Misplaced Pages

#307692

64-533: Pentangle may refer to: Pentagon , a five-sided polygon Pentagram , a five-pointed star drawn with five straight strokes Pentangle (band) , a British folk rock band The Pentangle (album) , a 1968 album by Pentangle Miss Pentangle, a character from The Worst Witch Pentangle Puzzles , a manufacturer and distributor of mechanical puzzles See also [ edit ] Pentacle (disambiguation) Pentagram (disambiguation) Topics referred to by

128-472: A : c a c + a − b : a b a + b − c . {\displaystyle {\frac {bc}{b+c-a}}:{\frac {ca}{c+a-b}}:{\frac {ab}{a+b-c}}.} An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two . Every triangle has three distinct excircles, each tangent to one of

192-508: A , y a ) {\displaystyle (x_{a},y_{a})} , ( x b , y b ) {\displaystyle (x_{b},y_{b})} , and ( x c , y c ) {\displaystyle (x_{c},y_{c})} , and the sides opposite these vertices have corresponding lengths a {\displaystyle a} , b {\displaystyle b} , and c {\displaystyle c} , then

256-457: A {\displaystyle a} be the length of B C ¯ {\displaystyle {\overline {BC}}} , b {\displaystyle b} the length of A C ¯ {\displaystyle {\overline {AC}}} , and c {\displaystyle c} the length of A B ¯ {\displaystyle {\overline {AB}}} . Now,

320-658: A {\displaystyle a} be the length of B C ¯ {\displaystyle {\overline {BC}}} , b {\displaystyle b} the length of A C ¯ {\displaystyle {\overline {AC}}} , and c {\displaystyle c} the length of A B ¯ {\displaystyle {\overline {AB}}} . Also let T A {\displaystyle T_{A}} , T B {\displaystyle T_{B}} , and T C {\displaystyle T_{C}} be

384-480: A {\displaystyle a} , b {\displaystyle b} , and c {\displaystyle c} are the side lengths of the original triangle. This is the same area as that of the extouch triangle . The three lines A T A {\displaystyle AT_{A}} , B T B {\displaystyle BT_{B}} and C T C {\displaystyle CT_{C}} intersect in

448-534: A ) = s − a . {\displaystyle d\left(A,T_{B}\right)=d\left(A,T_{C}\right)={\tfrac {1}{2}}(b+c-a)=s-a.} If the altitudes from sides of lengths a {\displaystyle a} , b {\displaystyle b} , and c {\displaystyle c} are h a {\displaystyle h_{a}} , h b {\displaystyle h_{b}} , and h c {\displaystyle h_{c}} , then

512-492: A + b + c . {\displaystyle \left({\frac {ax_{a}+bx_{b}+cx_{c}}{a+b+c}},{\frac {ay_{a}+by_{b}+cy_{c}}{a+b+c}}\right)={\frac {a\left(x_{a},y_{a}\right)+b\left(x_{b},y_{b}\right)+c\left(x_{c},y_{c}\right)}{a+b+c}}.} The inradius r {\displaystyle r} of the incircle in a triangle with sides of length a {\displaystyle a} , b {\displaystyle b} , c {\displaystyle c}

576-419: A , b , c , d , e {\displaystyle a,b,c,d,e} and diagonals d 1 , d 2 , d 3 , d 4 , d 5 {\displaystyle d_{1},d_{2},d_{3},d_{4},d_{5}} , the following inequality holds: A regular pentagon cannot appear in any tiling of regular polygons. First, to prove a pentagon cannot form

640-732: A r {\displaystyle {\tfrac {1}{2}}ar} . Since these three triangles decompose △ A B C {\displaystyle \triangle ABC} , we see that the area Δ  of  △ A B C {\displaystyle \Delta {\text{ of }}\triangle ABC} is: Δ = 1 2 ( a + b + c ) r = s r , {\displaystyle \Delta ={\tfrac {1}{2}}(a+b+c)r=sr,}      and      r = Δ s , {\displaystyle r={\frac {\Delta }{s}},} where Δ {\displaystyle \Delta }

704-519: A convex regular pentagon are in the golden ratio to its sides. Given its side length t , {\displaystyle t,} its height H {\displaystyle H} (distance from one side to the opposite vertex), width W {\displaystyle W} (distance between two farthest separated points, which equals the diagonal length D {\displaystyle D} ) and circumradius R {\displaystyle R} are given by: The area of

