{"id":26,"date":"2014-03-28T10:39:21","date_gmt":"2014-03-28T14:39:21","guid":{"rendered":"http:\/\/web.colby.edu\/thegeometricviewpoint\/?p=26"},"modified":"2014-05-15T10:53:31","modified_gmt":"2014-05-15T14:53:31","slug":"pythagoras-in-the-hyperbolic-plane","status":"publish","type":"post","link":"https:\/\/web.colby.edu\/thegeometricviewpoint\/2014\/03\/28\/pythagoras-in-the-hyperbolic-plane\/","title":{"rendered":"<wbr>Pythagoras in the hyperbolic plane"},"content":{"rendered":"<p>The Pythagorean theorem is surely the most famous of all mathematical theorems. The simplicity of its statement (that a right triangle with sides of length <img src='https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b' title='b' class='latex' \/> and hypotenuse of length <img src='https:\/\/s0.wp.com\/latex.php?latex=c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c' title='c' class='latex' \/> satisfies <img src='https:\/\/s0.wp.com\/latex.php?latex=a%5E2+%2B+b%5E2+%3D+c%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^2 + b^2 = c^2' title='a^2 + b^2 = c^2' class='latex' \/>) and the multiplicity of beautiful proofs (like the one shown at right from <a title=\"Byrne's edition of Euclid's Elements\" href=\"http:\/\/www.math.ubc.ca\/~cass\/euclid\/book1\/images\/bookI-prop47.html\" target=\"_blank\">Byrne&#8217;s edition<\/a> of Euclid&#8217;s Elements) contribute to its memorability.<\/p>\n<div id=\"attachment_28\" style=\"width: 160px\" class=\"wp-caption alignright\"><a href=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/Byrne-pythagorean.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-28\" class=\"size-thumbnail wp-image-28\" alt=\"Byrne's version of Euclid's proof of the Pythagorean theorem.\" src=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/Byrne-pythagorean-150x150.jpg\" width=\"150\" height=\"150\" \/><\/a><p id=\"caption-attachment-28\" class=\"wp-caption-text\">Byrne&#8217;s version of Euclid&#8217;s proof of the Pythagorean theorem.<\/p><\/div>\n<p>The fame of Pythagoras&#8217; theorem in Euclidean geometry makes it natural to ask if it holds in other geometries. For instance, is the Pythagorean theorem true in hyperbolic geometry, that radical challenger to Euclid&#8217;s throne?<\/p>\n<p>The naive answer is &#8220;No, the Pythagorean theorem does not hold in hyperbolic geometry&#8221;, as it is logically equivalent to Euclid&#8217;s 5th postulate (which is the defining difference between Euclidean and hyperbolic geometry). However, there <i>is<\/i> a theorem in hyperbolic geometry which is analogous to Pythagoras theorem:<\/p>\n<div style=\"border: 5px; padding-left: 5px; padding-top: 5px; padding-right: 5px; padding-bottom: 0px; border-style: ridge; border-color: #6A7F97; width: 70%;\"><b style=\"font-variant: small-caps; font-size: large;\">The hyperbolic Pythagorean theorem <\/b><br \/>\n<i>If a right triangle in the hyperbolic plane has sides of length <img src='https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b' title='b' class='latex' \/> and a hypotenuse of length <img src='https:\/\/s0.wp.com\/latex.php?latex=c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c' title='c' class='latex' \/>, then <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ccosh%28a%29%5Ccosh%28b%29+%3D+%5Ccosh+c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\cosh(a)\\cosh(b) = \\cosh c' title='\\cosh(a)\\cosh(b) = \\cosh c' class='latex' \/>.<br \/>\n<\/i><\/div>\n<div id=\"attachment_39\" style=\"width: 310px\" class=\"wp-caption alignleft\"><a href=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithLabels.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-39\" class=\"size-medium wp-image-39\" alt=\"Right triangle in the hyperbolic plane\" src=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithLabels-300x300.jpg\" width=\"300\" height=\"300\" \/><\/a><p id=\"caption-attachment-39\" class=\"wp-caption-text\">Right triangle in the hyperbolic plane<\/p><\/div>\n<p>As in the figure at left, the edges of the triangle are hyperbolic geodesics (we&#8217;ll review what those are below), with the sides of length <img src='https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b' title='b' class='latex' \/> adjacent to the right angle and the hypotenuse (of length <img src='https:\/\/s0.wp.com\/latex.php?latex=c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c' title='c' class='latex' \/>) is the edge across from the right angle. The function <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ccosh+x+%3D+%5Cfrac%7Be%5Ex+%2B+e%5E%7B-x%7D%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\cosh x = \\frac{e^x + e^{-x}}{2}' title='\\cosh x = \\frac{e^x + e^{-x}}{2}' class='latex' \/> is hyperbolic cosine.