In geometry , the CIRCUMSCRIBED CIRCLE or CIRCUMCIRCLE of a polygon is a circle which passes through all the vertices of the polygon. The center of this circle is called the CIRCUMCENTER and its radius is called the CIRCUMRADIUS. A polygon which has a circumscribed circle is called a CYCLIC POLYGON (sometimes a CONCYCLIC POLYGON, because the vertices are concyclic ). All regular simple polygons , all isosceles trapezoids , all triangles and all rectangles are cyclic. A related notion is the one of a minimum bounding circle , which is the smallest circle that completely contains the polygon within it. Not every polygon has a circumscribed circle, as the vertices of a polygon do not need to all lie on a circle, but every polygon has a unique minimum bounding circle, which may be constructed by a linear time algorithm. Even if a polygon has a circumscribed circle, it may not coincide with its minimum bounding circle; for example, for an obtuse triangle , the minimum bounding circle has the longest side as diameter and does not pass through the opposite vertex. CONTENTS * 1 Triangles * 1.1 Straightedge and compass construction * 1.2 Alternate construction * 1.3 Circumcircle equations * 1.3.1
* 1.4 Circumcenter coordinates * 1.4.1
* 1.5 Angles
* 1.6
* 2 Cyclic quadrilaterals * 3 Cyclic n-gons * 3.1 Point on the circumcircle
* 3.2
* 4 See also * 5 Notes * 6 References * 7 External links * 7.1
TRIANGLES All triangles are cyclic; i.e., every triangle has a circumscribed circle. This can be proven on the grounds that the general equation for a circle with center (a, b) and radius r in the Cartesian coordinate system is ( x a ) 2 + ( y b ) 2 = r 2 . {displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}.} Since this equation has three parameters (a, b, r) only three points' coordinate pairs are required to determine the equation of a circle. Since a triangle is defined by its three vertices, and exactly three points are required to determine a circle, every triangle can be circumscribed. STRAIGHTEDGE AND COMPASS CONSTRUCTION Construction of the circumcircle (red) and the circumcenter Q (red dot) The circumcenter of a triangle can be constructed by drawing any two of the three perpendicular bisectors . The center is the point where the perpendicular bisectors intersect, and the radius is the length to any of the three vertices. This is because the circumcenter is equidistant from any pair of the triangle's vertices, and all points on the perpendicular bisectors are equidistant from two of the vertices of the triangle. ALTERNATE CONSTRUCTION Alternate construction of the circumcenter (intersection of broken lines) An alternate method to determine the circumcenter is to draw any two lines each one departing from one of the vertices at an angle with the common side, the common angle of departure being 90° minus the angle of the opposite vertex. (In the case of the opposite angle being obtuse, drawing a line at a negative angle means going outside the triangle.) In coastal navigation , a triangle's circumcircle is sometimes used as a way of obtaining a position line using a sextant when no compass is available. The horizontal angle between two landmarks defines the circumcircle upon which the observer lies. CIRCUMCIRCLE EQUATIONS Cartesian Coordinates In the
