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.[2] 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 Cartesian coordinates 1.3.2 Parametric equation 1.3.3 Trilinear and barycentric coordinates 1.3.4 Higher dimensions 1.4 Circumcenter coordinates 1.4.1 Cartesian coordinates
1.4.2 Trilinear coordinates
1.4.3 Barycentric coordinates
1.4.4 Circumcenter vector
1.4.5
Cartesian coordinates
1.5 Angles
1.6
Triangle
2 Cyclic quadrilaterals 3 Cyclic ngons 3.1 Point on the circumcircle
3.2
Polygon
4 See also 5 Notes 6 References 7 External links 7.1 MathWorld 7.2 Interactive Triangles[edit] All triangles are cyclic; i.e., every triangle has a circumscribed circle. Straightedge and compass construction[edit] 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. For three noncollinear points, these two lines cannot be parallel, and the circumcenter is the point where they cross. Any point on the bisector is equidistant from the two points that it bisects, from which it follows that this point, on both bisectors, is equidistant from all three triangle vertices. The circumradius is the distance from it to any of the three vertices. Alternate construction[edit] 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[edit]
Cartesian coordinates[edit]
In the Euclidean plane, it is possible to give explicitly an equation
of the circumcircle in terms of the
Cartesian coordinates
A = ( A x , A y ) B = ( B x , B y ) C = ( C x , C y ) displaystyle begin aligned mathbf A &=(A_ x ,A_ y )\mathbf B &=(B_ x ,B_ y )\mathbf C &=(C_ x ,C_ y )end aligned are the coordinates of points A, B, and C. The circumcircle is then the locus of points v = (vx,vy) in the Cartesian plane satisfying the equations
v − u
2 = r 2
A − u
2 = r 2
B − u
2 = r 2
C − u
2 = r 2 displaystyle begin aligned mathbf v mathbf u ^ 2 &=r^ 2 \mathbf A mathbf u ^ 2 &=r^ 2 \mathbf B mathbf u ^ 2 &=r^ 2 \mathbf C mathbf u ^ 2 &=r^ 2 end aligned guaranteeing that the points A, B, C, and v are all the same distance r from the common center u of the circle. Using the polarization identity, these equations reduce to the condition that the matrix [
v
2 − 2 v x − 2 v y − 1
A
2 − 2 A x − 2 A y − 1
B
2 − 2 B x − 2 B y − 1
C
2 − 2 C x − 2 C y − 1 ] displaystyle begin bmatrix mathbf v ^ 2 &2v_ x &2v_ y &1\mathbf A ^ 2 &2A_ x &2A_ y &1\mathbf B ^ 2 &2B_ x &2B_ y &1\mathbf C ^ 2 &2C_ x &2C_ y &1end bmatrix has a nonzero kernel. Thus the circumcircle may alternatively be described as the locus of zeros of the determinant of this matrix: det [
v
2 v x v y 1
A
2 A x A y 1
B
2 B x B y 1
C
2 C x C y 1 ] = 0. displaystyle det begin bmatrix mathbf v ^ 2 &v_ x &v_ y &1\mathbf A ^ 2 &A_ x &A_ y &1\mathbf B ^ 2 &B_ x &B_ y &1\mathbf C ^ 2 &C_ x &C_ y &1end bmatrix =0. Using cofactor expansion, let S x = 1 2 det [
A
2 A y 1
B
2 B y 1
C
2 C y 1 ] , S y = 1 2 det [ A x
A
2 1 B x
B
2 1 C x
C
2 1 ] , a = det [ A x A y 1 B x B y 1 C x C y 1 ] , b = det [ A x A y
A
2 B x B y
B
2 C x C y
C
