Movatterモバイル変換

Jump to content

Circumcircle

From Wikipedia, the free encyclopedia

(Redirected fromCircumcenter)

Circle that passes through the vertices of a triangle

Ingeometry, thecircumscribed circle orcircumcircle of atriangle is acircle that passes through all threevertices. The center of this circle is called thecircumcenter of the triangle, and its radius is called thecircumradius. The circumcenter is the point ofintersection between the threeperpendicular bisectors of the triangle's sides, and is atriangle center.

More generally, ann-sidedpolygon with all its vertices on the same circle, also called the circumscribed circle, is called acyclic polygon, or in the special casen = 4, acyclic quadrilateral. Allrectangles,isosceles trapezoids,right kites, andregular polygons are cyclic, but not every polygon is.

Straightedge and compass construction

Construction of the circumcircle of triangle△ABC and the circumcenterQ

The circumcenter of a triangle can beconstructed by drawing any two of the threeperpendicular bisectors. For three non-collinear 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.

Alternative construction

Alternative construction of the circumcenter (intersection of broken lines)

An alternative 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.)

Incoastal navigation, a triangle's circumcircle is sometimes used as a way of obtaining aposition line using asextant when nocompass is available. The horizontal angle between two landmarks defines the circumcircle upon which the observer lies.

Circumcircle equations

Cartesian coordinates

In theEuclidean plane, it is possible to give explicitly an equation of the circumcircle in terms of theCartesian coordinates of the vertices of the inscribed triangle. Suppose that

{\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 pointsA, B, C. The circumcircle is then the locus of points $\mathbf {v} =(v_{x},v_{y})$ in the Cartesian plane satisfying the equations

{\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 pointsA,B,C,v are all the same distancer from the common center $\mathbf {u}$ of the circle. Using thepolarization identity, these equations reduce to the condition that thematrix

{\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}&-1\end{bmatrix}}

has a nonzerokernel. Thus the circumcircle may alternatively be described as thelocus of zeros of thedeterminant of this matrix:

\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}&1\end{bmatrix}}=0.

Usingcofactor expansion, let

{\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}&1\end{bmatrix}},\\[5pt]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}&1\end{bmatrix}},\\[5pt]a&=\det {\begin{bmatrix}A_{x}&A_{y}&1\\B_{x}&B_{y}&1\\C_{x}&C_{y}&1\end{bmatrix}},\\[5pt]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 $a|\mathbf {v} |^{2}-2\mathbf {Sv} -b=0$ where $\mathbf {S} =(S_{x},S_{y}),$ 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 withS at infinity) – $\left|\mathbf {v} -{\tfrac {\mathbf {S} }{a}}\right|^{2}={\tfrac {b}{a}}+{\tfrac {|\mathbf {S} |^{2}}{a^{2}}},$ giving the circumcenter ${\tfrac {\mathbf {S} }{a}}$ and the circumradius ${\sqrt {{\tfrac {b}{a}}+{\tfrac {|\mathbf {S} |^{2}}{a^{2}}}}}.$ A similar approach allows one to deduce the equation of thecircumsphere of atetrahedron.

Parametric equation

Aunit vector perpendicular to the plane containing the circle is given by

{\widehat {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,P_c, a point on the circle,P₀ and a unit normal of the plane containing the circle,⁠ ${\widehat {n}},$ ⁠ one parametric equation of the circle starting from the pointP₀ and proceeding in a positively oriented (i.e.,right-handed) sense about⁠ ${\widehat {n}}$ ⁠ is the following:

\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[{\widehat {n}}\times (P_{0}-P_{c})\right].