Hence, given the radius, r, center, Pc, a point on the circle, P0 and a unit normal of the plane containing the circle, n {displaystyle scriptstyle {hat {n}}} , one parametric equation of the circle starting from the point P0 and proceeding in a positively oriented (i.e., right-handed ) sense about n {displaystyle scriptstyle {hat {n}}} is the following: R ( s ) = P c + cos ( s r ) ( P 0 P c ) + sin ( s r ) . {displaystyle mathrm {R} (s)=mathrm {P_{c}} +cos left({frac {mathrm {s} }{mathrm {r} }}right)(P_{0}-P_{c})+sin left({frac {mathrm {s} }{mathrm {r} }}right)left.} Trilinear And Barycentric Coordinates An equation for the circumcircle in trilinear coordinates x : y : z is :p. 199 a/x + b/y + c/z = 0. An equation for the circumcircle in barycentric coordinates x : y : z is a2/x + b2/y + c2/z = 0. The isogonal conjugate of the circumcircle is the line at infinity, given in trilinear coordinates by ax + by + cz = 0 and in barycentric coordinates by x + y + z = 0. Higher Dimensions Additionally, the circumcircle of a triangle embedded in d dimensions can be found using a generalized method. Let A, B, and C be d-dimensional points, which form the vertices of a triangle. We start by transposing the system to place C at the origin: a = A C , b = B C . {displaystyle {begin{aligned}mathbf {a} &=mathbf {A} -mathbf {C} ,\mathbf {b} width:12.836ex; height:5.843ex;" alt="{displaystyle {begin{aligned}mathbf {a} &=mathbf {A} -mathbf {C} ,\mathbf {b} "> r = a b a b 2 a b = a b 2 sin = A B 2 sin , {displaystyle r={frac {leftmathbf {a} rightleftmathbf {b} rightleftmathbf {a} -mathbf {b} right}{2leftmathbf {a} times mathbf {b} right}}={frac {leftmathbf {a} -mathbf {b} right}{2sin theta }}={frac {leftmathbf {A} -mathbf {B} right}{2sin theta }},} where θ is the interior angle between A and B. The circumcenter, p0, is given by p 0 = ( a 2 b b 2 a ) ( a b ) 2 a b 2 + C . {displaystyle p_{0}={frac {(leftmathbf {a} right^{2}mathbf {b} -leftmathbf {b} right^{2}mathbf {a} )times (mathbf {a} times mathbf {b} )}{2leftmathbf {a} times mathbf {b} right^{2}}}+mathbf {C} .} This formula only works in three dimensions as the cross product is not defined in other dimensions, but it can be generalized to the other dimensions by replacing the cross products with following identities: ( a b ) c = ( a c ) b ( b c ) a , a ( b c ) = ( a c ) b ( a b ) c , a b = a 2 b 2 ( a b ) 2 . {displaystyle {begin{aligned}(mathbf {a} times mathbf {b} )times mathbf {c} &=(mathbf {a} cdot mathbf {c} )mathbf {b} -(mathbf {b} cdot mathbf {c} )mathbf {a} ,\mathbf {a} times (mathbf {b} times mathbf {c} )&=(mathbf {a} cdot mathbf {c} )mathbf {b} -(mathbf {a} cdot mathbf {b} )mathbf {c} ,\leftmathbf {a} times mathbf {b} right width:38.246ex; height:11.176ex;" alt="{displaystyle {begin{aligned}(mathbf {a} times mathbf {b} )times mathbf {c} &=(mathbf {a} cdot mathbf {c} )mathbf {b} -(mathbf {b} cdot mathbf {c} )mathbf {a} ,\mathbf {a} times (mathbf {b} times mathbf {c} )&=(mathbf {a} cdot mathbf {c} )mathbf {b} -(mathbf {a} cdot mathbf {b} )mathbf {c} ,\leftmathbf {a} times mathbf {b} right"> U = ( U x , U y ) {displaystyle U=(U_{x},U_{y})} are U x = 1 D U y = 1 D {displaystyle {begin{aligned}U_{x}&={frac {1}{D}}left\U_{y} margin-bottom: -0.251ex; width:80.303ex; height:10.843ex;" alt="{displaystyle {begin{aligned}U_{x}&={frac {1}{D}}left\U_{y}"> D = 2 . {displaystyle D=2left.,} Without loss of generality this can be expressed in a simplified form after translation of the vertex A to the origin of the Cartesian coordinate systems, i.e., when A′ = A − A = (A′x,A′y) = (0,0). In this case, the coordinates of the vertices B′ = B − A and C′ = C − A represent the vectors from vertex A′ to these vertices. Observe that this trivial translation is possible for all triangles and the circumcenter U = ( U x , U y ) {displaystyle U'=(U'_{x},U'_{y})} of the triangle A′B′C′ follow as U x = 1 D , U y = 1 D {displaystyle {begin{aligned}U'_{x}&={frac {1}{D'}}left,\U'_{y} margin-bottom: -0.251ex; width:45.084ex; height:10.843ex;" alt="{displaystyle {begin{aligned}U_{x}&={frac {1}{D}}left,\U_{y}&={frac {1}{D}}leftend{aligned}}}" /> with D = 2 ( B x C y B y C x ) . {displaystyle D'=2(B'_{x}C'_{y}-B'_{y}C'_{x}).,} Due to the translation of vertex A to the origin, the circumradius r can be computed as r = U = U x 2 + U y 2 {displaystyle r=U'={sqrt {{U'_{x}}^{2}+{U'_{y}}^{2}}}} and the actual circumcenter of ABC follows as U = U + A {displaystyle U=U'+A} Trilinear Coordinates The circumcenter has trilinear coordinates :p.19 cos α : cos β : cos γ where α, β, γ are the angles of the triangle. In terms of the side lengths a, b, c, the trilinears are a ( b 2 + c 2 a 2 ) : b ( c 2 + a 2 b 2 ) : c ( a 2 + b 2 c 2 ) . {displaystyle a(b^{2}+c^{2}-a^{2}):b(c^{2}+a^{2}-b^{2}):c(a^{2}+b^{2}-c^{2}).} Barycentric Coordinates The circumcenter has barycentric coordinates a 2 ( b 2 + c 2 a 2 ) : b 2 ( c 2 + a 2 b 2 ) : c 2 ( a 2 + b 2 c 2 ) , {displaystyle a^{2}(b^{2}+c^{2}-a^{2}):;b^{2}(c^{2}+a^{2}-b^{2}):;c^{2}(a^{2}+b^{2}-c^{2}),,} where a, b, c are edge lengths (BC, CA, AB respectively) of the triangle. In terms of the triangle's angles , , , {displaystyle alpha ,beta ,gamma ,} the barycentric coordinates of the circumcenter are sin 2 : sin 2 : sin 2 . {displaystyle sin 2alpha :sin 2beta :sin 2gamma .} Circumcenter Vector Since the
Here U is the vector of the circumcenter and A, B, C are the vertex vectors. The divisor here equals 16S 2 where S is the area of the triangle. Cartesian Coordinates From Cross- And Dot-products In
Then the radius of the circle is given by r = P 1 P 2 P 2 P 3 P 3 P 1 2 ( P 1 P 2 ) ( P 2 P 3 ) {displaystyle mathrm {r} ={frac {leftP_{1}-P_{2}rightleftP_{2}-P_{3}rightleftP_{3}-P_{1}right}{2leftleft(P_{1}-P_{2}right)times left(P_{2}-P_{3}right)right}}} The center of the circle is given by the linear combination P c = P 1 + P 2 + P 3 {displaystyle mathrm {P_{c}} =alpha ,P_{1}+beta ,P_{2}+gamma ,P_{3}} where = P 2 P 3 2 ( P 1 P 2 ) ( P 1 P 3 ) 2 ( P 1 P 2 ) ( P 2 P 3 ) 2 = P 1 P 3 2 ( P 2 P 1 ) ( P 2 P 3 ) 2 ( P 1 P 2 ) ( P 2 P 3 ) 2 = P 1 P 2 2 ( P 3 P 1 ) ( P 3 P 2 ) 2 ( P 1 P 2 ) ( P 2 P 3 ) 2 {displaystyle {begin{aligned}alpha ={frac {leftP_{2}-P_{3}right^{2}left(P_{1}-P_{2}right)cdot left(P_{1}-P_{3}right)}{2leftleft(P_{1}-P_{2}right)times left(P_{2}-P_{3}right)right^{2}}}\beta ={frac {leftP_{1}-P_{3}right^{2}left(P_{2}-P_{1}right)cdot left(P_{2}-P_{3}right)}{2leftleft(P_{1}-P_{2}right)times left(P_{2}-P_{3}right)right^{2}}}\gamma ={frac {leftP_{1}-P_{2}right^{2}left(P_{3}-P_{1}right)cdot left(P_{3}-P_{2}right)}{2leftleft(P_{1}-P_{2}right)times left(P_{2}-P_{3}right)right^{2}}}end{aligned}}} Location Relative To The Triangle The circumcenter's position depends on the type of triangle: *
* The circumcenter of an acute triangle is inside the triangle * The circumcenter of a right triangle is at the center of the hypotenuse * The circumcenter of an obtuse triangle is outside the triangle These locational features can be seen by considering the trilinear or barycentric coordinates given above for the circumcenter: all three coordinates are positive for any interior point, at least one coordinate is negative for any exterior point, and one coordinate is zero and two are positive for a non-vertex point on a side of the triangle. ANGLES The angles which the circumscribed circle forms with the sides of the triangle coincide with angles at which sides meet each other. The side opposite angle α meets the circle twice: once at each end; in each case at angle α (similarly for the other two angles). This is due to the ALTERNATE SEGMENT THEOREM, which states that the angle between the tangent and chord equals the angle in the alternate segment. TRIANGLE CENTERS ON THE CIRCUMCIRCLE OF TRIANGLE ABC In this section, the vertex angles are labeled A, B, C and all coordinates are trilinear coordinates : * Steiner point = bc / (b2 − c2) : ca / (c2 − a2) : ab / (a2 −
