Spatial Aptitude

Joining the centre to the vertices gives three kinds of triangles:

• 1, 4, 7: a 60° angle at the centre, so equilateral with side \(r\); area \(\frac{\sqrt{3}}{4}r^{2}\).
• 2, 3, 5, 6, 8, 9: sides \(r\) and \(r\) with 30° between them; area \(\frac{1}{2}r^{2}\sin 30^\circ = \frac{r^{2}}{4}\).
• 10, 11, 12: sides \(d\) and \(d\) with the dodecagon’s 150° angle; area \(\frac{d^{2}}{4} = \frac{(2 - \sqrt{3})r^{2}}{4}\), since \(d^{2} = 2r^{2}(1 - \cos 30^\circ)\).

The total area is \(3r^{2}\), so 3 squares of side \(r\). One equilateral, two 30° and one small triangle add up to exactly \(r^{2}\), and they fit: the equilateral triangle stands on one side of the square, the 30° triangles fill the spaces beside it, and the small triangle closes the gap at the opposite side. So each square uses 4 triangles.

Answer: A

The three sectors of the circle all touch one another, so they need three different colours: say 1 (upper left), 2 (bottom) and 3 (upper right).

In the band around the circle, the left part touches sectors 1 and 2, so it must be 3, and the right part touches sectors 2 and 3, so it must be 1. The part below the circle touches sector 2, the left part (3) and the right part (1), so it needs a fourth colour. Three colours are not enough.

Four are: colour the top band part 2 and the bottom band part 4, the outer top and bottom parts 1, the outer left and right parts 2, and all four outer corner parts 3.

Answer: A

Face 5 touches 4 and 1 (view 1) and 6 and 2 (view 3), so 3 is opposite 5. Face 4 touches 5 and 1 (view 1) and 6 and 3 (view 2), so 2 is opposite 4. That leaves 1 opposite 6.

In each net, the four squares of the vertical strip fold into a ring, so its 1st and 3rd squares are opposite, as are its 2nd and 4th; the two side squares are opposite too.

• (A): 1–6, 4–2 and 5–3, all as required.
• (B) and (C): 1 opposite 2, wrong.
• (D): 4 opposite 3 and 5 opposite 2, wrong.

Folding (A) also shows each view’s three faces in the same clockwise order, so it makes this dice.

Answer: A

Joined base to base, the two cones share one axis, which passes through both vertices and the centre of the common base. A plane through both vertices therefore contains this axis and cuts each cone in an isosceles triangle: its base is a diameter of the circle and its equal sides are slant heights.

The two triangles sit on opposite sides of the same diameter, so the outer boundary is a quadrilateral whose four sides are all equal slant heights: a rhombus. It cannot be an ellipse, because a plane through the axis meets a cone’s curved surface in straight lines.

Answer: A