SECTION 10

#1732902392308

768-470: A regular tiling (one in which all faces are congruent, thus requiring that all the polygons be pentagons), observe that 360° / 108° = 3 1 ⁄ 3 (where 108° Is the interior angle), which is not a whole number; hence there exists no integer number of pentagons sharing a single vertex and leaving no gaps between them. More difficult is proving a pentagon cannot be in any edge-to-edge tiling made by regular polygons: The maximum known packing density of

832-433: A convex regular pentagon with side length t {\displaystyle t} is given by If the circumradius R {\displaystyle R} of a regular pentagon is given, its edge length t {\displaystyle t} is found by the expression and its area is since the area of the circumscribed circle is π R 2 , {\displaystyle \pi R^{2},}

896-434: A given edge. This process was described by Euclid in his Elements circa 300 BC. The regular pentagon has Dih 5 symmetry , order 10. Since 5 is a prime number there is one subgroup with dihedral symmetry: Dih 1 , and 2 cyclic group symmetries: Z 5 , and Z 1 . These 4 symmetries can be seen in 4 distinct symmetries on the pentagon. John Conway labels these by a letter and group order. Full symmetry of

960-403: A point in the triangle is the ratio of all the distances to the triangle sides. Because the incenter is the same distance from all sides of the triangle, the trilinear coordinates for the incenter are   1 : 1 : 1. {\displaystyle \ 1:1:1.} The barycentric coordinates for a point in a triangle give weights such that the point is the weighted average of

1024-465: A range of sets of values, thus permitting it to form a family of pentagons. In contrast, the regular pentagon is unique up to similarity, because it is equilateral and it is equiangular (its five angles are equal). A cyclic pentagon is one for which a circle called the circumcircle goes through all five vertices. The regular pentagon is an example of a cyclic pentagon. The area of a cyclic pentagon, whether regular or not, can be expressed as one fourth

1088-444: A regular pentagon is ( 5 − 5 ) / 3 ≈ 0.921 {\displaystyle (5-{\sqrt {5}})/3\approx 0.921} , achieved by the double lattice packing shown. In a preprint released in 2016, Thomas Hales and Wöden Kusner announced a proof that this double lattice packing of the regular pentagon (known as the "pentagonal ice-ray" Chinese lattice design, dating from around 1900) has

1152-406: A right-angled triangle with one side equal to r {\displaystyle r} and the other side equal to r cot ⁡ A 2 {\displaystyle r\cot {\tfrac {A}{2}}} . The same is true for △ I B ′ A {\displaystyle \triangle IB'A} . The large triangle is composed of six such triangles and

1216-533: A single point called the Gergonne point , denoted as G e {\displaystyle G_{e}} (or triangle center X 7 ). The Gergonne point lies in the open orthocentroidal disk punctured at its own center, and can be any point therein. The Gergonne point of a triangle has a number of properties, including that it is the symmedian point of the Gergonne triangle. Trilinear coordinates for

1280-434: A triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle. The incircle radius is no greater than one-ninth the sum of the altitudes. The squared distance from the incenter I {\displaystyle I} to the circumcenter O {\displaystyle O}

1344-398: Is K T = K 2 r 2 s a b c {\displaystyle K_{T}=K{\frac {2r^{2}s}{abc}}} where K {\displaystyle K} , r {\displaystyle r} , and s {\displaystyle s} are the area, radius of the incircle , and semiperimeter of the original triangle, and

SECTION 20

#1732902392308

1408-529: Is a Fermat prime . A variety of methods are known for constructing a regular pentagon. Some are discussed below. One method to construct a regular pentagon in a given circle is described by Richmond and further discussed in Cromwell's Polyhedra . The top panel shows the construction used in Richmond's method to create the side of the inscribed pentagon. The circle defining the pentagon has unit radius. Its center

1472-536: Is defined by the three touchpoints of the incircle on the three sides. The touchpoint opposite A {\displaystyle A} is denoted T A {\displaystyle T_{A}} , etc. This Gergonne triangle, △ T A T B T C {\displaystyle \triangle T_{A}T_{B}T_{C}} , is also known as the contact triangle or intouch triangle of △ A B C {\displaystyle \triangle ABC} . Its area

