Glen Whitney
c99b51dafa
Began with incenter.html, the first one alphabetically. Needed one new point construction method, and a new option to see what was going on. Got the planar diagrams on that page working. The next step on #36 will be to get 3D diagrams as the theorem on this page generalizes to 3D. That will be a bigger task, so merging this now. Reviewed-on: #39 Co-authored-by: Glen Whitney <glen@studioinfinity.org> Co-committed-by: Glen Whitney <glen@studioinfinity.org>
393 lines
13 KiB
HTML
393 lines
13 KiB
HTML
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
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<html>
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<head>
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<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
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<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
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<title>The incenter via algebra</title>
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<link rel="stylesheet" type="text/css" media="screen" href="style.css">
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<script type="text/javascript"
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src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
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</script>
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</head>
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<body style="visibility:hidden">
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<h1>A triangle's incenter via algebra</h1>
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<h2>… and extension to tetrahedra</h2>
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<table class="centered">
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<tr><td align="center">
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<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
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<param name="background" value="ffffff">
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<param name="title" value="The incenter via algebra">
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<param name="e[1]" value="A;point;fixed;100,250">
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<param name="e[2]" value="B;point;fixed;300,250">
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<param name="e[3]" value="C;point;free;50,50;red;red">
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<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
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<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
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<param name="e[6]" value="C';point;proportion;A,pt1,A,C,A,B,A,B;none;none">
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<param name="e[7]" value="pt2;point;extend;B,A,A,C;none;none">
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<param name="e[8]" value="A';point;proportion;B,pt2,B,A,B,C,B,C;none;none">
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<param name="e[9]" value="pt3;point;extend;C,B,B,A;none;none">
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<param name="e[10]" value="B';point;proportion;C,pt3,C,B,C,A,C,A;none;none">
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<param name="e[11]" value="AA';line;connect;A,A';none;none;magenta">
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<param name="e[12]" value="BB';line;connect;B,B';none;none;magenta">
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<param name="e[13]" value="CC';line;connect;C,C';none;none;magenta">
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<param name="e[14]" value="O;point;intersection;AA',BB'">
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<param name="e[15]" value="H;point;foot;O,A,B;none;none">
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<param name="e[16]" value="IC;circle;radius;O,H;none;none;black;none">
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</applet>
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</td></tr>
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<tr><td>
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<b>
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Drag $C$ to change the geometry.<br>
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Press “r” to reset the diagram to its initial state.<br>
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The incenter lies at the intersection of angle bisectors.
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</b>
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</td></tr></table>
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<h2>Statement of the problem</h2>
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<p>
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It is a fact of elementary geometry that the center of
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a triangle's incircle—known as its <em>incenter</em>—lies
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at the point of intersection of the
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triangle's angle bisectors. This leads to an elegant expression for
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the location of the incenter as a linear combination of the
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triangle's vertices, as stated in:
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<p>
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<b>Proposition 1:</b>
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<i>
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Let $a$, $b$, $c$ be the lengths of the
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sides opposite to the vertices $A$, $B$, $C$ of the triangle $ABC$.
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Then the incenter, $O$, is expressed as a linear combination of
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the vertices as:
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\[
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O = \frac{a}{p}A + \frac{b}{p}B + \frac{c}{p}C,
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\]
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where $p = a + b + c$ is the triangle's perimeter.
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</i>
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<p>
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To make sense of this statement, you need to know something about
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the algebra of points. The following capsule summary is all that's needed:
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<blockquote>
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<p>
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<b>The algebra of points:</b>
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Consider the points $A$ and $B$ and a variable $t$ that takes
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values in the range 0 to 1.
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If $T$ is a point on the segment $AB$
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such that $AT/AB = t$, we write $T = (1-t)A + tB$. Note that when
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$t=0$ we get $T=A$, and when $t=1$ we get $T=B$.
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As the value of $t$ ranges from 0 to 1, the point $T$ slides
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from $A$ to $B$.
