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Every edge of a tetrahedron with length $p$ is extended through the vertices by $p$.

Now all 12 points create a new solid $J$ of which I seek the volume dependent on the volume of the tetrahedron in the centre. enter image description here

With some help the solution becomes clear:

The whole Volume of all pyramids is: $$V_P=4\cdot\frac{1}{6}\cdot\left(\frac{a}{\sqrt{2}}\right)^3 +4\cdot\frac{1}{6}\cdot\left(\sqrt{2}a\right)^3 =\frac{3}{\sqrt{2}}a^3 \tag{1}$$

The side of the large cube is $$s=\frac{3}{\sqrt{2}}a\tag{2}$$ and the Volume is respectively $$V_C=\left(\frac{3}{\sqrt{2}}a\right)^3=\frac{27\sqrt{2}}{4}\cdot a^3 \tag{3}$$

The last step is subtracting the Volume of the pyramids.

$$V_J=V_C-V_P= \frac{27\sqrt{2}a^3}{4}-\frac{3}{\sqrt{2}}a^3=\frac{21\sqrt{2}a^3}{4} \tag{4}$$

Thx for the help.

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  • $\begingroup$ You are using the term side - do you mean that every edge is extended on both sides by p? $\endgroup$
    – Moti
    Sep 4, 2019 at 20:50
  • $\begingroup$ @Moti Jup, i'll change it $\endgroup$ Sep 4, 2019 at 20:51
  • $\begingroup$ I do not see the hexagon. Note that extending the rectangles create a CUBE - could this help you? $\endgroup$
    – Moti
    Sep 4, 2019 at 21:04
  • $\begingroup$ @Moti Its rather a six-sided figure than a regular hexagon. Its in the middle and in the same plane like the base of the tetrahedron. $\endgroup$ Sep 4, 2019 at 21:11

4 Answers 4

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I will generalize the problem a little bit for a better understanding of the geometry.
The volume you seek is $$\frac{21\sqrt{2}}{4} p^3$$

Let $A_1,A_2,A_3,A_4$ be the vertices of a regular tetrahedron with side $p = \sqrt{2}\ell$. Choose a coordinate system to make its centroid the origin.

In this coordinate system, the eight points $\pm A_1,\pm A_2\ldots$ is forming a cube of side $\ell$. The tetrahedron can be recovered from this cube by removing 4 corners of side $\ell$ at $-A_1,-A_2,...$

By a corner of side $r$ at a point $P$, I'm referring to any right angled tetrahedron $PQRS$ with $\angle QPR = \angle RPS = \angle SPQ = 90^\circ$ and $|PQ| = |PR| = |PS| = r$. Since the volume of such a corner is $\frac16 r^3$, the volume of tetrahedron is $$ \left(1 - \frac{4}{6}\right)\ell^3 = \frac13 \ell^3 = \frac{\sqrt{2}}{12} p^3$$ This is what you already know.

Extend the six edges of tetrahedron on both directions by a factor $\lambda$. One obtain twelve points of the form:

$$B_{ij} \stackrel{def}{=} (1 + \lambda)A_i - \lambda A_j \quad\text{where}\quad 1 \le i, j \le 4, \quad i \ne j$$ Let $C(\lambda) = {\rm co}\left(\{ B_{ij} : 1 \le i, j \le 4, i \ne j \}\right)$ be their convex hull. The problem at hand can be rephrased as:

  • What is the volume of $C(\lambda)$ when $\lambda = 1$?

Let $A'_k = (1+2\lambda) A_k$ for $1 \le k \le 4$ and $\mu = \frac{\lambda}{1+2\lambda}$. Notice

$$B_{ij} = (1+\lambda)A_i - \lambda A_j = (1 - \mu)A'_i + \mu( -A'_j)$$

The point $B_{ij}$ is lying on the edge $A'_i \to -A'_j$ of a cube with vertices at $\pm A'_1,\pm A'_2,\ldots$

The convex hull $C(\lambda)$ can be obtained from this cube by removing

  • four corners of side $\mu |A'_i + A'_j| = \lambda\ell$ at $A'_1,A'_2,\ldots$
  • another four corners of side $(1 + \lambda)\ell$ at $-A'_1, -A'_2,\ldots$.

This leads to

$$\verb/Vol/(C(\lambda)) = \left[( 1 + 2\lambda)^3 - \frac{4}{6}(\lambda^3 + (1+\lambda)^3)\right] \ell^3$$ Substitute $\lambda$ by $1$, the volume we seek is $$\verb/Vol/(C(1)) = 21\ell^3 = \frac{21\sqrt{2}}{4} p^3$$

Following is a picture illustrating what the convex hull $C(1)$ looks like. Together with the $8$ semi-transparent right tetrahedra (four with side $\ell$, another four with side $2\ell$), they can be combined to form a cube of side $3\ell$.

