The most common statement known as Steiner's theorem (Casey 1893, p. 329) states that the Pascal lines of the hexagons 123456, 143652, and 163254 formed by interchanging the vertices at positions 2, 4, and 6 are concurrent (where the numbers denote the order in which the vertices of the hexagon are taken). The 20 points of concurrence so generated are known as Steiner points.Another theorem due to Steiner lets lines and join a variable point on a conic section to two fixed points on the same conic section. Then and are projectively related.A third "Steiner's theorem" states that if two opposite edges of a tetrahedron move on two fixed skew lines in any way whatsoever but remain fixed in length, then the volume of the tetrahedron remains constant (Altshiller-Court 1979, p. 87)...
The Lemoine hexagon is a cyclic hexagon with vertices given by the six concyclic intersections of the parallels of a reference triangle through its symmedian point . The circumcircle of the Lemoine hexagon is therefore the first Lemoine circle. There are two definitions of the hexagon that differ based on the order in which the vertices are connected.The first definition is the closed self-intersecting hexagon in which alternate sides , , and pass through the symmedian point (left figure). The second definition (Casey 1888, p. 180) is the hexagon formed by the convex hull of the first definition, i.e., the hexagon (right figure).The sides of this hexagon have the property that, in addition to , , and , the remaining sides , , and are antiparallel to , , and , respectively.For the self-intersecting Lemoine hexagon, the perimeter and area are(1)(2)and for the simple hexagon, they are given by(3)(4)(Casey 1888, p. 188), where is the..
Let the opposite sides of a convex cyclic hexagon be , , , , , and , and let the polygon diagonals , , and be so chosen that , , and have no common polygon vertex (and likewise for , , and ), thenThis is an extension of Ptolemy's theorem tothe hexagon.
The dual of Brianchon's theorem (Casey 1888, p. 146), discovered by B. Pascal in 1640 when he was just 16 years old (Leibniz 1640; Wells 1986, p. 69). It states that, given a (not necessarily regular, or even convex) hexagon inscribed in a conic section, the three pairs of the continuations of opposite sides meet on a straight line, called the Pascal line.In 1847, Möbius (1885) published the following generalization of Pascal's theorem: if all intersection points (except possibly one) of the lines prolonging two opposite sides of a -gon inscribed in a conic section are collinear, then the same is true for the remaining point.
In the arbelos, consider the semicircles and with centers and passing through . The Apollonius circle of , and the large semicircle of the arbelos is an Archimedean circle . This circle has radius(as it must), and centerThe line perpendicular to and passing through the center of is called the Schoch line.Now let and be two semicircles through with radii proportional to and respectively. The circle tangent to and with its center on the Schoch line is an Archimedean circle. These circles are called Woo circles.Let be the radical axis of the great semicircle of the arbelos and . From a point on consider the tangents to the circle on diameter . The circle with center on the Schoch line and tangent to these tangents is a Woo circle (Okumura and Watanabe 2004).An applet for investigating Woo circles and Schoch lines has been prepared by Schoch (2005)...
The circle through the cusp of the arbelos and the tangent points of the first Pappus circle, which is congruent to the two Archimedes' circles. If and , then the radius of the Bankoff circle is
Draw the perpendicular line from the intersection of the two small semicircles in the arbelos. The two circles and tangent to this line, the large semicircle, and each of the two semicircles are then congruent and known as Archimedes' circles.For an arbelos with outer semicircle of unit radius and parameter , Archimedes' circles have radii(1)and centers(2)(3)Circles that are constructed in a natural way using an arbelos and are congruent to Archimedes' circles are known as Archimedean circles.
An Archimedean circle is a circle defined in the arbelos in a natural way and congruent to Archimedes' circles, i.e., having radiusfor an arbelos with outer semicircle of unit radius and parameter .
The term "arbelos" means shoemaker's knife in Greek, and this term is applied to the shaded area in the above figure which resembles the blade of a knife used by ancient cobblers (Gardner 1979). Archimedes himself is believed to have been the first mathematician to study the mathematical properties of this figure. The position of the central notch is arbitrary and can be located anywhere along the diameter.The arbelos satisfies a number of unexpected identities (Gardner 1979, Schoch). 1. Call the diameters of the left and right semicircles and , respectively, so the diameter of the enclosing semicircle is 1. Then the arc length along the bottom of the arbelos is(1)so the arc length along the enclosing semicircle is the same as the arc length along the two smaller semicircles. 2. Draw the perpendicular from the tangent of the two semicircles to the edge of the large circle. Then the area of the arbelos is the same as the area of the circle with..
Let a convex polygon be inscribed in a circle and divided into triangles from diagonals from one polygon vertex. The sum of the radii of the circles inscribed in these triangles is the same independent of the polygon vertex chosen (Johnson 1929, p. 193).If a triangle is inscribed in a circle, another circle inside the triangle, a square inside the circle, another circle inside the square, and so on. Then the equation relating the inradius and circumradius of a regular polygon,(1)gives the ratio of the radii of the final to initial circles as(2)Numerically,(3)(OEIS A085365), where is the corresponding constant for polygon circumscribing. This constant is termed the Kepler-Bouwkamp constant by Finch (2003). Kasner and Newman's (1989) assertion that is incorrect, as is the value of 0.8700... given by Prudnikov et al. (1986, p. 757)...
Circumscribe a triangle about a circle, another circle around the triangle, a square outside the circle, another circle outside the square, and so on. The circumradius and inradius for an -gon are then related by(1)so an infinitely nested set of circumscribed polygons and circles has(2)(3)(4)Kasner and Newman (1989) and Haber (1964) state that , but this is incorrect, and the actual answer is(5)(OEIS A051762).By writing(6)it is possible to expand the series about infinity, change the order of summation, do the sum symbolically, and obtain the quickly converging series(7)where is the Riemann zeta function.Bouwkamp (1965) produced the following infinite productformulas for the constant,(8)(9)(10)where is the sinc function (cf. Prudnikov et al. 1986, p. 757), is the Riemann zeta function, and is the Dirichlet lambda function. Bouwkamp (1965) also produced the formula with accelerated convergence(11)where(12)(cited in Pickover..
A triangle is a 3-sided polygon sometimes (but not very commonly) called the trigon. Every triangle has three sides and three angles, some of which may be the same. The sides of a triangle are given special names in the case of a right triangle, with the side opposite the right angle being termed the hypotenuse and the other two sides being known as the legs. All triangles are convex and bicentric. That portion of the plane enclosed by the triangle is called the triangle interior, while the remainder is the exterior.The study of triangles is sometimes known as triangle geometry, and is a rich area of geometry filled with beautiful results and unexpected connections. In 1816, while studying the Brocard points of a triangle, Crelle exclaimed, "It is indeed wonderful that so simple a figure as the triangle is so inexhaustible in properties. How many as yet unknown properties of other figures may there not be?" (Wells 1991, p. 21).It is..
A Heronian triangle is a triangle having rational side lengths and rational area. The triangles are so named because such triangles are related to Heron's formula(1)giving a triangle area in terms of its side lengths , , and semiperimeter . Finding a Heronian triangle is therefore equivalent to solving the Diophantine equation(2)The complete set of solutions for integer Heronian triangles (the three side lengths and area can be multiplied by their least common multiple to make them all integers) were found by Euler (Buchholz 1992; Dickson 2005, p. 193), and parametric versions were given by Brahmagupta and Carmichael (1952) as(3)(4)(5)(6)(7)This produces one member of each similarity class of Heronian triangles for any integers , , and such that , , and (Buchholz 1992).The first few integer Heronian triangles sorted by increasing maximal side lengths, are ((3, 4, 5), (5, 5, 6), (5, 5, 8), (6, 8, 10), (10, 10, 12), (5, 12, 13), (10, 13,..
A tiling consisting of a rhombus such that 17 rhombuses fit around a point and a second tile in the shape of six rhombuses stuck together. These two tiles can fill the plane in exactly four different ways. Two tiles which tile the plane in ways can be constructed using a rhombus of a shape such that pack around a point together with a complex piece made by sticking rhombuses together (Wells 1991).
The Penrose tiles are a pair of shapes that tile the plane only aperiodically (when the markings are constrained to match at borders). These two tiles, illustrated above, are called the "kite" and "dart," respectively. In strict Penrose tiling, the tiles must be placed in such a way that the colored markings agree; in particular, the two tiles may not be combined into a rhombus (Hurd).Two additional types of Penrose tiles known as the rhombs (of which there are two varieties: fat and skinny) and the pentacles (or which there are six type) are sometimes also defined that have slightly more complicated matching conditions (McClure 2002).In 1997, Penrose sued the Kimberly Clark Corporation over their quilted toilet paper, which allegedly resembles a Penrose aperiodic tiling (Mirsky 1997). The suit was apparently settled out of court.To see how the plane may be tiled aperiodically using the kite and dart, divide the kite into..
The term diamond is another word for a rhombus. The term is also used to denote a square tilted at a angle.The diamond shape is a special case of the superellipse with parameter , giving it implicit Cartesian equation(1)Since the diamond is a rhombus with diagonals and , it has inradius(2)(3)Writing as an algebraic curve gives the quartic curve(4)which is a diamond curve with the diamond edges extended to infinity.When considered as a polyomino, the diamond of order can be considered as the set of squares whose centers satisfy the inequality . There are then squares in the order- diamond, which is precisely the centered square number of order . For , 2, ..., the first few values are 1, 5, 13, 25, 41, 61, 85, 113, 145, ... (OEIS A001844).The diamond is also the name given to the unique 2-polyiamond...
A rounded rectangle is the shape obtained by taking the convex hull of four equal circles of radius and placing their centers at the four corners of a rectangle with side lengths and .A filled rounded rectangle with (or ) is called a stadium.The rounded rectangle has perimeter(1)A filled rounded rectangle has area(2)For a rounded square with (center) length and width , the corner radius can be determined by measuring the generalized diameter from the edge of one rounded corner to the diagonally opposite corner. From the Pythagorean theorem, the diagonal distance from the rounded corner to the corner of the circumscribed square is(3)and the corner radius is related to the edge length of the square circumscribing the corner circle by(4)Combining gives(5)(6)
The salinon is the figure illustrated above formed from four connected semicircles. The word salinon is Greek for "salt cellar," which the figure resembles. If the radius of the large enclosing circle is and the radius of the small central circle is , then the radii of the two small side circles are .In his Book of Lemmas, Archimedes proved that the salinon has an area equal to the circle having the line segment joining the top and bottom points as its diameter (Wells 1991), namely
A lune is a plane figure bounded by two circular arcs of unequal radii, i.e., a crescent. (By contrast, a plane figure bounded by two circular arcs of equal radius is known as a lens.) For circles of radius and whose centers are separated by a distance , the area of the lune is given by(1)where(2)is the area of the triangle with side lengths , , and . The second of these can be obtained directly by subtracting the areas of the two half-lenses whose difference producing the colored region above.In each of the figures above, the area of the lune is equal to the area of the indicated triangle. Hippocrates of Chios squared the above left lune (Dunham 1990, pp. 19-20; Wells 1991, pp. 143-144), as well as two others, in the fifth century BC. Two more squarable lunes were found by T. Clausen in the 19th century (Shenitzer and Steprans 1994; Dunham 1990 attributes these discoveries to Euler in 1771). In the 20th century, N. G. Tschebatorew..
The term "vesica piscis," meaning "fish bladder" in Latin, is used for the particular symmetric lens formed by the intersection of two equal circles whose centers are offset by a distance equal to the circle radii (Pedoe 1995, p. xii). The height of the lens is given by letting in the equation for a circle-circle intersection(1)giving(2)The vesica piscis therefore has two equilateral triangles inscribed in it as illustrated above.The area of the vesica piscis is given by plugging into the circle-circle intersection area equation with ,(3)giving(4)(5)(OEIS A093731). Since each arcof the lens is precisely 1/3 of a circle, perimeter is given by(6)Renaissance artists frequently surrounded images of Jesus with the vesica piscis (Pedoe 1995, p. xii; Rawles 1997).
The perpendicular distance from an arc's midpoint to the chord across it, equal to the radius minus the apothem ,(1)For a regular polygon of side length ,(2)(3)(4)(5)(6)where is the circumradius, the inradius, is the side length, and is the number of sides.
A (general, asymmetric) lens is a lamina formed by the intersection of two offset disks of unequal radii such that the intersection is not empty, one disk does not completely enclose the other, and the centers of curvatures are on opposite sides of the lens. If the centers of curvature are on the same side, a lune results.The area of a general asymmetric lens obtained from circles of radii and and offset can be found from the formula for circle-circle intersection, namely(1)(2)Similarly, the height of such a lens is(3)(4)A symmetric lens is lens formed by the intersection of two equal disk. The area of a symmetric lens obtained from circles with radii and offset is given by(5)and the height by(6)A special type of symmetric lens is the vesica piscis (Latin for "fish bladder"), corresponding to a disk offset which is equal to the disk radii.A lens-shaped region also arises in the study of Bessel functions, is very important in the theory of..