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</blockquote>
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<h2>A preliminary proposition</h2>
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<p>
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The following elementary proposition is needed for the proof of
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Proposition 1. This one is a standard result and is
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likely to be found in Euclid's <em>Elements</em> but I haven't checked.
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<p>
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<i>PS:</i>
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After this web page was written,
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I found out that
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Proposition 2 appears in Euclid's <em>Elements</em> as
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<a href="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI3.html">Book VI, Proposition 3</a>
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with a simpler and more elegant proof!
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Nevertheless, I am retaining my original proof here as
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a not-so-elegant alternative.
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<p>
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<b>Proposition 2:</b>
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<i>
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Let the bisector of the angle $C$ in the triangle $ABC$ meet the side $AB$
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at point $P$. We have:
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\[
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\frac{PA}{PB} = \frac{b}{a},
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\]
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where $a$ and $b$ are as in Proposition 1.
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</i>
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<p>
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<b>Proof:</b>
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From $P$ drop perpendicular $PM$ and $PN$ onto the sides $AC$ and $BC$;
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see the diagram below. The right triangles $CMP$ and $CNP$ are congruent
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because they share a common hypotenuse and their angles at $C$ are equal.
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Therefore, $PM = PN$.
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<table class="centered">
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<tr><td align="center">
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<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
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<param name="background" value="ffffff">
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<param name="title" value="The incenter via algebra">
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<param name="e[1]" value="A;point;fixed;50,250">
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<param name="e[2]" value="B;point;fixed;350,250">
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<param name="e[3]" value="C;point;free;100,50;red;red">
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<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
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<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
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<param name="e[6]" value="P;point;proportion;A,pt1,A,C,A,B,A,B">
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<param name="e[7]" value="CP;line;connect;C,P;none;none;magenta">
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<param name="e[8]" value="M;point;foot;P,C,A">
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<param name="e[9]" value="N;point;foot;P,C,B">
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<param name="e[10]" value="PM;line;connect;P,M;none;none;green">
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<param name="e[11]" value="PN;line;connect;P,N;none;none;green">
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<param name="e[12]" value="H;point;foot;C,A,B">
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<param name="e[13]" value="CH;line;connect;C,H;none;none;cyan">
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<param name="e[14]" value="AH;line;connect;A,H;none;none">
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<param name="e[15]" value="BN;line;connect;B,N;none;none">
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<param name="e[16]" value="AM;line;connect;A,M;none;none">
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</applet>
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</td></tr>
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<tr><td>
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<b>
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Drag $C$ to change the geometry.<br>
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Press “r” to reset the diagram to its initial state.<br>
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</b>
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</td></tr></table>
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<p>
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Let $H$ be the foot of the altitude dropped from $C$. Let us note
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the area of the triangle $APC$ may be expressed in two different ways
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in terms of the altitudes $CH$ and $PM$:
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\[
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\frac{1}{2} AP \cdot CH = \frac{1}{2} AC \cdot PM.
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\]
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Similarly, the area of the triangle $BPC$ may be expressed in two
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different way in terms of the altitudes $CH$ and $PN$:
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\[
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\frac{1}{2} BP \cdot CH = \frac{1}{2} BC \cdot PN.
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\]
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Dividing these two equalities and recalling that $PM = PN$,
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we arrive at the desired assertion. <b>QED</b>
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<h2>The proof of Proposition 1</h2>
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<p>
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The diagram below shows the bisector $CC'$ of the angle $C$.
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It is known from elementary geometry that the incenter $O$ lies
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on the bisector.
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From Proposition 2 we have $C'A/C'B = b/a$. It
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follows that $AC'/AB = b/(a+b)$. In terms of the notation of
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<em>the algebra of points</em> introduced earlier in this page,
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this is expressed as:
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\[
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C' = \Big(1 - \frac{b}{a+b} \Big) A + \frac{b}{a+b} B
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= \frac{a}{a+b} A + \frac{b}{a+b} B.