A cube with 8 corners removed

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Here is an answer - complementing your 3d object to a cube with side - $3p/2^{0.5}$

enter image description here

Subtract 4 right angle pyramids with diagonal $p$ and 4 right angle pyramids with diagonal $2p$. The pyramids sides are, $p/2^{0.5}$ and $2^{0.5}p$.

The volume of the cube is given by:$(3p/2^{0.5})^3$

The volume of the pyramids - $4X(1/6)X(p/2^{0.5})^3+4X(1/6)X(2p/2^{0.5})^3$

I will leave you to subtract the volumes.

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$\textit{First solution:}$

Let the vertices of the tetrahedron $T$ be denoted by $A, B, C$ and $D$. For two different points $X$ and $Y$, let $X_{Y}$ be the uniquely defined point on the straight line $X Y$, which is obtained by extending the line $\overline{X Y}$ by $a$ beyond $Y.$ The vertices of the solid to be examined are therefore the points $B_{A}, C_{A}, D_{A}, A_{B}, C_{B}$, $D_{B}, A_{C}, B_{C}, D_{C}, A_{D}, B_{D}$ and $C_{D}$.

P1

Let $\varepsilon(X Y Z)$ denote the plane defined by three non-collinear points $X, Y$ and $Z$. Each of the four planes $\varepsilon(A B C), \varepsilon(A B D), \varepsilon(A C D)$ and $\varepsilon(B C D)$, in which the surface of the tetrahedron $T$ lies, decomposes the whole space into two half-spaces. Of these two, we call the one containing the tetrahedron the inner half-space. The half-space facing away from the tetrahedron is called the outer half-space.

We now consider the intersection of the body with the plane $\varepsilon(A B C)$. Figure P2 shows this intersection, with the point $D$ behind the plane.

P2

The part of the body that is in the outer half-space with respect to the plane $\varepsilon(A B C)$ is shown in Figure P3.

P3

This outer part of the body with respect to $\varepsilon(A B C)$ is further decomposed by the other planes $\varepsilon(A B D)$, $\varepsilon(A C D)$ and $\varepsilon(B C D)$, into a total of seven parts. These are shown below.

The part that lies on the inner side with respect to all three planes is the body with corners $D_{A} D_{B} D_{C} A B C$. In addition there are three parts, each of which is in exactly one other outer half-space and in two inner ones, namely $B_{A} B_{C} D_{C} D_{A} A C, A_{C} A_{B} D_{B} D_{C} C B$ and $C_{B} C_{A} D_{A} D_{B} B A$. Finally, there are three parts located in two more outer half-spaces and one inner, namely $B_{A} C_{A} D_{A} A, A_{C} B_{C} D_{C} C$ and $A_{B} D_{B} C_{B} B$.

These seven parts, when put together, now already give the entire outer part of the body under consideration with respect to $\varepsilon(A B C)$. The common outer space of $\varepsilon(A B D), \varepsilon(A C D)$ and $\varepsilon(B C D)$ is thus completely in the inner half-space with respect to $\varepsilon(A B C)$. In other words, there is no point that is in all four outer half-spaces. Because of the symmetry of the regular tetrahedron, the analogous figures are congruent with respect to the other three planes. The four planes $\varepsilon(A B C), \varepsilon(A B D), \varepsilon(A C D)$ and $\varepsilon(B C D)$ thus decompose the body under consideration into the following 15 parts:

  • the four parts which lie in the interior with respect to one plane and in the exterior with respect to the other three,
  • the six parts that lie inside with respect to two planes and outside with respect to the other two,
  • the four parts which are inside with respect to three planes and outside with respect to the others,
  • the interior of all the planes formed by the tetrahedron $A B C D$ itself.

We now calculate the volume of the parts of the body in a complete case distinction according to the position to the four planes.

$\textit{Case 1:}$

$1$ inner and $3$ outer half-spaces. So of these there are four parts, each congruent to the tetrahedron $A_{B} D_{B} C_{B} B$ (Figure P4) and are formed by point reflection at one of the four vertices from the tetrahedron $A B C D$. Together they have the volume $V_{1}=4 \mathrm{~V}$.

P4

$\textit{Case 2:}$

$2$ inner and $2$ outer half-spaces. Here, each of the six tetrahedral edges gives rise to a body congruent to $C_{B} C_{A} D_{A} D_{B} B A$ (Figure P5).