2 ] displaystyle begin aligned S_ x &= frac 1 2 det begin bmatrix mathbf A ^ 2 &A_ y &1\mathbf B ^ 2 &B_ y &1\mathbf C ^ 2 &C_ y &1end bmatrix ,\S_ y &= frac 1 2 det begin bmatrix A_ x &mathbf A ^ 2 &1\B_ x &mathbf B ^ 2 &1\C_ x &mathbf C ^ 2 &1end bmatrix ,\a&=det begin bmatrix A_ x &A_ y &1\B_ x &B_ y &1\C_ x &C_ y &1end bmatrix ,\b&=det begin bmatrix A_ x &A_ y &mathbf A ^ 2 \B_ x &B_ y &mathbf B ^ 2 \C_ x &C_ y &mathbf C ^ 2 end bmatrix end aligned we then have av2 − 2Sv − b = 0 and, assuming the three points were not in a line (otherwise the circumcircle is that line that can also be seen as a generalized circle with S at infinity), v − S/a2 = b/a + S2/a2, giving the circumcenter S/a and the circumradius √b/a + S2/a2. A similar approach allows one to deduce the equation of the circumsphere of a tetrahedron. Parametric equation[edit] A unit vector perpendicular to the plane containing the circle is given by n ^ = ( P 2 − P 1 ) × ( P 3 − P 1 )
( P 2 − P 1 ) × ( P 3 − P 1 )
. displaystyle hat n = frac (P_ 2 P_ 1 )times (P_ 3 P_ 1 ) (P_ 2 P_ 1 )times (P_ 3 P_ 1 ) . 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., righthanded) sense about n ^ displaystyle scriptstyle hat n is the following: R ( s ) = P c + cos ( s r ) ( P 0 − P c ) + sin ( s r ) [ n ^ × ( P 0 − P c ) ] . 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[ hat n times (P_ 0 P_ c )right]. Trilinear and barycentric coordinates[edit] An equation for the circumcircle in trilinear coordinates x : y : z is[1]: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[edit] Additionally, the circumcircle of a triangle embedded in d dimensions can be found using a generalized method. Let A, B, and C be ddimensional 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 &=mathbf B mathbf C .end aligned The circumradius, r, is then r = ‖ a ‖ ‖ b ‖ ‖ a − b ‖ 2 ‖ a × b ‖ = ‖ a − b ‖ 2 sin θ = ‖ A − B ‖ 2 sin θ , 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&= sqrt leftmathbf a right^ 2 leftmathbf b right^ 2 (mathbf a cdot mathbf b )^ 2 .end aligned Circumcenter coordinates[edit]
Cartesian coordinates[edit]
The
Cartesian coordinates
U = ( U x , U y ) displaystyle U=left(U_ x ,U_ y right) are U x = 1 D [ ( A x 2 + A y 2 ) ( B y − C y ) + ( B x 2 + B y 2 ) ( C y − A y ) + ( C x 2 + C y 2 ) ( A y − B y ) ] U y = 1 D [ ( A x 2 + A y 2 ) ( C x − B x ) + ( B x 2 + B y 2 ) ( A x − C x ) + ( C x 2 + C y 2 ) ( B x − A x ) ] displaystyle begin aligned U_ x &= frac 1 D left[left(A_ x ^ 2 +A_ y ^ 2 right)left(B_ y C_ y right)+left(B_ x ^ 2 +B_ y ^ 2 right)left(C_ y A_ y right)+left(C_ x ^ 2 +C_ y ^ 2 right)left(A_ y B_ y right)right]\U_ y &= frac 1 D left[left(A_ x ^ 2 +A_ y ^ 2 right)left(C_ x B_ x right)+left(B_ x ^ 2 +B_ y ^ 2 right)left(A_ x C_ x right)+left(C_ x ^ 2 +C_ y ^ 2 right)left(B_ x A_ x right)right]end aligned with D = 2 [ A x ( B y − C y ) + B x ( C y − A y ) + C x ( A y − B y ) ] . displaystyle D=2left[A_ x left(B_ y C_ y right)+B_ x left(C_ y A_ y right)+C_ x left(A_ y B_ y right)right]., 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 ′ [ C y ′ ( B x ′ 2 + B y ′ 2 ) − B y ′ ( C x ′ 2 + C y ′ 2 ) ] , U y ′ = 1 D ′ [ B x ′ ( C x ′ 2 + C y ′ 2 ) − C x ′ ( B x ′ 2 + B y ′ 2 ) ] displaystyle begin aligned U'_ x &= frac 1 D' left[C'_ y left( B'_ x ^ 2 + B'_ y ^ 2 right)B'_ y left( C'_ x ^ 2 + C'_ y ^ 2 right)right],\U'_ y &= frac 1 D' left[B'_ x left( C'_ x ^ 2 + C'_ y ^ 2 right)C'_ x left( B'_ x ^ 2 + B'_ y ^ 2 right)right]end aligned with D ′ = 2 ( B x ′ C y ′ − B y ′ C x ′ ) . displaystyle D'=2left(B'_ x C'_ y B'_ y C'_ x right)., 