of the square. The ratio of areas <math>\frac{S_1}{S_2}</math> is
  
 
<math> \mathrm{(A) \ } \frac{1}{3}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{1}{\sqrt{2}}\qquad \mathrm{(D) \ } \sqrt2-1\qquad \mathrm{(E) \ } \frac{\sqrt{2}}4</math>
 
<math> \mathrm{(A) \ } \frac{1}{3}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{1}{\sqrt{2}}\qquad \mathrm{(D) \ } \sqrt2-1\qquad \mathrm{(E) \ } \frac{\sqrt{2}}4</math>
 
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==Solution==
 
==Solution==

Revision as of 09:40, 8 May 2007

Problem

2007 CyMO-14.PNG

In square $ABCD$ the segment $KB$ equals a side of the square. The ratio of areas $\frac{S_1}{S_2}$ is

$\mathrm{(A) \ } \frac{1}{3}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{1}{\sqrt{2}}\qquad \mathrm{(D) \ } \sqrt2-1\qquad \mathrm{(E) \ } \frac{\sqrt{2}}4$

Solution

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See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 13
Followed by
Problem 15
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