<\/p>\n<h3>The hyperbolic plane<\/h3>\n<p>The hyperbolic plane was discovered by Bolyai and Lobachevsky when they investigated the effect of replacing Euclid&#8217;s parallel postulate with an alternative. They operated purely deductively: they had no graphical representation of hyperbolic geometry to work with. Later mathematicians, such as Klein and Poincar\u00e9, discovered ways of representing hyperbolic geometry inside of Euclidean geometry by giving new meanings to terms such as &#8220;line&#8221;. Using calculus, we can give a succinct description (called the Poincar\u00e9 disc model) of the hyperbolic plane as follows.<\/p>\n<p>The entirety of hyperbolic geometry will take place inside the open unit disc (the blue disc at left) in the plane <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cmathbb+R%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\mathbb R^2' title='\\mathbb R^2' class='latex' \/>. The unit circle (the boundary of the disc) is not part of the world in which we do hyperbolic geometry. We refer to it as <i>the infinity circle<\/i>. A path <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cgamma%28t%29+%3D+%28x%28t%29%2C+y%28t%29%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\gamma(t) = (x(t), y(t))' title='\\gamma(t) = (x(t), y(t))' class='latex' \/> for <img src='https:\/\/s0.wp.com\/latex.php?latex=t_0+%5Cleq+t+%5Cleq+t_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_0 \\leq t \\leq t_1' title='t_0 \\leq t \\leq t_1' class='latex' \/> in the disc has length defined by an integral similar to the integral defining path length in euclidean geometry. The length of <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cgamma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\gamma' title='\\gamma' class='latex' \/> in euclidean geometry is given by <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cint_%7Bt_0%7D%5E%7Bt_1%7D+%7C%7C%5Cgamma%27%28t%29%7C%7C%5Cthinspace+dt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\int_{t_0}^{t_1} ||\\gamma&#039;(t)||\\thinspace dt' title='\\int_{t_0}^{t_1} ||\\gamma&#039;(t)||\\thinspace dt' class='latex' \/>, where <img src='https:\/\/s0.wp.com\/latex.php?latex=%7C%7Cv%7C%7C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='||v||' title='||v||' class='latex' \/> denotes the magnitude of the vector <img src='https:\/\/s0.wp.com\/latex.php?latex=v&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v' title='v' class='latex' \/>. The length of <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cgamma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\gamma' title='\\gamma' class='latex' \/> in hyperbolic geometry on the other hand is given by the integral <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cint_%7Bt_0%7D%5E%7Bt_1%7D+%5Cfrac%7B%7C%7C%5Cgamma%27%28t%29%7C%7C%7D%7B1+-+%7C%7C%5Cgamma%28t%29%7C%7C%5E2%7D+%5Cthinspace+dt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\int_{t_0}^{t_1} \\frac{||\\gamma&#039;(t)||}{1 - ||\\gamma(t)||^2} \\thinspace dt' title='\\int_{t_0}^{t_1} \\frac{||\\gamma&#039;(t)||}{1 - ||\\gamma(t)||^2} \\thinspace dt' class='latex' \/>. A path between points <img src='https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A' title='A' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B' title='B' class='latex' \/> is a <i>geodesic<\/i> if it has length no greater than the length of any other path between <img src='https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A' title='A' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B' title='B' class='latex' \/>. Geodesics in hyperbolic geometry are the analogue of straight lines in euclidean geometry. If there were light in the hyperbolic plane, it would travel along geodesics.<\/p>\n<div id=\"attachment_43\" style=\"width: 310px\" class=\"wp-caption alignright\"><a href=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithGeodesics.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-43\" class=\"size-medium wp-image-43\" alt=\"Right triangle drawn with geodesics\" src=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithGeodesics-300x300.jpg\" width=\"300\" height=\"300\" \/><\/a><p id=\"caption-attachment-43\" class=\"wp-caption-text\">A triangle is formed by three geodesics intersecting pairwise.<\/p><\/div>\n<p>In the Poincar\u00e9 disc model of hyperbolic geometry it turns out that the geodesics are segments of diameters of the disc and portions of circles in <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cmathbb+R%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\mathbb R^2' title='\\mathbb R^2' class='latex' \/> which intersect the infinity circle at right angles. If three geodesics intersect in three points, not all lying on the same geodesic, then the three geodesics define a triangle. The image on the right shows our triangle arising from three geodesics. It may seem as though the triangle we&#8217;ve drawn is somewhat special in that two of the sides lie on diameters of the disc. However using hyperbolic isometries (the analogue of euclidean translations, rotations, and reflections) we may move (without changing lengths or angles) <i>any<\/i> hyperbolic triangle so that two of its sides lie on diameters, as we have indicated.