b2) = the nonvertex point of intersection of the circumcircle with the
Steiner ellipse. (The
OTHER PROPERTIES The diameter of the circumcircle, called the CIRCUMDIAMETER and equal to twice the CIRCUMRADIUS, can be computed as the length of any side of the triangle divided by the sine of the opposite angle : diameter = a sin A = b sin B = c sin C . {displaystyle {text{diameter}}={frac {a}{sin A}}={frac {b}{sin B}}={frac {c}{sin C}}.} As a consequence of the law of sines , it does not matter which side and opposite angle are taken: the result will be the same. The diameter of the circumcircle can also be expressed as diameter = a b c 2 area = A B B C C A 2 A B C = a b c 2 s ( s a ) ( s b ) ( s c ) = 2 a b c ( a + b + c ) ( a + b + c ) ( a b + c ) ( a + b c ) {displaystyle {begin{aligned}{text{diameter}}&{}={frac {abc}{2cdot {text{area}}}}={frac {ABBCCA}{2Delta ABC}}\&{}={frac {abc}{2{sqrt {s(s-a)(s-b)(s-c)}}}}\ width:61.041ex; height:20.176ex;" alt="{displaystyle {begin{aligned}{text{diameter}}&{}={frac {abc}{2cdot {text{area}}}}={frac {ABBCCA}{2Delta ABC}}\&{}={frac {abc}{2{sqrt {s(s-a)(s-b)(s-c)}}}}\"> s ( s a ) ( s b ) ( s c ) {displaystyle {sqrt {scriptstyle {s(s-a)(s-b)(s-c)}}}} above is the area of the triangle, by Heron\'s formula . Trigometric expressions for the diameter of the circumcircle include :p.379 diameter = 2 area sin A sin B sin C . {displaystyle {text{diameter}}={sqrt {frac {2cdot {text{area}}}{sin Asin Bsin C}}}.} The triangle's nine-point circle has half the diameter of the circumcircle. In any given triangle, the circumcenter is always collinear with the
centroid and orthocenter . The line that passes through all of them is
known as the
The isogonal conjugate of the circumcenter is the orthocenter . The useful minimum bounding circle of three points is defined either by the circumcircle (where three points are on the minimum bounding circle) or by the two points of the longest side of the triangle (where the two points define a diameter of the circle). It is common to confuse the minimum bounding circle with the circumcircle. The circumcircle of three collinear points is the line on which the three points lie, often referred to as a circle of infinite radius. Nearly collinear points often lead to numerical instability in computation of the circumcircle. Circumcircles of triangles have an intimate relationship with the
By Euler\'s theorem in geometry , the distance between the circumcenter O and the incenter I is O I = R ( R 2 r ) , {displaystyle OI={sqrt {R(R-2r)}},} where r is the incircle radius and R is the circumcircle radius; hence the circumradius is at least twice the inradius (Euler\'s triangle inequality ), with equality only in the equilateral case. :p. 198 The distance between O and the orthocenter H is :p. 449 O H = R 2 8 R 2 cos A cos B cos C = 9 R 2 ( a 2 + b 2 + c 2 ) . {displaystyle OH={sqrt {R^{2}-8R^{2}cos Acos Bcos C}}={sqrt {9R^{2}-(a^{2}+b^{2}+c^{2})}}.} For centroid G and nine-point center N we have I G a + b + c 9 R 2 a 2 + b 2 + c 2 27 4 R 2 m a 2 + m b 2 + m c 2 . {displaystyle {begin{aligned}3{sqrt {3}}R&geq a+b+c\9R^{2}&geq a^{2}+b^{2}+c^{2}\{frac {27}{4}}R^{2} width:25.671ex; height:11.843ex;" alt="{displaystyle {begin{aligned}3{sqrt {3}}R&geq a+b+c\9R^{2}&geq a^{2}+b^{2}+c^{2}\{frac {27}{4}}R^{2}">:p.122,#96 4 R 2 h 2 ( t 2 h 2 ) = t 4 ( m 2 h 2 ) . {displaystyle 4R^{2}h^{2}(t^{2}-h^{2})=t^{4}(m^{2}-h^{2}).