The areas of the regions illustrated above can be found from the equations(1)(2)Since we want to solve for three variables, we need a third equation. This can be taken as(3)where(4)(5)leading to(6)Combining the equations (1), (2), and (6) gives the matrix equation(7)which can be inverted to yield(8)(9)(10)
A portion of a disk whose upper boundary is a (circular) arc and whose lower boundary is a chord making a central angle radians (), illustrated above as the shaded region. The entire wedge-shaped area is known as a circular sector.Circular segments are implemented in the Wolfram Language as DiskSegment[x, y, r, q1, q2]. Elliptical segments are similarly implemented as DiskSegment[x, y, r1, r2, q1, q2].Let be the radius of the circle, the chord length, the arc length, the height of the arced portion, and the height of the triangular portion. Then the radius is(1)the arc length is(2)the height is(3)(4)(5)and the length of the chord is(6)(7)(8)(9)From elementary trigonometry, the angle obeys the relationships(10)(11)(12)(13)The area of the (shaded) segment is then simply given by the area of the circular sector (the entire wedge-shaped portion) minus the area of the bottom triangular portion,(14)Plugging in gives(15)(16)(17)(18)where..
Half a circle. The area of a semicircle of radius is given by(1)(2)(3)The weighted mean of is(4)(5)The semicircle is the cross section of a hemispherefor any plane through the z-axis.The perimeter of the curved boundary is given by(6)With , this gives(7)The perimeter of the semicircular lamina is then(8)The weighted value of of the semicircular curve is given by(9)(10)(11)so the geometric centroid is(12)The geometric centroid of the semicircularlamina is given by(13)(Kern and Bland 1948, p. 113).
A circular sector is a wedge obtained by taking a portion of a disk with central angle radians (), illustrated above as the shaded region. A sector with central angle of radians would correspond to a filled semicircle. Let be the radius of the circle, the chord length, the arc length, the sagitta (height of the arced portion), and the apothem (height of the triangular portion). Then(1)(2)(3)(4)(5)(6)(7)(8)(9)The angle obeys the relationships(10)(11)(12)(13)The area of the sector is(14)(15)(Beyer 1987). The area can also be found by direct integration as(16)It follows that the weighted mean of the is(17)(18)so the geometric centroid of the circular sectoris(19)(20)(21)(Gearhart and Schulz 1990). Checking shows that this obeys the proper limits for a semicircle () and for an isosceles triangle ()...
There are a number of meanings for the word "arc" in mathematics. In general, an arc is any smooth curve joining two points. The length of an arc is known as its arc length.In a graph, a graph arc isan ordered pair of adjacent vertices.In particular, an arc is any portion (other than the entire curve) of the circumference of a circle. An arc corresponding to the central angle is denoted . Similarly, the size of the central angle subtended by this arc (i.e., the measure of the arc) is sometimes (e.g., Rhoad et al. 1984, p. 421) but not always (e.g., Jurgensen 1963) denoted .The center of an arc is the center of the circle of whichthe arc is a part.An arc whose endpoints lie on a diameter of a circleis called a semicircle.For a circle of radius , the arc length subtended by a central angle is proportional to , and if is measured in radians, then the constant of proportionality is 1, i.e.,(1)The length of the chord connecting the arc's endpointsis(2)As..
An orthodiagonal quadrangle is a quadrangle whose diagonals are perpendicular to each other. If , , , and are the sides of a quadrangle, then this quadrangle is orthodiagonal iff .If is a cyclic orthodiagonal quadrangle, then the quadrangle formed by the tangents to the circumcircle through the vertices of form a bicentric quadrilateral . The circumcenters of and and the point of intersection of the diagonals of are collinear.
If , , and are three points on one line, , , and are three points on another line, and meets at , meets at , and meets at , then the three points , , and are collinear. Pappus's hexagon theorem is self-dual.The incidence graph of the configuration corresponding to the theorem is the Pappus graph.
The regular tessellation consisting of regular hexagons (i.e., a hexagonal grid).In general, the term honeycomb is used to refer to a tessellation in dimensions for . The only regular honeycomb in three dimensions is , which consists of eight cubes meeting at each polyhedron vertex. The only quasiregular honeycomb (with regular cells and semiregular vertex figures) has each polyhedron vertex surrounded by eight tetrahedra and six octahedra and is denoted .Ball and Coxeter (1987) use the term "sponge" for a solid that can be parameterized by integers , , and that satisfy the equationThe possible sponges are , , , , and .There are many semiregular honeycombs, such as , in which each polyhedron vertex consists of two octahedra and four cuboctahedra .
Given a triangle and the excentral triangle , define the -vertex of the hexyl triangle as the point in which the perpendicular to through the excenter meets the perpendicular to through the excenter , and similarly define and . Then is known as the hexyl triangle of , and forms a hexagon with parallel sides (Kimberling 1998 pp. 79 and 172).The hexyl triangle has trilinear vertex matrix(1)where , , and (Kimberling 1998, p. 172).It has side lengths(2)(3)(4)and area(5)(6)(7)where is the area of the reference triangle, is the circumradius, and is the inradius. It therefore has the same side lengths and area as the excentral triangle.The Cevians triangles with Cevian points corresponding to Kimberling centers with , 20, 21, 27, 63, and 84 are perspective to the hexyl triangle. That anticevian triangles and antipedal triangles corresponding to Kimberling centers for , 9, 19, 40, 57, 63, 84, 610, 1712, and 2184 are also perspective to the..
Take any triangle with polygon vertices , , and . Pick a point on the side opposite , and draw a line parallel to . Upon reaching the side at , draw the line parallel to . Continue (left figure). Then the line closes for any triangle. If is the midpoint of , then (right figure).Let be the ratio in which the sides of the reference triangle are divided i.e., , and define . Then the coordinates of the vertices of the figure are shown above.The six vertexes of Thomsen's figure lie on an ellipse having the triangle centroid as its center. The area of this ellipse iswhere is the area of the reference triangle. When (or ), the ellipse becomes the Steiner circumellipse, and when , it becomes the Steiner inellipse (M. Tarquini, pers. comm., Sep. 2, 2005).Thomsen's figure is similar to a Tucker hexagon. While Thomsen's hexagon closes after six parallels, a Tucker hexagon closes after alternately three parallels and three antiparallels...
Marion's theorem (Mathematics Teacher 1993, Maushard 1994, Morgan 1994) states that the area of the central hexagonal region determined by trisection of each side of a triangle and connecting the corresponding points with the opposite vertex is given by 1/10 the area of the original triangle.This can easily be shown using trilinear coordinates. In the above diagram, , , and, from the multisection formula, the trisection points have trilinear coordinates(1)(2)(3)(4)(5)(6)The other labeled points can then be computed as(7)(8)(9)(10)(11)(12)(13)(14)(15)(16)(17)(18)Using the trilinear equation for the area of a triangle then gives the following areas of the colored triangles illustrated above in terms of the area of the original triangle.(19)(20)(21)(22)Taking the remaining red portion then gives(23)(24)as originally stated.A generalization of Marion's theorem sometimes known as Morgan's theorem was found by Ryan Morgan,..
A hexagon tiling is a tiling of the planeby identical hexagons.The regular hexagon forms a regular tessellation,also called a hexagonal grid, illustrated above.There are at least three tilings of irregular hexagons,illustrated above.They are given by the following types:(1)(Gardner 1988). Note that the periodic hexagonal tessellationis a degenerate case of all three tilings with(2)and(3)Amazingly, the number of plane partitions contained in an box also gives the number of hexagon tilings by rhombi for a hexagon of side lengths , , , , , (David and Tomei 1989, Fulmek and Krattenthaler 2000). The asymptotic distribution of rhombi in a random hexagon tiling by rhombi was given by Cohn et al. (1998). A variety of enumerations for various explicit positions of rhombi are given by Fulmek and Krattenthaler (1998, 2000)...
After a half rotation of the coin on the left around the central coin (of the same radius), the coin undergoes a complete rotation. In other words, a coin makes two complete rotations when rolled around the boundary of an identical coin. This fact is readily apparent in the generation of the cardioid as one disk rolling on another.
In a given circle, find an isosceles triangle whose legs pass through two given points inside the circle. This can be restated as: from two points in the plane of a circle, draw lines meeting at the point of the circumference and making equal angles with the normal vector at that point.The problem is called the billiard problem because it corresponds to finding the point on the edge of a circular "billiard" table at which a cue ball at a given point must be aimed in order to carom once off the edge of the table and strike another ball at a second given point.The problem is equivalent to the determination of the point on a spherical mirror where a ray of light will reflect in order to pass from a given source to an observer. It is also equivalent to the problem of finding, given two points and a circle such that the points are both inside or outside the circle, the ellipse whose foci are the two points and which is tangent to the given circle.The problem was..
The lines joining the vertices , , and of a given triangle with the circumcenters of the triangles , , and (where is the circumcenter of ), respectively, are concurrent. Their point of concurrence is known as the Kosnita point.
A 16-sided polygon, sometimes also called a hexakaidecagon. The regular hexadecagon is a constructible polygon, and the inradius , circumradius , and area of the regular hexadecagon of side length 1 are(1)(2)(3)
Consider the plane figure obtained by drawing each diagonal in a regular polygon with vertices. If each point of intersection is associated with a node and diagonals are split ar each intersection to form segments associated with edges, the resulting figure is a planar graph here termed the polygon diagonal intersection graph and denoted .For , 2, ..., the vertex counts of are 1, 2, 3, 5, 10, 19, 42, 57, 135, 171, ... (OEIS A007569), which are given by a finite sum of(1)times polynomials in with , 4, 6, 12, 18, 24, 30, 42, 60, 84, 90, 120, and 210 (Poonen and Rubinstein 1998).For , 2, ..., the edge counts of are 0, 1, 3, 8, 20, 42, 91, 136, 288, ... (OEIS A135565), which are again given by a finite sum of polynomials times .Similarly, for , 2, ..., the numbers of regions into which the polygon is divided are given by 1, 4, 11, 24, 50, 80, 154, 220, 375, ... (OEIS A007678), where the th term is given in closed form by(2)(Griffiths 2010), where(3)For odd, all terms but the..
Let be a heptagon with seven vertices given in cyclic order inscribed in a conic. Then the Pascal lines of the seven hexagons obtained by omitting each vertex of in turn and keeping the remaining vertices in the same cyclic order are the sides of a heptagon which circumscribes a conic.Moreover, the Brianchon points of the seven hexagons obtained by omitting the sides of one at a time and keeping the remaining sides in the natural cyclic order are the vertices of the original heptagon.
A 30-sided polygon. The regular triacontagon with side length 1 has inradius , circumradius , and area given by(1)(2)(3)
The (signed) area of a planar non-self-intersecting polygon with vertices , ..., iswhere denotes a determinant. This can be writtenwhere the signs can be found from the diagram above.Note that the area of a convex polygon is defined to be positive if the points are arranged in a counterclockwise order, and negative if they are in clockwise order (Beyer 1987).
Flat polygons embedded in three-space can be transformed into a congruent planar polygon as follows. First, translate the starting vertex to (0, 0, 0) by subtracting it from each vertex of the polygon. Then find the normal to the polygon by taking the cross product of the first and last vertices. Now, let be the rotation matrix for Euler angles , , and , and solve(1)for and (after first expressing sines in terms of cosines using . The result is(2)(3)The signs are chosen as follows:(4)(5)Plugging these back in and applying to the original polygon then gives a polygon whose vertices all have one component zero. This component can then be dropped. The only special cases which need to be taken into account are , in which case the polygon is parallel to the -plane and the third components can be immediately dropped. The second occurs when , in which case there is no component of the normal vector along the x-axis, so the Euler rotation will not work. However, simply..
A skew polygon such that every two consecutive sides (but no three) belong to a face of a regular polyhedron. Every regular polyhedron can be orthogonally projected onto a plane in such a way that one Petrie polygon becomes a regular polygon with the remainder of the projection interior to it. The Petrie polygon of the polyhedron has sides, whereThe Petrie polygons shown above correspond to the Platonicsolids.
