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<h1>A triangle's incenter via algebra</h1>
<h2>… and extension to tetrahedra</h2>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
<param name="background" value="ffffff">
<param name="title" value="The incenter via algebra">
<param name="e[1]" value="A;point;fixed;100,250">
<param name="e[2]" value="B;point;fixed;300,250">
<param name="e[3]" value="C;point;free;50,50;red;red">
<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
<param name="e[6]" value="C';point;proportion;A,pt1,A,C,A,B,A,B;none;none">
<param name="e[7]" value="pt2;point;extend;B,A,A,C;none;none">
<param name="e[8]" value="A';point;proportion;B,pt2,B,A,B,C,B,C;none;none">
<param name="e[9]" value="pt3;point;extend;C,B,B,A;none;none">
<param name="e[10]" value="B';point;proportion;C,pt3,C,B,C,A,C,A;none;none">
<param name="e[11]" value="AA';line;connect;A,A';none;none;magenta">
<param name="e[12]" value="BB';line;connect;B,B';none;none;magenta">
<param name="e[13]" value="CC';line;connect;C,C';none;none;magenta">
<param name="e[14]" value="O;point;intersection;AA',BB'">
<param name="e[15]" value="H;point;foot;O,A,B;none;none">
<param name="e[16]" value="IC;circle;radius;O,H;none;none;black;none">
</applet>
</td></tr>
<tr><td>
<b>
Drag $C$ to change the geometry.<br>
Press “r” to reset the diagram to its initial state.<br>
The incenter lies at the intersection of angle bisectors.
</b>
</td></tr></table>
<h2>Statement of the problem</h2>
<p>
It is a fact of elementary geometry that the center of
a triangle's incircle—known as its <em>incenter</em>—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:
<p>
<b>Proposition 1:</b>
<i>
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.
</i>
<p>
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:
<blockquote>
<p>
<b>The algebra of points:</b>
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 $B$.
</blockquote>
<h2>A preliminary proposition</h2>
<p>
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 <em>Elements</em> but I haven't checked.
<p>
<i>PS:</i>
After this web page was written,
I found out that
Proposition 2 appears in Euclid's <em>Elements</em> as
<a href="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI3.html">Book VI, Proposition 3</a>
with a simpler and more elegant proof!
Nevertheless, I am retaining my original proof here as
a not-so-elegant alternative.
<p>
<b>Proposition 2:</b>
<i>
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.
</i>
<p>
<b>Proof:</b>
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, $PM = PN$.
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
<param name="background" value="ffffff">
<param name="title" value="The incenter via algebra">
<param name="e[1]" value="A;point;fixed;50,250">
<param name="e[2]" value="B;point;fixed;350,250">
<param name="e[3]" value="C;point;free;100,50;red;red">
<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
<param name="e[6]" value="P;point;proportion;A,pt1,A,C,A,B,A,B">
<param name="e[7]" value="CP;line;connect;C,P;none;none;magenta">
<param name="e[8]" value="M;point;foot;P,C,A">
<param name="e[9]" value="N;point;foot;P,C,B">
<param name="e[10]" value="PM;line;connect;P,M;none;none;green">
<param name="e[11]" value="PN;line;connect;P,N;none;none;green">
<param name="e[12]" value="H;point;foot;C,A,B">
<param name="e[13]" value="CH;line;connect;C,H;none;none;cyan">
<param name="e[14]" value="AH;line;connect;A,H;none;none">
<param name="e[15]" value="BN;line;connect;B,N;none;none">
<param name="e[16]" value="AM;line;connect;A,M;none;none">
</applet>
</td></tr>
<tr><td>
<b>
Drag $C$ to change the geometry.<br>
Press “r” to reset the diagram to its initial state.<br>
</b>
</td></tr></table>
<p>
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. <b>QED</b>
<h2>The proof of Proposition 1</h2>
<p>
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
<em>the algebra of points</em> 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.