} Carnot\'s theorem states that the sum of the distances from the circumcenter to the three sides equals the sum of the circumradius and the inradius . :p.83 Here a segment's length is considered to be negative if and only if the segment lies entirely outside the triangle. If a triangle has two particular circles as its circumcircle and incircle , there exist an infinite number of other triangles with the same circumcircle and incircle, with any point on the circumcircle as a vertex. (This is the n=3 case of Poncelet\'s porism ). A necessary and sufficient condition for such triangles to exist is the above equality O I = R ( R 2 r ) . {displaystyle OI={sqrt {R(R-2r)}}.} :p. 188 CYCLIC QUADRILATERALS Cyclic quadrilaterals Main article:
Quadrilaterals that can be circumscribed have particular properties including the fact that opposite angles are supplementary angles (adding up to 180° or π radians). CYCLIC N-GONS For a cyclic polygon with an odd number of sides, all angles are equal if and only if the polygon is regular. A cyclic polygon with an even number of sides has all angles equal if and only if the alternate sides are equal (that is, sides 1, 3, 5, ... are equal, and sides 2, 4, 6, ... are equal). A cyclic pentagon with rational sides and area is known as a Robbins pentagon ; in all known cases, its diagonals also have rational lengths. In any cyclic n-gon with even n, the sum of one set of alternate angles (the first, third, fifth, etc.) equals the sum of the other set of alternate angles. This can be proven by induction from the n=4 case, in each case replacing a side with three more sides and noting that these three new sides together with the old side form a quadrilateral which itself has this property; the alternate angles of the latter quadrilateral represent the additions to the alternate angle sums of the previous n-gon. Let one n-gon be inscribed in a circle, and let another n-gon be tangential to that circle at the vertices of the first n-gon. Then from any point P on the circle, the product of the perpendicular distances from P to the sides of the first n-gon equals the product of the perpendicular distances from P to the sides of the second n-gon. :p. 72 POINT ON THE CIRCUMCIRCLE Let a cyclic n-gon have vertices A1 , ..., An on the unit circle. Then for any point M on the minor arc A1An, the distances from M to the vertices satisfy :p.190,#332.10 { M A 1 + M A 3 + + M A n 2 + M A n 3 + M A n 1 n 2 if n is even . {displaystyle {begin{cases}MA_{1}+MA_{3}+dots +MA_{n-2}+MA_{n} 3 1 / cos ( n ) = 8.7000366 . {displaystyle prod _{ngeq 3}1/cos left({frac {pi }{n}}right)=8.7000366ldots .} (sequence A051762 in the OEIS ). The reciprocal of this constant is
the
SEE ALSO *
NOTES * ^ A B Whitworth, William Allen. Trilinear Coordinates and Other
Methods of Modern Analytical
* ^ Wolfram page on barycentric coordinates
* ^ Dörrie, Heinrich, 100 Great Problems of Elementary
Mathematics, Dover, 1965.
* ^ Nelson, Roger, "Euler's triangle inequality via proof without
words," Mathematics Magazine 81(1), February 2008, 58-61.
* ^ Dragutin Svrtan and Darko Veljan, "Non-Euclidean versions of
some classical triangle inequalities", Forum Geometricorum 12 (2012),
197–209. http://forumgeom.fau.edu/FG2012volume12/FG201217index.html
* ^ Marie-Nicole Gras, "Distances between the circumcenter of the
extouch triangle and the classical centers", Forum Geometricorum 14
(2014), 51-61.
http://forumgeom.fau.edu/FG2014volume14/FG201405index.html
* ^ Smith, Geoff, and Leversha, Gerry, "Euler and triangle
geometry",
REFERENCES * Coxeter, H.S.M. (1969). "Chapter 1". Introduction to geometry. Wiley. pp. 12–1 |