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Given the following object - box of which a rectangular pyramid is removed.

By means of unmarked ruler, draw lines on the surface of the object to guide cuts of the object into two objects with the same volume - half of the original object volume.

enter image description here

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  • $\begingroup$ May I know are point $E$ and point $G$ fixed points? $\endgroup$ – 00xxqhxx00 Dec 13 '20 at 5:50
  • $\begingroup$ Yes. ABCD a rectangle and EFGB a rectangle contained in ABCD. $\endgroup$ – Moti Dec 13 '20 at 5:53
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    $\begingroup$ Also, do you provide the ratio of $AE:EB$ and $BG:GC$? Or they could be randomised? $\endgroup$ – 00xxqhxx00 Dec 13 '20 at 5:56
  • $\begingroup$ No. It is not impacting the solution. $\endgroup$ – Moti Dec 14 '20 at 5:11
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General solution (what we need to construct)

It can be done with a single cut.

We will freely use the centers of any rectangle as they are trivial to construct. Let M be the center of BEFG. It is easy to verify that the plane given by D,M,I does the trick.
Indeed any plane through DI passes through the centroid and hence cuts the full box in half (mirroring at the three principal planes through the box centroid maps the box and any such plane onto themselves but swaps halfspaces).
Likewise any plane through MI cuts the pyramid in half. This can be seen by observing that it passes through the centroid of every horizonotal section (by intercept theorem).

1 The easy way, assuming we can use B

So all we need to do is find the antipode of the second intersection X of DM with ABCD.

One easy way of achieving that is to take advantage of the triplets of parallel lines AB,FG,CD and AD,EF,BC. Using those we can easily construct the midpoints of all the sides of ABCD. The midpoints in turn allow us to find the mirror image (on the opposite edge of ABCD) of X. Once we have that the rest is trivial.

Or, even simpler, only mark two sides which is sufficient to guide a single cut: enter image description here

2 The hard way, assuming we cannot use what's been cut off

Thanks @Bass for drawing my attention to this interpretation.

We will assume nondegeneracy, i.e $E \ne F \ne G$. We still can make use of the triplets of parallel lines as above to construct a half size copy of the "L" DAEFGC anchored at D. We can repeat this procedure as needed until we have a copy with the point corresponding to B on terra firma. We can now work on the shrunk copy as above the line DM will be identical to its shrunk copy. After mirroring the shrunk copy of X it will sit on a complete face, so we can blow it up to original scale (in terms of distance to D) and proceed as above. EDIT: Actually, the blow up can be done on the L, see picture bottom right.

showing not everything but a few steps to give a flavor. Top row: the shrink step, bottom left: constructing the cut line DM through the base, the shrunk copy and the full original were it available give the same result. I': this can be directly connected with I to give the cut on that surface. X' this must be put back to scale, right panel shows how, mirrored to the top edge and from there transferred to the edge antipodal to BC. enter image description here

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  • $\begingroup$ I was just about to ask OP if we are allowed to make marks in the air where the removed part used to be. (Or maybe there's another way to find those points?) $\endgroup$ – Bass Dec 13 '20 at 15:07
  • $\begingroup$ Good point @Bass. As long as there is a nondegenerate L_shaped bit left, we can work around using B. I'll update the answer. $\endgroup$ – Paul Panzer Dec 13 '20 at 15:17
  • $\begingroup$ @Paul You can not use B. Just drawing on surface. $\endgroup$ – Moti Dec 13 '20 at 19:12
  • $\begingroup$ @Paul I do not see two volume equal objects $\endgroup$ – Moti Dec 13 '20 at 19:31
  • $\begingroup$ Ok, I've filled in some more detail, rest really is trivial. $\endgroup$ – Paul Panzer Dec 13 '20 at 21:18

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