Heights in a tetrahedron I

 

Can all four heights be equally long in a not regular tetrahedron? Give an example that they can or prove that it is impossible. (First published on Facebook/Mathematics)

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YES it is possible for all four heights to be equally long in a not regular tetrahedron.

If alla the faces are congruent then they have equal area and so the corresponding heights are equal (since the volume V = Bh/3 where B is any face area and h the corresponding height). The problem is then, can four congruent, not equilateral triangles fit together to form a tetrahedron? – They can, and they need not even be isosceles; any acute-angled triangles will do. To see this look at this picture:

Three congruent, acute-angled triangles are placed in a plane as shown. They form together a larger triangle, similar to the smaller ones. The broken lines will, if extended be heights in the large triangle. Thus they coincide at the one point H. Then, if the triangles surronding the one in the middle are folded upwards their outer corners will, as seen from above, all reach the point H. But their distances to the other two corners are equal for adjacent triangles so they are also at the same altitude and the triangle sides will coincide.
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Trying the same with right-angled or obtuse-angled triangles will rather obviously not work. Another way to see that acute angles are necessary is to look at a formula for the volume of the tetrahedron. – Let the sides of each small triangle be a, b, c. Then, if vectors with lengths a, b, c from one corner of the tetrahedron along three edges will point at the other three vertices. The volume is then given by the scalar-cross product, V = (1/6)( a · (b × c)), which can be written as a determinant with row or column vectors a, b, c. (Scalar multiplikation *, cross product x, underscore _ denotes vector.) The square V² is then essentially the determinant of the matrix squared,

36V² = = det [ [a·a , a·b , a·c ] , [b·a , b·b , b·c ] , [c·a , c·b , c·c ] ].

Using the cosine theorem this can be expressed in the side lengths a, b, c:

72V² = (a² + c² – b²)(b² + c² – a²)(a² + b² – c²)

(see also http://en.wikipedia.org/wiki/Tetrahedron).

From this either all three factors on the right hand side are positive or one is positive and two negative. But by the cosine theorem, a² + b² – c² > 0 means that the angle opposite to c is acute, and a² + b² – c² < 0 that it is obtuse, and there cannot be two obtuse angles in a tringle, so all three factors are positive. So the triangles are acute-angled. Of course, they can also not be right-angled since that would make one factor zero.
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Using Herons formula and V = Bh/3 we can also find a nice expression for the height:

h² = 2(a² + b² – c²)(b² + c² – a²)(c² + a² - b²)/
((a + b + c)·(a + b – c)·(b + c – a)·(c + a – b))

By the above the numerator is positive so the denominator must also be positive. This means simply that the triangle inequalities are fulfilled so that the edges with lengths a, b, c actually do form triangles!
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This answers the question and more. However, as often happens new questions pop up. – Is it possible for the heights of the tetrahedron to be equal without the faces being congruent?

Expectation value

Problem from Facebook/Mathematics

Bhargav Gnv Find the mathematical expectation of the area of the projection of a cube with edge of length onto a plane with an isotropically distributed random direction of projection.

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Solution

Let the cube be positioned in a 3d coordinate system with centre at the origin and edges parallel to axes,and with edge length 1. Let the direction of projection be given by a unit vector n = (cos α, cos β, cos γ) where α, β, γ are the angles of the vector to the coordinate axes. Also, for simplicity we confine ourselves to the first octant. Because of symmetry that doesn’t affect the result.

Now, from any direction exactly three non-parallel faces of the cube can be seen, including the ”degenerate” cases when some face is seen from its own plane. Therefore the area of the projection is cos α + cos β + cos γ. Isotropically distributed random direction of projection means that every position of the tip of vector n on the unit sphere around the origin is equally probable. Since the area element of the sphere is sin θ dθ dφ in polar coordinates the probability density f(θ) for the tip’s position is proportional to sin θ (no φ-dependence by azimuthal symmetry). By normalization f(θ) = (2/π)*sin θ. Thus the expectation value
<A> = ∫∫ (cos α + cos β + cos γ)(2/(π))*sin θ dθ dφ
Σ
where Σ = [0,π/2]x[0,π/2].

To calculate this we must express the direction cosines cos α, cos β, cos γ in polar angles θ, φ. Let’s say that α, β, γ are the angles with the x-, y-, and x-axis respectively. Then the tip of the vector n is at (sin θ cos φ, sin θ sin φ, cos θ). Using the scalar product of n with the unit vetors e_x, e_y, e_z, these components are equal to the direction cosines. Thus,
<A> = ∫∫ (sin θ cos φ + sin θ sin φ + cos θ)(2/π)*sin θ dθ dφ =
= (2/π)*∫∫ sin²θ (cos φ + sin φ) + cos θ sin θ dθ dφ =
= (2/π) * ∫ [θ=0,π/2] sin²θ dθ * ∫ [θ=0,π/2] cos φ + sin φ dφ
+ (2/π) * (π/2) * ∫ [θ=0,π/2] cos θ sin θ dθ =
= (2/π)*(π/4)*2 + 1*1/2*2 = 1 + 1 = 2.

With edge a we get of course <A> = 2a². Seems resonable, since three faces are seen contracted and each face has area a², so <A> must be something between a² and 3a².

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Ooppss! ∫ [θ=0,π/2] cos θ sin θ dθ = 1/4*2 = 1/2 so <A> = 3a²/2.

Also, the estimate can be improved. The greatest projected area occurs in the direction (1/√3, 1/√3, 1/√3) and that area is 3a²/√3 = a²√3. So we must have a² < <A> < a²√3, which satisfied by <A> = 3a²/2. – In general the expaction value of a non-constant random variable must be strictly between the smallest and the greatest value.