In geometry, a pentagon (from pente, which is Greek for the number 5) is any five-sided polygon. A pentagon may be simple or self-intersecting. The sum of the internal angles in a simple pentagon is 540°. A pentagram is an example of a self-intersecting pentagon.
In a regular pentagon, all sides are equal in length and each interior angle is 108°. A regular pentagon has five lines of reflectional symmetry, and rotational symmetry of order 5 (through 72°, 144°, 216° and 288°). Its Schläfli symbol is {5}. The diagonals of a regular pentagon are in golden ratio to its sides.
The area of a regular convex pentagon with side length t is given by
- Failed to parse (Missing texvc executable; please see math/README to configure.): A = \frac{{t^2 \sqrt {25 + 10\sqrt 5 } }}{4} = \frac{5t^2 \tan(54^\circ)}{4} \approx 1.720477401 t^2.
A pentagram or pentangle is a regular star pentagon. Its Schläfli symbol is {5/2}. Its sides form the diagonals of a regular convex pentagon – in this arrangement the sides of the two pentagons are in the golden ratio.
When a regular pentagon is inscribed in a circle with radius R, its edge length t is given by the expression
- Failed to parse (Missing texvc executable; please see math/README to configure.): t = R\ {\sqrt { \frac {5-\sqrt{5}}{2}} } = 2R\sin 36^\circ = 2R\sin\frac{\pi}{5} \approx 1.17557050458 R.
The area of any regular polygon is:
- Failed to parse (Missing texvc executable; please see math/README to configure.): A = \frac{1}{2}Pa
where P is the perimeter of the polygon, a is the apothem. One can then substitute the respective values for P and a, which makes the formula:
- Failed to parse (Missing texvc executable; please see math/README to configure.): A = \frac{1}{2} \times \frac{5t}{1} \times \frac{t\tan(54^\circ)}{2}
with t as the given side length. Then we can then rearrange the formula as:
- Failed to parse (Missing texvc executable; please see math/README to configure.): A = \frac{1}{2} \times \frac{5t^2\tan(54^\circ)}{2}
and then, we combine the two terms to get the final formula, which is:
- Failed to parse (Missing texvc executable; please see math/README to configure.): A = \frac{5t^2\tan(54^\circ)}{4}.
The diagonals of a regular pentagon (hereby represented by D) can be calculated based upon the golden ratio φ and the known side T (see discussion of the pentagon in Golden ratio):
- Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {D}{T} = \varphi = \frac {1+ \sqrt {5} }{2} \ ,
Accordingly:
- Failed to parse (Missing texvc executable; please see math/README to configure.): D = T \times \varphi \ .
If a regular pentagon with successive vertices A, B, C, D, E is inscribed in a circle, and if P is any point on that circle between points B and C, then PA + PD = PB + PC + PE.
A variety of methods are known for constructing a regular pentagon. Some are discussed below.
A regular pentagon is constructible using a compass and straightedge, either by inscribing one in a given circle or constructing one on a given edge. This process was described by Euclid in his Elements circa 300 BC.[1]
One method to construct a regular pentagon in a given circle is described by Richmond[2] and further discussed in Cromwell's "Polyhedra."[3]
The top panel describes the construction used in the animation above to create the side of the inscribed pentagon. The circle defining the pentagon has unit radius. Its center is located at point C and a midpoint M is marked half way along its radius. This point is joined to the periphery vertically above the center at point D. Angle CMD is bisected, and the bisector intersects the vertical axis at point Q. A horizontal line through Q intersects the circle at point P, and chord PD is the required side of the inscribed pentagon.
To determine the length of this side, the two right triangles DCM and QCM are depicted below the circle. Using Pythagoras' theorem and two sides, the hypotenuse of the larger triangle is found as Failed to parse (Missing texvc executable; please see math/README to configure.): \scriptstyle \sqrt{5}/2 . Side h of the smaller triangle then is found using the half-angle formula:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \tan ( \phi/2) = \frac{1-\cos(\phi)}{\sin (\phi)} \ ,
where cosine and sine of ϕ are known from the larger triangle. The result is:
- Failed to parse (Missing texvc executable; please see math/README to configure.): h = \frac{\sqrt 5 - 1}{4} \ .