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=leftU'right= sqrt U'_ x ^ 2 + U'_ y ^ 2 and the actual circumcenter of ABC follows as U = U ′ + A displaystyle U=U'+A Trilinear coordinates[edit] The circumcenter has trilinear coordinates[1]: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[2] a ( b 2 + c 2 − a 2 ) : b ( c 2 + a 2 − b 2 ) : c ( a 2 + b 2 − c 2 ) . displaystyle aleft(b^ 2 +c^ 2 a^ 2 right):bleft(c^ 2 +a^ 2 b^ 2 right):cleft(a^ 2 +b^ 2 c^ 2 right). Barycentric coordinates[edit] 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 left(b^ 2 +c^ 2 a^ 2 right):;b^ 2 left(c^ 2 +a^ 2 b^ 2 right):;c^ 2 left(a^ 2 +b^ 2 c^ 2 right),, [3] where a, b, c are edge lengths (BC, CA, AB respectively) of the triangle. In terms of the triangle's angles α , β , γ , displaystyle alpha ,beta ,gamma , the barycentric coordinates of the circumcenter are[2] sin 2 α : sin 2 β : sin 2 γ . displaystyle sin 2alpha :sin 2beta :sin 2gamma . Circumcenter vector[edit]
Since the
Cartesian coordinates
U = a 2 ( b 2 + c 2 − a 2 ) A + b 2 ( c 2 + a 2 − b 2 ) B + c 2 ( a 2 + b 2 − c 2 ) C 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 U= frac a^ 2 left(b^ 2 +c^ 2 a^ 2 right)A+b^ 2 left(c^ 2 +a^ 2 b^ 2 right)B+c^ 2 left(a^ 2 +b^ 2 c^ 2 right)C a^ 2 left(b^ 2 +c^ 2 a^ 2 right)+b^ 2 left(c^ 2 +a^ 2 b^ 2 right)+c^ 2 left(a^ 2 +b^ 2 c^ 2 right) . 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
P 1 = [ x 1 y 1 z 1 ] , P 2 = [ x 2 y 2 z 2 ] , P 3 = [ x 3 y 3 z 3 ] displaystyle mathrm P_ 1 = begin bmatrix x_ 1 \y_ 1 \z_ 1 end bmatrix ,mathrm P_ 2 = begin bmatrix x_ 2 \y_ 2 \z_ 2 end bmatrix ,mathrm P_ 3 = begin bmatrix x_ 3 \y_ 3 \z_ 3 end bmatrix 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 = α P 1 + β P 2 + γ P 3 displaystyle mathrm P_ c =alpha ,P_ 1 +beta ,P_ 2 +gamma ,P_ 3 where α =
P 2 − P 3
2 ( P 1 − P 2 ) ⋅ ( P 1 − P 3 ) 2
( P 1 − P 2 ) × ( P 2 − P 3 )
2 β =
P 1 − P 3
2 ( P 2 − P 1 ) ⋅ ( P 2 − P 3 ) 2
( P 1 − P 2 ) × ( P 2 − P 3 )
2 γ =
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[edit] The circumcenter's position depends on the type of triangle:
If and only if
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 nonvertex point on a side of the triangle. Angles[edit]
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
Steiner point = bc / (b2 − c2) : ca / (c2 − a2) : ab /
(a2 − b2) = the nonvertex point of intersection of the circumcircle
with the Steiner ellipse. (The Steiner ellipse, with center =
centroid(ABC), is the ellipse of least area that passes through A, B,
and C. An equation for this ellipse is 1/(ax) + 1/(by) + 1/(cz) = 0.)
Tarry point
Other properties[edit] 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
Δ 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(sa)(sb)(sc) \& = frac 2abc sqrt (a+b+c)(a+b+c)(ab+c)(a+bc) end aligned where a, b, c are the lengths of the sides of the triangle and s = (a + b + c)/2 is the semiperimeter. The expression s ( s − a ) ( s − b ) ( s − c ) displaystyle sqrt scriptstyle s(sa)(sb)(sc) above is the area of the triangle, by Heron's formula.[3] Trigonometric expressions for the diameter of the circumcircle include[4]: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 ninepoint 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 Euler line.