<\/p>\n<h3>Proving the hyperbolic Pythagorean theorem<\/h3>\n<p>Here is a sketch of the proof of the hyperbolic Pythagorean theorem. It is an abbreviated version of the proof given by Martin Greenberg in his excellent text <a title=\"Euclidean and non-Euclidean Geometries\" href=\"http:\/\/www.whfreeman.com\/catalog\/Product\/euclideanandnon-euclideangeometries-fourthedition-greenberg\" target=\"_blank\"><i>Euclidean and non-Euclidean Geometries<\/i><\/a>.<\/p>\n<div id=\"attachment_49\" style=\"width: 310px\" class=\"wp-caption alignleft\"><a href=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithVerts.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-49\" class=\"size-medium wp-image-49\" alt=\"The triangle ABC\" src=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithVerts-300x300.jpg\" width=\"300\" height=\"300\" srcset=\"https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithVerts-300x300.jpg 300w, https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithVerts-150x150.jpg 150w, https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithVerts-1024x1024.jpg 1024w, https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/RightTrianglewithVerts.jpg 2000w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><p id=\"caption-attachment-49\" class=\"wp-caption-text\">The triangle ABC<\/p><\/div>\n<p>Let <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ctriangle+ABC&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\triangle ABC' title='\\triangle ABC' class='latex' \/> be a right triangle in the hyperbolic plane with <img src='https:\/\/s0.wp.com\/latex.php?latex=C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C' title='C' class='latex' \/> the right angle. Without loss of generality, we may assume that the vertex <img src='https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A' title='A' class='latex' \/> is the origin and that two of the edges, one of which is the hypotenuse, are portions of diameters, as in our picture. Let <img src='https:\/\/s0.wp.com\/latex.php?latex=d%28AB%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='d(AB)' title='d(AB)' class='latex' \/> be the hyperbolic distance from point <img src='https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A' title='A' class='latex' \/> to point <img src='https:\/\/s0.wp.com\/latex.php?latex=B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B' title='B' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Coverline%7BAB%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\overline{AB}' title='\\overline{AB}' class='latex' \/> the euclidean distance.<\/p>\n<p>We have already defined the hyperbolic cosine function <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ccosh&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\cosh' title='\\cosh' class='latex' \/>. The hyperbolic sine function is defined similarly: <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Csinh%28x%29+%3D+%28e%5Ex+-+e%5E%7B-x%7D%29%2F2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\sinh(x) = (e^x - e^{-x})\/2' title='\\sinh(x) = (e^x - e^{-x})\/2' class='latex' \/> and the hyperbolic tangent function is simply <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ctext%7Btanh+%7D%28x%29+%3D+%5Csinh%28x%29%2F%5Ccosh%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\text{tanh }(x) = \\sinh(x)\/\\cosh(x)' title='\\text{tanh }(x) = \\sinh(x)\/\\cosh(x)' class='latex' \/>.<\/p>\n<p>Using our path length formula, it is straightforward to verify that <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ctext%7Btanh%7D%28d%28AB%29%29+%3D+%5Cfrac%7B2%5Coverline%7BAB%7D%7D%7B1%2B%5Coverline%7BAB%7D%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\text{tanh}(d(AB)) = \\frac{2\\overline{AB}}{1+\\overline{AB}^2}' title='\\text{tanh}(d(AB)) = \\frac{2\\overline{AB}}{1+\\overline{AB}^2}' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ctext%7Btanh%7D%28d%28AC%29%29+%3D+%5Cfrac%7B2%5Coverline%7BAC%7D%7D%7B1%2B%5Coverline%7BAC%7D%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\text{tanh}(d(AC)) = \\frac{2\\overline{AC}}{1+\\overline{AC}^2}' title='\\text{tanh}(d(AC)) = \\frac{2\\overline{AC}}{1+\\overline{AC}^2}' class='latex' \/>.