<?xml version="1.0" encoding="UTF-8" standalone="yes"?><oembed><version><![CDATA[1.0]]></version><provider_name><![CDATA[Azimuth]]></provider_name><provider_url><![CDATA[https://johncarlosbaez.wordpress.com]]></provider_url><author_name><![CDATA[John Baez]]></author_name><author_url><![CDATA[https://johncarlosbaez.wordpress.com/author/johncarlosbaez/]]></author_url><title><![CDATA[Geometry Puzzles]]></title><type><![CDATA[link]]></type><html><![CDATA[<p>Here are four puzzles about <i>areas</i>, in approximate order of increasing difficulty. </p>
<h3>Mysteries of the equilateral triangle </h3>
<div align="center">
<a href="http://www.m-hikari.com/mccartin-2.pdf"><br />
<img src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/triangle_and_circles.jpg" /></a>
</div>
<p><b>Puzzle:</b> Show the area of the orange circle equals the total area of the two blue regions.</p>
<p>In case you&#8217;re wondering, the picture above shows an equilateral triangle with a small circle inscribed in it and a big circle circumscribed around it.  </p>
<p>The puzzle is easy if you think about it the right way.  I never knew this cool fact until last month, when I read this fun free book:</p>
<p>• Brian McCartin, <a href="http://www.m-hikari.com/mccartin-2.pdf">Mysteries of the equilateral triangle</a>.</p>
<p>I&#8217;ve given talks on the numbers <a href="http://math.ucr.edu/home/baez/numbers/#5">5</a>, <a href="http://math.ucr.edu/home/baez/numbers/#8">8</a>, and <a href="http://math.ucr.edu/home/baez/numbers/#24">24</a>.  I&#8217;ve imagined writing a book that had chapters on all my favorite natural numbers, starting with Chapter 0.  But it would be very hard to write Chapter 3, since this number has <i>too many</i> interesting properties!  McCartin&#8217;s book covers a lot of them.</p>
<h3> Square the lune to the tune of Claire de Lune </h3>
<div align="center">
<a href="https://en.wikipedia.org/wiki/Lune_of_Hippocrates"><br />
<img width="450" src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/lune_of_hippocrates.jpg" /></a>
</div>
<p><b>Puzzle:</b> Show the crescent has the same area as the triangle.</p>
<p>Greek mathematicians really wanted to <a href=""><b>square the circle</b></a>, by which I mean: use straightedge and compass to first draw a circle and then construct a square with the same area. </p>
<p>In 440 BC, Hippocrates of Chios figured out how to square the above crescent-shaped region, which lies between the circle centered at O and the smaller circle centered at D.  So, this region is called the <a href="https://en.wikipedia.org/wiki/Lune_of_Hippocrates"><b>Lune of Hippocrates</b></a>.</p>
<p>I&#8217;ve heard it said that this result gave some Greek geometers hope that the circle could be squared.  I&#8217;m not sure this is true; Hippocrates himself was probably too smart to be fooled.   But in any event, it would have been a false hope.  Much later, around 1885, Lindemann and Weierstrass proved that squaring the circle was impossible.</p>
<p>Any crescent-shaped region formed by two circular arcs is called a <b>lune</b>.   It&#8217;s widely believed that there are only 5 <b>squarable lunes</b>.  In other words, there are only 5 shapes of lune constructible by straightedge and compass whose area equals that of a square constructible using straightedge and compass.  (Obviously these lunes can come in many different <i>sizes</i>.)</p>
<p>Hippocrates discovered three squarable lunes.  Two more were discovered by Martin Johan Wallenius in 1766.  A proof that these are the only squarable lunes was given by Tchebatorew and Dorodnow, and summarized by the famous topologist Postnikov:</p>
<p>&bull; M. M. Postnikov, The problem of squarable Lunes, translated from the Russian by Abe Shenitzer, <i><a href="http://www.jstor.org/discover/10.2307/2589121?uid=3739560&amp;uid=2&amp;uid=4&amp;uid=3739256&amp;sid=21103281963343">American Mathematical Monthly</a></i>, <b>107</b> (Aug.-Sep. 2000), 645&#8211;651.</p>
<p>However, there may be a loophole in this proof: <a href="http://mathoverflow.net/a/151566/2893">Will Jagy claims</a> that these Russians assumed without proof that the ratio of two angles involved in the construction of the lune is rational.   With this assumption, finding a squarable lune amounts to finding a rational number <img src='https://s0.wp.com/latex.php?latex=u&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='u' title='u' class='latex' /> and a <a href="https://en.wikipedia.org/wiki/Constructible_number">constructible number</a> <img src='https://s0.wp.com/latex.php?latex=%5Csin+a&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;sin a' title='&#92;sin a' class='latex' /> such that</p>
<p><img src='https://s0.wp.com/latex.php?latex=%5Csin+%28+u+a%29+%3D+%5Csqrt%7Bu%7D+%5Csin+a+&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;sin ( u a) = &#92;sqrt{u} &#92;sin a ' title='&#92;sin ( u a) = &#92;sqrt{u} &#92;sin a ' class='latex' /></p>
<p><b>Puzzle:</b> Why should <img src='https://s0.wp.com/latex.php?latex=u&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='u' title='u' class='latex' /> be rational?  Do you know the true state of the art here?</p>
<p>(My puzzles include hard questions I don&#8217;t know the answer to, and this is one.)</p>
