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>
156 lines
5.4 KiB
HTML
156 lines
5.4 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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<!-- fix buggy IE8, especially for mathjax -->
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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>A geometric inequality</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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MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
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</script>
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</head>
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<body style="visibility:hidden">
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<h1>A geometric inequality</h1>
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<h4>…and its solution by
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<a href="mailto:haoyuep@aol.com">Dan Hoey</a></h4>
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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="400">
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<param name="background" value="ffffff">
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<param name="title" value="A geometric inequality">
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<param name="e[1]" value="A;point;fixed;100,300">
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<param name="e[2]" value="B;point;fixed;300,300">
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<param name="e[3]" value="ABC;polygon;equilateralTriangle;A,B">
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<param name="e[4]" value="C;point;vertex;ABC,3">
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<param name="e[5]" value="P;point;free;180,320;red;red">
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<param name="e[6]" value="PA;line;connect;P,A;none;none;green">
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<param name="e[7]" value="PB;line;connect;P,B;none;none;green">
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<param name="e[8]" value="PC;line;connect;P,C;none;none;magenta">
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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 the point $P$.<br>
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Press “r” to reset the diagram to its initial state.<br>
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Proposition: $PC \le PA + PB$.
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</b>
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</td></tr></table>
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<h2>The problem</h2>
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<p>
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<b>Proposition:</b> <i>Let $ABC$ be an equilateral triangle and $P$
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be an arbitrary point in its plane. Then $PC \le PA + PB$.</i>
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<p>
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This was brought up in
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<a href="http://mathforum.org/kb/message.jspa?messageID=1086018">a message </a>
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on the <code>geometry.puzzles</code> newsgroup on November 11, 2001.
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Go to that message and scroll to the bottom of the page to see the discussion
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thread.
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<p>
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On November 22, 2001
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<a href="mailto:haoyuep@aol.com">Dan Hoey</a>
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offered a particularly nice solution. He also commented that he
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had learned that problem may be related to the <em>Van Schooten Theorem</em>,
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which indeed it is. See
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<a href="http://www.cut-the-knot.org/Curriculum/Geometry/Pompeiu.shtml">Van
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Schooten's and Pompeiu's Theorems: What are these?</a> for much detail and
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historical background.
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<h2>The proof</h2>
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<p>
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Here is Dan Hoey's proof of the proposition as stated above.
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<p>
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On the line segment $AP$ construct the equilateral triangle
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$APD$, as shown in the diagram below, then add the line segment $DC$.
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<p>
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Let us show that the triangles $APB$ and $ADC$ are congruent. For this,
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Let us observe that the sides $AP$ and $AB$ in the triangle $APB$
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equal the sides $AD$ and $AC$ in the triangle $ADC$, by the construction.
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Moreover, the angles $BAP$ and $CAD$ are equal because each equals
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the difference of
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a 60 degree angle and the angle $DAB$. Therefore, the triangles $APB$ and
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$ADC$ are congruent by the side-angle-side equality.
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We conclude, in particular, that $PB = DC$.
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<p>
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In the triangle $PDC$ we have $PC \le PD + DC$. In this inequality
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replace $PD$ and $DC$ by their equivalents $PA$ and $PB$ to arrive at
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$PC \le PA + PB$. <b>QED</b>
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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="400">
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<param name="background" value="ffffff">
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<param name="title" value="A geometric inequality">
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<param name="e[1]" value="A;point;fixed;100,300">
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<param name="e[2]" value="B;point;fixed;300,300">
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<param name="e[3]" value="ABC;polygon;equilateralTriangle;A,B">
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<param name="e[4]" value="C;point;vertex;ABC,3">
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<param name="e[5]" value="P;point;free;180,320;red;red">
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<param name="e[6]" value="PA;line;connect;P,A;none;none;green">
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<param name="e[7]" value="PB;line;connect;P,B;none;none;green">
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<param name="e[8]" value="PC;line;connect;P,C;none;none;magenta">
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<param name="e[9]" value="APD;polygon;equilateralTriangle;A,P;none;none;none">
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<param name="e[10]" value="D;point;vertex;APD,3;">
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<param name="e[11]" value="AD;line;connect;A,D;none;none;lightGray">
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<param name="e[12]" value="PD;line;connect;P,D;none;none;lightGray">
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<param name="e[13]" value="CD;line;connect;C,D;none;none;lightGray">
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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 the point $P$.<br>
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Press “r” to reset the diagram to its initial state.<br>
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Observation: The triangles $APB$ and $ADC$ are congruent.
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</b>
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</td></tr></table>
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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 November 23, 2001.<br>
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Cosmetic revisions on June 23, 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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<a href="http://validator.w3.org/check?uri=referer">
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<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
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<a href="http://jigsaw.w3.org/css-validator/check/referer">
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<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
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</td></tr>
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</table>
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</body>
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</html>
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