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<h1>A geometric inequality</h1>
<h4>…and its solution by
<a href="mailto:haoyuep@aol.com">Dan Hoey</a></h4>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="400" height="400">
<param name="background" value="ffffff">
<param name="title" value="A geometric inequality">
<param name="e[1]" value="A;point;fixed;100,300">
<param name="e[2]" value="B;point;fixed;300,300">
<param name="e[3]" value="ABC;polygon;equilateralTriangle;A,B">
<param name="e[4]" value="C;point;vertex;ABC,3">
<param name="e[5]" value="P;point;free;180,320;red;red">
<param name="e[6]" value="PA;line;connect;P,A;none;none;green">
<param name="e[7]" value="PB;line;connect;P,B;none;none;green">
<param name="e[8]" value="PC;line;connect;P,C;none;none;magenta">
</applet>
</td></tr>
<tr><td>
<b>
Drag the point $P$.<br>
Press “r” to reset the diagram to its initial state.<br>
Proposition: $PC \le PA + PB$.
</b>
</td></tr></table>
<h2>The problem</h2>
<p>
<b>Proposition:</b> <i>Let $ABC$ be an equilateral triangle and $P$
be an arbitrary point in its plane. Then $PC \le PA + PB$.</i>
<p>
This was brought up in
<a href="http://mathforum.org/kb/message.jspa?messageID=1086018">a message </a>
on the <code>geometry.puzzles</code> newsgroup on November 11, 2001.
Go to that message and scroll to the bottom of the page to see the discussion
thread.
<p>
On November 22, 2001
<a href="mailto:haoyuep@aol.com">Dan Hoey</a>
offered a particularly nice solution. He also commented that he
had learned that problem may be related to the <em>Van Schooten Theorem</em>,
which indeed it is. See
<a href="http://www.cut-the-knot.org/Curriculum/Geometry/Pompeiu.shtml">Van
Schooten's and Pompeiu's Theorems: What are these?</a> for much detail and
historical background.
<h2>The proof</h2>
<p>
Here is Dan Hoey's proof of the proposition as stated above.
<p>
On the line segment $AP$ construct the equilateral triangle
$APD$, as shown in the diagram below, then add the line segment $DC$.
<p>
Let us show that the triangles $APB$ and $ADC$ are congruent. For this,
Let us observe that the sides $AP$ and $AB$ in the triangle $APB$
equal the sides $AD$ and $AC$ in the triangle $ADC$, by the construction.
Moreover, the angles $BAP$ and $CAD$ are equal because each equals
the difference of
a 60 degree angle and the angle $DAB$. Therefore, the triangles $APB$ and
$ADC$ are congruent by the side-angle-side equality.
We conclude, in particular, that $PB = DC$.
<p>
In the triangle $PDC$ we have $PC \le PD + DC$. In this inequality
replace $PD$ and $DC$ by their equivalents $PA$ and $PB$ to arrive at
$PC \le PA + PB$. <b>QED</b>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="400" height="400">
<param name="background" value="ffffff">
<param name="title" value="A geometric inequality">
<param name="e[1]" value="A;point;fixed;100,300">
<param name="e[2]" value="B;point;fixed;300,300">
<param name="e[3]" value="ABC;polygon;equilateralTriangle;A,B">
<param name="e[4]" value="C;point;vertex;ABC,3">
<param name="e[5]" value="P;point;free;180,320;red;red">
<param name="e[6]" value="PA;line;connect;P,A;none;none;green">
<param name="e[7]" value="PB;line;connect;P,B;none;none;green">
<param name="e[8]" value="PC;line;connect;P,C;none;none;magenta">
<param name="e[9]" value="APD;polygon;equilateralTriangle;A,P;none;none;none">
<param name="e[10]" value="D;point;vertex;APD,3;">
<param name="e[11]" value="AD;line;connect;A,D;none;none;lightGray">
<param name="e[12]" value="PD;line;connect;P,D;none;none;lightGray">
<param name="e[13]" value="CD;line;connect;C,D;none;none;lightGray">
</applet>
</td></tr>
<tr><td>
<b>
Drag the point $P$.<br>
Press “r” to reset the diagram to its initial state.<br>
Observation: The triangles $APB$ and $ADC$ are congruent.
</b>
</td></tr></table>
<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 November 23, 2001.<br>
Cosmetic revisions on June 23, 2010.
</em>
<p>
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