With this side known, attention turns to the lower diagram to find the side s of the regular pentagon. First, side a of the right-hand triangle is found using Pythagoras' theorem again:
- Failed to parse (Missing texvc executable; please see math/README to configure.): a^2 = 1-h^2 \ ; \ a = \frac{1}{2}\sqrt { \frac {5+\sqrt 5}{2}} \ .
Then s is found using Pythagoras' theorem and the left-hand triangle as:
- Failed to parse (Missing texvc executable; please see math/README to configure.): s^2 = (1-h)^2 + a^2 = (1-h)^2 + 1-h^2 = 1-2h+h^2 + 1-h^2 = 2-2h=2-2\left(\frac{\sqrt 5 - 1}{4}\right) \
-
- Failed to parse (Missing texvc executable; please see math/README to configure.): =\frac {5-\sqrt 5}{2} \ .
The side s is therefore:
- Failed to parse (Missing texvc executable; please see math/README to configure.): s = \sqrt{ \frac {5-\sqrt 5}{2}} \ ,
a well established result.[4] Consequently, this construction of the pentagon is valid.
An alternative method is this:
- Draw a circle in which to inscribe the pentagon and mark the center point O. (This is the green circle in the diagram to the right).
- Choose a point A on the circle that will serve as one vertex of the pentagon. Draw a line through O and A.
- Construct a line perpendicular to the line OA passing through O. Mark its intersection with one side of the circle as the point B.
- Construct the point C as the midpoint of O and B.
- Draw a circle centered at C through the point A. Mark its intersection with the line OB (inside the original circle) as the point D.
- Draw a circle centered at A through the point D. Mark its intersections with the original (green) circle as the points E and F.
- Draw a circle centered at E through the point A. Mark its other intersection with the original circle as the point G.
- Draw a circle centered at F through the point A. Mark its other intersection with the original circle as the point H.
- Construct the regular pentagon AEGHF.
Animation that is almost the same as this alternative method
Method using Carlyle circles
The Carlyle circle was invented as a geometric method to find the roots of a quadratic equation.[5] This methodology leads to a procedure for constructing a regular pentagon. The steps are as follows:[6]
- Draw a circle in which to inscribe the pentagon and mark the center point O.
- Draw a horizontal line through the center of the circle. Mark one intersection with the circle as point B.
- Construct a vertical line through the center. Mark one intersection with the circle as point A.
- Construct the point M as the midpoint of O and B.
- Draw a circle centered at M through the point A. Mark its intersection with the horizontal line (inside the original circle) as the point W and its intersection outside the circle as the point V.
- Draw a circle of radius OA and center W. It intersects the original circle at two of the vertices of the pentagon.
- Draw a circle of radius OA and center V. It intersects the original circle at two of the vertices of the pentagon.
- The fifth vertex is the intersection of the horizontal axis with the original circle.
A direct method using degrees follows:
- Draw a circle and choose a point to be the pentagon's (e.g. top center)
- Choose a point A on the circle that will serve as one vertex of the pentagon. Draw a line through O and A.
- Draw a guideline through it and the circle's center
- Draw lines at 54° (from the guideline) intersecting the pentagon's point
- Where those intersect the circle, draw lines at 18° (from parallels to the guideline)
- Join where they intersect the circle
After forming a regular convex pentagon, if one joins the non-adjacent corners (drawing the diagonals of the pentagon), one obtains a pentagram, with a smaller regular pentagon in the center. Or if one extends the sides until the non-adjacent sides meet, one obtains a larger pentagram. The accuracy of this method depends on the accuracy of the protractor used to measure the angles.
Overhand knot of a paper strip
- A regular pentagon may be created from just a strip of paper by tying an overhand knot into the strip and carefully flattening the knot by pulling the ends of the paper strip. Folding one of the ends back over the pentagon will reveal a pentagram when backlit.
- Construct a regular hexagon on stiff paper or card. Crease along the three diameters between opposite vertices. Cut from one vertex to the center to make an equilateral triangular flap. Fix this flap underneath its neighbor to make a pentagonal pyramid. The base of the pyramid is a regular pentagon.
