Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 14"
I_like_pie (talk | contribs) |
(img) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| + | <div style="float:right"> | ||
| + | [[Image:2007 CyMO-14.PNG|250px]] | ||
| + | </div> | ||
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> | ||
| − | |||
| − | |||
==Solution== | ==Solution== | ||
Revision as of 10:40, 8 May 2007
Problem
In square 
the segment 
equals a side of the square. The ratio of areas 
is
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
| 2007 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