<\/p>\n<p>We begin by showing:<\/p>\n<p><b>Lemma<\/b>: <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Csin+A+%3D+%5Cfrac%7B%5Csinh+a%7D%7B%5Csinh+c%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\sin A = \\frac{\\sinh a}{\\sinh c}' title='\\sin A = \\frac{\\sinh a}{\\sinh c}' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ccos+A+%3D+%5Cfrac%7B%5Ctext%7Btanh+%7D+a%7D%7B%5Ctext%7Btanh+%7D+c%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\cos A = \\frac{\\text{tanh } a}{\\text{tanh } c}' title='\\cos A = \\frac{\\text{tanh } a}{\\text{tanh } c}' class='latex' \/><\/p>\n<p>Once we have those equations, the hyperbolic Pythagorean theorem can be derived from the equality <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Csin%5E2+A+%2B+%5Ccos%5E2+A+%3D+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\sin^2 A + \\cos^2 A = 1' title='\\sin^2 A + \\cos^2 A = 1' class='latex' \/> by applying identities for <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Csinh&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\sinh' title='\\sinh' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ccosh&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\cosh' title='\\cosh' class='latex' \/> analogous to the identities involving <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Csin&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\sin' title='\\sin' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ccos&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\cos' title='\\cos' class='latex' \/>. We leave it as a pleasant challenge to the reader to work out those details.<\/p>\n<div id=\"attachment_65\" style=\"width: 310px\" class=\"wp-caption alignright\"><a href=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean1.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-65\" class=\"size-medium wp-image-65\" alt=\"Extending geodesics\" src=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean1-300x229.jpg\" width=\"300\" height=\"229\" srcset=\"https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean1-300x229.jpg 300w, https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean1-1024x783.jpg 1024w, https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean1.jpg 1511w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><p id=\"caption-attachment-65\" class=\"wp-caption-text\">Extend the geodesics into the plane outside the disc<\/p><\/div>\n<p>We now set about proving the Lemma, by considering the image in <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cmathbb+R%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\mathbb R^2' title='\\mathbb R^2' class='latex' \/> at right. Extend the geodesics making up the sides of <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ctriangle+ABC&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\triangle ABC' title='\\triangle ABC' class='latex' \/> into <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cmathbb+R%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\mathbb R^2' title='\\mathbb R^2' class='latex' \/>. This means that the geodesic <img src='https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' \/> is now part of a circle <img src='https:\/\/s0.wp.com\/latex.php?latex=H&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='H' title='H' class='latex' \/> centered at a point <img src='https:\/\/s0.wp.com\/latex.php?latex=O&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='O' title='O' class='latex' \/>.<\/p>\n<p>The circle <img src='https:\/\/s0.wp.com\/latex.php?latex=H&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='H' title='H' class='latex' \/> intersects the infinity circle in two points. Join those two points by a line (drawn in red in the figure) and let <img src='https:\/\/s0.wp.com\/latex.php?latex=P&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='P' title='P' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=Q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q' title='Q' class='latex' \/> be the points where that line intersects the extensions of the other two edges of <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ctriangle+ABC&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\triangle ABC' title='\\triangle ABC' class='latex' \/>, as in the picture. (The points <img src='https:\/\/s0.wp.com\/latex.php?latex=P&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='P' title='P' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=Q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q' title='Q' class='latex' \/> have a special relationship to the points <img src='https:\/\/s0.wp.com\/latex.php?latex=B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B' title='B' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=Q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q' title='Q' class='latex' \/>: they are the images of <img src='https:\/\/s0.wp.com\/latex.php?latex=P&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='P' title='P' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=Q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q' title='Q' class='latex' \/> under the conversion from the Poincar\u00e9 disc model of the hyperbolic plane, to the Klein model of the hyperbolic plane. But that is a story for a different day.)