Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 14"
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==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== | |
| + | <math>\triangle KBC</math> and <math>\triangle KCD</math> have the same heights (<math>\perp BD</math>), so the ratio of their areas is simply the ratio of <math>KD:KB</math>. | ||
| − | = | + | Let the length of <math>KB</math> be <math>x</math>. Then <math>KD=x\sqrt{2}-x</math>, and the ratio of <math>S_1:S_2</math> is <math>x\sqrt{2}-x:x</math>, or <math>\sqrt{2}-1:1\Longrightarrow\mathrm{ D}</math>. |
| − | {{ | ||
==See also== | ==See also== | ||
{{CYMO box|year=2007|l=Lyceum|num-b=13|num-a=15}} | {{CYMO box|year=2007|l=Lyceum|num-b=13|num-a=15}} | ||
Latest revision as of 22:02, 8 May 2007
Problem
In square 
the segment 
equals a side of the square. The ratio of areas 
is
Solution
and 
have the same heights (
), so the ratio of their areas is simply the ratio of 
.
Let the length of be 
. Then 
, and the ratio of 
is 
, or 
.
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 | ||
