Show that each tetrahedron in this sequence has the same altitude to its equilateral base (see details)?

Question

For non-negative integers n define

a(0) = -1, a(1) = 1, a(n) = 4a(n-1) - a(n-2) and 
b(0) = 1, b(1) = 1, b(n) = 4b(n-1) - b(n-2).

Each positive integer n, let T(n) be the tetrahedron whose base is the equilateral triangle each of whose sides is b(n), and whose other three edges each equal a(n). Show that each of these tetrahedra has the same length of altitude to its base.

[note -- I deleted an earlier version of this question, which was misstated]

Indica · Accepted Answer

Need to show that h(n) = 3b(n)²−a(n)² is constant (this is a multiple of altitude²)

General solution of the recurrence is c₁αⁿ+c₂βⁿ where c₁,c₂ depend on initial conditions and α,β solve the auxiliary equation λ²−4λ+1=0. Hence α=2+√3, β=2−√3

Write h(n) = (√3.b(n)−a(n))(√3.b(n)+a(n)) = c(n)*d(n)

Since a(n), b(n) satisfy the same linear recurrence so will the linear combinations c(n) & d(n)

For the sequence c, c(0)=√3+1, c(1)=√3−1 so we get the following equations for c₁,c₂
√3+1 = c₁+c₂ and √3−1=(2+√3)c₁+(2−√3)c₂   →   c₁=0, c₂=√3+1

For the sequence d, d(0)=√3−1, d(1)=√3+1 so we get the following equations for c₁,c₂
√3−1 = c₁+c₂ and √3+1=(2+√3)c₁+(2−√3)c₂   →   c₁=√3−1, c₂=0

∴ h(n) = (√3+1)(2−√3)ⁿ * (√3−1)(2+√3)ⁿ = (3−1) * 1ⁿ = 2

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Show that each tetrahedron in this sequence has the same altitude to its equilateral base (see details)?

1 Answer