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.


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).
