A triangle's incenter via algebra

… and extension to tetrahedra

Statement of the problem

It is a fact of elementary geometry that the center of a triangle's incircle—known as its incenter—lies at the point of intersection of the triangle's angle bisectors. This leads to an elegant expression for the location of the incenter as a linear combination of the triangle's vertices, as stated in:

Proposition 1: Let $a$, $b$, $c$ be the lengths of the sides opposite to the vertices $A$, $B$, $C$ of the triangle $ABC$. Then the incenter, $O$, is expressed as a linear combination of the vertices as: \[ O = \frac{a}{p}A + \frac{b}{p}B + \frac{c}{p}C, \] where $p = a + b + c$ is the triangle's perimeter.

To make sense of this statement, you need to know something about the algebra of points. The following capsule summary is all that's needed:

The algebra of points: Consider the points $A$ and $B$ and a variable $t$ that takes values in the range 0 to 1. If $T$ is a point on the segment $AB$ such that $AT/AB = t$, we write $T = (1-t)A + tB$. Note that when $t=0$ we get $T=A$, and when $t=1$ we get $T=B$. As the value of $t$ ranges from 0 to 1, the point $T$ slides from $A$ to $B$.

A preliminary proposition

The following elementary proposition is needed for the proof of Proposition 1. This one is a standard result and is likely to be found in Euclid's Elements but I haven't checked.

PS: After this web page was written, I found out that Proposition 2 appears in Euclid's Elements as Book VI, Proposition 3 with a simpler and more elegant proof! Nevertheless, I am retaining my original proof here as a not-so-elegant alternative.

Proposition 2: Let the bisector of the angle $C$ in the triangle $ABC$ meet the side $AB$ at point $P$. We have: \[ \frac{PA}{PB} = \frac{b}{a}, \] where $a$ and $b$ are as in Proposition 1.

Proof: From $P$ drop perpendicular $PM$ and $PN$ onto the sides $AC$ and $BC$; see the diagram below. The right triangles $CMP$ and $CNP$ are congruent because they share a common hypotenuse and their angles at $C$ are equal. Therefore, $PM = PN$.

Drag $C$ to change the geometry. Press “r” to reset the diagram to its initial state.

Let $H$ be the foot of the altitude dropped from $C$. Let us note the area of the triangle $APC$ may be expressed in two different ways in terms of the altitudes $CH$ and $PM$: \[ \frac{1}{2} AP \cdot CH = \frac{1}{2} AC \cdot PM. \] Similarly, the area of the triangle $BPC$ may be expressed in two different way in terms of the altitudes $CH$ and $PN$: \[ \frac{1}{2} BP \cdot CH = \frac{1}{2} BC \cdot PN. \] Dividing these two equalities and recalling that $PM = PN$, we arrive at the desired assertion.   QED

The proof of Proposition 1

The diagram below shows the bisector $CC'$ of the angle $C$. It is known from elementary geometry that the incenter $O$ lies on the bisector. From Proposition 2 we have $C'A/C'B = b/a$. It follows that $AC'/AB = b/(a+b)$. In terms of the notation of the algebra of points introduced earlier in this page, this is expressed as: \[ C' = \Big(1 - \frac{b}{a+b} \Big) A + \frac{b}{a+b} B = \frac{a}{a+b} A + \frac{b}{a+b} B. \] Since $O = (1-t)C + tC'$ for some $t$, we arrive at: \[ O = (1-t)C + t\Big[ \frac{a}{a+b} A + \frac{b}{a+b} B \Big]. \] This expresses the triangle's incenter $O$ in terms of its vertices, sides lengths, and a yet unknown quantity $t$ that takes values in the range 0 to 1.

Drag $C$ to change the geometry. Press “r” to reset the diagram to its initial state.

If we repeat the same calculation by replacing the vertex $C$ by vertex $A$, we arrive at: \[ O = (1-s)A + s\Big[ \frac{b}{b+c} B + \frac{c}{b+c} C \Big], \] where $s$ is yet another unknown that takes values in the range from 0 to 1.

Equating the two expressions for $O$ and collecting the terms, we arrive at: \[ \Big[\frac{ta}{a+b} + s - 1\Big] A + \Big[\frac{tb}{a+b} - \frac{sb}{b+c} \Big] B + \Big[1 - t - \frac{sc}{b+c} \Big] C = 0. \] In this equations, if the coefficient of $C$ is nonzero, then we may solve for $C$, obtaining $C = \alpha A + \beta B$ for some $\alpha$ and $\beta$. But this would mean that $C$ lies along the line $AB$ which would mean the the three points $A$, $B$ and $C$ are collinear, therefore the triangle $ABC$ is degenerate. We conclude that if the triangle is non-degenerate, then the coefficient of $C$ is zero. Similar arguments show that the coefficients of $A$ and $B$ are zero. Thus, the following system of three equations hold: \[ \frac{ta}{a+b} + s - 1 = 0, \quad \frac{tb}{a+b} - \frac{sb}{b+c} = 0, \quad 1 - t - \frac{sc}{b+c} = 0. \] Solving this system for the unknowns $t$ and $s$ we obtain: \[ t = \frac{a+b}{a+b+c}, \quad s = \frac{b+c}{a+b+c}. \] Substituting these in either of the expressions for $O$ we arrive at: \[ O = \frac{a}{a+b+C}A + \frac{b}{a+b+C}B + \frac{c}{a+b+C}C, \] which is equivalent to Proposition 1's assertion.   QED

Extension to tetrahedra

Propositions 1 extends to tetrahedra:

Proposition 3 Let $a$, $b$, $c$, $d$ be the areas of the faces opposite to the vertices $A$, $B$, $C$, $D$ of the tetrahedron $ABCD$. The the tetrahedron's incenter $O$ is given by: \[ O = \frac{a}{\mathcal{A}} A + \frac{b}{\mathcal{A}} B + \frac{c}{\mathcal{A}} C + \frac{d}{\mathcal{A}} D, \] where $\mathcal{A} = a + b + c + d$ is the tetrahedron's surface area.

This is proved with the aid of the following extension of Proposition 2:

Proposition 4 Let $a$, $b$, $c$, $d$ be the areas of the faces opposite to the vertices $A$, $B$, $C$, $D$ of the tetrahedron $ABCD$. Let the bisector plane of the (internal) dihedral angle of edge $CD$ intersect edge $AB$ at point $P$. The we have: \[ \frac{PA}{PB} = \frac{b}{a}. \]

Drag $D$ to change the geometry. Press “r” to reset the diagram to its initial state. The plane $CDP$ bisects the dihedral angle of edge $CD$.

I leave the proofs to you, the diligent reader. You will find useful information in a discussion in the sci.math newsgroup from January 17, 2006.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on June 26, 2010.

Go to Geometry Problems and Puzzles