A star polygon , with positive integers, is a figure formed by connecting with straight lines every th point out of regularly spaced points lying on a circumference. The number is called the polygon density of the star polygon. Without loss of generality, take . The star polygons were first systematically studied by Thomas Bradwardine.The circumradius of a star polygon with and unit edge lengths is given by(1)and its characteristic polynomial is a factor of the resultant with respect to of the polynomials(2)(3)where is a Chebyshev polynomial of the first kind (Gerbracht 2008).The usual definition (Coxeter 1969) requires and to be relatively prime. However, the star polygon can also be generalized to the star figure (or "improper" star polygon) when and share a common divisor (Savio and Suryanaroyan 1993). For such a figure, if all points are not connected after the first pass, i.e., if , then start with the first unconnected point..
The Star of Lakshmi is the star figure , that is used in Hinduism to symbolize Ashtalakshmi, the eight forms of wealth. This symbol appears prominently in the Lugash national museum portrayed in the fictional film The Return of the Pink Panther.The interior of a Star of Lakshmi with edges of length is a regular octagon with side lengths(1)The areas of the intersection and union of the two constituent squares are(2)(3)
The pentagram, also called the five-point star, pentacle, pentalpha, or pentangle, is the star polygon .It is a pagan religious symbol that is one of the oldest symbols on Earth and is known to have been used as early as 4000 years B.C. It represents the "sacred feminine" or "divine goddess" (Brown 2003, pp. 35-37). However, in modern American pop culture, it more commonly represents devil worship. In the novel The Da Vinci Code, dying Louvre museum curator Jacque Saunière draws a pentagram on his abdomen with his own blood as a clue to identify his murderer (Brown 2003, p. 35).In the above figure, let the length from one tip to another connected tip be unity, the length from a tip to an adjacent dimple be , the edge lengths of the inner pentagon be , the inradius of the inner pentagon be , the circumradius of the inner pentagon be , the circumradius of the pentagram be , and the additional horizontal and vertical..
A star polygon-like figure for which and are not relatively prime. Examples include the hexagram , star of Lakshmi , and nonagram .
A 15-sided polygon, sometimes also called the pentakaidecagon. For a regular pentadecagon with side length 1, the inradius , circumradius , and area are(1)(2)(3)
A principal vertex of a simple polygon is called an ear if the diagonal that bridges lies entirely in . Two ears and are said to overlap ifThe two-ears theorem states that, except for triangles, every simple polygon has at least two nonoverlapping ears.
A polygon is said to be simple (or a Jordan polygon) if the only points of the plane belonging to two polygon edges of are the polygon vertices of . Such a polygon has a well-defined interior and exterior. Simple polygons are topologically equivalent to a disk.The breaking up of self-intersecting polygons into simple polygons (illustrated above) is also called polygon tessellation (Woo et al. 1999).
Given a polygon with an even number of sides, the derived polygon is obtained by joining the points which are a fractional distance along each side. If , then the derived polygons are called midpoint polygons and tend to a shape with opposite sides parallel and equal in length. Furthermore, alternate polygons have approximately the same length, and the original and all derived polygons have the same centroid.Amazingly, if , the derived polygons still approach a shape with opposite sides parallel and equal in length, and all have the same centroid. The above illustrations show 20 derived polygons for ratios , 0.5, 0.7, and 0.9. More amazingly still, if the original polygon is skew, a plane polygonal is approached which has these same properties.
A cyclic polygon is a polygon with vertices upon which a circle can be circumscribed. Since every triangle has a circumcircle, every triangle is cyclic. It is conjectured that for a cyclic polygon of sides, (where is the area) satisfies a monic polynomial of degree , where(1)(2)(Robbins 1995). It is also conjectured that a cyclic polygon with sides satisfies one of two polynomials of degree . The first few values of are 1, 7, 38, 187, 874, ... (OEIS A000531).For triangles , the polynomial is Heron's formula, which may be written(3)and which is of order in . For a cyclic quadrilateral, the polynomial is Brahmagupta's formula, which may be written(4)which is of order in . Robbins (1995) gives the corresponding formulas for the cyclic pentagon and cyclic hexagon.
A cyclic pentagon is a not necessarily regular pentagon on whose polygon vertices a circle may be circumscribed. Let such a pentagon have edge lengths , ..., , and area , and let(1)denote the th-order symmetric polynomial on the five variables consisting of the squares of the pentagon side lengths , so(2)(3)(4)(5)(6)In addition, also define(7)(8)(9)(10)(11)Then the area of the pentagon satisfies(12)a seventh order polynomial in (Robbins 1995). This is also times the polynomial discriminant of the cubic equation(13)(Robbins 1995).
The star figure composed of three equilateral triangles rotated at angles , , and . It has been called the star of Goliath by analogy with the star of David (i.e., hexagram).
A hexagon (not necessarily regular) on whose polygon vertices a circle may be circumscribed. Let(1)denote the th-order symmetric polynomial on the six variables consisting of the squares of the hexagon side lengths , so(2)(3)(4)(5)(6)(7)Then let be the area of the hexagon and define(8)(9)(10)(11)(12)The area of the hexagon then satisfies(13)or this equation with replaced by , a seventh-order polynomial in . This is times the polynomial discriminant of the cubic equation(14)
The closed cyclic self-intersecting hexagon formed by joining the adjacent antiparallels in the construction of the cosine circle. The sides of this hexagon have the property that, in addition to , , and being antiparallel to , , , the remaining sides , , and . The cosine hexagon is a special case of a Tucker hexagon.
The regular pentagon is the regular polygon withfive sides, as illustrated above.A number of distance relationships between vertices of the regular pentagon can be derived by similar triangles in the above left figure,(1)where is the diagonal distance. But the dashed vertical line connecting two nonadjacent polygon vertices is the same length as the diagonal one, so(2)(3)Solving the quadratic equation and taking the plus sign (since the distance must be positive) gives the golden ratio(4)The coordinates of the vertices of a regular pentagon inscribed in a unit circle relative to the center of the pentagon are given as shown in the above figures, with(5)(6)(7)(8)The circumradius, inradius, sagitta, and area of a regular pentagon of side length are given by(9)(10)(11)(12)(13)(14)(15)where is the golden ratio. The height of a regular pentagon of side length is given by(16)(17)Five regular pentagons can be arranged around an identical..
A planar polygon is convex if it contains all the line segments connecting any pair of its points. Thus, for example, a regular pentagon is convex (left figure), while an indented pentagon is not (right figure). A planar polygon that is not convex is said to be a concave polygon.Let a simple polygon have vertices for , 2, ..., , and define the edge vectors as(1)where is understood to be equivalent to . Then the polygon is convex iff all turns from one edge vector to the next have the same sense. Therefore, a simple polygon is convex iff(2)has the same sign for all , where denotes the perp dot product (Hill 1994). However, a more efficient test that doesn't require a priori knowledge that the polygon is simple is known (Moret and Shapiro 1991).The happy end problem considers convex -gons and the minimal number of points (in the general position) in which a convex -gon can always be found. The answers for , 4, 5, and 6 are 3, 5, 9, and 17. It is conjectured that , but..
The regular octagon is the regular polygon witheight sides, as illustrated above.The inradius , circumradius , and area of the regular octagon can be computed directly from the formulas for a general regular polygon with side length and sides as(1)(2)(3)(4)(5)(6)The vertex angle , central angle , and exterior angle are given by(7)(8)(9)The octagon with alternate sides parallel to the - or -axes as illustrated above is the shape used for traffic stop signs in the United States.An octagon with inradius 1 in this orientation can bespecified by the inequalities(10)The octagon may also be orientated so that its vertices lie along the - and -axes and at degree angles to them, as illustrated above. An octagon with circumradius in this orientation can be specified by the inequality(11)The Castel del Monte in Southern Italy consists of a central octagonal core with an inner octagonal courtyard. This, in turn, is surrounded by eight perfectly octagonal..
A principal vertex of a simple polygon is called a mouth if the diagonal is an extremal diagonal (i.e., the interior of lies in the exterior of ).
The regular nonagon is the regular polygon with nine sides and Schläfli symbol .The regular nonagon cannot be constructed using the classical Greek rules of geometric construction, but Conway and Guy (1996) give a Neusis construction based on angle trisection. Madachy (1979) illustrates how to construct a nonagon by folding and knotting a strip of paper. Although the regular nonagon is not a constructible polygon, Dixon (1991) gives constructions for several angles which are close approximations to the nonagonal angle , including angles of and .Given a regular nonagon, let be the midpoint of one side, be the mid-arc point of the arc connecting an adjacent side, and the midpoint of . Then, amazingly, (Karst, quoted in Bankoff and Garfunkel 1973).
A concave polygon is a polygon that is not convex.A simple polygon is concave iff at least one of its internal angles is greater than . An example of a non-simple (self-intersecting) polygon is a star polygon.A concave polygon must have at least four sides.
The regular hexagon is the regular polygon withsix sides, as illustrated above.The inradius , circumradius , sagitta , and area of a regular hexagon can be computed directly from the formulas for a general regular polygon with side length and sides,(1)(2)(3)(4)(5)(6)(7)(8)Therefore, for a regular hexagon,(9)so(10)In proposition IV.15, Euclid showed how to inscribe a regular hexagon in a circle. To construct a regular hexagon with a compass and straightedge, draw an initial circle . Picking any point on the circle as the center, draw another circle of the same radius. From the two points of intersection, draw circles and . Finally, draw centered on the intersection of circles and . The six circle-circle intersections then determine the vertices of a regular hexagon.A plane perpendicular to a axis of a cube (Gardner 1960; Holden 1991, p. 23), octahedron (Holden 1991, pp. 22-23), and dodecahedron (Holden 1991, pp. 26-27)..
A derived polygon with side ratios chosen as so that inscribed polygons are constructed by connecting the midpoints of the base polygon. For a triangle , the midpoint-inscribed polygons , , ... are similar triangles. For a quadrilateral , the midpoint-inscribed polygon is a parallelogram known as the Varignon parallelogram, and , , , ... are similar parallelograms, as are , , , ....
The regular heptagon is the seven-sided regular polygon illustrated above, which has Schläfli symbol . According to Bankoff and Garfunkel (1973), "since the earliest days of recorded mathematics, the regular heptagon has been virtually relegated to limbo." Nevertheless, Thébault (1913) discovered many beautiful properties of the heptagon, some of which are discussed by Bankoff and Garfunkel (1973).Although the regular heptagon is not a constructible polygon using the classical rules of Greek geometric construction, it is constructible using a Neusis construction (Johnson 1975; left figure above). To implement the construction, place a mark on a ruler , and then build a square of side length . Then construct the perpendicular bisector at to , and draw an arc centered at of radius . Now place the marked ruler so that it passes through , lies on the arc, and falls on the perpendicular bisector. Then , and two such triangles..
If a plane cuts the sides , , , and of a skew quadrilateral in points , , , and , thenboth in magnitude and sign (Altshiller-Court 1979, p. 111).More generally, if , , ..., are the polygon vertices of a finite polygon with no "minimal sides" and the side meets a curve in the points and , thenwhere denotes the distance from point to .
The regular hendecagon is the regular polygon with 11 sides, as illustrated above, and has Schläfli symbol .The regular hendecagon cannot be constructed using the classical Greek rules of geometric construction.The rim of the U.S. Susan B. Anthony dollar coin was a regular hendecagon, as illustrated above.The inradius, circumradius, and area of the regular hendecagon with unit side lengths are(1)(2)(3)These can be written exactly as the polynomial roots(4)(5)(6)
The regular dodecagon, illustrated above, is the constructible 12-sided regular polygon that can be denoted using the Schläfli symbol .The Australian 50-cent piece is dodecagonal, as illustrated above.The inradius , circumradius , and area can be computed directly from the formulas for a general regular polygon with side length and sides,(1)(2)(3)(4)(5)(6)Kűrschák's theorem gives the area of the dodecagon inscribed in a unit circle with ,(7)(Wells 1991, p. 137).A plane perpendicular to a axis of a dodecahedron or icosahedron cuts the solid in a regular decagonal cross section (Holden 1991, pp. 24-25).The Greek, Latin, andMaltese crosses are all irregular dodecagons.
A polygon which has both a circumcircle (which touches each vertex) and an incircle (which is tangent to each side). All triangles are bicentric with(1)where is the circumradius, is the inradius, and is the separation of centers. For bicentric quadrilaterals, a result sometimes known as Fuss's problem, the circles satisfy(2)(Dörrie 1965, Salazar 2006) or, in another form,(3)(Davis; Durége 1861; Casey 1888, pp. 109-110; Johnson 1929; Dörrie 1965).If the circles permit successive tangents around the incircle which close the polygon for one starting point on the circumcircle, then they do so for all points on the circumcircle, a result known as Poncelet's porism.
The regular decagon is constructible 10-sided regular polygon with Schläfli symbol . The inradius , circumradius , and area can be computed directly from the formulas for a general regular polygon with side length and sides,(1)(2)(3)(4)(5)(6)Here, is the golden ratio.
