Cube A is 1 unit on each side, with a body diagonal connecting points p & q. A cube B is then constructed with edge pq. As cube A spins along edge pq, does the volume of the intersecting cubes remain constant (at 1/4 unit cubed) or does it vary? And if it does vary, what are the maximum and minimum.Naraht (talk) 02:47, 25 February 2025 (UTC)[reply]

The rotating cube B cuts the cube A along two triangles T and S, both with fixed vertices p and q, and both with a third vertex moving along 8 edges of the cube A (the edges which are adjacent to neither p nor q). The variation of the volume of the intersection is proportional to the difference of the areas of T and S (the sign being given by the sense of rotation). Since T and S have fixed bases, their area is proportional to their heights, let's call them t and s. If we project the cubes onto the plane orthogonal to the diagonal pq, we see a hexagon PA and a rotating square PB with a fixed vertex on the center of the hexagon, cutting the hexagon along two rotating orthogonal segments of length t and s. The variation of the area of the intersection is proportional to the s,t.So the volume of the intersection of A and B is proportional to the area of the intersection of PA and PB. It follows that it is maximum when a face of B meets a vertex of A, and it is minimum when s=t and the intersection is symmetric pma 22:49, 3 March 2025 (UTC)[reply]

Nice argument, clearly explained! So the intersection of cube A with a wedge with edge pq has a constant volume iff the wedge angle is multiple of 60°? catslash (talk) 23:09, 3 March 2025 (UTC)[reply]

True, nice remark! pma 01:02, 4 March 2025 (UTC)[reply]

Actually my argument is not correct: the variation of volume is not proportional to t-s, but to t² - s² (for an angle dw the volume gains ⅓area(T)tdw, and looses ⅓area(S)sdw ). The conclusion is the same though... pma 01:22, 4 March 2025 (UTC)[reply]

Yes, I get

${\displaystyle {\begin{aligned}{\frac {\partial V}{\partial \theta }}&=\int _{0}^{t}{\sqrt {3}}{\frac {t-u}{t}}u\partial u-\int _{0}^{s}{\sqrt {3}}{\frac {s-u}{s}}u\partial u\\&=\left[{\sqrt {3}}{\frac {u^{2}}{t}}{\frac {3t-2u}{6}}\right]_{u=0}^{u=t}-\left[{\sqrt {3}}{\frac {u^{2}}{s}}{\frac {3s-2u}{6}}\right]_{u=0}^{u=s}\\&={\frac {t^{2}-s^{2}}{2{\sqrt {3}}}}\end{aligned}}}$

but to determine that the volume varies, it is only necessary to realize that

${\displaystyle {\frac {\partial V}{\partial \theta }}=f(t)-f(s)}$

which is the key insight which eluded me. catslash (talk) 22:42, 5 March 2025 (UTC)[reply]