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
<param name="background" value="ffffff">
<param name="title" value="The incenter via algebra">
<param name="e[1]" value="A;point;fixed;100,250">
<param name="e[2]" value="B;point;fixed;300,250">
<param name="e[3]" value="C;point;free;50,50;red;red">
<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
<param name="e[6]" value="C';point;proportion;A,pt1,A,C,A,B,A,B">
<param name="e[7]" value="pt2;point;extend;B,A,A,C;none;none">
<param name="e[8]" value="A';point;proportion;B,pt2,B,A,B,C,B,C;none;none">
<param name="e[9]" value="pt3;point;extend;C,B,B,A;none;none">
<param name="e[10]" value="B';point;proportion;C,pt3,C,B,C,A,C,A;none;none">
<param name="e[11]" value="AA';line;connect;A,A';none;none;none">
<param name="e[12]" value="BB';line;connect;B,B';none;none;none">
<param name="e[13]" value="CC';line;connect;C,C';none;none;magenta">
<param name="e[14]" value="O;point;intersection;AA',BB'">
<param name="e[15]" value="H;point;foot;O,A,B;none;none">
<param name="e[16]" value="IC;circle;radius;O,H;none;none;black;none">
</applet>
</td></tr>
<tr><td>
<b>
Drag $C$ to change the geometry.<br>
Press “r” to reset the diagram to its initial state.<br>
</b>
</td></tr></table>
<p>
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.
<p>
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. <b>QED</b>
<h2>Extension to tetrahedra</h2>
<p>
Propositions 1 extends to tetrahedra:
<p>
<b>Proposition 3</b>
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.
<p>
This is proved with the aid of the following extension of Proposition 2:
<p>
<b>Proposition 4</b>
<i>
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:
\[
\frac{PA}{PB} = \frac{b}{a}.
\]
</i>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
<param name="background" value="ffffff">
<param name="title" value="Triangle's incenter">
<param name="e[1]" value="A;point;fixed;100,250,300">
<param name="e[2]" value="B;point;fixed;300,250,300">
<param name="e[3]" value="C;point;fixed;350,200,-400">
<param name="e[4]" value="D;point;free;200,40;red;red">
<param name="e[5]" value="ABCD;polyhedron;tetrahedron;A,B,C,D">
<param name="e[6]" value="ABC;plane;3points;A,B,C;none;none;none;none">
<!-- construct the bisector of the dihedral angle with edge DC -->
<param name="e[7]" value="pl1;plane;perpendicular;D,C;none;none;none;none">
<param name="e[8]" value="Apl1;point;foot;A,pl1;none;none">
<param name="e[9]" value="Bpl1;point;foot;B,pl1;none;none">
<param name="e[10]" value="pt1;point;angleBisector;Apl1,D,Bpl1,pl1;none;none">
<param name="e[11]" value="tmp1;plane;3points;D,C,pt1;none;none;none;none">
<param name="e[12]" value="P;point;intersection;A,B,tmp1">
<param name="e[13]" value="DP;line;connect;D,P;none;none;cyan">
<param name="e[14]" value="CP;line;connect;C,P;none;none;cyan">
</applet>
</td></tr>
<tr><td>
<b>
Drag $D$ to change the geometry.<br>
Press “r” to reset the diagram to its initial state.<br>
The plane $CDP$ bisects the dihedral angle of edge $CD$.
</b>
</td></tr></table>
<p>
I leave the proofs to you, the diligent reader. You will find
useful information in a
<a
href="http://groups.google.com/group/sci.math/browse_thread/thread/56c187e018a85111/436f448dfab35a83">discussion
in the sci.math newsgroup</a> from January 17, 2006.
<hr width="60%">
<p>
<em>This applet was created by
<a href="http://userpages.umbc.edu/~rostamia">Rouben Rostamian</a>
using
<a href="http://aleph0.clarku.edu/~djoyce/home.html">David Joyce</a>'s
<a href="http://aleph0.clarkU.edu/~djoyce/java/Geometry/Geometry.html">Geometry
Applet</a>
on June 26, 2010.
</em>
<p>
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