Also called Chvátal's art gallery theorem. If the walls of an art gallery are made up of straight line segments, then the entire gallery can always be supervised by watchmen placed in corners, where is the floor function. This theorem was proved by Chvátal (1975). It was conjectured that an art gallery with walls and holes requires watchmen, which has now been proven by Bjorling-Sachs and Souvaine (1991, 1995) and Hoffman et al. (1991).In the Season 2 episode "Obsession" (2006) of the television crime drama NUMB3RS, Charlie mentions the art gallery theorem while building an architectural model.
The apeirogon is an extension of the definition of regular polygon to a figure with an infinite number of sides. Its Schläfli symbol is .The apeirogon can produce a regular tiling on the hyperbolic plane. This is achieved by letting each edge of a regular polygon have length , and each internal angle of the polygon be . Then construct triangle , where is the midpoint of an edge, is an adjacent vertex, and is the center of the polygon. This is a right triangle, with the right angle at . But the length of side is , and the angle is . The length of side , the radius of the circumscribed circle, can the be determined using the standard formula for a right triangle on a surface that has a constant curvature of ,(1)(2)The value of is never less than one, while the value of increases from zero to one as increases. Only for small values of is less than one, which it has to be for any real value of . Thus, by making large enough, the figure has an infinite number of sides and is..
A polygon vertex of a simple polygon is a principal polygon vertex if the diagonal intersects the boundary of only at and .
A regular polygram is generalization of a (regular) polygon on sides (i.e., an -gon) obtained by connecting every th vertex around a circle with every th, "picking up" the pencil as needed to repeat the procedure after traversing the circle until none of the vertices remain unconnected.Lachlan (1893) defines polygram to be a figure consisting of straight lines.The best-known polygrams are the pentagram and hexagram(a.k.a. Star of David). The following table summarizes some named polygrams.symbolpolygram5pentagram6hexagram7heptagram8octagramstar of Lakshmi9nonagram10decagram
A 24-sided polygon. The regular icositetragon is constructible. For side length 1, the inradius , circumradius , and area are given by(1)(2)(3)
A 32-sided polygon. The regular icosidodecagon is a constructible polygon, and the regular icosidodecahedron of side length 1 has inradius , circumradius , and area (1)(2)(3)
A 20-sided polygon. The regular icosagon is a constructible polygon, and the regular icosagon of unit side length has inradius , circumradius , and area given by(1)(2)(3)The swastika is an irregular icosagon.
In general, the internal similitude center of two circles and with centers given in Cartesian coordinates is given by(1)In trilinear coordinates, the internal center of similitude is given by , where(2)(3)(4)The incircle and circumcircle of a triangle have two similitude centers, namely the internal center of similitude Si and the external similitude center Se. The internal center of similitude of these two circles Si is the isogonal conjugate of the Gergonne point of . It is Kimberling center and has equivalent triangle center functions(5)(6)(7)The two points Si and Se share certain similar properties, but there seems to be no straightforward analogy between the two. For instance, Si is the homothetic center of the tangential, intangents, and extangents triangles of triangle taken pairwise, but the only comparable property of the external similitude center Se is more complicated: Se is the homothetic center of the tangential triangle..
Let and denote two directly similar figures in the plane, where corresponds to under the given similarity. Let , and define . Then is also directly similar to .
In general, the external similitude center of two circles and with centers given in Cartesian coordinates is given by(1)In trilinear coordinates, the external center of similitude is given by , where(2)(3)(4)The incircle and circumcircle of a triangle have two similitude centers, namely the internal similitude center Si and the external center of similitude Se. The external center of similitude of the circumcircle and incircle Se is the isogonal conjugate of the Nagel point of . It is Kimberling center and has equivalent triangle center functions(5)(6)(7)The two points Si and Se share certain similar properties, but there seems to be no straightforward analogy between the two. For instance, the internal similitude center Si is the homothetic center of the tangential, intangents, and extangents triangles of triangle taken pairwise, but the only comparable property of Se is more complicated: Se is the homothetic center of the tangential..
If two similar figures lie in the plane but do not have parallel sides (i.e., they are similar but not homothetic), there exists a center of similitude, also called a self-homologous point, which occupies the same homologous position with respect to the two figures (Johnson 1929, p. 16).The similitude center of two triangles and can be constructed by extending each pair of corresponding sides of the triangles and locating their intersection, then drawing the circumcircle passing through two corresponding vertices of the triangles and the point of intersection of the pair of lines through corresponding sides that contain these points. Repeating for each of the three vertices gives three circles that intersect in a unique point, as illustrated above. This point is the similitude center (Johnson 1929).The locus of similitude centers of two nonconcentric circlesis another circle having the line joining the two homothetic centers as..
A transformation that preserves angles and changes all distances in the same ratio, called the ratio of magnification. A similarity can also be defined as a transformation that preserves ratios of distances.A similarity therefore transforms figures into similar figures. When written explicitly in terms of transformation matrices in three dimensions, similarities are commonly referred to as similarity transformations.Examples of similarities include the following. 1. Central dilation: a transformation of lines to parallel lines that is not merely a translation. 2. Geometric contraction: a transformationin which the scale is reduced. 3. Dilation: a transformation taking each line to a parallel line whose length is a fixed multiple of the length of the original line. 4. Expansion: a transformation in which the scale isincreased. 5. Isometry: a transformation that preserves distances.6. Reflection: a transformation in which all..
Two figures are said to be similar when all corresponding angles are equal and all distances are increased (or decreased) in the same ratio, called the ratio of magnification (Coxeter and Greitzer 1967, p. 94). A transformation that takes figures to similar figures is called a similarity.Two figures are directly similar when all corresponding angles are equal and described in the same rotational sense. This relationship is written . (The symbol is also used to mean "is the same order of magnitude as" and "is asymptotic to.") Two figures are inversely similar when all corresponding angles are equal and described in the opposite rotational sense.
The meeting point of lines that connect corresponding points from homothetic figures. In the above figure, is the homothetic center of the homothetic figures and . For figures which are similar but do not have parallel sides, a similitude center exists (Johnson 1929, pp. 16-20).Given two nonconcentric circles, draw radii parallel and in the same direction. Then the line joining the extremities of the radii passes through a fixed point on the line of centers which divides that line externally in the ratio of radii. This point is called the external homothetic center, or external center of similitude (Johnson 1929, pp. 19-20 and 41).If radii are drawn parallel but instead in opposite directions, the extremities of the radii pass through a fixed point on the line of centers which divides that line internally in the ratio of radii (Johnson 1929, pp. 19-20 and 41). This point is called the internal homothetic center, or internal..
Two figures are homothetic if they are related by an expansion or geometric contraction. This means that they lie in the same plane and corresponding sides are parallel; such figures have connectors of corresponding points which are concurrent at a point known as the homothetic center. The homothetic center divides each connector in the same ratio , known as the similitude ratio. For figures which are similar but do not have parallel sides, a similitude center exists.
The Maltese cross is a symbol identified with the Christian warrior whose outward points form an octagon (left figure). Another class of cross sometimes (incorrectly) known as the Maltese cross is the cross pattée (from the French word meaning "paw," which each arm of the cross resembles). The TeX macro gives the form of the cross pattée illustrated in the middle figure. Around 1901, Dudeney published a seven-piece dissection of what he termed a "Maltese cross" (but which is actually a variant of the cross pattée) to a square (right figure) due to A. E. Hill (Gardner 1991, p. 46).
An irregular dodecagonal cross in the shape of a dagger . The six faces of a cube can be cut along seven edges and unfolded into a Latin cross (i.e., the Latin cross is the net of the cube). Similarly, eight hypersurfaces of a hypercube can be cut along 17 squares and unfolded to form a three-dimensional Latin cross.Another cross also called the Latin cross is illustrated above. It is a Greekcross with flared ends, and is also known as the crux immissa or cross patée.
An -uniform tessellation is a tessellation than has transitivity classes of vertices. The 1-uniform tessellations are sometimes known as Archimedean tessellations.The number of -uniform tessellations for , 2, ... are 11, 20, 61, 151, 332, 673, ... (OEIS A068599).
A rectangle which cannot be built up of squares all of different sizes is called an imperfect rectangle. A rectangle which can be built up of squares all of different sizes is called perfect. The number of perfect rectangles of orders 8, 9, 10, ... are 0, 2, 6, 22, 67, 213, 744, 2609, ... (OEIS A002839) and the corresponding numbers of imperfect rectangles are 0, 1, 0, 0, 9, 34, 103, 283, ... (OEIS A002881).Anderson maintains an online database of perfect rectangles at https://www.squaring.net/.
There are at least 15 classes of convex pentagonal tilings, as illustrated above. The first five were discovered during investigations of German mathematician Karl Reinhardt in 1918. After a gap of 50 years, R. B. Kershner found three more in 1968. Richard James subsequently discovered a ninth type of pentagonal tiling in 1975 and over the next few years, Marjorie Rice discovered another four types. Rolf Stein found a 14th tiling in 1985. The most recently discovered 15th tiling was found by Casey Mann, Jennifer McLoud and David Von Derau of the University of Washington Bothell in 2015 using a computer to exhaustively search through a large but finite set of possibilities (Bellos 2015).It has not been proven whether these 15 cases exhaust all possible tilings, but no others are known.Note that the tile in the 14th tiling is essentially different from the others because it is unique (up to similarity), while all the others form families..
A triangle tiling is a tiling of the plane by identical triangles. Any triangle tiles the plane (Wells 1991, p. 208).The total number of triangles (including inverted ones) in the above figures are given by(1)The first few values are 1, 5, 13, 27, 48, 78, 118, 170, 235, 315, 411, 525, 658,812, 988, 1188, 1413, 1665, ... (OEIS A002717).
If the three straight lines joining the corresponding vertices of two triangles and all meet in a point (the perspector), then the three intersections of pairs of corresponding sides lie on a straight line (the perspectrix). Equivalently, if two triangles are perspective from a point, they are perspective from a line.The 10 lines and 10 3-line intersections form a configuration sometimes called Desargues' configuration.Desargues' theorem is self-dual.
A regular polygon is an -sided polygon in which the sides are all the same length and are symmetrically placed about a common center (i.e., the polygon is both equiangular and equilateral). Only certain regular polygons are "constructible" using the classical Greek tools of the compass and straightedge.The terms equilateral triangle and square refer to the regular 3- and 4-polygons, respectively. The words for polygons with sides (e.g., pentagon, hexagon, heptagon, etc.) can refer to either regular or non-regular polygons, although the terms generally refer to regular polygons in the absence of specific wording.A regular -gon is implemented in the Wolfram Language as RegularPolygon[n], or more generally as RegularPolygon[r, n], RegularPolygon[x, y, rspec, n], etc.The sum of perpendiculars from any point to the sides of a regular polygon of sides is times the apothem.Let be the side length, be the inradius, and the circumradius..
A Mrs. Perkins's quilt is a dissection of a square of side into a number of smaller squares. The name "Mrs. Perkins's Quilt" comes from a problem in one of Dudeney's books, where he gives a solution for . Unlike a perfect square dissection, however, the smaller squares need not be all different sizes. In addition, only prime dissections are considered so that patterns which can be dissected into lower-order squares are not permitted.The smallest numbers of squares needed to create relatively prime dissections of an quilt for , 2, ... are 1, 4, 6, 7, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, ... (OEIS A005670), the first few of which are illustrated above.On October 9-10, L. Gay (pers. comm. to E. Pegg, Jr.) discovered 18-square quilts for side lengths 88, 89, and 90, thus beating all previous records. The following table summarizes the smallest numbers of squares known to be needed for various side lengths , with those for (and possibly..
The altitudes of a triangle are the Cevians that are perpendicular to the legs opposite . The three altitudes of any triangle are concurrent at the orthocenter (Durell 1928). This fundamental fact did not appear anywhere in Euclid's Elements.The triangle connecting the feet of the altitudes is known as the orthic triangle.The altitudes of a triangle with side length , , and and vertex angles , , have lengths given by(1)(2)(3)where is the circumradius of . This leads to the beautiful formula(4)Other formulas satisfied by the altitude include(5)where is the inradius, and(6)(7)(8)where are the exradii (Johnson 1929, p. 189). In addition,(9)(10)(11)where is again the circumradius.The points , , , and (and their permutations with respect to indices; left figure) all lie on a circle, as do the points , , , and (and their permutations with respect to indices; right figure).Triangles and are inversely similar.Additional properties involving..