1536-497: Is different from Wikidata All article disambiguation pages All disambiguation pages Pentagon A pentagon may be simple or self-intersecting . A self-intersecting regular pentagon (or star pentagon ) is called a pentagram . A regular pentagon has Schläfli symbol {5} and interior angles of 108°. A regular pentagon has five lines of reflectional symmetry , and rotational symmetry of order 5 (through 72°, 144°, 216° and 288°). The diagonals of

1600-549: Is given by O I ¯ 2 = R ( R − 2 r ) = a b c a + b + c [ a b c ( a + b − c ) ( a − b + c ) ( − a + b + c ) − 1 ] {\displaystyle {\overline {OI}}^{2}=R(R-2r)={\frac {a\,b\,c\,}{a+b+c}}\left[{\frac {a\,b\,c\,}{(a+b-c)\,(a-b+c)\,(-a+b+c)}}-1\right]} and

1664-411: Is given by r = ( s − a ) ( s − b ) ( s − c ) s , {\displaystyle r={\sqrt {\frac {(s-a)(s-b)(s-c)}{s}}},} where s = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} is the semiperimeter. The tangency points of

1728-412: Is located at point C and a midpoint M is marked halfway along its radius. This point is joined to the periphery vertically above the center at point D . Angle CMD is bisected, and the bisector intersects the vertical axis at point Q . A horizontal line through Q intersects the circle at point P , and chord PD is the required side of the inscribed pentagon. To determine the length of this side,

1792-400: Is the area of △ A B C {\displaystyle \triangle ABC} and s = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} is its semiperimeter . For an alternative formula, consider △ I T C A {\displaystyle \triangle IT_{C}A} . This is

1856-495: Is the radius r of the inscribed circle, of a regular pentagon is related to the side length t by Like every regular convex polygon, the regular convex pentagon has a circumscribed circle . For a regular pentagon with successive vertices A, B, C, D, E, if P is any point on the circumcircle between points B and C, then PA + PD = PB + PC + PE. For an arbitrary point in the plane of a regular pentagon with circumradius R {\displaystyle R} , whose distances to

1920-529: Is truly the side of a regular pentagon, m ∠ C D P = 54 ∘ {\displaystyle m\angle \mathrm {CDP} =54^{\circ }} , so DP = 2 cos(54°), QD = DP cos(54°) = 2cos (54°), and CQ = 1 − 2cos (54°), which equals −cos(108°) by the cosine double angle formula . This is the cosine of 72°, which equals ( 5 − 1 ) / 4 {\displaystyle \left({\sqrt {5}}-1\right)/4} as desired. The Carlyle circle

1984-1024: The Law of sines in the triangle △ I A B {\displaystyle \triangle IAB} . We get A I ¯ sin ⁡ B 2 = c sin ⁡ ∠ A I B {\displaystyle {\frac {\overline {AI}}{\sin {\frac {B}{2}}}}={\frac {c}{\sin \angle AIB}}} . We have that ∠ A I B = π − A 2 − B 2 = π 2 + C 2 {\displaystyle \angle AIB=\pi -{\frac {A}{2}}-{\frac {B}{2}}={\frac {\pi }{2}}+{\frac {C}{2}}} . It follows that A I ¯ = c   sin ⁡ B 2 cos ⁡ C 2 {\displaystyle {\overline {AI}}=c\ {\frac {\sin {\frac {B}{2}}}{\cos {\frac {C}{2}}}}} . The equality with

Pentangle - Misplaced Pages Continue

2048-426: The g5 subgroup has no degrees of freedom but can be seen as directed edges . A pentagram or pentangle is a regular star pentagon. Its Schläfli symbol is {5/2}. Its sides form the diagonals of a regular convex pentagon – in this arrangement the sides of the two pentagons are in the golden ratio . An equilateral pentagon is a polygon with five sides of equal length. However, its five internal angles can take

2112-424: The Gergonne point are given by sec 2 ⁡ A 2 : sec 2 ⁡ B 2 : sec 2 ⁡ C 2 , {\displaystyle \sec ^{2}{\tfrac {A}{2}}:\sec ^{2}{\tfrac {B}{2}}:\sec ^{2}{\tfrac {C}{2}},} or, equivalently, by the Law of Sines , b c b + c −

