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>
131 lines
4.4 KiB
HTML
131 lines
4.4 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>An equilateral triangle inscribed in a rectangle</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>An equilateral triangle inscribed in a rectangle</h1>
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<p>
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The following solution to the
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<a href="http://userpages.umbc.edu/~rostamia/Geometry/inscribed-equilateral.html">inscribed triangle puzzle</a>
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is due to <b>Peter Renz</b>
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who communicated it to me on December 2016.
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<p>
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For the sake of completeness, let's begin with the statement of the puzzle.
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The figure below depicts a equilateral triangle $AEF$ inscribed in a rectangle
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in such a way that the two share a vertex. We wish to show that the
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area of the pink triangle ($ECF$) is the sum of the areas of the other two
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colored triangles.
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<div class="centered">
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<img src="./inscribed-equilateral-solution/frame000.png" width=500
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height=417 alt="[frame000.png]">
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</div>
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<p>
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As noted in the referring page (see the link above), a solution with
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the aid of trigonometry is quite straightforward. The purpose of this
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page is to present a solution in the style of <i>Euclid</i>, without appeal
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to trigonometry.
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<p>
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The following animation encapsulates Peter Renz's solution in its
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entirety.
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<div class="centered">
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<img src="./inscribed-equilateral-solution/waterwheel.gif" width=500
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height=417 alt="[animation.gif]">
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</div>
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The animation should be self-explanatory if you stare at it long enough.
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Nevertheless, I will now proceed to point out the reasoning through
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several still images extracted from that animation.
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<ol>
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<li>
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Construct a circle on the diameter $FE$, and then subdivide its boundary
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into six 60-degree wedges starting at the vertex $C$. Mark the division
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points $C$, $P$, $Q'$, $C'$, $P'$, and $Q$, as shown.
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<div class="centered">
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<img src="./inscribed-equilateral-solution/frame007.png" width=500
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height=417 alt="[frame007.png]">
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</div>
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<li>
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Rotate the triangle $FDA$ about $F$ by 60 degrees to bring the edge $FA$ to
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coincide with $FE$.
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<div class="centered">
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<img src="./inscribed-equilateral-solution/frame073.png" width=500
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height=417 alt="[frame073.png]">
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</div>
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The key observation is that the rotation moves the vertex $D$ into $P'$.
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<li>
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Further rotate the triangle about the circle's center by 180 degrees to
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place it in the $FPE$ position.
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<div class="centered">
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<img src="./inscribed-equilateral-solution/frame111.png" width=500
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height=417 alt="[frame111.png]">
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</div>
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<li>
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Rotate the triangle $EBA$ about $E$ by 60 degrees to bring the edge $EA$ to
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coincide with $EF$. Then the vertex $B$ will move onto $Q'$ for reasons
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similar to those explained above.
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<div class="centered">
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<img src="./inscribed-equilateral-solution/frame177.png" width=500
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height=417 alt="[frame177.png]">
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</div>
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<li>
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The two red line segments in the figure below are parallel and of equal lengths
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by virtue of being the side and the “radius” of the
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regular hexagon (not shown) inscribed in the circle.
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<div class="centered">
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<img src="./inscribed-equilateral-solution/frame180.png" width=500
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height=417 alt="[frame180.png]">
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</div>
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The altitudes of the three triangles, dropped from the vertices $C$,
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$P$, and $Q'$ are shown in dashed lines. It should be clear that the
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altitude dropped from $C$ is equal in length to the sum of those of
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the other two altitudes. Since the three triangles share a common base,
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the area of one is the sum of the areas of the other two. Q.E.D.
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</ol>
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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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