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<h1>Triangles with common base</h1>
<h4>An interesting problem proposed by
<a href="mailto:stevebg@adelphia.net">Steve Gray</a></h4>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="650" height="250">
<param name="background" value="ffffff">
<param name="title" value="Triangles with common base">
<param name="e[1]" value="A;point;fixed;50,215">
<param name="e[2]" value="B;point;fixed;250,215">
<param name="e[3]" value="T0;polygon;equilateralTriangle;A,B">
<param name="e[4]" value="C;point;vertex;T0,3">
<param name="e[5]" value="D;point;free;115,173;red;red">
<param name="e[6]" value="DC;line;connect;D,C;none;none;orange">
<param name="e[7]" value="DA;line;connect;D,A;none;none;blue">
<param name="e[8]" value="DB;line;connect;D,B;none;none;green">
<param name="e[9]" value="P;point;fixed;300,215">
<param name="e[10]" value="Q;point;fixed;600,215">
<param name="e[11]" value="PQ;line;connect;P,Q">
<param name="e[12]" value="B';point;similar;P,Q,C,A,D">
<param name="e[13]" value="PB';line;connect;P,B';none;none;green">
<param name="e[14]" value="QB';line;connect;Q,B';none;none;green">
<param name="e[15]" value="C';point;similar;P,Q,A,B,D">
<param name="e[16]" value="PC';line;connect;P,C';none;none;orange">
<param name="e[17]" value="QC';line;connect;Q,C';none;none;orange">
<param name="e[18]" value="A';point;similar;P,Q,B,C,D">
<param name="e[19]" value="PA';line;connect;P,A';none;none;blue">
<param name="e[20]" value="QA';line;connect;Q,A';none;none;blue">
<param name="e[21]" value="T1;polygon;triangle;B',C',A';none;none;lightGray">
<!-- construct the centroid -->
<param name="e[22]" value="midA'B';point;midpoint;A',B';none;none">
<param name="e[23]" value="li1;line;connect;C',midA'B';none;none;none">
<param name="e[24]" value="midB'C';point;midpoint;B',C';none;none">
<param name="e[25]" value="li2;line;connect;A',midB'C';none;none;none">
<param name="e[26]" value="M;point;intersection;li1,li2">
</applet>
</td></tr>
<tr><td>
<b>
Drag the point $D$.<br>
Press “r” to reset the diagram to its initial state.<br>
Proposition: $A'B'C'$ is equilateral and its centroid $M$ is fixed.
</b>
</td></tr></table>
<h2>The construction</h2>
<p>
This problem was proposed by
<a href="mailto:stevebg@adelphia.net">Steve Gray</a>
in the <code>geometry.puzzles</code> newsgroup
(<a href="http://mathforum.org/kb/message.jspa?messageID=1088552">see
the original message</a>) on July 26, 2002.
Scroll to the bottom of that page for a link to the solution.
<p>
Consider an equilateral triangle $ABC$, a line segment $PQ$,
and an arbitrary point $D$, as seen in the diagram above.
On the segment $PQ$ construct three triangles
$PC'Q$,
$PA'Q$,
$PB'Q$,
similar to the triangles
$ADB$,
$BDC$,
$CDA$,
respectively.
<p>
<b>Proposition 1:</b>
The triangle $A'B'C'$ is equilateral.
<p>
<b>Proposition 2:</b>
The centroid of $A'B'C'$ is independent of $D$.
<p>
Steve adds:
<blockquote>
<p>
Now generalize this for a regular $n$-gon. The new points form a
regular $n$-gon whose centroid is independent of $D$.
This problem is original so far as I know. I am interested in the
simplest synthetic solution; no algebra, please.
</blockquote>
<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>
July 26, 2002.<br>
Cosmetic revisions on June 17, 2010.
</em>
<p>
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