There's an old problem I was once asked to solve. It is to determine the length of a brace when you know a frame width, height and a brace width. An old method of bracing a frame was to put a weld at the corners of the brace where it touched the frame. The problem is, someone had to draw that and calculate the length of the square cut brace. In reality it wasn't a big issue because a little clearance isn't the end of the world.
I think I have solved the problem however I am struggling to simplify it.
Is there anyone out there who can provide an alternative solution or simplify my solution (or prove it incorrect).
In my solution "BL" is unknown, my question mark looks like a 2....
My Solution:
I am sorry to tell you that, as far as I am concerned, there are no more elegant solutions to this problem other than the one based on the three equations you have already derived yourself. If you want to, you can approach the problem from a different angle, but you will end up with same set of equations. Furthermore, I am not aware of any geometrical construction that could help us to determine the measure of $BL$.
We begin by deriving the first of your three equations in order to show to the visitors to this page that it is correct. Then, we transform (i.e. simplify) the very equation into a form that can be solved using a technique known to mathematicians.
The shape of the brace is a rectangle, the length and width of which equal to $BL$ and $W$ respectively. $\mathrm{Fig.\space 2}$ shows the bottom half of the brace. Pay your attention to the half-diagonal 1 and 2 of the brace, which are equal in length. We apply Pythagoras theorem to the two large colored right angle triangles, of which the half-diagonals are the respective hypotenuses, to obtain, $$ \left(\frac{L}{2}-B\right)^2+\left(\frac{H}{2}\right)^2 = \left(\frac{H}{2}-A\right)^2+\left(\frac{L}{2}\right)^2. \tag{1} $$
We apply Pythagoras theorem to the small colored right angle triangle, of which the width of the brace is the hypotenuse, to obtain, $$W^2 = A^2 + B^2. \tag{2}$$
When we replace $A$ from (1) by substituting its value obtained from (2), we get, $$ \left(\frac{L}{2}-B\right)^2+ \left(\frac{H}{2}\right)^2 = \left(\frac{H}{2}-\sqrt{W^2 – B^2}\right)^2 + \left(\frac{L}{2}\right)^2. \tag{3} $$
Once we expand (3), we get, $$ H\sqrt{W^2 – B^2} =W^2 + LB -2B^2. $$
Now, we square both sides of this equation and, then, simplify it as shown below. $$\text{Squared:}\quad H^2\left(W^2 – B^2\right) = W^4 + 2W^2LB – 4W^2B^2 + L^2B^2 – 4LB^3 +4B^4\qquad$$ $$\text{Simplified:}\quad 4B^4 – 4LB^3 + \left(L^2+H^2-4W^2\right)B^2 + 2LW^2B + W^2\left(W^2-H^2\right)=0$$
This is the quartic equation we must solve for $B$. Since $L$, $W$, and $H$ are real values, this equation has 0, 2, or 4 real roots. We need to use numerical methods, such as Newton-Raphson method, to determine the value of $B$.
When we substitute the given values of $W (=10)$, $L (=120)$, and $H (=70)$ in this quartic equation, we get the following equation. $$B^4-120B^3+4725B^2+6000B -120000=0$$
This equation has only two real roots and one of them is a negative value. The only positive real root is the value of $B$, i.e. $$B = \frac{5}{2}\left(12-7\sqrt{3}+\sqrt{56\sqrt{3}-93}\right) \approx 4.68588809382830997.$$