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A cube has four diagonal planes. Let them be $P_1, P_2, P_3, P_4$. $P_1$ and $P_2$ intersect at exactly two corners. The cube is cut by $P_1$ and $P_2$ diagonal planes. What are the volumes of the four pieces of the cube?

By diagonal planes, I'm referring to the plane that goes through four corners of the cube and cuts the cube in half.

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    $\begingroup$ Why 4 diagonal planes not 6? If the cube has corners $(±1,±1,±1)$, then I'd expect the planes $x=y,x=-y,x=z,x=-z,y=z,y=-z$ to meet your requirements. $\endgroup$
    – MvG
    Commented Feb 16, 2016 at 9:47

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Let's assume the cube has corners $(±1,±1,±1)$ and you have $P_1:x=y$ and $P_2:x=z$. So the common corners are $(1,1,1)$ and $(-1,-1,-1)$.

One of the pieces has $x<y\wedge x<z$. Its volume is therefore

$$\int_{-1}^1\mathrm dx\int_x^1\mathrm dy\int_x^1\mathrm dz= \int_{-1}^1(1-x)^2\,\mathrm dx=\left[\frac13x^3-x^2+x\right]_{-1}^1=\frac83$$

Another piece has $y<x\wedge x<z$. Its volume is

$$\int_{-1}^1\mathrm dx\int_{-1}^x\mathrm dy\int_x^1\mathrm dz= \int_{-1}^1(1+x)(1-x)\,\mathrm dx=\left[x-\frac13x^3\right]_{-1}^1=\frac43$$

The piece $x>y\wedge x>z$ is just like $x<y\wedge x<z$ so it has volume $\frac83$. The piece $z<x\wedge x<y$ is just like $y<x\wedge x<z$ with volume $\frac43$. Together these pieces have volume $2\left(\frac83+\frac43\right)=8=2^3$ which matches the volume of the whole cube. If your cube has a different edge length than $2$, scale these figures accordingly.

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