The points of intersection of the adjacent angle trisectors of the angles of any triangle are the polygon vertices of an equilateral triangle known as the first Morley triangle. Taylor and Marr (1914) give two geometric proofs and one trigonometric proof.A line is parallel to a side of the first Morley triangle if and only ifin directed angles modulo (Ehrmann and Gibert 2001).An even more beautiful result is obtained by taking the intersections of the exterior, as well as interior, angle trisectors, as shown above. In addition to the interior equilateral triangle formed by the interior trisectors, four additional equilateral triangles are obtained, three of which have sides which are extensions of a central triangle (Wells 1991).A generalization of Morley's theorem was discovered by Morley in 1900 but first published by Taylor and Marr (1914). Each angle of a triangle has six trisectors, since each interior angle trisector has two associated..
An obtuse triangle is a triangle in which one of the angles is an obtuse angle. (Obviously, only a single angle in a triangle can be obtuse or it wouldn't be a triangle.) A triangle must be either obtuse, acute, or right.From the law of cosines, for a triangle with side lengths , , and ,(1)with the angle opposite side . For an angle to be obtuse, . Therefore, an obtuse triangle satisfies one of , , or .An obtuse triangle can be dissected into no fewer than seven acutetriangles (Wells 1986, p. 71).A famous problem is to find the chance that three points picked randomly in a plane are the polygon vertices of an obtuse triangle (Eisenberg and Sullivan 1996). Unfortunately, the solution of the problem depends on the procedure used to pick the "random" points (Portnoy 1994). In fact, it is impossible to pick random variables which are uniformly distributed in the plane (Eisenberg and Sullivan 1996). Guy (1993) gives a variety of solutions to the..
The interior of the triangle is the set of all points inside a triangle, i.e., the set of all points in the convex hull of the triangle's vertices.The simplest way to determine if a point lies inside a triangle is to check the number of points in the convex hull of the vertices of the triangle adjoined with the point in question. If the hull has three points, the point lies in the triangle's interior; if it is four, it lies outside the triangle.To determine if a given point lies in the interior of a given triangle, consider an individual vertex, denoted , and let and be the vectors from to the other two vertices. Expressing the vector from to in terms of and then gives(1)where and are constants. Solving for and gives(2)(3)where(4)is the determinant of the matrix formed from the column vectors and . Then the point lies in the interior of the triangle if and .If the convex hull of the triangle vertices plus the point is bounded by four points, the point lies outside..
The hexagram is the star polygon , also known as the star of David or Solomon's seal, illustrated at left above.It appears as one of the clues in the novel TheDa Vinci Code (Brown 2003, p. 455).For a hexagram with circumradius (red circle), the inradius (green circle) is(1)and the circle passing through the intersections of the triangles has radius(2)The interior of a hexagram is a regular hexagon with side lengths equal to 1/3 that of the original hexagram. Given a hexagram with line segments of length , the areas of the intersection and union of the two constituent triangles are(3)(4)There is a "nonregular" hexagram that can be obtained by spacing the integers 1 to 6 evenly around a circle and connecting . The resulting figure is called a "unicursal hexagram" and was evidently discovered in the 19th century. It is not regular because there are some edges going from to (mod 6) and some edges going from to (mod 6). However,..
A hexagon is a six-sided polygon. Several special types of hexagons are illustrated above. In particular, a hexagon with vertices equally spaced around a circle and with all sides the same length is a regular polygon known as a regular hexagon.Given an arbitrary hexagon, take each three consecutive vertices, and mark the fourth point of the parallelogram sharing these three vertices. Taking alternate points then gives two congruent triangles, as illustrated above (Wells 1991).Given an arbitrary hexagon, connecting the centroids of each consecutive three sides gives a hexagon with equal and parallel sides known as the centroid hexagon (Wells 1991).
A Greek cross, also called a square cross, is a cross inthe shape of a plus sign. It is a non-regular dodecagon.A square cross appears on the flag of Switzerland, and also on the key to the Swiss Bank deposit box in D. Brown's novel The Da Vinci Code (Brown 2003, pp. 146 and 171-172).Greek crosses can tile the plane, as noted by the protagonist Christopher in The Curious Incident of the Dog in the Night-Time (Haddon 2003, pp. 203-204).
A tiling of regular polygons (in two dimensions), polyhedra (three dimensions), or polytopes ( dimensions) is called a tessellation. Tessellations can be specified using a Schläfli symbol.The breaking up of self-intersecting polygons into simple polygons is also called tessellation (Woo et al. 1999), or more properly, polygon tessellation.There are exactly three regular tessellationscomposed of regular polygons symmetrically tiling the plane.Tessellations of the plane by two or more convex regular polygons such that the same polygons in the same order surround each polygon vertex are called semiregular tessellations, or sometimes Archimedean tessellations. In the plane, there are eight such tessellations, illustrated above (Ghyka 1977, pp. 76-78; Williams 1979, pp. 37-41; Steinhaus 1999, pp. 78-82; Wells 1991, pp. 226-227).There are 14 demiregular (or polymorph) tessellations which are..
The combination of a central dilation and a rotation about the same center. However, the combination of a central dilation and a rotation whose centers are distinct is also a spiral symmetry. In fact, any two directly similar figures are related either by a translation or by a spiral symmetry (Coxeter and Greitzer 1967, p. 97).A spiral similarity tessellation of any ordinary tessellation can be constructed by placing a series of polygonal tiles of decreasing size on an equilateral spiral.
An attractive tiling of the square composed of two types of triangular tiles. It consists of 16 equilateral triangles and 32 -- isosceles triangles arranged in the shape of a dodecagon.The composition of Kürschák's tile is motivated by drawing inward-pointing equilateral triangles on each side of a unit square and then connecting adjacent vertices to form a smaller square rotated with respect to the original square. Joining the midpoints of the square together with the intersections of the equilateral triangles then gives a dodecagon (Wells 1991) with circumradius
A demiregular tessellation, also called a polymorph tessellation, is a type of tessellation whose definition is somewhat problematical. Some authors define them as orderly compositions of the three regular and eight semiregular tessellations (which is not precise enough to draw any conclusions from), while others defined them as a tessellation having more than one transitivity class of vertices (which leads to an infinite number of possible tilings).The number of demiregular tessellations is commonly given as 14 (Critchlow 1970, pp. 62-67; Ghyka 1977, pp. 78-80; Williams 1979, p. 43; Steinhaus 1999, pp. 79 and 81-82). However, not all sources apparently give the same 14. Caution is therefore needed in attempting to determine what is meant by "demiregular tessellation."A more precise term of demiregular tessellations is 2-uniform tessellations (Grünbaum and Shephard 1986, p. 65)...
Regular tessellations of the plane by two or more convex regular polygons such that the same polygons in the same order surround each polygon vertex are called semiregular tessellations, or sometimes Archimedean tessellations. In the plane, there are eight such tessellations, illustrated above (Ghyka 1977, pp. 76-78; Williams 1979, pp. 37-41; Steinhaus 1999, pp. 78-82; Wells 1991, pp. 226-227). Williams (1979, pp. 37-41) also illustrates the dual tessellations of the semiregular tessellations. The dual tessellation of the tessellation of squares and equilateral triangles is called the Cairo tessellation (Williams 1979, p. 38; Wells 1991, p. 23).
Let be the group of symmetries which map a monohedral tiling onto itself. The transitivity class of a given tile T is then the collection of all tiles to which T can be mapped by one of the symmetries of . If has transitivity classes, then is said to be -isohedral. Berglund (1993) gives examples of -isohedral tilings for , 2, and 4.The numbers of isohedral -polyomino tilings (more specifically, the number of polyominoes with cells that tile the plane by rotation but not by translation) for , 8, 9, ... are 3, 11, 60, 199, 748, ... (OEIS A075201), the first few of which are illustrated above (Myers).The numbers of -polyomino tilings that tile the plane isohedrally without additional restriction for , 2, ... are 1, 1, 2, 5, 12, 35, 104, ... (OEIS A075205).
Consider a two-dimensional tessellation with regular -gons at each polygon vertex. In the plane,(1)(2)so(3)(Ball and Coxeter 1987), and the only factorizations are(4)(5)(6)Therefore, there are only three regular tessellations (composed of the hexagon, square, and triangle), illustrated above (Ghyka 1977, p. 76; Williams 1979, p. 36; Wells 1991, p. 213).There do not exist any regular star polygon tessellations in the plane. Regular tessellations of the sphere by spherical triangles are called triangular symmetry groups.
An aperiodic tiling is a non-periodic tiling in which arbitrarily large periodic patches do not occur. A set of tiles is said to be aperiodic if they can form only non-periodic tilings. The most widely known examples of aperiodic tilings are those formed by Penrose tiles.The Federation Square buildings in Melbourne, Australia feature an aperiodic pinwheel tiling attributed to Charles Radin. The tiling is illustrated above in a pair of photographs by P. Bourke.
A plane tiling is said to be isohedral if the symmetry group of the tiling acts transitively on the tiles, and -isohedral if the tiles fall into n orbits under the action of the symmetry group of the tiling. A -anisohedral tiling is a tiling which permits no -isohedral tiling with .The numbers of anisohedral polyominoes with , 9, 10, ... are 1, 9, 44, 108, 222, ... (OEIS A075206), the first few of which are illustrated above (Myers).
A triangle formed by three circular arcs. By extending the arcs into complete circles, the points of intersection , , and are obtained. This gives the three circular triangles, , , , and , which are called the associated triangles to .The circular triangle and its associated circles have a total of eight incircles and six circumcircles. These systems of circles have some remarkable properties, including the Hart circle, which is an analog of the nine-point circle in Feuerbach's theorem.A closed-form set of formulas for the area of circular triangles like is given by Fewell (2006).
The three circular triangles , , , and obtained by extending the arcs of a circular triangle into complete circles.
There are two types of squares inscribing reference triangle in the sense that all vertices lie on the sidelines of . The first type has two adjacent vertices of the square on one side, the second type has two opposite vertices on one side.Casey (1888, pp. 10-11) illustrates how to inscribe a square in an arbitrary triangle . Construct the perpendicular and the line segment . Bisect , and let be the intersection of the bisector with . Then draw and through , perpendicular to and parallel to , respectively. Let be the intersection of and , and then construct and through and perpendicular to . Then is an inscribed square. Permuting the order in which the vertices are taken gives an additional two congruent squares. These squares, however, are not necessarily the largest inscribed squares. Calabi's triangle is the only triangle (besides the equilateral triangle) for which the largest inscribed square can be inscribed in three different ways.An..
There are a number of interesting results related to the tiling of squares. For example, M. Laczkovich has shown that there are exactly three shapes of non-right triangles that tile the square with similar copies, corresponding to angles , , and (Stein and Szabó 1994). In particular, given triangles of shape with no two the same size, tile the square. The best known solution has 8 triangles (Berlekamp 1999).The total number of squares contained in a grid of unit square is the square pyramidal number
Gardner showed how to dissect a square into eight and nine acute scalene triangles.W. Gosper discovered a dissection of a unit square into 10 acute isosceles triangles, illustrated above (pers. comm. to Ed Pegg, Jr., Oct 25, 2002). The coordinates can be found from solving the four simultaneous equations (1)(2)(3)(4)for the four unknowns and picking the solutions for which . The solutions are roots of 12th order polynomials with numerical values given approximately by(5)(6)(7)(8)Pegg has constructed a dissection of a square into 22 acute isosceles triangles.Guy (1989) asks if it is possible to triangulate a square with integer side lengths such that the resulting triangles have integer side lengths (Trott 2004, p. 104).
The number of different triangles which have integer side lengths and perimeter is(1)(2)(3)where is the partition function giving the number of ways of writing as a sum of exactly terms, is the nearest integer function, and is the floor function (Andrews 1979, Jordan et al. 1979, Honsberger 1985). A slightly complicated closed form is given by(4)The values of for , 2, ... are 0, 0, 1, 0, 1, 1, 2, 1, 3, 2, 4, 3, 5, 4, 7, 5, 8, 7, 10, 8, 12, 10, 14, 12, 16, ... (OEIS A005044), which is also Alcuin's sequence padded with two initial 0s.The generating function for is given by(5)(6)(7) also satisfies(8)It is not known if a triangle with integer sides, triangle medians, and area exists (although there are incorrect proofs of the impossibility in the literature). However, R. L. Rathbun, A. Kemnitz, and R. H. Buchholz have shown that there are infinitely many triangles with rational sides (Heronian triangles) with two rational..
In general, a cross is a figure formed by two intersecting line segments. In linear algebra, a cross is defined as a set of mutually perpendicular pairs of vectors of equal magnitude from a fixed origin in Euclidean -space.The word "cross" is also used to denote the operation of the cross product, so would be pronounced " cross ."
