Vincent's question?

Question

Well that didn't take you long did it! What you didn't say at the start of your answer is that the radii of sphere and circle not being included implied that they didn't matter so you could choose circle radius = 0.

I don't think that you will find this one so easy.

An oil company has three wells A, B, C which form an acute angled triangle with side lengths BC = a, AC = b and AB = c. Thecompany wishes to put a storage depot D somewhere inside the triangle, joined by straight pipes to each well. Thecompany wants to minimise the total length of piping used.

It is fairly easy to discover empirically that the point D is where DA, DB, and DC make 120 degrees with one another and this is not too difficult to prove. However the relationship between min(DA + DB + DC) and a, b, c is another matter.

I feel confident that you won't get the answer to this one so quickly because as far as I'm aware there is no easy way to do it.

Quadrill, I'm really impressed that you got it so quickly and so easily. I first did this some years ago upon the instigation of one of my students. After much, much more algebra than you, I got the formula M = 0.5(a^2 + b^2 + c^2) + 0.5*sqrt(3*(a^2 + b^2 + c^2)^2 - 6*(a^4 + b^4 + c^4)). I tried both our formulas with a = 7, b = 8, c = 9 and got identical results so they are most probably equivalent. I still prefer mine as it gives M directly in terms of a, b, c.

Answer 1

Let's call the minimum DA=α, minimum DB=β, and minimum DC=γ (that's supposed to be a gamma).

Then the law of cosines tells us that:
c² = α² + β² - 2αβ cos(120) = α² + β² + αβ Similarly,
b² = α² + γ² - 2αγ cos(120) = α² + γ² + αγ and
a² = β² + γ² - 2βγ cos(120) = β² + γ² + βγ

Adding these gives us:
a² + b² + c² = 2(α² + β² + γ²) + (αβ + αγ + βγ)
so that
α² + β² + γ² = (a² + b² + c²)/2 - (αβ + αγ + βγ)/2

By the law of sines we have that:
ΔADB = αβ sin(120)/2 = αβ√3/4
ΔCDB = βγ sin(120)/2 = βγ√3/4
ΔADC = αγ sin(120)/2 = αγ√3/4
Summing these gives:
Δ = ΔABC = (αβ + αγ + βγ)√3/4

We are looking for M = min(α + β + γ) so putting it all together,
M² = (α + β + γ)² =
α² + β² + γ² + 2((αβ + αγ + βγ) =
(a² + b² + c²)/2 - (αβ + αγ + βγ)/2 + 2((αβ + αγ + βγ) =
(a² + b² + c²)/2 + (3/2)(αβ + αγ + βγ) =
(a² + b² + c²)/2 + (2√3)Δ

where Δ² = s(s-a)(s-b)(s-c) and s=(a+b+c)/2

Answer 2

Previous answer looks good. This is now called the Steiner point, first explored in 1826.

http://faculty.evansville.edu/ck6/tcenters/class/steiner.html