<\/p>\n<p>The key to the whole business is to apply the definition of cosine to the <i>euclidean<\/i> triangle <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ctriangle+PAQ&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\triangle PAQ' title='\\triangle PAQ' class='latex' \/>. Doing so, we obtain: <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ccos+A+%3D+%5Coverline%7BAQ%7D%2F%5Coverline%7BAP%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\cos A = \\overline{AQ}\/\\overline{AP}' title='\\cos A = \\overline{AQ}\/\\overline{AP}' class='latex' \/>. Converting to hyperbolic distances, we arrive at<\/p>\n<div style=\"border: 2px; padding-left: 5px; padding-top: 5px; padding-right: 5px; padding-bottom: 0px; border-style: ridge; border-color: #6A7F97; width: 45%;\">(*) \u00a0 \u00a0 <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Ccos+A+%3D+%5Ctext%7Btanh+%7Db%2F%5Ctext%7Btanh+%7Dc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\cos A = \\text{tanh }b\/\\text{tanh }c' title='\\cos A = \\text{tanh }b\/\\text{tanh }c' class='latex' \/>.<\/div>\n<p>Let <img src='https:\/\/s0.wp.com\/latex.php?latex=R&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R' title='R' class='latex' \/> be the point, other than <img src='https:\/\/s0.wp.com\/latex.php?latex=B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B' title='B' class='latex' \/>, where the line <img src='https:\/\/s0.wp.com\/latex.php?latex=AB&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='AB' title='AB' class='latex' \/> intersects the circle <img src='https:\/\/s0.wp.com\/latex.php?latex=H&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='H' title='H' class='latex' \/>. The point <img src='https:\/\/s0.wp.com\/latex.php?latex=R&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R' title='R' class='latex' \/> is the result of applying the inversion <img src='https:\/\/s0.wp.com\/latex.php?latex=%28x%2Cy%29+%5Cto+%5Cfrac%7B1%7D%7Bx%5E2+%2B+y%5E2%7D%28x%2Cy%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(x,y) \\to \\frac{1}{x^2 + y^2}(x,y)' title='(x,y) \\to \\frac{1}{x^2 + y^2}(x,y)' class='latex' \/> to the point <img src='https:\/\/s0.wp.com\/latex.php?latex=B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B' title='B' class='latex' \/>. This implies that <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Coverline%7BAR%7D+%3D+1%2F%5Coverline%7BAB%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\overline{AR} = 1\/\\overline{AB}' title='\\overline{AR} = 1\/\\overline{AB}' class='latex' \/>. Hence,<\/p>\n<div style=\"border: 2px; padding-left: 5px; padding-top: 5px; padding-right: 5px; padding-bottom: 0px; border-style: ridge; border-color: #6A7F97; width: 45%;\">(**) \u00a0 \u00a0 \u00a0<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Coverline%7BBR%7D+%3D+%5Coverline%7BAR%7D+-+%5Coverline%7BAB%7D+%3D+2%2F%5Csinh+d%28AB%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\overline{BR} = \\overline{AR} - \\overline{AB} = 2\/\\sinh d(AB)' title='\\overline{BR} = \\overline{AR} - \\overline{AB} = 2\/\\sinh d(AB)' class='latex' \/>.<\/div>\n<p>Recalling that <img src='https:\/\/s0.wp.com\/latex.php?latex=c+%3D+d%28AB%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c = d(AB)' title='c = d(AB)' class='latex' \/>, we have <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Coverline%7BBR%7D+%3D+2%2Fc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\overline{BR} = 2\/c' title='\\overline{BR} = 2\/c' class='latex' \/>. Letting <img src='https:\/\/s0.wp.com\/latex.php?latex=S&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='S' title='S' class='latex' \/> be the other intersection point between the line <img src='https:\/\/s0.wp.com\/latex.php?latex=AC&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='AC' title='AC' class='latex' \/> and the circle <img src='https:\/\/s0.wp.com\/latex.php?latex=H&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='H' title='H' class='latex' \/>, we also have <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Coverline%7BCS%7D+%3D+2%2F%5Csinh+b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\overline{CS} = 2\/\\sinh b' title='\\overline{CS} = 2\/\\sinh b' class='latex' \/>.<\/p>\n<div id=\"attachment_67\" style=\"width: 310px\" class=\"wp-caption alignleft\"><a href=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean2.