Origami Fun

Note that all triangles in the figure are equilateral triangles. Denote by $A_1$ , $A_2$ the areas of the equilateral triangles of side $s_1$ and $s_2$ respectively.
Then, we have $\frac{{{A}_{1}}}{{{A}_{2}}}=2\Rightarrow {{\left( \frac{{{s}_{1}}}{{{s}_{2}}} \right)}^{2}}=2\Rightarrow {{s}_{1}}^{2}=2{{s}_{2}}^{2}\Rightarrow {{s}_{1}}={{s}_{2}}\sqrt{2} \ \ \ \ \ (1)$ Moreover,

${{s}_{1}}+{{s}_{2}}=1\overset{\left( 1 \right)}{\mathop{\Rightarrow }}\,{{s}_{2}}\sqrt{2}+{{s}_{2}}=1\Rightarrow {{s}_{2}}\left( \sqrt{2}+1 \right)=1\Rightarrow {{s}_{2}}=\frac{1}{\sqrt{2}+1}\Rightarrow {{s}_{2}}=\sqrt{2}-1$ Hence, $\left( 1 \right)\Rightarrow {{s}_{1}}=\left( \sqrt{2}-1 \right)\sqrt{2}\Rightarrow {{s}_{1}}=2-\sqrt{2}$ Finally,

${{s}_{1}}+2{{s}_{2}}=2-\sqrt{2}+2\left( \sqrt{2}-1 \right)=\sqrt{2}\approx 1,4143$ For the answer, $\left\lfloor 1000\left( \left( {{s}_{1}}+2{{s}_{2}} \right) \right) \right\rfloor =\boxed{1414}$ .

Note that $s_1 + s_2 = 1$ , the side length of the equilateral triangular paper. For figure of the same shape, in this case the equilateral triangle, area is directly proportional to the square of the linear dimension, in this case the side length. This means that $\dfrac {s_1^2}{s_2^2} = 2 \implies s_1 = \sqrt 2 s_2$ . Then we have:

$\begin{aligned} s_1 + s_2 & = 1 \\ (\sqrt 2 + 1)s_2 & = 1 \\ \implies s_2 & = \frac 1{\sqrt 2 + 1} \\ s_1 + 2s_2 & = (\sqrt 2 + 2) s_2 = \frac {2+\sqrt 2}{\sqrt 2 +1} = \frac {\sqrt 2(\sqrt 2 +1)}{\sqrt 2+1} = \sqrt 2 \\ \implies \left \lfloor 1000(s_1+2s_2)\right \rfloor & = \lfloor 1000\sqrt 2 \rfloor = \boxed{1414} \end{aligned}$

Since the large triangle has side length 1 and is equilateral, its height is given by sin(60 degrees) = sqrt(3) / 2. Let h be the height of the folded triangle. We then see that the height of the little triangle is sqrt(3)/2 - h. Hence, we have:

Area(folded triangle) = A(fold) = (1/2) h s1, and

Area(little triangle) = A(little) = (1/2) (s2) (sqrt(3)/2 - h) = (1/2) (1-s1) (sqrt(3)/2 - h)

(the last equality follows since s1 + s2 = 1 = length of the side of the big triangle).

Utilizing the relationship A(fold) = 2*A(little) we find:

(1/2) (s1) h = (1-s1)*(sqrt(3)/2 - h) (equation 1)

But, from the folded triangle, we find:

tan(60 degrees) = sqrt(3) = h/((s1)/2), so that:

h = (s1)*sqrt(3)/2

Substituting this into equation 1 and simplifying yields the equation:

(s1)^2 -4*s1 +2 = 0 or

s1 = 2 - sqrt(2) (we reject the negative root).

Hence, s2 = 1-s1 = sqrt(2) - 1, so that

1000 (s1 + 2 s2) = 1000*sqrt(2)

The greatest integer function of this last quantity is 1414.