Question

A paper shown in Panel I is folded along the dashed lines (- - -) to construct a cube. The shaded regions shown in Panel I appear on the outer surface of the cube. Referring to the cubes shown in Panel II, which one of the options is correct?

• A. Only (i) can correspond to the unfolded cube in Panel I.
• B. Only (ii) can correspond to the unfolded cube in Panel I.
• C. Both (i) and (ii) can correspond to the unfolded cube in Panel I.
• D. Neither (i) nor (ii) can correspond to the unfolded cube in Panel I.

Hint:

Only the diamond face is symmetric under reflection. Check whether the wedge and sliver next to it point into their shared corner from the same side that the net's diagonal cut would produce, or from the mirrored side.

Answer 1

The Correct Option is A

Step 1: Read the net face by face.
Panel I is a strip of four squares stacked one below the other, with one extra square glued to the right side of the topmost square and one extra square glued to the left side of the bottommost square. That is six squares in all, one for each face of the cube once every dashed line is folded through a right angle. Going through them in order: the topmost square carries a grey right-angled triangle tucked into its lower-left corner, not quite reaching the far corner. The square glued to its right is shaded completely grey from edge to edge, with no white left on it anywhere. The second square of the vertical strip carries a small grey square sitting exactly in its upper-right corner. The third square of the strip carries a large grey diamond whose four points touch the midpoints of the four sides, a shape that looks identical no matter which way it is turned. The fourth, bottom square of the strip is cut in half along the diagonal running from its upper-right corner to its lower-left corner, with the lower-right half shaded grey. The square glued to the left of this last square is cut along the same diagonal direction, but here it is the upper-left half that is shaded grey, the exact complement of its neighbour.
Step 2: Fix the one idea the question is actually testing.
The diamond is the only marking on the whole net with full symmetry. It looks identical after any rotation and even after a mirror flip, so by itself it cannot decide anything. What decides the question is the pair of triangles on either side of the diamond in the strip, the lower-right triangle on the fourth square and its complementary upper-left triangle on the flap glued to it, together with one physical fact: folding paper is a rigid motion. A rigid fold can spin a shape around in space any way at all, but it can never turn a shape into its own mirror image, since that would need the ink lifted off one side of the sheet and printed on the other. So once the strip is folded into a cube with every grey mark kept facing outward, the diamond face and its two triangular neighbours can only ever meet each other, around their shared cube corner, in one fixed rotational sense. Swap that sense for its mirror and the resulting picture cannot be produced by folding this particular net, however the finished solid is later turned in the hand.
Step 3: Fold the diamond square and its two triangular neighbours and see how they close up.
Hold the strip with the diamond square facing you. Its neighbour carrying the diagonal cut folds back to sit on one side of the diamond face, and the second diagonal-cut square, attached beyond it, wraps around to sit on the third visible side. Because the shaded lower-right triangle and the shaded upper-left triangle are complementary halves of the very same diagonal cut, when the shared edge between them is folded to a right angle, the grey portions come together only at the end of that shared cube edge nearest the diamond face, while the far end of the same edge stays white on both faces. This fixes the picture completely: the diamond sits in the middle, a broad grey wedge appears on one neighbouring face with its grey portion reaching right into the corner shared with the diamond, and a narrow grey sliver appears on the other neighbouring face, also reaching into that same shared corner, both fading to white as they move toward the side of the cube away from the diamond.
Step 4: Check cube (i) against this picture.
Cube (i) shows exactly this arrangement. The diamond sits centred on the front face, unchanged as it must be. The grey wedge on the top face reaches down into the corner it shares with the front and right faces, and the grey sliver on the right face also reaches up into that same shared corner, both fading to white on the side away from the front face. This is precisely the fixed rotational relationship produced by folding the net with every shaded region kept outward, so cube (i) is achievable by an actual fold.
Step 5: Check cube (ii) against the same picture.
Cube (ii) is built from the same family of shapes, a wedge on the top face and a sliver on the right face, but every one of them has slid to the opposite side of its face. The front face no longer shows the diamond at all, only a small triangular patch sitting in a corner away from the shared edge, and the top wedge and the right sliver both sit toward the back of the cube, away from the shared corner, instead of reaching into it. That is the mirror image of the arrangement obtained in Step 3. The only way every one of these markings could swap sides simultaneously is to flip the whole net over before folding, which would put the grey ink on the inside of the finished cube instead of the outside. Since the question fixes the shaded regions to the outer surface, cube (ii) cannot be produced by any actual folding of Panel I.
Step 6: Conclude.
Only cube (i) matches a genuine folding of the net in Panel I; cube (ii) is its mirror image and cannot be built by folding alone. So only (i) can correspond to the unfolded cube in Panel I. \[ \boxed{\text{Only (i) can correspond to the unfolded cube, so the correct option is (A).}} \]