Can You Slice The Ice?

Riddler Classic

Winter is almost upon us, at least for those of us in the Northern Hemisphere. Official ice master and deliverer Kristoff and his trusty pal Sven are preparing ice cubes that will be sent to the hot springs, where Kristoff’s extended family lives.

At the moment, they are studying a cube whose side lengths measure 1 meter, which means its surface-area-to-volume ratio is 6. (Kristoff doesn’t particularly care about the units here, but knows they are inverse meters.) Sven is concerned that much of the ice will melt before they reach the hot springs, so he suggests that Kristoff cut the ice along a single plane to minimize the surface-area-to-volume ratio of the larger resulting piece.

Along which plane should Kristoff cut the ice? And what will the resulting surface-area-to-volume ratio be?

Solution

I looked at a few different cases. Images created and acquired from here. In all cases, let the reduction be of length $x$, so each reduced side is now $1-x$.

One Corner (Triangular Pyramid) Cut Off

Surface Area

Three isosceles right triangles, each of area $\dfrac{x^2}{2}$ are missing from the cube. A new equilateral triangle of area $\dfrac{x^2\sqrt{3}}{4}$ is added.

$$SA = 6 - \dfrac{3x^2}{2} + \dfrac{x^2\sqrt{3}}{4}$$

$$SA = \dfrac{24 + \left(\sqrt{3} - 6\right)x^2}{4}$$

Volume

The volume of the missing triangular pyramid is $\dfrac{x^3}{6}$.

$$V = 1 - \dfrac{x^3}{6}$$

$$= \dfrac{6-x^3}{6}$$

Ratio

$$\dfrac{SA}{V} = \dfrac{\dfrac{24 + \left(\sqrt{3} - 6\right)x^2}{4}}{\dfrac{6-x^3}{6}}$$

$$= \dfrac{72 + \left(3\sqrt{3} - 18\right)x^2}{12-2x^3}$$
Taking the derivative,
$$0 = \dfrac{\left(12-2x^3\right)\left(6\sqrt{3}x-36x\right) - \left(72 + \left(3\sqrt{3} - 18\right)x^2\right)\left(-6x^2\right)}{\left(12-2x^3\right)^2}$$

Using WolframAlpha yields $x \approx 0.73485$, resulting in a minimum ratio of $5.8079$.

Two Corners (Triangular Prism) Cut Off

Surface Area

Two isosceles right triangles, each of area $\dfrac{x^2}{2}$ and two rectangles, each of area $x$ are missing from the cube. A new recatangle of area $x\sqrt{2}$ is added.

$$SA = 6 - \dfrac{2x^2}{2} -2x + x\sqrt{2}$$

$$SA = 6 - x^2 - \left(2 - \sqrt{2}\right)x$$

Volume

The volume of the missing triangular prism is $\dfrac{x^2}{2}$.

$$V = 1 - \dfrac{x^2}{2}$$

$$= \dfrac{2-x^2}{2}$$

Ratio

$$\dfrac{SA}{V} = \dfrac{6 - x^2 - \left(2 - \sqrt{2}\right)x}{\dfrac{2-x^2}{2}}$$

$$= \dfrac{12 - 2x^2 - \left(4 - 2\sqrt{2}\right)x}{2-x^2}$$
Taking the derivative,
$$0 = \dfrac{\left(2-x^2\right)\left(-4x - 4 + 2\sqrt{2}\right) - \left(12 - 2x^2 - \left(4 - 2\sqrt{2}\right)x\right)(-2x)}{\left(2-x^2\right)^2}$$

Using WolframAlpha yields $x \approx 0.148052$, resulting in a minimum ratio of $5.9566$.

Four Corners (Rectangular Prism) Cut Off

Surface Area

Four rectangles, each of area $x$, and one square of area $1$, are missing from the cube. A new square of area $1$ is added.

$$SA = 6 - 4x$$

Volume

The volume of the missing rectangular prism is $x$.

$$V = 1 - x$$

Ratio

$$\dfrac{SA}{V} = \dfrac{6 - 4x}{1-x} = 4 + \dfrac{2}{1-x}$$

The ratio is always increasing, which means the inital cube has the minimum ratio of 6.

Answer

Comparing all three cases, cutting off a triangular prism of base side length $0.73485$ at one corner yields the minimum ratio of $5.8079$

Rohan Lewis

2021.12.05