A Cube and an Octahedron

The volume of the composite 3D figure is given by $V=V_o + 8V_p$ , where $V_o$ is the volume of the octahedron and $V_p$ is the volume of one of the pyramids of the corners of cube.

Let the shared center of the cube and octahedron be the origin $O(0,0,0)$ . The $x$ - and $y$ -axes is each perpendicular to a horizontal edge of the octahedron and the $z$ -axis is vertical and passes through the top vertex of the octahedron $A$ .

The top-half of the octahedron has a $\dfrac 32 \times \dfrac 32$ square base, whose diagonals are $\dfrac {3\sqrt 2}2$ , and an altitude $AO = \sqrt{\left(\dfrac 32\right)^2 - \left( \dfrac {3\sqrt 2}4 \right)^2}$ $= \dfrac {3\sqrt 2}4$ . Then it volume is $V_o = 2 \times \dfrac 13 \left(\dfrac 32\right)^2 AO$ $= \dfrac {9\sqrt 2}8$ .

Let us look at the 3D figure along the $x$ -axis. Then the intercept of $y$ -axis by its perpendicular horizontal octahedron edge is $B \left(0, \dfrac 34, 0\right)$ . The straight line $AB$ is given by $\dfrac {z-0}{y-\frac 34} = - \dfrac {\frac {3\sqrt 2}4}{\frac 34}$ $\implies z = \dfrac {3\sqrt 2}4 - \sqrt 2 y$ . We note that the top-right vertex of cube is $D\left(0, \dfrac 1{\sqrt 2}, \dfrac 12\right)$ . Then $C \left(0, y_c, \dfrac 12\right)$ and $E \left(0, \dfrac 1{\sqrt 2}, z_e\right)$ . And $\dfrac 12 = \dfrac {3\sqrt 2}4 - \sqrt 2 y_c$ , $\implies y_c = \dfrac {3-\sqrt 2}4$ and $z_e = \dfrac {3\sqrt 2}4 - \sqrt 2 \times \dfrac 1{\sqrt 2}$ $= \dfrac {3\sqrt 2-4}4$ . Let $CD = a$ and $DE = b$ . Then $a = y_d - y_c$ $= \dfrac 1{\sqrt 2} - \dfrac {3-\sqrt 2}4$ $= \dfrac {3\sqrt 2-3}4$ and $b = d_e - z_e$ $= \dfrac 12 - \dfrac {3\sqrt 2-4}4$ $=\dfrac {6-3\sqrt 2}4$ .

Now, we note that the inverted pyramid of vertex $D$ has a isosceles right triangle base with a hypotenuse of $2a$ and a height of $b$ . Therefore $V_p = \dfrac 13 \times a^2 \times b$ $=\dfrac {90-63\sqrt 2}{64}$ .

Therefore $V = V_o + 8V_p = \dfrac {9\sqrt 2}8 + 8 \times \dfrac {90-63\sqrt 2}{64}$ $= \dfrac {90-54\sqrt 2}8$ $\approx \boxed{1.704}$ .