<p>For a nice discussion of the 5 squarable lunes, with pictures, see this page:</p>
<p>• <a href="http://www.mathpages.com/home/kmath171/kmath171.htm">The five squarable lunes</a>, MathPages.</p>
<h3> Twice in a blue lune </h3>
<div align="center">
<a href="https://en.wikipedia.org/wiki/Lune_of_Hippocrates"><br />
<img src="https://upload.wikimedia.org/wikipedia/commons/3/3d/Lunules-better.png" /></a>
</div>
<p><b>Puzzle:</b> Show the area of this right triangle equals the total area inside the blue lunes.  The outside of each lune is a semicircle.  The inside of each lune is part of the circle containing the points A, B, and C.</p>
<p>The circle through all 3 corners of a triangle is called its <b>circumcircle</b>.  You can <a href="http://www.mathopenref.com/constcircumcircle.html">construct the circumcircle using a straightedge and compass</a>, if you want.</p>
<p>Again this is a famous old problem.  The two blue lunes here are called the <b>Lunes of Alhazen</b>.  This problem was later posed and solve by Leonardo da Vinci!  </p>
<p>This puzzle is a lot of fun, so I urge you not to give up&#8212;but if you do, you can see da Vinci&#8217;s solution <a href="http://www.hypatiamaze.org/davinci/leo_lune2.html">here</a>.</p>
<h3> The arbelos</h3>
<div align="center">
<a href="http://commons.wikimedia.org/wiki/File:Arbelos-tikz.svg"><br />
<img width="300" src="https://i1.wp.com/math.ucr.edu/home/baez/mathematical/arbelos.jpg" /></a>
</div>
<p><b>Puzzle:</b> show the area of the green shape equals the area of the circle.</p>
<p>The green shape is called an arbelos, which means &#8216;shoemaker&#8217;s knife&#8217; in Greek, since it looks a bit like that.  It shows up in Propositions 4-8 of the <a href="https://en.wikipedia.org/wiki/Book_of_Lemmas"><i>Book of Lemmas</i></a>.  This book goes back at least to <a href="https://en.wikipedia.org/wiki/Th%C4%81bit_ibn_Qurra">Thābit ibn Qurra</a>, a mathematician, astronomer and physician who live in Baghdad from 826 to 901 AD.  Ibn Qurra said the book was written by Archimedes!  But nobody is sure.</p>
<p>So, when you do this puzzle, you may be matching wits with Archimedes.  But you&#8217;ll certainly be sharing some thoughts with Thābit ibn Qurra.</p>
<p>Math has been called the world&#8217;s longest conversation.  Perhaps this is an exaggeration: people have been passing on stories for a long time.  But Babylonians were computing the square root of two back in 1700 BC, probably using techniques that are still interesting today&#8230; so it really is remarkable how old mathematics still makes good sense, unlike the old creation myths.</p>
<p>For more on the arbelos try:</p>
<p>&bull; <a href="http://en.wikipedia.org/wiki/Arbelos">Arbelos</a>, Wikipedia.</p>
<p>and</p>
<p>&bull; Thomas Schoch, <a href="http://www.retas.de/thomas/arbelos/references.html">Arbelos references</a>, 2013.</p>
<p>For the 15 propositions in the Book of Lemmas, see:</p>
<p>&bull; <a href="http://en.wikipedia.org/wiki/Book_of_Lemmas"><i>Book of Lemmas</i></a>,  Wikipedia.</p>
<h3> Two semicircles is half a circle </h3>
<div align="center">
<a href="http://www.cut-the-knot.org/proofs/Semicircles.shtml"><br />
<img width="300" src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/semicircle_puzzle.png" /></a>
</div>
<p><b>Puzzle:</b> show the total area of the two semicircles is half the area of the large circle.</p>
<p>Amazingly, it seems this fact was noticed only in 2011:</p>
<p>• Andrew K. Jobbings, Two semicircles fill half a circle, <i>The Mathematical Gazette</i> <b>95</b> (Nov. 2011), 538&#8211;540.</p>
<p>However, Andrew Jobbings is a genius when it comes to &#8216;recreational mathematics&#8217;.  For more of his work, check out this page:</p>
<p>&bull; Andrew Jobbings, <a href="http://www.arbelos.co.uk/papers.html">Arbelos</a>.</p>
<p>I learned about this puzzle from Alexander Bogomolny, who has a wonderful website full of Javascript geometry demonstrations:</p>
<p>&bull; Alexander Bogomolny, <a href="http://www.cut-the-knot.org/proofs/Semicircles.shtml">A property of semicircles</a>.</p>
<p>You&#8217;ll get a scary warning asking &#8220;Do you want to run this application?&#8221;  Say yes.  You&#8217;ll get an applet that lets you slide the point where the semicircles touch: no matter where it is, the semicircles have the same total area!  Click &#8220;hint&#8221; and you&#8217;ll get a hint.  If you&#8217;re still stuck, and too impatient to solve the puzzle yourself, scroll down and see a proof!</p>
<p>However, that proof is long and confusing.  With geometry I like demonstrations where after some thought you can simply <i>see</i> that the result is true.  For this puzzle, such a demonstration was provided by Greg Egan.  </p>
<p><a href="http://math.ucr.edu/home/baez/mathematical/semicircle_puzzle_egan.gif">Click here</a> if you give up on this puzzle and want to see Egan&#8217;s solution.   You may need to ponder it a bit, but when you&#8217;re done you should say &#8220;yes, this result is <i>clear!</i>&#8221;</p>
<p>As a separate hint: the answers to this puzzle and the previous one are similar, in a very nice way!</p>
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