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<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>
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