Take the third case.
Suppose Arctos, Cuspis, and Dos are each cut half-way. This figure is an equilateral triangle, whose sides are the diagonal of a half-unit squared. The figure Z1 . X1 . Y1 is also an equilateral triangle, and the figure Z11⁄2 . X11⁄2 . Y11⁄2 is an equilateral hexagon.
It is easy for us to see what these shapes are, and also, what the figures of any other set would be, as ZI . XII . YII or ZI . XII . YIII but we must learn them as a two-dimensional being would, so that we may see how to learn the three-dimensional sections of a tessaract.
It is evident that the resulting figures are the same whether we fix the cube, and then turn the sectional plane to the required position, or whether we fix the sectional plane, and then turn the cube. Thus, in the first case we might have fixed the plane, and then so placed the cube that one plane side coincided with the sectional plane, and then have drawn the cube half-way through, in a direction at right angles to the plane, when we should have seen the square first mentioned. In the second case (ZO . XI . YI) we might have put the cube with Arctos coinciding with the plane and with Cuspis and Dos equally inclined to it, and then have drawn the cube through the plane at right angles to it until the lines (Cuspis and Dos) were cut at the required distances from Corvus. In the third case we might have put the cube with only Corvus coinciding with the plane and with Cuspis, Dos, and Arctos equally inclined to it (for any of the shapes in the set ZI . XI . YI) and then have drawn it through as before. The resulting figures are exactly the same as those we got before; but this way is the best to use, as it would probably be easier for a two-dimensional being to think of a cube passing through his space than to imagine his whole space turned round, with regard to the cube.
We have already seen ([p. 117]) how a two-dimensional being would observe the sections of a cube when it is put with one plane side coinciding with his space, and is then drawn partly through.
Now, suppose the cube put with the line Arctos coinciding with his space, and the lines Cuspis and Dos equally inclined to it. At first he would only see Arctos. If the cube were moved until Dos and Cuspis were each cut half-way, Arctos still being parallel to the plane, Arctos would disappear at once; and to find out what he would see he would have to take the square sections of the cube, and find on each of them what lines are given by the new set of sections. Thus he would take Moena itself, which may be regarded as the first section of the square set. One point of the figure would be the middle point of Cuspis, and since the sectional plane is parallel to Arctos, the line of intersection of Moena with the sectional plane will be parallel to Arctos. The required line then cuts Cuspis half-way, and is parallel to Arctos, therefore it cuts Callis half-way.
Fig. 21.
Next he would take the square section half-way between Moena and Murex. He knows that the line Alvus of this section is parallel to Arctos, and that the point Dos at one of its ends is half-way between Corvus and Cista, so that this line itself is the one he wants (because the sectional plane cuts Dos half-way between Corvus and Cista, and is parallel to Arctos). In [Fig. 21] the two lines thus found are shown. a b is the line in Moena, and c d the line in the section. He must now find out how far apart they are. He knows that from the middle point of Cuspis to Corvus is half-a-unit, and from the middle point of Dos to Corvus is half-a-unit, and Cuspis and Dos are at right angles to each other; therefore from the middle point of Cuspis to the middle point of Dos is the diagonal of a square whose sides are half-a-unit in length. This diagonal may be written d (1⁄2)2. He would also see that from the middle point of Callis to the middle point of Via is the same length; therefore the figure is a parallelogram, having two of its sides, each one unit in length, and the other two each d (1⁄2)2.
He could also see that the angles are right, because the lines a c and b d are made up of the X and Y directions, and the other two, a b and d, are purely Z, and since they have no tendency in common, they are at right angles to each other.