P5

Adding $C_{B} C_{A} D_{A} D_{B} B A$ by the tetrahedron $A_{B} D_{B} C_{B} B$, we obtain a skew prism with a base congruent to that of the tetrahedron and with twice the height. Since a prism with the same base area and the same height has three times the volume of a corresponding pyramid, the volume of the skew prism is $3 \cdot 2 \mathrm{~V}=6 \mathrm{~V}$ and for the body $C_{B} C_{A} D_{A} D_{B} B A$ the volume $6 \mathrm{~V}-V=5 \mathrm{~V}$. In total, we thus find $V_{2}=6 \cdot 5 \mathrm{~V}=30 \mathrm{~V}$.

$\textit{Case 3:}$

$3$ inner and $1$ outer halfspaces. Over each tetrahedral face, a truncated pyramid is obtained, e.g. $D_{A} D_{B} D_{C} A B C$ (Forming P6).

P6

Adding the tetrahedron $A B C D$ to this body to form the complete pyramid results in a regular tetrahedron with edge length $2 a$, i.e. with volume $8 \mathrm{~V}$. The truncated pyramid itself therefore has the volume $8 \mathrm{~V}-V=7 \mathrm{~V}$. Together this gives $V_{1}=4 \cdot 7 \mathrm{~V}=28 \mathrm{~V}$ for this case.

$\textit{Case 4:}$

$4$ inner half-spaces. This is the tetrahedron itself with volume $V_{4}=V$. Since each two of the described sub-bodies have no common inner points, the total volume is $$ V_{1}+V_{2}+V_{3}+V_{4}=(4+30+28+1) V=63 \mathrm{~V} . $$

$\textit{Result:}$

The volume of the body spanned by the twelve points is $63 \mathrm{~V}$.


$\textit{Second Solution:}$

We consider a cube $A B C D E F G H$ with edge length $b=\sqrt{2} / 2 \cdot a$. Its face diagonals are then of length $\sqrt{2} \cdot b=a$. Thus the regular tetrahedron $A C F H$ inscribed in the cube has edges of length $a$ and is consequently congruent to $T$ (see Figure P7 ).

P7

From now on we work with the tetrahedron $A C F H$ instead of $T$ and use designations analogous to those of the first solution. The body to be considered is thus $C_{A} F_{A} H_{A} A_{C} F_{C} H_{C} A_{F} C_{F} H_{F} A_{H} C_{H} F_{H}$.

Let $M$ be the centre of the cube $A B C D E F G H$. By centric stretching with the centre $M$ by the stretching factor 3 the cube $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime} G^{\prime} H^{\prime}$ arises from $A B C D E F G H$.

Figure P8 shows a plane section through both cubes, namely the one passing through the tetrahedral edge $\overline{A C}$ and the centre $M$.

P8

In addition to $M$, $A$ and $C$, the vertices $G$ and $E$ of the cube $A B C D E F G H$ lie in the plane of intersection, because the points $M, A$ and $G$ are collinear and so are $M, C$ and $E$. In the section plane $A C G E$ is a rectangle with side lengths $a$ and $b$. Since the point $M$ is equidistant from each of the four vertices, it is the diagonal intersection of $A C G E$.

Furthermore, the point $A^{\prime}$ also lies in the intersection plane because $M, A$ and $A^{\prime}$ are collinear. Finally, an analogous argument shows that the vertices $C^{\prime}, G^{\prime}$ and $E^{\prime}$ of the cube $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime} G^{\prime} H^{\prime}$ also lie in the cutting plane.

For the centre $M_{1}$ of the tetrahedral edge $\overline{A C}$ now $\left|A M_{1}\right|=a / 2$ and $\left|M M_{1}\right|=b / 2$ hold. Furthermore, $\angle M M_{1} A$ is a right angle. If one now extends the line $\overline{A C}$ beyond $A$ by the length $a$ to the point $C_{A}$, the following applies $$ \frac{\left|C_{A} A\right|}{\left|A M_{1}\right|}=\frac{a}{a / 2}=2 . $$ Similarly we have $$ \frac{\left|A^{\prime} A\right|}{|A M|}=\frac{\left|A^{\prime} M\right|-|A M|}{|A M|}=\frac{3|A M|-|A M|}{|A M|}=2, $$ which shows that the triangle $A C_{A} A^{\prime}$ arises from the triangle $A M_{1} M$ by centric stretching with centre $A$ by a factor (-2). So, in particular, $\left|A^{\prime} C_{A}\right|=b$ holds, and furthermore $\angle A^{\prime} C_{A} A=\angle A^{\prime} C_{A} C$ is a right angle. Finally, because of $\left|A^{\prime} M\right| /|A M|=\left|C^{\prime} M\right| /|C M|=3$, we know after the inversion of the ray theorem that $\overline{A^{\prime} C^{\prime}}$ is parallel to $\overline{A C}$. Consequently, $\angle C^{\prime} A^{\prime} C_{A}$ is also a right angle, showing that the point $C_{A}$ lies on the cube edge $\overline{A^{\prime} E^{\prime}}$ and divides it in the ratio $1: 2$. The distances are $\left|A^{\prime} C_{A}\right|=b$ and $\left|E^{\prime} C_{A}\right|=2 b$. The cube $A B C D E F G H$, and hence the tetrahedron $A C F H$, is now symmetrical with respect to the rotations by $120^{\circ}$ about the spatial diagonal $\overline{A G}$ of the cube. Consequently, the points $F_{A}$ and $H_{A}$ lie on the other two edges of the cube $\overline{A^{\prime} D^{\prime}}$ and $\overline{A^{\prime} B^{\prime}}$ bounded by the corner $A^{\prime}$ and also divide them in the ratio $1: 2$, again the distance to the corner $A^{\prime}$ being the shorter in each case.