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
Delaunay triangulation
O I = R ( R − 2 r ) , displaystyle OI= sqrt R(R2r) , 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.[5][6]:p. 198 The distance between O and the orthocenter H is[7][8]: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 ninepoint center N we have I G < I O , 2 I N < I O , O I 2 = 2 R ⋅ I N . displaystyle begin aligned IG&<IO,\2IN&<IO,\OI^ 2 &=2Rcdot IN.end aligned The product of the incircle radius and the circumcircle radius of a triangle with sides a, b, and c is[9]: p. 189, #298(d) r R = a b c 2 ( a + b + c ) . displaystyle rR= frac abc 2(a+b+c) . With circumradius R, sides a, b, c, and medians ma, mb, and mc, we have[10]:p.289–290 3 3 R ≥ 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 &geq m_ a ^ 2 +m_ b ^ 2 +m_ c ^ 2 .end aligned If median m, altitude h, and internal bisector t all emanate from the same vertex of a triangle with circumradius R, then[11]: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
O I = R ( R − 2 r ) . displaystyle OI= sqrt R(R2r) . [9]:p. 188 Cyclic quadrilaterals[edit] Cyclic quadrilaterals Main article: Cyclic quadrilateral 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 ngons[edit] 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).[12] A cyclic pentagon with rational sides and area is known as a Robbins pentagon; in all known cases, its diagonals also have rational lengths.[13] In any cyclic ngon 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 ngon. Let one ngon be inscribed in a circle, and let another ngon be tangential to that circle at the vertices of the first ngon. Then from any point P on the circle, the product of the perpendicular distances from P to the sides of the first ngon equals the product of the perpendicular distances from P to the sides of the second ngon.[9]:p. 72 Point on the circumcircle[edit] Let a cyclic ngon 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[14]:p.190,#332.10 M A 1 + M A 3 + ⋯ + M A n − 2 + M A n < n 2 if n is odd ; M A 1 + M A 3 + ⋯ + M A n − 3 + M A n − 1 ≤ n 2 if n is even . displaystyle begin cases MA_ 1 +MA_ 3 +dots +MA_ n2 +MA_ n < frac n sqrt 2 quad text if ,n, text is odd ;\MA_ 1 +MA_ 3 +dots +MA_ n3 +MA_ n1 leq frac n sqrt 2 quad text if ,n, text is even .end cases
Polygon
A sequence of circumscribed polygons and circles. Any regular polygon is cyclic. Consider a unit circle, then circumscribe a regular triangle such that each side touches the circle. Circumscribe a circle, then circumscribe a square. Again circumscribe a circle, then circumscribe a regular 5gon, and so on. The radii of the circumscribed circles converge to the socalled polygon circumscribing constant ∏ n ≥ 3 1 / cos ( π 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 Kepler–Bouwkamp constant. See also[edit] Inscribed circle
Jung's theorem, an inequality relating the diameter of a point set to
the radius of its minimum bounding circle
Lester's theorem
Circumscribed sphere
Circumgon
Triangle
Notes[edit] ^ a b Whitworth, William Allen. Trilinear Coordinates and Other
Methods of Modern Analytical
Geometry
References[edit] Coxeter, H.S.M. (1969). "Chapter 1". Introduction to geometry. Wiley.
pp. 12–13. ISBN 0471504580.
Megiddo, N. (1983). "Lineartime algorithms for linear programming in
R3 and related problems". SIAM Journal on Computing. 12 (4):
759–776. doi:10.1137/0212052.
Kimberling, Clark (1998). "
Triangle
External links[edit] Derivation of formula for radius of circumcircle of triangle at
Mathalino.com
Semiregular anglegons and sidegons: respective generalizations of
rectangles and rhombi at Dynamic
Geometry
MathWorld[edit] Weisstein, Eric W. "Circumcircle". MathWorld. Weisstein, Eric W. "Cyclic Polygon". MathWorld. Weisstein, Eric W. "Steiner circumellipse". MathWorld. Interactive[edit]
Triangle