2176-445: The angles at the three vertices. The Cartesian coordinates of the incenter are a weighted average of the coordinates of the three vertices using the side lengths of the triangle relative to the perimeter (that is, using the barycentric coordinates given above, normalized to sum to unity) as weights. The weights are positive so the incenter lies inside the triangle as stated above. If the three vertices are located at ( x

2240-442: The centroid of the regular pentagon and its five vertices are L {\displaystyle L} and d i {\displaystyle d_{i}} respectively, we have If d i {\displaystyle d_{i}} are the distances from the vertices of a regular pentagon to any point on its circumcircle, then The regular pentagon is constructible with compass and straightedge , as 5

2304-425: The distance from the incenter to the center N {\displaystyle N} of the nine point circle is I N ¯ = 1 2 ( R − 2 r ) < 1 2 R . {\displaystyle {\overline {IN}}={\tfrac {1}{2}}(R-2r)<{\tfrac {1}{2}}R.} The incenter lies in the medial triangle (whose vertices are

2368-457: The excircles are called the exradii . The exradius of the excircle opposite A {\displaystyle A} (so touching B C {\displaystyle BC} , centered at J A {\displaystyle J_{A}} ) is r a = r s s − a = s ( s − b ) ( s − c ) s −

2432-500: The incenter is at ( a x a + b x b + c x c a + b + c , a y a + b y b + c y c a + b + c ) = a ( x a , y a ) + b ( x b , y b ) + c ( x c , y c )

2496-745: The incenter of △ A B C {\displaystyle \triangle ABC} as I {\displaystyle I} . The distance from vertex A {\displaystyle A} to the incenter I {\displaystyle I} is: A I ¯ = d ( A , I ) = c sin ⁡ B 2 cos ⁡ C 2 = b sin ⁡ C 2 cos ⁡ B 2 . {\displaystyle {\overline {AI}}=d(A,I)=c\,{\frac {\sin {\frac {B}{2}}}{\cos {\frac {C}{2}}}}=b\,{\frac {\sin {\frac {C}{2}}}{\cos {\frac {B}{2}}}}.} Use

2560-447: The incircle divide the sides into segments of lengths s − a {\displaystyle s-a} from A {\displaystyle A} , s − b {\displaystyle s-b} from B {\displaystyle B} , and s − c {\displaystyle s-c} from C {\displaystyle C} . See Heron's formula . Denote

2624-1121: The incircle is tangent to A B ¯ {\displaystyle {\overline {AB}}} at some point T C {\displaystyle T_{C}} , and so ∠ A T C I {\displaystyle \angle AT_{C}I} is right. Thus, the radius T C I {\displaystyle T_{C}I} is an altitude of △ I A B {\displaystyle \triangle IAB} . Therefore, △ I A B {\displaystyle \triangle IAB} has base length c {\displaystyle c} and height r {\displaystyle r} , and so has area 1 2 c r {\displaystyle {\tfrac {1}{2}}cr} . Similarly, △ I A C {\displaystyle \triangle IAC} has area 1 2 b r {\displaystyle {\tfrac {1}{2}}br} and △ I B C {\displaystyle \triangle IBC} has area 1 2

Pentangle - Misplaced Pages Continue

2688-494: The incircle radius r {\displaystyle r} and the circumcircle radius R {\displaystyle R} of a triangle with sides a {\displaystyle a} , b {\displaystyle b} , and c {\displaystyle c} is r R = a b c 2 ( a + b + c ) . {\displaystyle rR={\frac {abc}{2(a+b+c)}}.} Some relations among

2752-402: The inradius r {\displaystyle r} is one-third of the harmonic mean of these altitudes; that is, r = 1 1 h a + 1 h b + 1 h c . {\displaystyle r={\frac {1}{{\dfrac {1}{h_{a}}}+{\dfrac {1}{h_{b}}}+{\dfrac {1}{h_{c}}}}}.} The product of

2816-908: The internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system . While the incenter of △ A B C {\displaystyle \triangle ABC} has trilinear coordinates 1 : 1 : 1 {\displaystyle 1:1:1} , the excenters have trilinears J A = − 1 : 1 : 1 J B = 1 : − 1 : 1 J C = 1 : 1 : − 1 {\displaystyle {\begin{array}{rrcrcr}J_{A}=&-1&:&1&:&1\\J_{B}=&1&:&-1&:&1\\J_{C}=&1&:&1&:&-1\end{array}}} The radii of