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-67\" class=\"size-medium wp-image-67\" alt=\"The angle BOT\" src=\"http:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean2-300x232.jpg\" width=\"300\" height=\"232\" srcset=\"https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean2-300x232.jpg 300w, https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean2-1024x793.jpg 1024w, https:\/\/web.colby.edu\/thegeometricviewpoint\/files\/2014\/03\/HyperbolicPythagorean2.jpg 1520w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><p id=\"caption-attachment-67\" class=\"wp-caption-text\">The angle BOT is equal to the angle at B of triangle ABC.<\/p><\/div>\n<p>Finally, as in the figure at left, let <img src='https:\/\/s0.wp.com\/latex.php?latex=T&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T' title='T' class='latex' \/> be the orthogonal projection of <img src='https:\/\/s0.wp.com\/latex.php?latex=O%27&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='O&#039;' title='O&#039;' class='latex' \/>, the center of <img src='https:\/\/s0.wp.com\/latex.php?latex=H&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='H' title='H' class='latex' \/> onto the line <img src='https:\/\/s0.wp.com\/latex.php?latex=AB&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='AB' title='AB' class='latex' \/>. Some rather easy arithmetic, using the fact that the angles of a euclidean triangle sum to <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cpi&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\pi' title='\\pi' class='latex' \/>, shows that the angle <img src='https:\/\/s0.wp.com\/latex.php?latex=BOT&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='BOT' title='BOT' class='latex' \/> is equal to the angle <img src='https:\/\/s0.wp.com\/latex.php?latex=B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B' title='B' class='latex' \/>. Combining this fact with (*) and (**), we conclude that <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Csin+B+%3D+%5Csinh+b%2F+%5Csinh+c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\sin B = \\sinh b\/ \\sinh c' title='\\sin B = \\sinh b\/ \\sinh c' class='latex' \/>. Interchanging the roles of <img src='https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A' title='A' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B' title='B' class='latex' \/> in the preceding argument, concludes the proof of the lemma.<\/p>\n<h3>Final Thoughts<\/h3>\n<p>The euclidean and hyperbolic planes are certainly the most important of the two-dimensional geometries. The third most important geometry is spherical geometry. There is also a version of the Pythagorean theorem for triangles on the sphere. Thurston, in his famous book <i><a title=\"Three-dimensional geometry and topology\" href=\"http:\/\/press.princeton.edu\/titles\/6086.html\" target=\"_blank\">Three-dimensional Geometry and Topology<\/a><\/i>,\u00a0sketches a strategy for giving a combined proof of the law of cosines in the hyperbolic plane and in the sphere. The corresponding Pythagorean theorems follow from that. For the hyperbolic law of cosines, Thurston uses the hyperboloid model of the hyperbolic plane, which gives a unification of the Poincar\u00e9 disc model and the Klein model alluded to earlier.<\/p>\n<p>Finally, many of the beautiful proofs of the Pythagorean theorem make use of dissections of a square and the fact that <img src='https:\/\/s0.wp.com\/latex.php?latex=a%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^2' title='a^2' class='latex' \/> is the area of a square with sides of length <img src='https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' \/>. Is there a dissection proof of the hyperbolic Pythagorean theorem?<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Pythagorean theorem is surely the most famous of all mathematical theorems. The simplicity of its statement (that a right triangle with sides of length and and hypotenuse of length satisfies ) and the multiplicity of beautiful proofs (like the one shown at right from Byrne&#8217;s edition of Euclid&#8217;s Elements) contribute to its memorability. The [&hellip;]<\/p>\n","protected":false},"author":2290,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ngg_post_thumbnail":0,"footnotes":""},"categories":[149567,149568],"tags":[],"_links":{"self":[{"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/posts\/26"}],"collection":[{"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/users\/2290"}],"replies":[{"embeddable":true,"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/comments?post=26"}],"version-history":[{"count":51,"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/posts\/26\/revisions"}],"predecessor-version":[{"id":849,"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/posts\/26\/revisions\/849"}],"wp:attachment":[{"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/media?parent=26"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/categories?post=26"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/web.colby.edu\/thegeometricviewpoint\/wp-json\/wp\/v2\/tags?post=26"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}