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\]
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Since $O = (1-t)C + tC'$ for some $t$, we arrive at:
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\[
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O = (1-t)C + t\Big[ \frac{a}{a+b} A + \frac{b}{a+b} B \Big].
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\]
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This expresses the triangle's incenter $O$ in terms of its vertices,
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sides lengths, and a yet unknown quantity $t$ that takes values in the range
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0 to 1.
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<table class="centered">
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<tr><td align="center">
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<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
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<param name="background" value="ffffff">
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<param name="title" value="The incenter via algebra">
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<param name="e[1]" value="A;point;fixed;100,250">
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<param name="e[2]" value="B;point;fixed;300,250">
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<param name="e[3]" value="C;point;free;50,50;red;red">
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<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
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<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
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<param name="e[6]" value="C';point;proportion;A,pt1,A,C,A,B,A,B">
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<param name="e[7]" value="pt2;point;extend;B,A,A,C;none;none">
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<param name="e[8]" value="A';point;proportion;B,pt2,B,A,B,C,B,C;none;none">
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<param name="e[9]" value="pt3;point;extend;C,B,B,A;none;none">
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<param name="e[10]" value="B';point;proportion;C,pt3,C,B,C,A,C,A;none;none">
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<param name="e[11]" value="AA';line;connect;A,A';none;none;none">
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<param name="e[12]" value="BB';line;connect;B,B';none;none;none">
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<param name="e[13]" value="CC';line;connect;C,C';none;none;magenta">
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<param name="e[14]" value="O;point;intersection;AA',BB'">
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<param name="e[15]" value="H;point;foot;O,A,B;none;none">
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<param name="e[16]" value="IC;circle;radius;O,H;none;none;black;none">
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</applet>
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</td></tr>
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<tr><td>
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<b>
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Drag $C$ to change the geometry.<br>
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Press “r” to reset the diagram to its initial state.<br>
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</b>
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</td></tr></table>
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<p>
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If we repeat the same calculation by replacing the vertex $C$ by vertex $A$,
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we arrive at:
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\[
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O = (1-s)A + s\Big[ \frac{b}{b+c} B + \frac{c}{b+c} C \Big],
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\]
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where $s$ is yet another unknown that takes values in the range from 0 to 1.
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<p>
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Equating the two expressions for $O$ and collecting the terms, we arrive at:
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\[
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\Big[\frac{ta}{a+b} + s - 1\Big] A
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+
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\Big[\frac{tb}{a+b} - \frac{sb}{b+c} \Big] B
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+
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\Big[1 - t - \frac{sc}{b+c} \Big] C = 0.
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\]
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In this equations,
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if the coefficient of $C$ is nonzero, then we may solve for $C$,
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obtaining $C = \alpha A + \beta B$ for some $\alpha$
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and $\beta$. But this would
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mean that $C$ lies along the line $AB$ which would mean the the
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three points $A$, $B$ and $C$ are collinear, therefore the
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triangle $ABC$ is degenerate. We conclude that if the triangle is
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non-degenerate, then the coefficient of $C$ is zero. Similar
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arguments show that the coefficients of $A$ and $B$ are zero.
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Thus, the following system of three equations hold:
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\[
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\frac{ta}{a+b} + s - 1 = 0,
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\quad
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\frac{tb}{a+b} - \frac{sb}{b+c} = 0,
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\quad
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1 - t - \frac{sc}{b+c} = 0.
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\]
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Solving this system for the unknowns $t$ and $s$ we obtain:
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\[
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t = \frac{a+b}{a+b+c}, \quad s = \frac{b+c}{a+b+c}.
