### Cutting a Cube

A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical?

# Covering Cups

##### Stage: 3 Challenge Level:

Many thanks to Andrei of School 205 Bucharest for the inspiration for this solution. Well done Andrei. One might have expected the triangular solution to be best. I wonder what happens if you chop off the corners?

Using the following notation:

$h$ - for the height of the cup and

$d$ - for the diameter of the cup.

Test each of the two possible forms of the box, one with a rectangular base and one with a triangle as its base:

Cuboidal box

As there are six cups they could be ordered in two ways:

$1 \times 6$ cups

and

$2 \times 3$ cups

For any cuboid box, the surface area is: 2 times the top area + 2 times the side area + 2 times the front area:

The total area of a $1 \times 6$ cups box is:

$2 \times(h \times 6 \times d + d \times 6 \times d + d \times h) = 185470$ mm$^2$

The total area of a $2 \times3$ cups box is:

$2 \times(h \times2 \times d + h \times3 \times d + 2 \times d \times3 \times h) = 157250$ mm$^2$

For a triangular prism box

This case is illustrated in the figure of the problem.

The base is an equilateral triangle. The surface area is 2 times the area of the equilateral triangle + 3 times the area of each of the faces.

First calculate the side of the equilateral triangle $ABC$.

This is $2 \times BM + 2 \times85$ mm.

$BM$ is a tangent to the circumference of a corner cup (centre $O$, radius $d/2 = 85/2$ mm).

Also the angle $MOB = 60^{\circ}$

Therefore angle $MBO = 30^{\circ}$

\eqalign{ BM &= \frac{85}{2}\tan60^{\circ} \\ \; &= \frac{85}{2}\sqrt{3} \\ \; &= \frac{85\sqrt{3}}{2}\\ \mbox{The side of triangle ABC is} &= 2 \times d + 2 \times BM \\ \; &= 2 \times85 + 2 \times\frac{85\sqrt{3}}{2} \\ \; &= 85(2 + \sqrt{3}) \\ \mbox{The altitude of the equilateral triangle ABC is} &= AB \times\sin60^{\circ} \\ \; &= 85(2 + \sqrt{3}) \times\frac{\sqrt{3}}{2} \\ \; &= \frac{85\sqrt{3}(2 + \sqrt{3})}{2} \\ \mbox{Area of the two equilateral triangles} &= 85(2 + \sqrt{3}) \times\frac{85\sqrt{3}(2 + \sqrt{3})}{2} \\ \; &= 87149 \; \mbox{mm}^2 \\ \mbox{Area of the three faces} &= 3 \times83 \times85(2 + \sqrt{3}) \\ \; &= 78989 \; \mbox{mm}^2 \\ \mbox{The total surface area is } &= 166138 \; \mbox{mm}^2}

So the best box is a box of 2 x 3 cups.