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We are given a regular tetrahedron $ABCD$ ($ABC$ is its` base and $D$ is its vertex) and we reflect it through the middle of its height (point reflexion) - and thus we obtain a congruent regular tetrahedron $A'B'C'D'$.

$D'$ lies in the center of $ABC$, and $D$ in the center of $A'B'C'$.

Planes $\pi (ABC) \ || \ \pi (A'B'C'), \ \ \ \pi (ABD) \ || \ \pi (A'B'D'), \ \ \ \pi (B'C'D') \ || \ \pi (BCD)$, $ \ \ \ \pi (A'C'D') \ || \ \pi (ACD)$.

I drew a picture and I think that the intersection of the two tetrahedrons is a parallelepiped, but I don't know how to prove it more formally (I mean, I know that the respective sides of the tetrahedrons are parallel, because we reflect $ABCD$ in a point, but I am not sure if that's enough).

Secondly, how can we calculate the volume of the intersection?

Could you help me with that?

Thank you!

enter image description here

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  • $\begingroup$ Where did the comments go? $\endgroup$ – Hagrid May 25 '14 at 13:07
  • $\begingroup$ Well, anyway. Do you think it's the right intersection, anf if it is, how can one calculate its volume? $\endgroup$ – Hagrid May 25 '14 at 13:22
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    $\begingroup$ Sorry, when I woke up, I came to my senses and saw that my first answer was wildly incorrect, so I deleted it. I now have another answer. $\endgroup$ – Peter Woolfitt May 25 '14 at 15:49
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Note that the region of intersection has $6$ faces - one for each face not parallel to the base of $ABCD$. Additionally, by construction we know that this region has $3$ pairs of parallel faces.

This is sufficient to conclude that the region of intersection is a parallelepiped. Here or here you can see a set of equivalent definitions of parallelepiped.

For the volume, we will use the observation that if we can calculate the length $x$ where $x$ is an edge of the parallelepiped, we can calculate the volume of the parallelepiped with $V=6\frac{x^3}{6\sqrt{2}}=\frac{x^3}{\sqrt{2}}$. To see this consider the $3$ vectors of length $x$ from $D$ in the directions of $A$, $B$, and $C$. They form a regular tetrahedron, and the volume of a parallelepiped formed from $3$ vectors is $6$ times the volume of the tetrahedron formed from those $3$ vectors.

To calculate $x$, we use a little convoluted approach. First, note that the projection of each of these vectors onto the line segment $DD'$ is the same, but the sum of the 3 projections is just $|DD'|$, so the projection of any one of them onto $DD'$ is $\frac{1}{3}|DD'|$. By similar triangles we have $$\frac{x}{|DA|}=\frac{\frac{1}{3}|DD'|}{|DD'|}\rightarrow\frac{x}{|DA|}=\frac{1}{3}\rightarrow x=\frac{1}{3},$$

which results in a final total volume of $V=\frac{1}{27\sqrt{2}}$.

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  • $\begingroup$ Thank you. Could you tell me why the sum of the 3 projections is $|DD'|$? $\endgroup$ – Hagrid May 25 '14 at 17:12
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    $\begingroup$ Yeah, note that the sum of all the vectors is the vector from $D$ to $D'$, and their angles relative to the line $DD'$ are the same, so the sum of their projections is $|DD'|$. $\endgroup$ – Peter Woolfitt May 25 '14 at 17:23
  • $\begingroup$ Indeed, I see that now. From $D$ we go down towards $A$, then right, and then left and we're in $D'$. Thanks. $\endgroup$ – Hagrid May 25 '14 at 17:27
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The intersection is bounded by three pairs of parallel planes (pairs of face planes from the respective tetrahedra), so you're right that the intersection is a parallelipiped.

To find the volume in terms of the edge length $3\ell$ of one tetrahedron, note that each face of the intersection is a rhombus comprising two equilateral triangles of edge length $\ell$, so the area of the base is $(\sqrt{3}/2)\ell^{2}$, and that the "height" of the intersection (measured orthogonally to a face) is $(\sqrt{2}/\sqrt{3})\ell$, the height of a regular tetrahedron of side length $\ell$. The volume is consequently $\ell^{3}/\sqrt{2}$.

Edges and face of the intersection of two tetrahedra

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