2880-646: The internal bisector of one angle (at vertex A , for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex A , or the excenter of A . Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system . Suppose △ A B C {\displaystyle \triangle ABC} has an incircle with radius r {\displaystyle r} and center I {\displaystyle I} . Let

2944-588: The midpoints of the sides). The radius of the incircle is related to the area of the triangle. The ratio of the area of the incircle to the area of the triangle is less than or equal to π / 3 3 {\displaystyle \pi {\big /}3{\sqrt {3}}} , with equality holding only for equilateral triangles . Suppose △ A B C {\displaystyle \triangle ABC} has an incircle with radius r {\displaystyle r} and center I {\displaystyle I} . Let

3008-410: The optimal density among all packings of regular pentagons in the plane. There are no combinations of regular polygons with 4 or more meeting at a vertex that contain a pentagon. For combinations with 3, if 3 polygons meet at a vertex and one has an odd number of sides, the other 2 must be congruent. The reason for this is that the polygons that touch the edges of the pentagon must alternate around

3072-441: The pentagon, which is impossible because of the pentagon's odd number of sides. For the pentagon, this results in a polygon whose angles are all (360 − 108) / 2 = 126° . To find the number of sides this polygon has, the result is 360 / (180 − 126) = 6 2 ⁄ 3 , which is not a whole number. Therefore, a pentagon cannot appear in any tiling made by regular polygons. There are 15 classes of pentagons that can monohedrally tile

3136-446: The plane . None of the pentagons have any symmetry in general, although some have special cases with mirror symmetry. Inradius In geometry , the incircle or inscribed circle of a triangle is the largest circle that can be contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter . An excircle or escribed circle of

3200-446: The regular form is r10 and no symmetry is labeled a1 . The dihedral symmetries are divided depending on whether they pass through vertices ( d for diagonal) or edges ( p for perpendiculars), and i when reflection lines path through both edges and vertices. Cyclic symmetries in the middle column are labeled as g for their central gyration orders. Each subgroup symmetry allows one or more degrees of freedom for irregular forms. Only

3264-435: The regular pentagon fills approximately 0.7568 of its circumscribed circle. The area of any regular polygon is: where P is the perimeter of the polygon, and r is the inradius (equivalently the apothem ). Substituting the regular pentagon's values for P and r gives the formula with side length t . Similar to every regular convex polygon, the regular convex pentagon has an inscribed circle . The apothem , which

SECTION 50

#1732902392308

3328-415: The same term [REDACTED] This disambiguation page lists articles associated with the title Pentangle . If an internal link led you here, you may wish to change the link to point directly to the intended article. Retrieved from " https://en.wikipedia.org/w/index.php?title=Pentangle&oldid=1233704009 " Category : Disambiguation pages Hidden categories: Short description

3392-1441: The second expression is obtained the same way. The distances from the incenter to the vertices combined with the lengths of the triangle sides obey the equation I A ¯ ⋅ I A ¯ C A ¯ ⋅ A B ¯ + I B ¯ ⋅ I B ¯ A B ¯ ⋅ B C ¯ + I C ¯ ⋅ I C ¯ B C ¯ ⋅ C A ¯ = 1. {\displaystyle {\frac {{\overline {IA}}\cdot {\overline {IA}}}{{\overline {CA}}\cdot {\overline {AB}}}}+{\frac {{\overline {IB}}\cdot {\overline {IB}}}{{\overline {AB}}\cdot {\overline {BC}}}}+{\frac {{\overline {IC}}\cdot {\overline {IC}}}{{\overline {BC}}\cdot {\overline {CA}}}}=1.} Additionally, I A ¯ ⋅ I B ¯ ⋅ I C ¯ = 4 R r 2 , {\displaystyle {\overline {IA}}\cdot {\overline {IB}}\cdot {\overline {IC}}=4Rr^{2},} where R {\displaystyle R} and r {\displaystyle r} are

3456-512: The sides, incircle radius, and circumcircle radius are: a b + b c + c a = s 2 + ( 4 R + r ) r , a 2 + b 2 + c 2 = 2 s 2 − 2 ( 4 R + r ) r . {\displaystyle {\begin{aligned}ab+bc+ca&=s^{2}+(4R+r)r,\\a^{2}+b^{2}+c^{2}&=2s^{2}-2(4R+r)r.\end{aligned}}} Any line through

