Tetrahedron Incenter

Moving on from the solution below, I claim that the position vector of the incentre $I$ of a tetrahedron $ABCD$ is expressible in terms of the position vectors of the vertices $A,B,C,D$ and of the areas of its four faces as follows: $\mathbf{p} \; = \; \frac{BCD \mathbf{a} + ACD \mathbf{b} + ABD \mathbf{c} + ABC \mathbf{d}}{BCD + ACD + ABD + ABC}$

This is the three-dimensional analogue of the better-known result for the incentre of a triangle. Here is the proof. Splitting the tetrahedron $ABCD$ into the four tetrahedra $ABCI$ , $ABDI$ , $ACDI$ , $BCDI$ , all with bases one of the faces of the original tetrahedron and with height $r$ , the inradius, we deduce that $\tfrac13(ABC + ABD + ACD + BCD)r \; = \; V$ where $V$ is the volume of the tetrahedron $ABCD$ . The equation of the plane $ABC$ (for example) is $\big(\mathbf{r} - \mathbf{a}\big) \cdot \big(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}\big) \; = \; 0$ (note that $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})$ ) and so the distance of $I$ from the face $ABC$ is $\left|\frac{(\mathbf{p} - \mathbf{a})\cdot(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a})}{|\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}|}\right| \; = \; \frac{|(\mathbf{p} - \mathbf{a})\cdot(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a})|}{2ABC}$ Now $\begin{aligned} \mathbf{p} \cdot(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}) & = \; \frac{BCD + ABD + ACD}{ABC + ABD + ACD + BCD}\mathbf{a}\cdot(\mathbf{b} \times \mathbf{c}) + \frac{ABC}{ABC + ABD + ACD + BCD}\mathbf{d} \cdot(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}) \\ & = \; \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) + \frac{ABC}{ABC+ABD+ACD+BCD}(\mathbf{d}-\mathbf{a})\cdot(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}) \\ (\mathbf{p}-\mathbf{a})\cdot(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}) & = \; \frac{ABC}{ABD + ABD + ACD + BCD} (\mathbf{d}-\mathbf{a})\cdot(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}) \\ |(\mathbf{p}-\mathbf{a})\cdot(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a})| & = \; \frac{6ABC\ V}{ABC + ABD + ACD + BCD} \; = \; 2rABC \end{aligned}$ which shows that the distance from $I$ to $ABC$ is $r$ . Similar calculations show that the distance of $I$ to the other three faces is also $r$ , and hence that $I$ is indeed the incentre and $r$ the inradius.

Calculating the coordinates of the incentre is now elementary. (and it turns out the the formulae obtained below can be substantially simplified). We have $\mathbf{p} \; =\; \left(\frac{5\sqrt{233} + 10\sqrt{249} - 5\sqrt{313}}{17 + \sqrt{233}+ \sqrt{249} + \sqrt{313}}\,,\,\frac{12\sqrt{233}+2\sqrt{313}}{17 + \sqrt{233}+ \sqrt{249} + \sqrt{313}}\,,\,\frac{170}{17 + \sqrt{233}+ \sqrt{249} + \sqrt{313}}\right)$ from which we calculate $OI = |\mathbf{p}| = \boxed{4.759398680}$ .

