Teresa solved this problem, using this hints we gave:

The area of the equilateral triangle is

1 = \frac{1}{2} \times (2 t)^{2} \times \sin 60^{\circ}

t^{2} = \frac{1}{\sqrt{3}}

, so

t = \frac{1}{\sqrt[4]{3}} \approx 0.760

We know that $p = t$ and $q = 1 - t$ , so $p \approx 0.760$ and $q \approx 0.240$ .

The height of the equilateral triangle is $1 + s = t \sqrt{3}$ so $s \approx 0.316$ .

We can use Pythagoras' Theorem to find $m$ : $m^{2} = s^{2} + (1 - t)^{2} \approx 0.156$ , so $m \approx 0.397$ .

Now we can work out $θ$ : $\tan θ = \frac{1 - t}{s} \approx 0.760$ , so $θ \approx 37 . 2^{\circ}$ .

From the sine rule, we have $\frac{u}{\sin 30^{\circ}} = \frac{1}{\sin (150 - θ)}$ , so $u = \frac{1}{2 \sin (150 - θ)} \approx 0.542$ .

Also from the sine rule, $n = \frac{\sin θ}{\sin (150 - θ)} \approx 0.656$ .

By looking at the longest line across the square, we have $\frac{1}{m + u + v} = \cos θ$ , so $v = \frac{1}{\cos θ} - m - u \approx 0.317$ .

By looking at the rectangle and considering the diagonal, we see that $(r + n)^{2} = (1 + s)^{2} + t^{2}$ , so $r = \sqrt{(1 + s)^{2} + t^{2}} - n \approx 0.864$ .