Finally, since all four corners of the tetrahedron are equal, we find near the corners $C^{\prime}$, $F^{\prime}$ and $H^{\prime}$ of the cube $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime} G^{\prime} H^{\prime}$ an analogous position. In particular, therefore, on each edge of the cube $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime} G^{\prime} H^{\prime}$ there is exactly one vertex of the body under consideration $C_{A} F_{A} H_{A} A_{C} F_{C} H_{C} A_{F} C_{F} H_{F} A_{H} C_{H} F_{H}$. In each case the edges are divided in the ratio $1: 2$, in each case so that the distance to $A^{\prime}, C^{\prime}, F^{\prime}$ or $H^{\prime}$ is the shorter of the two (P9).

P9

The body $C_{A} F_{A} H_{A} A_{C} F_{C} H_{C} A_{F} C_{F} H_{F} A_{H} C_{H} F_{H}$ is thus obtained from the cube $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime} G^{\prime} H^{\prime}$ of edge length $3 b$ by removing a three-sided pyramid at each corner of the cube.

Each of these pyramids has a corner from which three paired right-angled edges of equal length extend. The edge length is equal to $b$ at the four corners $A^{\prime}, C^{\prime}, F^{\prime}$ and $H^{\prime}$ and equal to $2 b$ at the four corners $B^{\prime}, D^{\prime}, E^{\prime}$ and $G^{\prime}$.

The volume of a three-sided pyramid with one corner from which three pairwise right-angled edges of length $c$ extend is calculated to be $V_{c}=1 / 3 \cdot\left(c^{2} / 2\right) \cdot c=c^{3} / 6$. To see this, it is sufficient to choose one of the bounding isosceles right triangles as the base and note that the height is $c$. For the body under consideration, this gives a volume of $$ (3 b)^{3}-4 \cdot \frac{b^{3}}{6}-4 \cdot \frac{(2 b)^{3}}{6}=\left(27-\frac{2}{3}-\frac{16}{3}\right) b^{3}=21 \cdot b^{3} . $$ Finally, we note that the tetrahedron $A C F H$ arises from the cube $A B C D E F G H$ in quite an analogous way, namely by removing a three-sided pyramid at each of the vertices $B, D, E$ and $G$ of the cube, which, starting from $B, D, E$ and $G$ respectively, has three pairwise right-angled edges of length $b$. The tetrahedron $A C F H$ thus has a volume of $$ V=b^{3}-4 \cdot b^{3} / 6=b^{3} / 3 . $$ Thus $b^{3}=3 V$ holds, and the body $C_{A} F_{A} H_{A} A_{C} F_{C} H_{C} A_{F} C_{F} H_{F} A_{H} C_{H} F_{H}$ has, compared to the tetrahedron, the volume $$ 21 \cdot b^{3}=21 \cdot 3 \mathrm{~V}=63 \mathrm{~V} . $$

$\textit{Result:}$

Thus it follows again that the volume of the body spanned by the twelve points is exactly $63 \mathrm{~V}$.

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There are only three types of faces for the new solid. They are either equilateral triangles of side lenghth $p$ that are $\dfrac{7}{4} h$ away from the center, where $h = p \sqrt{\dfrac{2}{3}} $, or equilateral triangles of side $2p$ that are $ \frac{5}{4} h $ from the centroid, or rectangles of sides $ p $ and $ 2 p $ that are $ \dfrac{3 \sqrt{2}}{4 } p $ from the center.

$V = (1/3) \left(4 (\dfrac{\sqrt{3}}{4}) ( p^2 h (\dfrac{7}{4}) + 5 p^2 h ) + 6 (2 p^2)( \dfrac{3\sqrt{2}}{4} p ) \right) = \sqrt{3} ( \dfrac{9}{4} p^2 h ) + 3 \sqrt{2} p^3 $

$V = \sqrt{2} p^3 ( \dfrac{21}{4} ) = 63 V_0$

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