The hypotenuse of a right triangle is the triangle's longest side, i.e., the side opposite the right angle. The word derives from the Greek hypo- ("under") and teinein ("to stretch").The length of the hypotenuse of a right trianglecan be found using the Pythagorean theorem.Among his many other talents, Major General Stanley in Gilbert and Sullivan's operetta The Pirates of Penzance impresses the pirates with his knowledge of the hypotenuse in "The Major General's Song" as follows: "I am the very model of a modern Major-General, I've information vegetable, animal, and mineral, I know the kings of England, and I quote the fights historical, From Marathon to Waterloo, in order categorical; I'm very well acquainted too with matters mathematical, I understand equations, both the simple and quadratical, About binomial theorem I'm teeming with a lot o' news-- With many cheerful facts about the square of..
The Kakeya needle problems asks for the plane figure of least area in which a line segment of width 1 can be freely rotated (where translation of the segment is also allowed). Surprisingly, there is no minimum area (Besicovitch 1928). Another iterative construction which tends to as small an area as desired is called a Perron tree (Falconer 1990, Wells 1991).When the figure is restricted to be convex, the smallest region is an equilateral triangle of unit height. Wells (1991) states that Kakeya discovered this, while Falconer (1990) attributes it to Pál.If convexity is replaced by the weaker assumption of simply-connectedness, then the area can still be arbitrarily small, but if the set is required to be star-shaped, then is a known lower bound (Cunningham 1965).The smallest simple convex domain in which one can put a segment of length 1 which will coincide with itself when rotated by has area(OEIS A093823; Le Lionnais 1983). ..
In the above figure, let be the intersection of and and specify that . ThenA beautiful related theorem due to H. Stengel can be stated as follows. In the above figure, let lie on the side and lie on the side . Now let intersect the line at a point , and construct points , , , and so that . Then
Given two crossed ladders resting against two buildings, what is the distance between the buildings? Let the height at which they cross be and the lengths of the ladders and . The height at which touches the building is then obtained by simultaneously solving the equations(1)(2)and(3)the latter of which follows either immediately from the crossed ladders theorem or from similar triangles with , , and . Eliminating gives the equations(4)(5)These quartic equations can be solved for and given known values of , , and .There are solutions in which not only , , , , and are all integers, but so are , and . One example is .The problem can also be generalized to the situation in which the ends of the ladders are not pinned against the buildings, but propped fixed distances and away.
Let be the midpoint of the arc . Pick at random and pick such that (where denotes perpendicular). Then
Two points which are collinear with respect to a similitude center but are not homologous points. Four interesting theorems from Johnson (1929) follow. 1. Two pairs of antihomologous points form inversely similar triangles with the homothetic center. 2. The product of distances from a homotheticcenter to two antihomologous points is a constant. 3. Any two pairs of points which are antihomologous with respect to a similitudecenter lie on a circle. 4. The tangents to two circles at antihomologous points make equal angles with the line through the points.
The triangulation point of a reference triangle for which triangles , , and have congruent incircles. It is a special case of an Elkies point. Kimberling and Elkies (1987) showed that a unique such point exists for any triangle, but did not provide explicit constructions for this point.
Let five circles with concyclic centers be drawn such that each intersects its neighbors in two points, with one of these intersections lying itself on the circle of centers. By joining adjacent pairs of the intersection points which do not lie on the circle of center, an (irregular) pentagram is obtained each of whose five vertices lies on one of the circles with concyclic centers.Let the circle of centers have radius and let the five circles be centered and angular positions along this circle. The radii of the circles and their angular positions along the circle of centers can then be determined by solving the ten simultaneous equations (1)(2)for , ..., 5, where and .
Four or more points , , , , ... which lie on a circle are said to be concyclic. Three points are trivially concyclic since three noncollinear points determine a circle (i.e., every triangle has a circumcircle). Ptolemy's theorem can be used to determine if four points are concyclic.The number of the lattice points which can be picked with no four concyclic is (Guy 1994).A theorem states that if any four consecutive points of a polygon are not concyclic, then its area can be increased by making them concyclic. This fact arises in some proofs that the solution to the isoperimetric problem is the circle.
Given the incircle and circumcircle of a bicentric polygon of sides, the centroid of the tangent points on the incircle is a fixed point , known as the Weill point, independent of the particular polygon.More generally, the locus of the centroid of any number of the points is a circle (Casey 1888).
The minimal enclosing circle problem, sometimes also known as the bomb problem, is the problem of finding the circle of smallest radius that contains a given set of points in its interior or on its boundary. This smallest circle is known as the minimal enclosing circle.Jung's theorem states that every finite set of points with geometric span has an enclosing circle with radius no greater than .
Concentric circles are circles with a common center. The region between two concentric circles of different radii is called an annulus. Any two circles can be made concentric by inversion by picking the inversion center as one of the limiting points.Given two concentric circles with radii and , what is the probability that a chord chosen at random from the outer circle will cut across the inner circle? Depending on how the "random" chord is chosen, 1/2, 1/3, or 1/4 could all be correct answers. 1. Picking any two points on the outer circle and connecting them gives 1/3. 2. Picking any random point on a diagonal and then picking the chordthat perpendicularly bisects it gives 1/2. 3. Picking any point on the large circle, drawing a line to the center, and thendrawing the perpendicularly bisected chord gives 1/4. So some care is obviously needed in specifying what is meant by "random" in this problem.Given an arbitrary chord to the..
Four circles may be drawn through an arbitrary point on a torus. The first two circles are obvious: one is in the plane of the torus and the second perpendicular to it. The third and fourth circles (which are inclined with respect to the torus) are much more unexpected and are known as the Villarceau circles (Villarceau 1848, Schmidt 1950, Coxeter 1969, Melzak 1983).To see that two additional circles exist, consider a coordinate system with origin at the center of torus, with pointing up. Specify the position of by its angle measured around the tube of the torus. Define for the circle of points farthest away from the center of the torus (i.e., the points with ), and draw the x-axis as the intersection of a plane through the z-axis and passing through with the -plane. Rotate about the y-axis by an angle , where(1)In terms of the old coordinates, the new coordinates are(2)(3)So in coordinates, equation (◇) of the torus becomes(4)Expanding the left..
The three circumcircles through the triangle centroid of a given triangle and the pairs of the vertices of the second Brocard triangle are called the McCay circles (Johnson 1929, p. 306).The circumcircle of their centers (i.e., of the second Brocard triangle) is therefore the Brocard circle.The -McCay circle has center functionand radius,1/3 that of the Neuberg circle, where is the Brocard angle (Johnson 1929, p. 307).If the polygon vertex of a triangle describes a Neuberg circle , then its triangle centroid describes one of the McCay circles (Johnson 1929, p. 290). In the above figure, the inner triangle is the second Brocard triangle of , whose two indicated edges are concyclic with on the McCay circle.
A disk with radius 1.The (open) unit disk can also be considered to be the region in the complex plane defined by , where denotes the complex modulus. (The closed unit disk is similarly defined as .
Let , , , and be four circles of general position through a point . Let be the second intersection of the circles and . Let be the circle . Then the four circles , , , and all pass through the point . Similarly, let be a fifth circle through . Then the five points , , , and all lie on one circle . And so on.
The triquetra is a geometric figure consisting of three mutually intersecting vesica piscis lens shapes, as illustrated above. The central region common to all three lenses is a Reuleaux triangle.The triquetra has perimeterand its interior has area
While some authors define "circumference" as distance around an arbitrary closed object (sometimes restricted to a closed curved object), in the work, the term "perimeter" is used for this purpose and "circumference" is restricted to mean the perimeter of a circle.For radius or diameter ,where is pi.The term "circumference" is also sometimes used to refer to the "enclosingboundary" itself of a curved lamina or disk.
The circumcenter is the center of a triangle's circumcircle. It can be found as the intersection of the perpendicular bisectors. The trilinear coordinates of the circumcenter are(1)and the exact trilinear coordinatesare therefore(2)where is the circumradius, or equivalently(3)The circumcenter is Kimberling center .The distance between the incenter and circumcenter is , where is the circumradius and is the inradius.Distances to a number of other named triangle centers are given by(4)(5)(6)(7)(8)(9)(10)(11)(12)where is the triangle triangle centroid, is the orthocenter, is the incenter, is the symmedian point, is the nine-point center, is the Nagel point, is the de Longchamps point, is the circumradius, is Conway triangle notation, and is the triangle area.If the triangle is acute, the circumcenter is in the interior of the triangle. In a right triangle, the circumcenter is the midpoint of the hypotenuse.For an acute triangle,(13)where..
The area of the dodecagon () inscribed in a unit circle with is
In the figure above with tangent line and secant line ,(1)(Jurgensen et al. 1963, p. 346).The line tangent to a circle of radius centered at (2)(3)through can be found by solving the equation(4)giving(5)Two of these four solutions give tangent lines, as illustrated above, and the lengths of these lines are equal (Casey 1888, p. 29).
Let three equal circles with centers , , and intersect in a single point and intersect pairwise in the points , , and . Then the circumcircle of the reference triangle is congruent to the original three.Furthermore, the points , , , and form an orthocentric system.Here, the original three circles are known as Johnson circles and the triangle formed by their centers is known as the Johnson triangle. Amazingly, the Johnson triangle circumcircle is also congruent to the circumcircle of the reference triangle and centered at the orthocenter .A "triquetra" is a figure consisting of three circular arcs of equal radius, and has seen extensive use in heraldry (i.e., coats of arms), specifically in the case of the so-called Borromean rings. The term "Triquetra theorem" was used by Mackenzie (1992) to describe Johnson's theorem.Mackenzie (1992) generalized this theorem to the case where the three circles do not coincide. In this..
The radial curve of a unit circle from a radial point and parametric equations(1)(2)is another circle with parametricequations(3)(4)
The Johnson midpoint is the point of concurrence of the line segments joining the vertices of a reference triangle with the centers of a certain set of circles (that resemble but are not the Johnson circles). It also is the midpoint of each of these segments, as well as perspector of the reference triangle and the triangle determined by the centers of these circles.It has triangle center functionand is Kimberling center .
Johnson's theorem states that if three equal circles mutually intersect one another in a single point, then the circle passing through their other three pairwise points of intersection is congruent to the original three circles. If the pairwise intersections are taken as the vertices of a reference triangle , then the Johnson circles that are congruent to the circumcircle of have centers(1)(2)(3)where , , , and are Conway triangle notation.The centers of the Johnson circles form the Johnson triangle which, together with , form an orthocentric system.The point of threefold concurrence of the Johnson circles is the orthocenter of the reference triangle .Note also that intersections of the directed lines from the orthocenter of the reference triangle through the centers of the Johnson circles intersect the Johnson circles at the vertices of the anticomplementary triangle . The anticomplementary circle, with center and radius (where is..
The pedal curve of a unitcircle with parametric equation(1)(2)with pedal point is(3)(4)The pedal curve with respect to the center is thecircle itself (Gray 1997, pp. 119 and 124-135).If the pedal point is taken on the circumference (and in particular at the point ), the pedal curve is the cardioid(5)(6)and otherwise is a limaçon.
The first and second isodynamic points of a triangle can be constructed by drawing the triangle's angle bisectors and exterior angle bisectors. Each pair of bisectors intersects a side of the triangle (or its extension) in two points and , for , 2, 3. The three circles having , , and as diameters are the Apollonius circles , , and . The points and in which the three Apollonius circles intersect are the first and second isodynamic points, respectively.The two isodynamic points of a reference triangle are mutually inverse with respect to the circumcircle of (Gallatly 1913, p. 103). and have triangle center functionsrespectively. The antipedal triangles of bothpoints are equilateral and have areaswhere is the Brocard angle.The isodynamic points are isogonal conjugates of the Fermat points. They lie on the Brocard axis. The distances from either isodynamic point to the polygon vertices are inversely proportional to the sides. The pedal..
External (or positive) and internal (or negative) similarity points of two circles with centers and and radii and are the points and on the lines such thator
Draw an initial circle, and arrange six circles tangent to it such that they touch both the original circle and their two neighbors. Then the three lines joining opposite points of tangency are concurrent in a point. The figures above show several possible configurations (Evelyn et al. 1974, pp. 31-37).Letting the radii of three of the circles approach infinity turns three of the circles into the straight sides of a triangle and the central circle into the triangle's incircle. As illustrated above, the three lines connecting opposite points of tangency (with those along the triangle edges corresponding to the vertices of the contact triangle) concur (Evelyn et al. 1974, pp. 39 and 42).
Given triangle , there are four lines simultaneously tangent to the incircle (with center ) and the -excircle (with center ). Of these, three correspond to the sidelines of the triangle, and the fourth is known as the -intangent (Kimberling 1998, p. 161), illustrated above.The intangents intersect one another pairwise, and their points of intersection formthe so-called intangents triangle.