3520-441: The square root of one of the roots of a septic equation whose coefficients are functions of the sides of the pentagon. There exist cyclic pentagons with rational sides and rational area; these are called Robbins pentagons . It has been proven that the diagonals of a Robbins pentagon must be either all rational or all irrational, and it is conjectured that all the diagonals must be rational. For all convex pentagons with sides

3584-455: The total area is: Δ = r 2 ( cot ⁡ A 2 + cot ⁡ B 2 + cot ⁡ C 2 ) . {\displaystyle \Delta =r^{2}\left(\cot {\tfrac {A}{2}}+\cot {\tfrac {B}{2}}+\cot {\tfrac {C}{2}}\right).} The Gergonne triangle (of △ A B C {\displaystyle \triangle ABC} )

3648-633: The touchpoints where the incircle touches B C ¯ {\displaystyle {\overline {BC}}} , A C ¯ {\displaystyle {\overline {AC}}} , and A B ¯ {\displaystyle {\overline {AB}}} . The incenter is the point where the internal angle bisectors of ∠ A B C , ∠ B C A ,  and  ∠ B A C {\displaystyle \angle ABC,\angle BCA,{\text{ and }}\angle BAC} meet. The trilinear coordinates for

3712-403: The triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two . Every triangle has three distinct excircles, each tangent to one of the triangle's sides. The center of the incircle, called the incenter , can be found as the intersection of the three internal angle bisectors . The center of an excircle is the intersection of

3776-767: The triangle vertex positions. Barycentric coordinates for the incenter are given by   a : b : c {\displaystyle \ a:b:c} where a {\displaystyle a} , b {\displaystyle b} , and c {\displaystyle c} are the lengths of the sides of the triangle, or equivalently (using the law of sines ) by sin ⁡ A : sin ⁡ B : sin ⁡ C {\displaystyle \sin A:\sin B:\sin C} where A {\displaystyle A} , B {\displaystyle B} , and C {\displaystyle C} are

3840-516: The triangle's circumradius and inradius respectively. The collection of triangle centers may be given the structure of a group under coordinate-wise multiplication of trilinear coordinates; in this group, the incenter forms the identity element . The distances from a vertex to the two nearest touchpoints are equal; for example: d ( A , T B ) = d ( A , T C ) = 1 2 ( b + c −

3904-426: The triangle's sides. The center of an excircle is the intersection of the internal bisector of one angle (at vertex A {\displaystyle A} , for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex A {\displaystyle A} , or the excenter of A {\displaystyle A} . Because

SECTION 60

#1732902392308

3968-430: The two right triangles DCM and QCM are depicted below the circle. Using Pythagoras' theorem and two sides, the hypotenuse of the larger triangle is found as 5 / 2 {\displaystyle \scriptstyle {\sqrt {5}}/2} . Side h of the smaller triangle then is found using the half-angle formula : where cosine and sine of ϕ are known from the larger triangle. The result is: If DP

4032-947: The vertices of the intouch triangle are given by T A = 0 : sec 2 ⁡ B 2 : sec 2 ⁡ C 2 T B = sec 2 ⁡ A 2 : 0 : sec 2 ⁡ C 2 T C = sec 2 ⁡ A 2 : sec 2 ⁡ B 2 : 0. {\displaystyle {\begin{array}{ccccccc}T_{A}&=&0&:&\sec ^{2}{\frac {B}{2}}&:&\sec ^{2}{\frac {C}{2}}\\[2pt]T_{B}&=&\sec ^{2}{\frac {A}{2}}&:&0&:&\sec ^{2}{\frac {C}{2}}\\[2pt]T_{C}&=&\sec ^{2}{\frac {A}{2}}&:&\sec ^{2}{\frac {B}{2}}&:&0.\end{array}}} Trilinear coordinates for

4096-401: Was invented as a geometric method to find the roots of a quadratic equation . This methodology leads to a procedure for constructing a regular pentagon. The steps are as follows: Steps 6–8 are equivalent to the following version, shown in the animation: A regular pentagon is constructible using a compass and straightedge , either by inscribing one in a given circle or constructing one on

#307692