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?
Show 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.
Only (i) can correspond to the unfolded cube in Panel I.
Only (ii) can correspond to the unfolded cube in Panel I.
Both (i) and (ii) can correspond to the unfolded cube in Panel I.
Neither (i) nor (ii) can correspond to the unfolded cube in Panel I.
Show Solution
The Correct Option isA
Solution and Explanation
Label the six unit squares of Panel I. Call $F$ the square carrying the diamond, call $T$ and $R$ its two neighbours in the strip that are each cut by a diagonal into a grey half and a white half, and call the remaining three squares $H_1,H_2,H_3$, since they end up hidden on the far side of the cube once $F$ is turned to face the viewer.
Notice that $T$ and $R$ are cut by the same diagonal line but shaded on opposite halves of it. Picture that diagonal as a single arrow drawn across the flattened strip, running from a point on $T$ to a point on $R$. When the shared edge between $T$ and $R$ is creased to $90^\circ$, this arrow does not break, it simply turns the corner and keeps pointing the same way around the shared edge, so it ends up pointing into the common corner of the folded solid from one specific side only. Test this with an actual strip of paper and a pencil line drawn straight across a single crease: whichever way you spin the finished right angle afterward, the arrow always points into the corner from that same side. The only way to make it point in from the opposite side is to unfold the paper and crease it the other way, which puts the pencil mark on the inside of the fold instead of the outside.
Now compare this with the two pictures. In cube (i), the grey wedge on the top face and the grey sliver on the right face both point into their shared corner from the same side, exactly like the undisturbed arrow from $T$ and $R$. In cube (ii), the same two shapes have both been moved to the far side of their faces, away from that shared corner, which is only possible if the diagonal on $T$ and $R$ had run the other way. Since the direction of that diagonal is fixed by how Panel I is actually drawn, this reversed version is not something folding can produce, so cube (ii) is a mirror reflection of a valid folding rather than a valid folding itself.
This also disposes of options (B), (C) and (D) directly. Option (B) requires (ii) alone to work, but (ii) is the unreachable mirror case just shown. Option (C) requires both (i) and (ii) to work, which fails the moment (ii) is ruled out. Option (D) requires neither to work, but (i) matches the fold exactly, so (i) does work. That leaves exactly one option standing.
Hence only (i) can be produced by folding Panel I, so the answer is option (A).