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The planes $ABC$ , $ABD$ , $ACD$ and $BCD$ have inward-pointing unit normals $\begin{array}{rclcrcl} \mathbf{n}_{ABC} & =& \displaystyle \frac{(\mathbf{b}-\mathbf{a}) \times (\mathbf{c} - \mathbf{a})}{|(\mathbf{b}-\mathbf{a}) \times (\mathbf{c} - \mathbf{a})|} \; =\; \mathbf{k}& \hspace{2cm} & \mathbf{n}_{ABD} & =& \displaystyle \frac{(\mathbf{d}-\mathbf{a}) \times (\mathbf{b} - \mathbf{a})}{|(\mathbf{d}-\mathbf{a}) \times (\mathbf{b} - \mathbf{a})|} \\[2ex] \mathbf{n}_{ACD} & =& \displaystyle \frac{(\mathbf{c}-\mathbf{a}) \times (\mathbf{d} - \mathbf{a})}{|(\mathbf{c}-\mathbf{a}) \times (\mathbf{d} - \mathbf{a})|} && \mathbf{n}_{BCD} & =& \displaystyle\frac{(\mathbf{d}-\mathbf{b}) \times (\mathbf{c} - \mathbf{b})}{|(\mathbf{d}-\mathbf{b}) \times (\mathbf{c} - \mathbf{b})|} \end{array}$ and so the plane $\Pi_1$ through $AB$ that internally bisects the angle between the planes $ABC$ and $ABD$ has equation $\big(\mathbf{r} - \mathbf{a}\big) \cdot \big(\mathbf{n}_{ABC} - \mathbf{n}_{ABD}\big) \; = \; 0$ Similarly the plane $\Pi_2$ through $AC$ that internally bisects the angle between the planes $ABC$ and $ACD$ has equation $\big(\mathbf{r} - \mathbf{a}\big) \cdot \big(\mathbf{n}_{ABC} - \mathbf{n}_{ACD}\big) \; = \; 0$ and the plane $\Pi_3$ through $CD$ that intenally bisects the angle between the planes $ACD$ and $BCD$ has equation $\big(\mathbf{r} - \mathbf{c}\big) \cdot \big(\mathbf{n}_{ACD} - \mathbf{n}_{BCD}\big) \; = \; 0$ The incentre $I$ of the tetrahedron will be the intersection of the three planes $\Pi_1,\,\Pi_2,\,\Pi_3$ . The actual coordinates of $I$ can be determined exactly, but the end result is horrendous. For example, the $x$ -component is $5\frac{5976 \sqrt{233} + 2796 \sqrt{249} - 313 \sqrt{58017} + 498 \sqrt{72929} + 233 \sqrt{77937} - 12 \sqrt{18159321}}{ 2988 \sqrt{233} + 2796 \sqrt{249} + 517 \sqrt{58017} + 249 \sqrt{72929} + 233 \sqrt{77937} + 29 \sqrt{18159321}}$ Being content with decimal approximations, the coordinates of $I$ are $(2.21584, 3.32475, 2.58611)$ , which makes $OI = \boxed{4.759398680}$ .

$\vec{a}=(10,0,0)$ , $\vec{b}=(5,12,0)$ , $\vec{c}=(-5,2,0)$ , $\vec{d}=(0,0,10)$

By subtracting we get vectors along the edges from d: $\vec{e_1}=\vec{a}-\vec{d}=(10,0,-10)$ $\vec{e_2}=\vec{b}-\vec{d}=(5,12,-10)$ $\vec{e_3}=\vec{c}-\vec{d}=(-5,2,-10)$

Between these edges, we have three planes, with unit vectors perpendicular to them, pointing inward: $\vec{v_1}=\vec{e2}×\vec{e1}=(-12,5,-12)/\sqrt{313}$ $\vec{v_2}=\vec{e1}×\vec{e3}=(2,15,2)/\sqrt{233}$ $\vec{v_3}=\vec{e3}×\vec{e2}=(10,-10,-7)/\sqrt{249}$

Now of course the point $\vec{d}$ has distance $0$ to each of its adjacent planes. If we move away from $\vec{d}$ in the right direction, we can increase the distance to these 3 planes with an equal amount. Therefore we are looking for a vector $\vec{u}$ such that $\vec{u}\cdot\vec{v_1}=\vec{u}\cdot\vec{v_2}=\vec{u}\cdot\vec{v_3}$ . We find such a vector by calculating $\vec{u}=(\vec{v_1}-\vec{v_2})×(\vec{v_1}-\vec{v_3})$

(From this point, I used excel for the calculations)

Now the incentre must be at $\vec{p}=\vec{d}+k\vec{u}$ where $k$ is chosen such that p is equidistant to the plane ABC and the other three planes. Therefore $10+ku_3=k\vec{u}\cdot\vec{v_1}$ , or $k=\frac{10}{\vec{u}\cdot\vec{v_1}-u_3}$

We find $\vec{p}=(2.21583..., 3,32475..., 2.58610....) \text { and } |\vec{p}| = 4.7593986...$