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>
132 lines
4.4 KiB
HTML
132 lines
4.4 KiB
HTML
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
|
|
<html>
|
|
<head>
|
|
<!-- fix buggy IE8, especially for mathjax -->
|
|
<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
|
|
|
|
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
|
|
<title>An equilateral triangle inscribed in a rectangle</title>
|
|
<link rel="stylesheet" type="text/css" media="screen" href="style.css">
|
|
|
|
<script type="text/javascript"
|
|
src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
|
|
MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
|
|
</script>
|
|
|
|
</head>
|
|
<body style="visibility:hidden">
|
|
|
|
<h1>An equilateral triangle inscribed in a rectangle</h1>
|
|
|
|
<p>
|
|
The following solution to the
|
|
<a href="http://userpages.umbc.edu/~rostamia/Geometry/inscribed-equilateral.html">inscribed triangle puzzle</a>
|
|
is due to <b>Peter Renz</b>
|
|
who communicated it to me on December 2016.
|
|
|
|
<p>
|
|
For the sake of completeness, let's begin with the statement of the puzzle.
|
|
The figure below depicts a equilateral triangle $AEF$ inscribed in a rectangle
|
|
in such a way that the two share a vertex. We wish to show that the
|
|
area of the pink triangle ($ECF$) is the sum of the areas of the other two
|
|
colored triangles.
|
|
|
|
<div class="centered">
|
|
<img src="./inscribed-equilateral-solution/frame000.png" width=500
|
|
height=417 alt="[frame000.png]">
|
|
</div>
|
|
|
|
<p>
|
|
As noted in the referring page (see the link above), a solution with
|
|
the aid of trigonometry is quite straightforward. The purpose of this
|
|
page is to present a solution in the style of <i>Euclid</i>, without appeal
|
|
to trigonometry.
|
|
|
|
<p>
|
|
The following animation encapsulates Peter Renz's solution in its
|
|
entirety.
|
|
|
|
<div class="centered">
|
|
<img src="./inscribed-equilateral-solution/waterwheel.gif" width=500
|
|
height=417 alt="[animation.gif]">
|
|
</div>
|
|
|
|
The animation should be self-explanatory if you stare at it long enough.
|
|
Nevertheless, I will now proceed to point out the reasoning through
|
|
several still images extracted from that animation.
|
|
|
|
<ol>
|
|
|
|
<li>
|
|
Construct a circle on the diameter $FE$, and then subdivide its boundary
|
|
into six 60-degree wedges starting at the vertex $C$. Mark the division
|
|
points $C$, $P$, $Q'$, $C'$, $P'$, and $Q$, as shown.
|
|
|
|
<div class="centered">
|
|
<img src="./inscribed-equilateral-solution/frame007.png" width=500
|
|
height=417 alt="[frame007.png]">
|
|
</div>
|
|
|
|
<li>
|
|
Rotate the triangle $FDA$ about $F$ by 60 degrees to bring the edge $FA$ to
|
|
coincide with $FE$.
|
|
|
|
<div class="centered">
|
|
<img src="./inscribed-equilateral-solution/frame073.png" width=500
|
|
height=417 alt="[frame073.png]">
|
|
</div>
|
|
|
|
The key observation is that the rotation moves the vertex $D$ into $P'$.
|
|
|
|
<li>
|
|
Further rotate the triangle about the circle's center by 180 degrees to
|
|
place it in the $FPE$ position.
|
|
|
|
<div class="centered">
|
|
<img src="./inscribed-equilateral-solution/frame111.png" width=500
|
|
height=417 alt="[frame111.png]">
|
|
</div>
|
|
|
|
<li>
|
|
Rotate the triangle $EBA$ about $E$ by 60 degrees to bring the edge $EA$ to
|
|
coincide with $EF$. Then the vertex $B$ will move onto $Q'$ for reasons
|
|
similar to those explained above.
|
|
|
|
<div class="centered">
|
|
<img src="./inscribed-equilateral-solution/frame177.png" width=500
|
|
height=417 alt="[frame177.png]">
|
|
</div>
|
|
|
|
<li>
|
|
The two red line segments in the figure below are parallel and of equal lengths
|
|
by virtue of being the side and the “radius” of the
|
|
regular hexagon (not shown) inscribed in the circle.
|
|
|
|
<div class="centered">
|
|
<img src="./inscribed-equilateral-solution/frame180.png" width=500
|
|
height=417 alt="[frame180.png]">
|
|
</div>
|
|
|
|
The altitudes of the three triangles, dropped from the vertices $C$,
|
|
$P$, and $Q'$ are shown in dashed lines. It should be clear that the
|
|
altitude dropped from $C$ is equal in length to the sum of those of
|
|
the other two altitudes. Since the three triangles share a common base,
|
|
the area of one is the sum of the areas of the other two. Q.E.D.
|
|
|
|
</ol>
|
|
|
|
<table width="100%">
|
|
<tr>
|
|
<td valign="top">Go to <a href="index.html">Geometry Problems and Puzzles</a></td>
|
|
<td align="right" style="width:200px;">
|
|
<a href="http://validator.w3.org/check?uri=referer">
|
|
<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
|
|
<a href="http://jigsaw.w3.org/css-validator/check/referer">
|
|
<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
|
|
</td></tr>
|
|
</table>
|
|
|
|
</body>
|
|
</html>
|
|
|