# Volume of a solid created by an extended tetrahedron

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.

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.

• You are using the term side - do you mean that every edge is extended on both sides by p?
– Moti
Sep 4, 2019 at 20:50
• @Moti Jup, i'll change it Sep 4, 2019 at 20:51
• I do not see the hexagon. Note that extending the rectangles create a CUBE - could this help you?
– Moti
Sep 4, 2019 at 21:04
• @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. Sep 4, 2019 at 21:11

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 $$-A'_1, -A'_2,\ldots$$.

$$\verb/Vol/(C(\lambda)) = \left[( 1 + 2\lambda)^3 - \frac{4}{6}(\lambda^3 + (1+\lambda)^3)\right] \ell^3$$ 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 $$3\ell$$.

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

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.

$$\textit{First solution:}$$

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 $$C_{D}$$.

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.

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.

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 $$V_{1}=4 \mathrm{~V}$$.

$$\textit{Case 2:}$$

$$2$$ 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).

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).

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 ).

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 $$M$$.

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).

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 $$63 \mathrm{~V}$$.

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.

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

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