Pâté de Foie Gua

Let the vertices of the tetrahedron be $(0,0,0),\;\;(x,0,0),\;\;(0,y,0),\;\;(0,0,z)$

The three faces meeting at the origin are all right-angled triangles. Their areas are $\frac{yz}{2},\;\;\frac{zx}{2},\;\;\frac{xy}{2}$

We can use the fact that the area of a triangle in 3D space is given by the cross-product of the vectors representing two of its edges to find the fourth face has area $\frac12 \sqrt{y^2 z^2 +z^2 x^2+x^2 y^2}$

(note how this is related to the areas of the projections of the faces onto the coordinate planes - that's not a coincidence!) Other ways to get this formula are Heron's formula, or using the dot product to work out one of the angles.

Taking $(x,y,z)=(1,2,8)$ gives areas $1,4,8,9$ ; so solutions are possible.

Note that $(x,y,z)=(1,8,30)$ works, as does $(1,30,112)$ and $(1,112,418)$ . Finding a few more terms suggests we should look at the sequence defined by $a_1=8$ , $a_2=30$ and $a_{n+2}=4a_{n+1}-a_n$ .

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I've got a very messy proof that this sequence works; so there are infinitely many solutions, but the neater proof will have to wait.

Such a tetrahedron is a trirectangular tetrahedron , and by De Gua's Theorem , the areas of the triangular faces are related by:

$A_{ABO}^2 + A_{ACO}^2 + A_{BCO}^2 = A_{ABC}^2$

One way to show that there are infinitely many solutions to the above equation is to let $A_{ABO}^2 + A_{ACO}^2 = k^2$ so that $k^2 + A_{BCO}^2 = A_{ABC}^2$ , two sets of Pythagorean triples.

Using Euclid's Pythagorean triple generator on $A_{ABO}^2 + A_{ACO}^2 = k^2$ , we obtain $A_{ABO} = m^2 - n^2$ , $A_{ACO} = 2mn$ , and $k = m^2 + n^2$ for some positive integers $m > n$ .

If we further require $m$ and $n$ to have opposite parity, then $k$ will be odd, and will have odd factors $1$ and $k$ , so that there exists integers $p$ and $q$ such that $p - q = 1$ and $p + q = k$ , so that $k = k \cdot 1 = (p + q)(p - q) = p^2 - q^2$ .

Then using Euclid's Pythagorean triple generator again on $k^2 + A_{BCO}^2 = A_{ABC}^2$ , we obtain $k = p^2 - q^2$ , $A_{BCO} = 2pq$ , and $A_{ABC} = p^2 + q^2$ .

Since $p - q = 1$ and $p + q = k = m^2 + n^2$ , these two equations solve to $p = \frac{1}{2}(m^2 + n^2 + 1)$ and $q = \frac{1}{2}(m^2 + n^2 - 1)$ .

Putting this all together and simplifying gives us the generator:

$(m^2 - n^2)^2 + (2mn)^2 + (\frac{1}{2}((m^2 + n^2)^2 - 1))^2 = (\frac{1}{2}((m^2 + n^2)^2 + 1))^2$

for positive integers $m > n$ with opposite parity, and this shows us that there are infinitely many solutions .

We can further simplify (but reduce the number of solution sets) by letting $m = 2x$ and $n = 1$ , to obtain the generator:

$(4x^2 - 1)^2 + (4x)^2 + (4x^2(2x^2 + 1))^2 = (4x^2(2x^2 + 1) + 1)^2$

for any positive integer $x$ , whose first few solutions are $(3, 4, 12, 13), (15, 8, 144, 145), (35, 12, 684, 685), ...$ .