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In a tetrahedron (which is not necessarily regular) two opposite edges have the same length $a$ and they are perpendicular to each other. Moreover they are each perpendicular to a line of length $b$ which joins their midpoints. Express the volume of the tetrahedron in terms of $a$ and $b$ and prove your answer.

Hint: Can you imagine a more accessible related problem?.

To simplify this problem, what's the unknown here?

To re-examine this problem, we need to find the volume of the tetrahedron and to do this, we need to compute the base and height.

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  • $\begingroup$ What are "opposite edges" of a tetrahedron? Does it mean the pairs of edges that don't meet? Can those even be perpendicular? $\endgroup$ Commented Mar 29, 2019 at 12:49
  • $\begingroup$ @DanielMcLaury Depends on what you mean by "perpendicular". Their directions can certainly be perpendicular. $\endgroup$
    – Arthur
    Commented Mar 29, 2019 at 12:51
  • $\begingroup$ "In a tetrahedron (which is not necessarily regular) two opposite edges have the same length a and they are perpendicular to each other. Moreover they are each perpendicular to a line of length b which joins their midpoints." No, that's impossible. If the two edges "have the same length a and they are perpendicular to each other" then the line joining their midpoints has angle $\pi/4$ radians (45 degrees). $\endgroup$
    – user247327
    Commented Mar 29, 2019 at 12:53
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    $\begingroup$ @user247327 Consider the segment from $(-1, 0, 0)$ to $(1, 0, 0)$ and the segment from $(0, -1, 1)$ to $(0, 1, 1)$. The segment joining their midpoints goes from $(0,0,0)$ to $(0,0,1)$, and the three segments are all mutually perpendicular. At least under the definition of "perpendicular" which OP is clearly using (which doesn't require the lines to intersect). $\endgroup$
    – Arthur
    Commented Mar 29, 2019 at 13:00
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    $\begingroup$ @blue interesting analogy gives a way to visualise the problem. $\endgroup$
    – base 10
    Commented Mar 30, 2019 at 9:51

3 Answers 3

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Let $ABCD$ be the tetrahedron, $AB=CD=a$, $M$ and $N$ midpoints of $AB$ and $CD$, $MN=b$. We know $AB\perp CD$ and $AB,CD\perp MN$. Consider planes $\alpha,\beta$ such that $\alpha\supseteq AB$, $\alpha\parallel CD$ and $\beta\supseteq CD$, $\beta\parallel AB$. Then $\alpha\parallel\beta$ and $MN$ is their common normal.

Project $C,D$ on $\alpha$ to $C',D'$, and $A,B$ on $\beta$ to $A',B'$. Note $AA'=BB'=CC'=DD'=MN=b$. Since $AB\perp CD$, we see that $AB\perp C'D'$. Also $AB$ and $C'D'$ intersect in $M$, and moreover $M$ is the midpoint of both $AB$ and $C'D'$. Thus $AC'BD'$ is a square of diagonal $a$, so of edge $a/\sqrt 2$. Similarly, $A'CB'D$ is a square of edge $a/\sqrt 2$. $AB'CD'A'BC'D$ is a cuboid with edges $a/\sqrt 2$, $a/\sqrt2$ and $b$, hence its volume is $a^2b/2$.

Now the volume of $ABCD$ is $a^2b/2$ minus volumes of $ABCC'$, $ABDD'$, $CDAA'$ and $CDBB'$. Each od these tetrahedrons have volume $a^2b/12$, since their bases are half of square of side $a/\sqrt 2$ and the height is $b$. So the volume of $ABCD$ is $a^2b/2-4a^2b/12= a^2b/6$.


This is the picture: enter image description here

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We may assume that we have a vertical edge connecting the points $\left(0,0,\pm{1 \over2}a\right)$ and a horizontal edge connecting the points $\left(b,\pm{1\over2}a,0\right)$. The symmetry plane $z=0$ cuts the tetrahedron into two pyramids of base area ${1\over2} a b$ and height ${1\over2}a$. It follows that the volume of the tetrahedron is ${1\over6} a^2 b$.

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The other answers give appropriate approaches for this simple case, but I thought I'd share a little-known generalization.

We know that, for a tetrahedron, $$\text{volume} = \frac13 \cdot \text{area of base}\cdot\text{altitude}$$ And we know from trig that we can write the area of a triangle in terms of two base sides and their included angle; thus, we have

$$\text{volume} = \frac16\cdot \text{side}\cdot\text{side}\cdot\sin(\text{included angle}) \cdot \text{altitude} \tag{$\star$}$$

Fun Fact: $(\star)$ works even for opposite "$\text{side}$"s, provided "$\text{included angle}$" is taken to mean the angle determined by the direction vectors of those sides, and "$\text{altitude}$" is the extent of the tetrahedron in the direction perpendicular to the two "$\text{sides}$"s. (For opposite edges, "altitude" is equivalent to the shortest distance between the lines containing those edges.)

In the problem at hand, the opposites sides both have length $a$, the included angle is $90^\circ$ (whose sine is $1$), and the altitude is $b$, so that the volume is $\frac16a\cdot a\cdot 1\cdot b = \frac16a^2b$. Easy-peasy.


We can prove $(\star)$ using coordinates and a determinant (which would have been overkill for the particular problem, and may be well outside the context in which the problem was posed to OP; nevertheless ...). For instance, we can take our tetrahedron's vertices to be

$$A=(0,0,a) \qquad B = (0,0,b) \qquad C = (h,c\sin\theta,c\cos\theta) \qquad D = (h,d\sin\theta,d\cos\theta)$$

Here, we have opposite side-lengths $|AB|=|a-b|$ and $|CD|=|d-c|$, and "included angle" $\theta$. Moreover, those edges lie in planes $x=0$ and $x=h$, so that the tetrahedron lies entirely between those planes, giving a corresponding "altitude" of $h$. Then we can compute $$\text{volume} = \frac16\,|\Delta|$$ where $$\Delta := \left|\begin{array}{ccc} B_x-A_x & B_y-A_y & B_z-A_z \\ C_x-A_x & C_y-A_y & C_z-A_z \\ D_x-A_x & D_y-A_y & D_z-A_z \end{array}\right| = \left|\begin{array}{ccc} 0 & 0 & b-a \\ h & c\sin\theta & c\cos\theta-a \\ h & d\sin\theta & d\cos\theta-a \end{array}\right| = h (b-a) (d-c)\sin\theta$$ In absolute value, this gives $(\star)$. $\square$

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  • $\begingroup$ why assume adjacent sides of the base triangle are of length a? $\endgroup$
    – base 10
    Commented Mar 31, 2019 at 20:55
  • $\begingroup$ @base10: Um ... I don't assume adjacent sides of the base triangle are of length $a$. I write specifically: "In the problem at hand, the opposites sides both have length $a$, ...". The point of the Fun Fact is that the side-side-sine-altitude formula for volume, which typically applies to any pair of adjacent sides, actually applies to any pair of opposite sides (provided the angle and altitude are interpreted appropriately). $\endgroup$
    – Blue
    Commented Mar 31, 2019 at 22:43

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