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\]
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Substituting these in either of the expressions for $O$ we arrive at:
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\[
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O = \frac{a}{a+b+C}A + \frac{b}{a+b+C}B + \frac{c}{a+b+C}C,
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\]
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which is equivalent to Proposition 1's
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assertion. <b>QED</b>
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<h2>Extension to tetrahedra</h2>
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<p>
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Propositions 1 extends to tetrahedra:
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<p>
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<b>Proposition 3</b>
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Let
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$a$, $b$, $c$, $d$
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be the areas of the faces opposite to the vertices
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$A$, $B$, $C$, $D$ of the tetrahedron $ABCD$. The the tetrahedron's
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incenter $O$ is given by:
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\[
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O
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= \frac{a}{\mathcal{A}} A
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+ \frac{b}{\mathcal{A}} B
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+ \frac{c}{\mathcal{A}} C
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+ \frac{d}{\mathcal{A}} D,
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\]
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where $\mathcal{A} = a + b + c + d$ is the tetrahedron's surface area.
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<p>
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This is proved with the aid of the following extension of Proposition 2:
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<p>
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<b>Proposition 4</b>
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<i>
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Let
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$a$, $b$, $c$, $d$
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be the areas of the faces opposite to the vertices
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$A$, $B$, $C$, $D$ of the tetrahedron $ABCD$.
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Let the bisector plane of the (internal) dihedral angle of edge $CD$
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intersect edge $AB$ at point $P$.
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The we have:
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\[
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\frac{PA}{PB} = \frac{b}{a}.
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\]
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</i>
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<table class="centered">
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<tr><td align="center">
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<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
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<param name="background" value="ffffff">
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<param name="title" value="Triangle's incenter">
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<param name="e[1]" value="A;point;fixed;100,250,300">
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<param name="e[2]" value="B;point;fixed;300,250,300">
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<param name="e[3]" value="C;point;fixed;350,200,-400">
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<param name="e[4]" value="D;point;free;200,40;red;red">
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<param name="e[5]" value="ABCD;polyhedron;tetrahedron;A,B,C,D">
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<param name="e[6]" value="ABC;plane;3points;A,B,C;none;none;none;none">
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<!-- construct the bisector of the dihedral angle with edge DC -->
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<param name="e[7]" value="pl1;plane;perpendicular;D,C;none;none;none;none">
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<param name="e[8]" value="Apl1;point;foot;A,pl1;none;none">
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<param name="e[9]" value="Bpl1;point;foot;B,pl1;none;none">
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<param name="e[10]" value="pt1;point;angleBisector;Apl1,D,Bpl1,pl1;none;none">
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<param name="e[11]" value="tmp1;plane;3points;D,C,pt1;none;none;none;none">
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<param name="e[12]" value="P;point;intersection;A,B,tmp1">
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<param name="e[13]" value="DP;line;connect;D,P;none;none;cyan">
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<param name="e[14]" value="CP;line;connect;C,P;none;none;cyan">
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</applet>
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</td></tr>
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<tr><td>
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<b>
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Drag $D$ to change the geometry.<br>
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Press “r” to reset the diagram to its initial state.<br>
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The plane $CDP$ bisects the dihedral angle of edge $CD$.
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</b>
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</td></tr></table>
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<p>
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I leave the proofs to you, the diligent reader. You will find
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useful information in a
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<a
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href="http://groups.google.com/group/sci.math/browse_thread/thread/56c187e018a85111/436f448dfab35a83">discussion
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in the sci.math newsgroup</a> from January 17, 2006.
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<hr width="60%">
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<p>
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<em>This applet was created by
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<a href="http://userpages.umbc.edu/~rostamia">Rouben Rostamian</a>
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using
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<a href="http://aleph0.clarku.edu/~djoyce/home.html">David Joyce</a>'s
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<a href="http://aleph0.clarkU.edu/~djoyce/java/Geometry/Geometry.html">Geometry
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Applet</a>
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on June 26, 2010.
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</em>
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<p>
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<table width="100%">
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<tr>
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<td valign="top">Go to <a href="index.html">Geometry Problems and Puzzles</a></td>
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<td align="right" style="width:200px;">
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