For a unit circle with parametricequations(1)(2)the negative pedal curve with respect to the pedal point is(3)(4)Therefore if the point is inside the circle (), the negative pedal is an ellipse, if , it is a single point, if the point is outside the circle (), the negative pedal is a hyperbola.
The incenter is the center of the incircle for a polygon or insphere for a polyhedron (when they exist). The corresponding radius of the incircle or insphere is known as the inradius.The incenter can be constructed as the intersection of angle bisectors. It is also the interior point for which distances to the sides of the triangle are equal. It has trilinear coordinates 1:1:1, i.e., triangle center function(1)and homogeneous barycentric coordinates . It is Kimberling center .For a triangle with Cartesian vertices , , , the Cartesian coordinates of the incenter are given by(2)The distance between the incenter and circumcenter is , where is the circumradius and is the inradius, a result known as the Euler triangle formula.The incenter lies on the Nagel line and Soddy line, and lies on the Euler line only for an isosceles triangle. The incenter is the center of the Adams' circle, Conway circle, and incircle. It lies on the Darboux cubic, M'Cay cubic,..
The pedal curve of circleinvolute(1)(2)with the center as the pedal point is the Archimedes'spiral(3)(4)
Any one of the eight Apollonius circles of three given circles is tangent to a circle known as a Hart circle, as are the other three Apollonius circles having (1) like contact with two of the given circles and (2) unlike contact with the third.
The circle which touches the incircles , , , and of a circular triangle and its associated triangles. It is either externally tangent to and internally tangent to incircles of the associated triangles , , and (as in the above figure), or vice versa. The Hart circle has several properties which are analogous to the properties on the nine-point circle of a linear triangle. There are eight Hart circles associated with a given circular triangle.The Hart circle of any circular triangle and the Hart circles of the three associated triangles have a common tangent circle which touches the former in the opposite sense to that which it touches the latter (Lachlan 1893, p. 254). In addition, the circumcircle of any circular triangle is the Hart circle of the circular triangle formed by the circumcircles of the inverse associated triangles (Lachlan 1893, p. 254)...
The radical circle of three given circles is the circle having center at the radical center of the three circles and is orthogonal to all of them. (A circle with center at the radical center that is orthogonal to one of the original circles is always orthogonal to all three.)The following table summarizes the radical circle for some circle triplets.circlesradical circleexcirclesexcircles radical circleLucas circlesLucas circles radical circleMcCay circlesMcCay circles radical circlemixtilinear incirclesmixtilinear incircles radical circleNeuberg circlesNeuberg circles radical circlepower circlesde Longchamps circleStammler circlesStammler circles radical circletangent circlesincircle
Pick any two relatively prime integers and , then the circle of radius centered at is known as a Ford circle. No matter what and how many s and s are picked, none of the Ford circles intersect (and all are tangent to the x-axis). This can be seen by examining the squared distance between the centers of the circles with and ,(1)Let be the sum of the radii(2)then(3)But , so and the distance between circle centers is the sum of the circle radii, with equality (and therefore tangency) iff . Ford circles are related to the Farey sequence (Conway and Guy 1996).If , , and are three consecutive terms in a Farey sequence, then the circles and are tangent at(4)and the circles and intersect in(5)Moreover, lies on the circumference of the semicircle with diameter and lies on the circumference of the semicircle with diameter (Apostol 1997, p. 101)...
Let , , and be the lengths of the tangents to a circle from the vertices of a triangle with sides of lengths , , and . Then the condition that is tangent to the circumcircle of the triangle is thatThe theorem was discovered by Casey prior to Purser's independent discovery.
In Homogeneous coordinates , the equation of a circle isThe discriminant of this circle is defined asand the quadratic form is the basic invariant.
Given two circles, draw the tangents from the center of each circle to the sides of the other. Then the line segments and are of equal length.The theorem can be proved by brute force by setting up the nine equations(1)(2)(3)(4)(5)(6)(7)(8)(9)and using Gröbner basis to determine the polynomial equations satisfied by and while eliminate , , , , , , and . The resulting eighth-degree polynomials satisfied by and are identical, proving that .
Given triangle , there are four lines simultaneously tangent to the - and -excircles (with centers and , respectively). Of these, three correspond to the sidelines of the triangle, and the fourth is known as the -extangent (Kimberling 1998, p. 162), illustrated above.The extangents intersect one another pairwise, and their points of intersection formthe so-called extangents triangle.
The radius of an excircle. Let a triangle have exradius (sometimes denoted ), opposite side of length and angle , area , and semiperimeter . Then(1)(2)(3)(Johnson 1929, p. 189), where is the circumradius. Let be the inradius, then(4)(5)(Casey 1888, p. 65) and(6)Some fascinating formulas due to Feuerbach are(7)(Johnson 1929, pp. 190-191).
Members of a coaxal system satisfyfor values of . Picking then gives the two circlesof zero radius, known as point circles. The two point circles , real or imaginary, are called the limiting points of the coaxal system.
If the tangents at and to the circumcircle of a triangle intersect in a point , then the circle with center and which passes through and is called the excosine circle, and cuts and in two points which are extremities of a diameter.
Consider a unit circle and a radiant point located at . There are four different regimes of caustics, illustrated above.For radiant point at , the catacaustic is the nephroid(1)(2)(Trott 2004, p. 17, mistakenly states that the catacaustic for parallel light falling on any concave mirror is a nephroid.)For radiant point a finite distance , the catacaustic is the curve(3)(4)which is apparently incorrectly described as a limaçonby Lawrence (1972, p. 207).For radiant point on the circumference of the circle (), the catacaustic is the cardioid(5)(6)with Cartesian equation(7)For radiant point inside the circle, the catacausticis a discontinuous two-part curve.If the radiant point is the origin, then the catacaustic degenerates to a single point at the origin since all rays reflect upon themselves back through the origin...
A circle is the set of points in a plane that are equidistant from a given point . The distance from the center is called the radius, and the point is called the center. Twice the radius is known as the diameter . The angle a circle subtends from its center is a full angle, equal to or radians.A circle has the maximum possible area for a given perimeter,and the minimum possible perimeter for a given area.The perimeter of a circle is called the circumference, and is given by(1)This can be computed using calculus using the formula for arc length in polar coordinates,(2)but since , this becomes simply(3)The circumference-to-diameter ratio for a circle is constant as the size of the circle is changed (as it must be since scaling a plane figure by a factor increases its perimeter by ), and also scales by . This ratio is denoted (pi), and has been proved transcendental.Knowing , the area of the circle can be computed either geometrically or using calculus. As the..
Given a parabola with parametricequations(1)(2)the negative pedal curve for a pedal point has equation(3)(4)Taking the pedal point at the origin gives(5)(6)which is a semicubical parabola. Similarly, taking gives(7)(8)which is a Tschirnhausen cubic.
The center of an excircle. There are three excenters for a given triangle, denoted , , . The incenter and excenters of a triangle are an orthocentric system.where is the circumcenter, are the excenters, and is the circumradius (Johnson 1929, p. 190). Denote the midpoints of the original triangle , , and . Then the lines , , and intersect in a point known as the mittenpunkt.
Given a circle expressed in trilinearcoordinates bya central circle is a circle such that is a triangle center and is a homogeneous function that is symmetric in the side lengths , , and (Kimberling 1998, p. 226).The following table summarizes the triangle centers whose trilinears correspond to a circle with (for some appropriate value of ). In the table, indicated a circle function that is known but which does not appear among the list of Kimberling centers. Note also that the circumcircle is not actually a central circle, since its trilinears 0:0:0 are not those of a triangle center.circleKimberlingKimberlingcenterAdams' circle*incenter anticomplementary circlethird power pointorthocenter Apollonius circleBevan circleincenter Bevan point Brocard circletriangle centroid midpoint of the Brocard diametercircumcircle-0circumcenter Conway circleincenter cosine circlesymmedian point de Longchamps circlethird power..
Given a chord of a circle, draw any other two chords and passing through its midpoint. Call the points where and meet and . Then is also the midpoint of . There are a number of proofs of this theorem, including those by W. G. Horner, Johnson (1929, p. 78), and Coxeter (1987, pp. 78 and 144). The latter concise proof employs projective geometry.The following proof is given by Coxeter and Greitzer (1967, p. 46). In the figure at right, drop perpendiculars and from and to , and and from and to . Write , , and , and then note that by similar triangles(1)(2)(3)so(4)(5)so . Q.E.D.
Orthogonal circles are orthogonal curves, i.e., they cut one another at right angles. By the Pythagorean theorem, two circles of radii and whose centers are a distance apart are orthogonal if(1)Two circles with Cartesian equations(2)(3)are orthogonal if(4)A theorem of Euclid states that, for the orthogonal circles in the above diagram,(5)(Dixon 1991, p. 65).The radical lines of three given circles concur in the radical center . If a circle with center cuts any one of the three circles orthogonally, it cuts all three orthogonally. This circle is called the orthogonal circle (or radical circle) of the system. The orthogonal circle is the locus of a point whose polars with respect to the three given circles are concurrent (Lachlan 1893, p. 237).The following table lists circles orthogonal to various named circle.circleorthogonal circle(s)Apollonius circleStevanović circleBevan circleStevanović circleBrocard..
The diameter of a circle is the distance from a point on the circle to a point radians away, and is the maximum distance from one point on a circle to another. The diameter of a sphere is the maximum distance between two antipodal points on the surface of the sphere.If is the radius of a circle or sphere, then . The ratio of the circumference of a circle or great circle of a sphere to the diameter is pi,
There are four completely different definitions of the so-called Apollonius circles: 1. The set of all points whose distances from two fixed points are in a constant ratio (Durell 1928, Ogilvy 1990). 2. One of the eight circles that is simultaneously tangent to three given circles (i.e., a circle solving Apollonius' problem for three circles). 3. One of the three circles passing through a vertex and both isodynamic points and of a triangle (Kimberling 1998, p. 68). 4. The circle that touches all three excircles of atriangle and encompasses them (Kimberling 1998, p. 102). Given one side of a triangle and the ratio of the lengths of the other two sides, the locus of the third polygon vertex is the Apollonius circle (of the first type) whose center is on the extension of the given side. For a given triangle, there are three circles of Apollonius. Denote the three Apollonius circles (of the first type) of a triangle by , , and , and their centers..
Draw a circle that cuts three given circles perpendicularly. The solution is known as the radical circle of the given three circles. If it lies outside the three circles, then the circle with center and radius formed by the tangent from to one of the given circles intersects the given circles perpendicularly. Otherwise, if lies inside one of the circles, the problem is unsolvable.
If three circles , , and are taken in pairs, the external similarity points of the three pairs lie on a straight line. Similarly, the external similarity point of one pair and the two internal similarity points of the other two pairs lie upon a straight line, forming a similarity axis of the three circles.
Draw three circles in the plane, none of which lies completely inside another, and the common external tangent lines for each pair. Then points of intersection of the three pairs of tangent lines lie on a straight line.Monge's theorem has a three-dimensional analog which states that the apexes of the cones defined by four spheres, taken two at a time, lie in a plane (when the cones are drawn with the spheres on the same side of the apex; Wells 1991).
The angular twist of a shaft with given cross section is given by(1)(Roark 1954, p. 174), where is the twisting moment (commonly measured in units of inch-pounds-force), is the length (inches), is the modulus of rigidity (pounds-force per square inch), and (sometimes also denoted ) is the torsional rigidity multiplier for a given geometric cross section (inches to the fourth power). Note that the quantity is sometimes denoted (e.g., Timoshenko and Goodier 1951, p. 264).Values of are known exactly only for a small number of cross sections, and in closed form for even fewer. The following table lists approximate values for some common shapes (Timoshenko and Goodier 1951, pp. 258-280; Roark 1954, pp. 174-179).cross section approxOEIScircle1.570796...A019669equilateral triangle0.021650...A180317half-disk0.297556...A180310isosceles right triangle0.026089...A180314quarter-disk0.0825...sliced..
The area moment of inertia is a property of a two-dimensional plane shape which characterizes its deflection under loading. It is also known as the second moment of area or second moment of inertia. The area moment of inertia has dimensions of length to the fourth power. Unfortunately, in engineering contexts, the area moment of inertia is often called simply "the" moment of inertia even though it is not equivalent to the usual moment of inertia (which has dimensions of mass times length squared and characterizes the angular acceleration undergone by a solids when subjected to a torque).The second moment of area about the about the -axis is defined by(1)while more generally, the "product" moment of area is defined by(2)Here, the positive sign convention is used (e.g., Pilkey 2002, p. 15).More generally, the area moment of inertia tensor is given by(3)(4)by analogy with the moment of inertia tensor, which has negative..
