Length of edges of a Tetrahedron

[damon354]

Hello, the problem is the following:

ABCD is a tetrahedron such that the edges [AB], [AC] and [AD] are two to two orthogonal. Considering that AB = 3 and CD = √2, determine the minimum value of BC⁶ + BD⁶ - AC⁶ - AD⁶.

What I say:

To have the minimum value of BC⁶ + BD⁶ - AC⁶ - AD⁶, we have to determine the maximum value of AC + AD that we get if the CAD triangle is isosceles in A so that AC = AD. So we have: AD=1 and AC=1 and BD² = AB² +AD² = 3²+1² = 10 and BC = AB² + AC² = 3² + 1² + 10. So BC⁶ + BD⁶ - AC⁶ - AD⁶ = 10³ + 10³ - 1 - 1 = 1998.

I found this answer intuitively and I think it's correct because I checked with GeoGebra. But actually the problem is that I don't know how to prove this value properly.

 

[Dr.Peterson]

I assume "two to two orthogonal" means what I would call "pairwise orthogonal".

Your previous question was about vectors, but this one is not, on its surface at least. Can you suggest what you are learning that might be useful here? For instance, can you use calculus?

 

[damon354]

This exercise is also in the Chapter "Vectors and Dot Product" of the maths book. We learned four ways to calculate the dot product:

u.v = 1/2(||u+v||² -||u||² -||v||²) = 1/2(-||u-v||² +||u||² +||v||²)
u.v = ||u|| · ||v|| · cos(u, v)
u.v = xx' + yy' + zz' (coordinates)
And with an orthogonal projection

But in some exercises we have to use other theorems (Pytagoras, Thales, Al-Kashi...) and trigonometry.

What does mean "use calculus" ?

 

[Dr.Peterson]

What I ended up doing (not knowing that this is intended as a vector problem) was to express the four lengths in terms of one, which I called x, so that the required expression was a function of x, which turned out to be a quadratic in x^3. I then found the minimum of that quadratic, either algebraically or using the derivative (which is calculus). I did get your answer.

 

[damon354]

Thank you, the problem can be marked as solved.