Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 8"
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==Solution== | ==Solution== | ||
| − | {{ | + | In the quadrilateral <math>A\Gamma IZ</math>, we have three isosceles triangles <math>A\Gamma\Delta</math>, <math>AZE</math>, and <math>\Gamma \Delta I</math>. Those are congruent to each other, as well as <math>HAB</math>, <math>B\Gamma\Theta</math>, and <math>EK\Delta</math>. Also, <math>AE\Delta</math> is congruent to <math>AB\Gamma</math>. Thus we have two figures of equal area: <math>A\Gamma IZ</math> and a combination of two figures: <math>HB\Theta\Gamma A</math> and <math>EK\Delta</math>. Since the area of the whole star is 1, the area of <math>AZI\Gamma</math> is <math>\frac{1}{2}\mathrm{(B)}</math>. |
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=7|num-a=9}} | {{CYMO box|year=2006|l=Lyceum|num-b=7|num-a=9}} | ||
Latest revision as of 08:18, 12 August 2008
Problem
In the figure 
is a regular 5-sided polygon and 
, 
, 
, 
, 
are the points of intersections of the extensions of the sides.
If the area of the "star" 
is 1, then the area of the shaded quadrilateral 
is
Solution
In the quadrilateral , we have three isosceles triangles 
, 
, and 
. Those are congruent to each other, as well as 
, 
, and 
. Also, 
is congruent to 
. Thus we have two figures of equal area: and a combination of two figures: 
and . Since the area of the whole star is 1, the area of 
is 
.
See also
| 2006 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