The plane figure formed by a sequence of circles , , , ... that are all tangent to each other at the same point and such that the sequence of radii converges to zero. In the figure above, is chosen to be the circle with center and radius .The topology of this set and of its generalizations to higher dimensions has been intensively studied in recent years (Eda 2000, Eda and Kawamura 2002ab). This research was motivated by the following striking discovery: although the fundamental group of the circle is , the fundamental group of the figure eight is , where denotes the free product, and in general the fundamental group of the -petalled rose is (Massey 1989, pp. 123-125), the fundamental group of the Hawaiian ring is not a free group (Higman 1952, de Smit 1992, Black 1996).
An infinite sequence of circles such that every four consecutive circles are mutually tangent, and the circles' radii ..., , ..., , , , , , , ..., , , ..., are in geometric progression with ratiowhere is the golden ratio (Gardner 1979ab). Coxeter (1968) generalized the sequence to spheres.
A quadrilateral, sometimes also known as a tetragon or quadrangle (Johnson 1929, p. 61) is a four-sided polygon. If not explicitly stated, all four polygon vertices are generally taken to lie in a plane. (If the points do not lie in a plane, the quadrilateral is called a skew quadrilateral.) There are three topological types of quadrilaterals (Wenninger 1983, p. 50): convex quadrilaterals (left figure), concave quadrilaterals (middle figure), and crossed quadrilaterals (or butterflies, or bow-ties; right figure).A quadrilateral with two sides parallel is called a trapezoid, whereas a quadrilateral with opposite pairs of sides parallel is called a parallelogram.For a planar convex quadrilateral (left figure above), let the lengths of the sides be , , , and , the semiperimeter , and the polygon diagonals and . The polygon diagonals are perpendicular iff .An equation for the sum of the squares of side lengths is(1)where is the..
For a cyclic quadrilateral, the sum of theproducts of the two pairs of opposite sides equals the product of the diagonals(1)(Kimberling 1998, p. 223).This fact can be used to derive the trigonometryaddition formulas.Furthermore, the special case of the quadrilateral being a rectangle gives the Pythagorean theorem. In particular, let , , , , , and , so the general result is written(2)For a rectangle, , , and , so the theorem gives(3)
The figure formed when the midpoints of the sides of a convex quadrilateral are joined in order is a parallelogram. Equivalently, the bimedians bisect each other.The area of the Varignon parallelogram of a convex quadrilateral is half that of the quadrilateral, and the perimeter is equal to the sum of the diagonals of the original quadrilateral.
For a quadrilateral which is not cyclic,Ptolemy's theorem becomes an inequality:The Ptolemy inequality is still valid when is a triangular pyramid (Boomstra 1956-1957).
The figure formed when the midpoints of adjacent sides of a quadrilateral are joined. Varignon's theorem demonstrated that this figure is a parallelogram. The center of the Varignon parallelogram is the geometric centroid of four point masses placed on the vertices of the quadrilateral.
If a cyclic quadrilateral is inscribed in a circle of a coaxal system such that one pair of connectors touches another circle of the system at , then each pair of opposite connectors will touch a circle of the system ( at on , at on , at on , at on , and at on ), and the six points of contact , , , , , and will be collinear.The general theorem states that if , , ..., are any number of points taken in order on a circle of a given coaxal system so that , , ..., touch respectively fixed circles , , ..., of the system, then must touch a fixed circle of the system. Further, if , , ..., touch respectively any of the circles , , ..., , then must touch the remaining circle.
A quadrilateral in which a pair of opposite sides have the same length and are inclined at to each other (or equivalently, satisfy ). Some interesting theorems hold for such quadrilaterals. Let be an equilic quadrilateral with and . Then 1. The midpoints , , and of the diagonals and the side always determine an equilateral triangle. 2. If equilateral triangle is drawn outwardly on , then is also an equilateral triangle. 3. If equilateral triangles are drawn on , , and away from , then the three new graph vertices , , and are collinear. See Honsberger (1985) for additional theorems.
Given an arbitrary planar quadrilateral, place a square outwardly on each side, and connect the centers of opposite squares. Then van Aubel's theorem states that the two lines are of equal length and cross at a right angle.van Aubel's theorem is related to Napoleon's theorem and is a special case of the Petr-Neumann-Douglas theorem. It is sometimes (incorrectly) known simply as Aubel's theorem (Casey 1888; Wells 1991, p. 11; Kimberling 2003, p. 23).A second theorem sometimes known as van Aubel's theorem states that if is the Cevian triangle of a point , then
A parallelogram is a quadrilateral with opposite sides parallel (and therefore opposite angles equal). A quadrilateral with equal sides is called a rhombus, and a parallelogram whose angles are all right angles is called a rectangle. And, since a square is a degenerate case of a rectangle, both squares and rectangles are special types of parallelograms.The polygon diagonals of a parallelogram bisecteach other (Casey 1888, p. 2).The angles of a parallelogram satisfy the identities(1)(2)and(3)A parallelogram of base and height has area(4)The height of a parallelogram is(5)and the polygon diagonals and are(6)(7)(8)(9)(Beyer 1987).The sides , , , and diagonals , of a parallelogram satisfy(10)(Casey 1888, p. 22).The area of the parallelogram with sides formed by the vectors and is(11)(12)(13)where is the two-dimensional cross product and is the determinant.As shown by Euclid, if lines parallel to the sides are drawn through..
A trapezoid is a quadrilateral with two sides parallel. The trapezoid is equivalent to the British definition of trapezium (Bronshtein and Semendyayev 1977, p. 174). An isosceles trapezoid is a trapezoid in which the base angles are equal so . A right trapezoid is a trapezoid having two right angles.The area of the trapezoid is(1)(2)(3)The geometric centroid lies on the median between the base and top, and if the lower left-hand corner of the trapezoid is at the original, lies at(4)(5)(6)(cf. Harris and Stocker 1998, p. 83, who give but not ).The trapezoid depicted has central median(7)If vertical lines are extended from the endpoints of the upper side, the bases of the triangles formed on the left and right are(8)(9)respectively. This gives the vertex angles as(10)(11)(12)(13)from the lower left corner proceeding counterclockwise.In terms of the side length, the diagonals of the trapezoid are given by(14)(15)and the height..
The orthopoles of a line with respect to the four triangles formed by three out of four vertices of any quadrilateral lie on a straight line known as the orthopolar line of for the given quadrilateral (Servais 1923, McBrien 1942, Goormaghtigh 1947).
There are two common definitions of the trapezium. The American definition is a quadrilateral with no parallel sides; the British definition is a quadrilateral with two sides parallel (e.g., Bronshtein and Semendyayev 1977, p. 174)--which Americans call a trapezoid.Definitions for trapezoid and trapezium have causedcontroversy for more than two thousand years.Euclid (Book 1, Definition 22) stated, "Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has opposite sides and angles equal to one another but is neither equilateral nor right angled. And let quadrilaterals other than these be called trapezia."Proclus (also Heron and Posidonius) divided quadrilaterals into parallelograms and non-parallelograms. For the latter, Proclus assigned..
A quadrilateral which has an incircle, i.e., one for which a single circle can be constructed which is tangent to all four sides. Opposite sides of such a quadrilateral satisfy(1)where(2)is the semiperimeter, and the areais(3)where is the inradius. Using Bretschneider's formula together with (1) and (3) then gives the beautiful formula(4)(5)where and are the diagonal lengths.A rhombus is a special case of a tangential quadrilateral.
Given four points, , , , and , let be the orthocenter of . Then is the orthocenter , is the orthocenter of , and is the orthocenter of . The configuration is called an orthocentric quadrangle.
A cyclic quadrilateral is a quadrilateral for which a circle can be circumscribed so that it touches each polygon vertex. A quadrilateral that can be both inscribed and circumscribed on some pair of circles is known as a bicentric quadrilateral.The area of a cyclic quadrilateral is the maximum possible for any quadrilateral with the given side lengths. The opposite angles of a cyclic quadrilateral sum to radians (Euclid, Book III, Proposition 22; Heath 1956; Dunham 1990, p. 121). There exists a closed billiards path inside a cyclic quadrilateral if its circumcenter lies inside the quadrilateral (Wells 1991, p. 11).The area is then given by a special case of Bretschneider's formula. Let the sides have lengths , , , and , let be the semiperimeter(1)and let be the circumradius. Then(2)(3)the first of which is known as Brahmagupta's formula. Solving for the circumradius in (2) and (3) gives(4)The diagonals of a cyclic quadrilateral..
Let , , , and be four points on a circle, and , , , the orthocenters of triangles , etc. If, from the eight points, four with different subscripts are chosen such that three are from one set and the fourth from the other, these points form an orthocentric system. There are eight such systems, which are analogous to the six sets of orthocentric systems obtained using the feet of the angle bisectors, orthocenter, and polygon vertices of a generic triangle.On the other hand, if all the points are chosen from one set, or two from each set, with all different subscripts, the four points lie on a circle. There are four pairs of such circles, and eight points lie by fours on eight equal circles.The Simson line of with regard to triangle is the same as that of with regard to the triangle .
The figure determined by four lines, no three of which are concurrent, and their six points of intersection (Johnson 1929, pp. 61-62). Note that this figure is different from a complete quadrangle. A complete quadrilateral has three diagonals (compared to two for an ordinary quadrilateral). The midpoints of the diagonals of a complete quadrilateral are collinear on a line (Johnson 1929, pp. 152-153).A theorem due to Steiner (Mention 1862ab, Johnson 1929, Steiner 1971) states that in a complete quadrilateral, the bisectors of angles are concurrent at 16 points which are the incenters and excenters of the four triangles. Furthermore, these points are the intersections of two sets of four circles each of which is a member of a conjugate coaxal system. The axes of these systems intersect at the point common to the circumcircles of the quadrilateral.Newton proved that, if a conic section is inscribed in a complete quadrilateral,..
The term "square" can be used to mean either a square number (" is the square of ") or a geometric figure consisting of a convex quadrilateral with sides of equal length that are positioned at right angles to each other as illustrated above. In other words, a square is a regular polygon with four sides.When used as a symbol, denotes a square geometric figure with given vertices, while is sometimes used to denote a graph product (Clark and Suen 2000).A square is a special case of an isosceles trapezoid, kite, parallelogram, quadrilateral, rectangle, rhombus, and trapezoid.The diagonals of a square bisect one another and are perpendicular (illustrated in red in the figure above). In addition, they bisect each pair of opposite angles (illustrated in blue).The perimeter of a square with side length is(1)and the area is(2)The inradius , circumradius , and area can be computed directly from the formulas for a general regular polygon..
If the four points making up a quadrilateral are joined pairwise by six distinct lines, a figure known as a complete quadrangle results. A complete quadrangle is therefore a set of four points, no three collinear, and the six lines which join them. Note that a complete quadrilateral is different from a complete quadrangle.The midpoints of the sides of any complete quadrangle and the three diagonal points all lie on a conic known as the nine-point conic. If it is an orthocentric quadrilateral, the conic reduces to a circle.
A four-sided quadrilateral not contained in a plane. The lines connecting the midpoints of opposite sides of a skew quadrilateral intersect (and bisect) each other (Steinhaus 1999).The problem of finding the minimum bounding surface of a skew quadrilateral was solved by Schwarz (Schwarz 1890, Wells 1991) in terms of Abelian integrals and has the shape of a saddle. It is given by solving
An equilateral parallelogram whose acute angles are . Sometimes, the restriction to is dropped, and it is required only that two opposite angles are acute and the other two obtuse. The term rhombus is commonly used for an arbitrary equilateral parallelogram.The area of a lozenge of side length is(1)its diagonals have lengths(2)(3)and it has inradius(4)
Given a general quadrilateral with sides of lengths , , , and , the area is given by(1)(2)(Coolidge 1939; Ivanov 1960; Beyer 1987, p. 123) where and are the diagonal lengths and is the semiperimeter. While this formula is termed Bretschneider's formula in Ivanoff (1960) and Beyer (1987, p. 123), this appears to be a misnomer. Coolidge (1939) gives the second form of this formula, stating "here is one [formula] which, so far as I can find out, is new," while at the same time crediting Bretschneider (1842) and Strehlke (1842) with "rather clumsy" proofs of the related formula(3)(Bretschneider 1842; Strehlke 1842; Coolidge 1939; Beyer 1987, p. 123), where and are two opposite angles of the quadrilateral."Bretschneider's formula" can be derived by representing the sides of the quadrilateral by the vectors , , , and arranged such that and the diagonals by the vectors and arranged so that and . The..