Cyclic Bisecting

First consider an equilateral triangle with the same conditions, labelled as follows:

By rotational symmetry, $A_1 = A_2 = A_3$ , $B_1 = B_2 = B_3$ , and $C_1 = C_2 = C_3$ .

Since $PX = XU$ , $A_1 + B_1 = A_2 + C_1$ and $A_3 + B_3 + C_3 = D + B_2 + C_2$ . Using the above equations, we can deduce that $B_1 = C_1$ and $A_3 = D$ (so $B_1 = B_2 = B_3 = C_1 = C_2 = C_3$ and $A_1 = A_2 = A_3 = D$ ).

Examining triangles with a side ratio of $\frac{PX}{XY}$ , we have $\frac{A_1 + B_1}{C_1} = \frac{C_3}{D}$ , and using the above equalities this becomes $\frac{D + B_1}{B_1} = \frac{B_1}{D}$ , which simplifies to $B_1 = \frac{1 + \sqrt{5}}{2}D$ (the golden ratio!)

Since the area of the triangle is $1$ , we have $A_1 + A_2 + A_3 + B_1 + B_2 + B_3 + C_1 + C_2 + C_3 = 1$ , or $4D + 6B_1 = 1$ .

Solving $B_1 = \frac{1 + \sqrt{5}}{2}D$ and $4D + 6B_1 = 1$ gives $D = \boxed{\frac{7 - 3\sqrt{5}}{4}}$ .

Since any triangle with the same area can be transformed to this equilateral triangle by shear mapping , which also preserves areas and length ratios, this value of $D$ is true for any triangle with an area of $1$ .