A cyclic pentagon is one for which a circle called the circumcircle goes through all five vertices. The regular pentagon is an example of a cyclic pentagon. The area of a cyclic pentagon, whether regular or not, can be expressed as one fourth the square root of one of the roots of a septic equation whose coefficients are functions of the sides of the pentagon.[7][8][9]
There exist cyclic pentagons with rational sides and rational area; these are called Robbins pentagons. In a Robbins pentagon, either all diagonals are rational or all are irrational, and it is conjectured that all the diagonals must be rational.[10]
The K5 complete graph is often drawn as a regular pentagon with all 10 edges connected. This graph also represents an orthographic projection of the 5 vertices and 10 edges of the 5-cell. The rectified 5-cell, with vertices at the mid-edges of the 5-cell is projected inside a pentagon.
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Pentagonal cross-section of okra.
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Starfruit is another fruit with fivefold symmetry.
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An illustration of brittle stars, also echinoderms with a pentagonal shape.
The best known packing of equal-sized regular pentagons on a plane is a dual lattice structure which covers 92.131% of the plane.
A pentagon cannot appear in any tiling made by regular polygons. To prove a pentagon cannot form a regular tiling (one in which all faces are congruent), observe that 360 / 108 = 31⁄3, which is not a whole number. More difficult is proving a pentagon cannot be in any edge-to-edge tiling made by regular polygons:
There are no combinations of regular polygons with 4 or more meeting at a vertex that contain a pentagon. For combinations with 3, if 3 polygons meet at a vertex and one has an odd number of sides, the other 2 must be congruent. The reason for this is that the polygons that touch the edges of the pentagon must alternate around the pentagon, which is impossible because of the pentagon's odd number of sides. For the pentagon, this results in a polygon whose angles are all (360 − 108) / 2 = 126°. To find the number of sides this polygon has, the result is 360 / (180 − 126) = 62⁄3, which is not a whole number. Therefore, a pentagon cannot appear in any tiling made by regular polygons.
- ^ George Edward Martin (1998). Geometric constructions. Springer. p. 6. ISBN 0-387-98276-0. http://books.google.com/books?id=ABLtD3IE_RQC&pg=PA6.
- ^ Herbert W Richmond (1893). "Pentagon". http://mathworld.wolfram.com/Pentagon.html.
- ^ Peter R. Cromwell. Polyhedra. p. 63. ISBN 0-521-66405-5. http://books.google.com/books?id=OJowej1QWpoC&pg=PA63.
- ^ This result agrees with Herbert Edwin Hawkes, William Arthur Luby, Frank Charles Touton (1920). "Exercise 175". Plane geometry. Ginn & Co.. p. 302. http://books.google.com/books?id=eOdHAAAAIAAJ&pg=PA302.
- ^ Eric W. Weisstein (2003). CRC concise encyclopedia of mathematics (2nd ed.). CRC Press. p. 329. ISBN 1-58488-347-2. http://books.google.com/books?id=Zg1_QZsylysC&pg=PA329.
- ^ Duane W DeTemple (1991). "Carlyle Circles and the Lemoine Simplicity of Polygon Constructions". The American Mathematical Monthly 98 (2): 97–108. http://apollonius.math.nthu.edu.tw/d1/ne01/jyt/linkjstor/regular/1.pdf. JSTOR link
- ^ Weisstein, Eric W. "Cyclic Pentagon." From MathWorld--A Wolfram Web Resource. [1]
- ^ Robbins, D. P. "Areas of Polygons Inscribed in a Circle." Discr. Comput. Geom. 12, 223-236, 1994.
- ^ Robbins, D. P. "Areas of Polygons Inscribed in a Circle." Amer. Math. Monthly 102, 523-530, 1995.
- ^ *Buchholz, Ralph H.; MacDougall, James A. (2008), "Cyclic polygons with rational sides and area", Journal of Number Theory 128 (1): 17–48, DOI:10.1016/j.jnt.2007.05.005, MR 2382768, http://docserver.carma.newcastle.edu.au/785/ .
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