{"id":1264,"date":"2015-02-01T01:00:19","date_gmt":"2015-02-01T01:00:19","guid":{"rendered":"http:\/\/blogs.ams.org\/visualinsight\/?p=1264"},"modified":"2015-07-29T00:47:44","modified_gmt":"2015-07-29T00:47:44","slug":"pentagon-decagon-packing","status":"publish","type":"post","link":"https:\/\/blogs.ams.org\/visualinsight\/2015\/02\/01\/pentagon-decagon-packing\/","title":{"rendered":"Pentagon-Decagon Packing"},"content":{"rendered":"
\n
\"Pentagon-Decagon<\/a>

Pentagon-Decagon Packing – Greg Egan<\/p><\/div>\n<\/div>\n

Two regular pentagons and a regular decagon meet snugly at a vertex: their interior angles sum to 360°. However, they can’t tile the plane. However, they come fairly close, as shown in this picture by Greg Egan<\/a>. <\/p>\n

Say we have three regular polygons with \\(p, q,\\) and \\(r\\) sides, respectively. Their interior angles will sum to 360° if and only if<\/p>\n

$$ \\frac{1}{p} + \\frac{1}{q} + \\frac{1}{r} = \\frac{1}{2} $$<\/p>\n

So, two regular pentagons and a regular decagon can meet snugly at a vertex because<\/p>\n

$$ \\frac{1}{5} + \\frac{1}{5} + \\frac{1}{10} = \\frac{1}{2} $$<\/p>\n

In fact there are 10 solutions of the equation <\/p>\n

$$ \\frac{1}{p} + \\frac{1}{q} + \\frac{1}{r} = \\frac{1}{2} $$<\/p>\n

for natural numbers \\(p, q, \\) and \\(r\\), not counting the order. Of these, only 4 give tilings of the plane by regular polygons. The other 6 are ‘forbidden’.<\/p>\n

Here are all 10 solutions:<\/p>\n

    \n
  1. \n$$ \\frac{1}{6} + \\frac{1}{6} + \\frac{1}{6} = \\frac{1}{2} $$
    \ngives the regular     hexagonal tiling<\/a>.\n<\/li>\n

  2. \n$$ \\frac{1}{5} + \\frac{1}{5} + \\frac{1}{10} = \\frac{1}{2} $$
    \ndoes not give a tiling by regular polygons, but it gives the packing shown above, and three more described by     Robert Fathauer<\/a>.<\/p>\n
    \n\"Regular<\/a>\n<\/div>\n<\/li>\n

  3. \n$$ \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{8} = \\frac{1}{2} $$
    \ngives the semiregular     truncated square tiling<\/a>.\n<\/li>\n

  4. \n$$ \\frac{1}{4} + \\frac{1}{6} + \\frac{1}{12} = \\frac{1}{2} $$
    \ngives the semiregular     truncated trihexagonal tiling<\/a>.\n<\/li>\n

  5. \n$$ \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{20} = \\frac{1}{2} $$
    \ndoes not give a tiling by regular polygons, but it gives a packing described by     Robert Fathauer<\/a>.<\/p>\n
    \n\"Regular<\/a>\n<\/div>\n<\/li>\n

  6. \n$$ \\frac{1}{3} + \\frac{1}{12} + \\frac{1}{12} = \\frac{1}{2} $$
    \ngives the semiregular     truncated hexagonal tiling<\/a>.\n<\/li>\n

  7. \n$$ \\frac{1}{3} + \\frac{1}{10} + \\frac{1}{15} = \\frac{1}{2} $$
    \ndoes not give a tiling by regular polygons, but it gives two packings described by     Robert Fathauer<\/a> and Kevin Jardine<\/a>.<\/p>\n
    \n\"Regular<\/a><\/div>\n<\/li>\n

  8. \n$$ \\frac{1}{3} + \\frac{1}{9} + \\frac{1}{18} = \\frac{1}{2} $$
    \ndoes not give a tiling by regular polygons, but it gives a packing described by     Robert Fathauer<\/a>.<\/p>\n
    \n\"Regular<\/a>\n<\/div>\n<\/li>\n

  9. \n$$ \\frac{1}{3} + \\frac{1}{8} + \\frac{1}{24} = \\frac{1}{2} $$
    \ndoes not give a tiling by regular polygons, but it gives two packings discovered by     Robert Fathauer<\/a> and Kevin Jardine<\/a>.<\/p>\n
    \n\"Regular<\/a>\n<\/div>\n<\/li>\n

  10. \n$$ \\frac{1}{3} + \\frac{1}{7} + \\frac{1}{42} = \\frac{1}{2} $$
    \ndoes not give a tiling by regular polygons, but it gives a packing discovered by     Robert Fathauer<\/a>.<\/p>\n
    \n\"Regular<\/a>\n<\/div>\n<\/li>\n<\/ol>\n

    Because 42 is the largest number appearing in the above list of solutions, this number—or twice this number—appears in Hurwitz’s automorphism theorem<\/a>. This says that the number of orientation-preserving conformal transformations of a compact Riemann surface of genus \\(g > 1\\) is at most \\(84(g \u2212 1)\\). For more on this, see:<\/p>\n

    • John Baez, 42<\/a>.<\/p>\n

    For more on the 6 ‘forbidden tilings’—the 6 solutions on the list above that do not give tilings of the plane by regular polygons—see this set of webpages:<\/p>\n

    • Kevin Jardine, Imperfect congruence: Kepler, Dürer and the mystery of the forbidden tilings<\/a>.<\/p>\n

    Also see:<\/p>\n

    • Tiling by regular polygons<\/a>, Wikipedia.<\/p>\n

    The above pictures of the 6 forbidden tilings were created by dllu<\/a> and placed on Wikimedia Commons<\/a> under a Creative Commons CC0 1.0 Universal Public Domain Dedication<\/a> copyright.<\/p>\n

    \n

    Visual Insight<\/i> is a place to share striking images that help explain advanced topics in mathematics. I\u2019m always looking for truly beautiful images, so if you know about one, please drop a comment here<\/a> and let me know!<\/p>\n

    <\/div>","protected":false},"excerpt":{"rendered":"

    Two regular pentagons and a regular decagon meet snugly at a vertex: their interior angles sum to 360°. However, they can’t tile the plane. However, they come fairly close, as shown in this picture by Greg Egan<\/a>. <\/p>\n

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