Find the side length of a square with line segments of length 1, 2, and 3 extending from each corner and intersecting at their tips I know trigonometry should be involved in this somehow but am stuck at where to construct the triangles.

 A: Let the origin be at the bottom left corner of the square, and let the point where the line segments meet be $(x,y)$, then
$ x^2 + (y - a)^2 = 1 $
$ (x - a)^2 + (y - a)^2 = 4$
$ x^2 + y^2  = 9 $
These are three quadratic equations in three unknowns with a special structure.  Subtracting (1) from (3), gives us,
$ 2 a y - a^2 = 8 $
Subtracting (2) from (3) gives us
$ 2 a x + 2 a y - 2 a^2 = 5 $
Substracting the last two equations, yields
$ - 2 a x + a^2 = 3 $
Therefore, $ x = \dfrac{ a^2 - 3 }{2 a} $
Plug this into the equation involving $x$ and $y$, and solve for $y$ you get
$ a^2 - 3 + 2 a y - 2 a^2 = 5 $
Therefore,
$ y = \dfrac{ a^2 + 8 }{2 a} $
Now we have expressions for $x$ and $y$ in terms of the side length $a$.  Substitute this into the third equation from the top, you get
$  (a^2 - 3 )^2 + (a^2 + 8)^2 = 36 a^2 $
Let $ u = a^2 $ , then this last equation becomes
$ 2 u^2 - 26 u + 73 = 0$
For which the solutions are
$ u = \dfrac{1}{4} ( 26 \pm \sqrt{92} )  $
From which the two possible side lengths are
$ a_1 \approx 2.02536 $ and $a_2 = 2.9829374 $
Out of which, only the second solution is valid (by checking the values of $x$ and $y$ from the above equations).
