# Geometric mean of areas of tetrahedron

Consider a tetrahedron ABCD such that the areas of the triangles ABC, BCD, ACD, ABD are $k,l,m,n$ respectively. The volume of tetrahedron is $(1/6)$ cubic units . If each face ACB, ACD, BCD of the tetrahedron subtends a right angle at C, then we have to find the geometric mean of $k,l,m$.

I remember that $\frac{1}{3}(base)(height)=\frac{1}{6}$.

in this base area is know , I am confused how to find height

• What does it mean that a face of a tetrahedron subtends a right angle at some point? A tetrahedron is a solid, hence we should talk about solid angles. Did you just mean that $ACB,ACD,BCD$ are perpendicular to each other? Additionally, please use Mathjax. – Jack D'Aurizio Apr 21 '17 at 11:15
• If such planes are orthogonal and $x,y,z$ are the lengths of the edges through $C$, the volume is just $\frac{xyz}{6}$ and the areas of the previous three triangles are $\frac{xy}{2},\frac{xz}{2},\frac{yz}{2}$. Since $xyz=1$, it is pretty easy to find the geometric mean of the wanted areas. We may also notice that the area of $ABD$ is completely irrelevant. – Jack D'Aurizio Apr 21 '17 at 11:20

## 2 Answers

Your question can be written more clearly. There is no need to involve side areas and geometric means to find the tetrahedron height $H.$

Let $C$ be origin, $CA=x, CB=y, CD=z$ be the intercepts of a plane cutting the three axes

$$Vol = \frac {xyz}{6}$$

which can be found by integration of slices parallel to coordinate planes.

Area of the triangle $ABD$ is $\Delta$ where tetrahedron slant lengths through vertex $C$ are

$$Z= \sqrt{x^2+y^2},\,X= \sqrt{y^2+z^2}, \,Y= \sqrt{z^2+x^2}\,;$$

$$2s= (X+Y+Z),\,\Delta = \sqrt{s (s-X)(s-Y)(s-Z)};$$

$$Vol = \frac13 \cdot \Delta \cdot H; \quad H =\frac{xyz}{2 \Delta}.$$

EDIT2:

Or perhaps are you hinting about square root of lateral triangles area product divided by base area $\Delta$ is valid up to a constant even if angle at $C$ is not a right angle, as a formula for H? I mean for a scalene tetrahedron?

$$\sqrt{{\frac{xy}{2}\cdot \frac{yz}{2}\cdot \frac{zx}{2}}}= \frac{xyz}{2\sqrt2}$$

Take a coordinate system such that $C$ is the origin, and axes $Ox, Oy, Oz$ are defined by $\vec{CA}, \vec{CB}, \vec{CD}$ resp. Let us denote:

$$\tag{1}a:=CA, \ \ \ \ b:=CB, \ \ \ \ c:=CD.$$

the lengths of the sides issued from $C$. Thus, by definition of $k,l,m$, using the classical formula for the area of a right triangle:

$$\tag{2}k=\tfrac{1}{2}ab, \ \ \ \ l=\tfrac{1}{2}bc, \ \ \ \ m=\tfrac{1}{2}ca$$

Multyplying together the equalities (2), we get:

$$\tag{3}klm=\dfrac{a^2b^2c^2}{8} \ \ \ \iff \ \ \ abc=\sqrt{8klm}.$$

Besides, the base plane $(P)$ of the tetrahedron has equation

$$\tag{4}\tfrac{x}{a}+\tfrac{y}{b}+\tfrac{z}{c}-1=0$$

(Proof: equation (4) is verified for the 3 points $(x,y,z)=\underbrace{(a,0,0)}_{\text{point A}}, \underbrace{(0,b,0)}_{\text{point B}}, \underbrace{(0,0,c)}_{\text{point D}}$).

The height $h$ can be described as being the distance of the origin to the plane $(P)$. There is a classical formula for the distance of a point to a plane given by its cartesian equation (see formula (9) in (http://mathworld.wolfram.com/Point-PlaneDistance.html)); when we apply it with $(x_0,y_0,z_0)=(0,0,0)$ and equation (3), we obtain:

$$h=\dfrac{1}{\sqrt{\tfrac{1}{a^2}+\tfrac{1}{b^2}+\tfrac{1}{c^2}}}=\dfrac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\dfrac{\sqrt{8klm}}{\sqrt{4k^2+4l^2+4m^2}}$$

with the final simplification

$$h=\dfrac{\sqrt{2klm}}{\sqrt{k^2+l^2+m^2}}$$

Remark: as said by Jack D'Aurizio, area $n$ is of no use.