# Question 3

Given an acute angled triangle $$T$$, show that there is a unique tetrahedron whose faces are all congruent to $$T$$ and find the volume of this tetrahedron.

## Official Solution by Fionntan Roukema

### Existence

Let $$T$$ be an acute angled triangle with side lengths $$a,b,c$$. We suppose with out loss of generality that $$a\leq b\leq c$$. We consider the cuboid $$C$$ with side lengths equal to $$x,y,z$$ where $x=\sqrt{\frac{a^2+b^2-c^2}2},\qquad y=\sqrt{\frac{a^2+c^2-b^2}2}, \qquad z=\sqrt{\frac{c^2+b^2-a^2}2}.$ As $$a\leq b\leq c$$ we can see that the cube is non-degenerate provided that $$a^2+b^2-c^2>0$$. The law of cosines guarantees that this is the case when the angle opposite the side with length $$c$$ is acute. Using Pythagoras's Theorem we can see that the diagonals of the cuboid have lengths $$a$$, $$b$$, and $$c$$ (see the diagram below).

The convex hull of the highlighted diagonals in the cuboid above is a tetrahedron $$\Delta$$ whose faces consist of four congruent triangles with side lengths $$a$$, $$b$$, $$c$$. A triangle is uniquely determine by three edge lengths. Therefore, the tetrahedron $$\Delta$$ has four faces congruent to $$T$$.

### Uniqueness

We will now show that if there exists a tetrahedron $$\Delta$$ in $$\mathbb{R}^3$$ whose faces are congruent to $$T$$ then it is unique. We may assume that three of the vertices have coordinates $$v_1=(0,0,0)$$, $$v_2=(c,0,0)$$ and $$v_3=(\alpha,\beta,0)$$ where the distance from $$v_1$$ to $$v_3$$ is $$b$$ and the distance between $$v_2$$ and $$v_3$$ is $$a$$.

As each vertex is connected to three edges of lengths $$a,b,c$$, any point $$v_4=(x,y,z)$$ in $$\mathbb{R}^3$$ that is the fourth vertex of a tetrahedron whose faces are congruent to $$T$$ satisfies: $x^2+y^2+z^2=a^2$ $(x-c)^2+y^2+z^2=b^2$ $(x-\alpha)^2+(y-\beta)^2+z^2=c^2$ Note that the first two equations determine $$x$$ uniquely. The value of $$x$$ in the second two equations determines $$y$$ uniquely.

Substituting the values of $$x$$ and $$y$$ into (1) determines $$z$$ up to sign. The two constructed tetrahedra are congruent via a reflection in the $$xy$$-plane.

### Volume

We now calculate the volume of $$\Delta$$. The cuboid $$C$$ consists of the tetrahedron $$\Delta$$, and four corners each of which has volume $$\frac{xyz}6$$ (as it is well-known that the volume of a cone with height $$h$$ over a domain $$D$$ is $$\frac{h}3\textrm{area}(D)$$). Thus $\text{vol}(C)=\textrm{vol}(\Delta)+4\left(\frac{xyz}6\right)\qquad\text{and so}\qquad \textrm{vol}(\Delta)=\frac{xyz}3.$ Substituting back in the values we know, we get, $\textrm{vol}(T)=\sqrt{\frac{(a^2+b^2-c^2)(a^2+c^2-b^2)(b^2+c^2-a^2)}{72}}.$