Trilinear and barycentric coordinates

An equation for the circumcircle intrilinear coordinatesx :y :z is^[1] ${\tfrac {a}{x}}+{\tfrac {b}{y}}+{\tfrac {c}{z}}=0.$ An equation for the circumcircle inbarycentric coordinatesx :y :z is ${\tfrac {a^{2}}{x}}+{\tfrac {b^{2}}{y}}+{\tfrac {c^{2}}{z}}=0.$

Theisogonal conjugate of the circumcircle is the line at infinity, given intrilinear 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 three dimensions can be found using a generalized method. LetA,B,C be three-dimensional points, which form the vertices of a triangle. We start by transposing the system to placeC at the origin:

{\begin{aligned}\mathbf {a} &=\mathbf {A} -\mathbf {C} ,\\\mathbf {b} &=\mathbf {B} -\mathbf {C} .\end{aligned}}

The circumradiusr is then

r={\frac {\left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\left\|\mathbf {a} -\mathbf {b} \right\|}{2\left\|\mathbf {a} \times \mathbf {b} \right\|}}={\frac {\left\|\mathbf {a} -\mathbf {b} \right\|}{2\sin \theta }}={\frac {\left\|\mathbf {A} -\mathbf {B} \right\|}{2\sin \theta }},

whereθ is the interior angle betweena andb. The circumcenter,p₀, is given by

p_{0}={\frac {(\left\|\mathbf {a} \right\|^{2}\mathbf {b} -\left\|\mathbf {b} \right\|^{2}\mathbf {a} )\times (\mathbf {a} \times \mathbf {b} )}{2\left\|\mathbf {a} \times \mathbf {b} \right\|^{2}}}+\mathbf {C} .

This formula only works in three dimensions as thecross product is not defined in other dimensions, but it can be generalized to the other dimensions by replacing the cross products with following identities:

{\begin{aligned}\mathbf {u} \times (\mathbf {v} \times \mathbf {w} )&=(\mathbf {u} \cdot \mathbf {w} )\mathbf {v} -(\mathbf {u} \cdot \mathbf {v} )\mathbf {w} ,\\\left\|\mathbf {u} \times \mathbf {v} \right\|^{2}&=\left\|\mathbf {u} \right\|^{2}\left\|\mathbf {v} \right\|^{2}-(\mathbf {u} \cdot \mathbf {v} )^{2}.\end{aligned}}

This gives us the following equation for the circumradiusr:

r={\frac {\left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\left\|\mathbf {a} -\mathbf {b} \right\|}{2{\sqrt {\left\|\mathbf {a} \right\|^{2}\left\|\mathbf {b} \right\|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2}}}}}

and the following equation for the circumcenterp₀:

p_{0}={\frac {((\left\|\mathbf {a} \right\|^{2}\mathbf {b} -\left\|\mathbf {b} \right\|^{2}\mathbf {a} )\cdot \mathbf {b} )\mathbf {a} -((\left\|\mathbf {a} \right\|^{2}\mathbf {b} -\left\|\mathbf {b} \right\|^{2}\mathbf {a} )\cdot \mathbf {a} )\mathbf {b} }{2(\left\|\mathbf {a} \right\|^{2}\left\|\mathbf {b} \right\|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2})}}+\mathbf {C}

which can be simplified to:

p_{0}={\frac {\left\|\mathbf {a} \right\|^{2}\left\|\mathbf {b} \right\|^{2}(\mathbf {a} +\mathbf {b} )-(\mathbf {a} \cdot \mathbf {b} )(\left\|\mathbf {a} \right\|^{2}\mathbf {b} +\left\|\mathbf {b} \right\|^{2}\mathbf {a} )}{2(\left\|\mathbf {a} \right\|^{2}\left\|\mathbf {b} \right\|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2})}}+\mathbf {C}

Circumcenter coordinates

Cartesian coordinates

TheCartesian coordinates of the circumcenter $U=\left(U_{x},U_{y}\right)$ are

{\begin{aligned}U_{x}&={\frac {1}{D}}\left[(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})\right]\\[5pt]U_{y}&={\frac {1}{D}}\left[(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})\right]\end{aligned}}

with

D=2\left[A_{x}(B_{y}-C_{y})+B_{x}(C_{y}-A_{y})+C_{x}(A_{y}-B_{y})\right].\,

Without loss of generality this can be expressed in a simplified form after translation of the vertexA 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 vertexA' to these vertices. Observe that this trivial translation is possible for all triangles and the circumcenter $U'=(U'_{x},U'_{y})$ of the triangle△A'B'C' follow as

{\begin{aligned}U'_{x}&={\frac {1}{D'}}\left[C'_{y}({B'_{x}}^{2}+{B'_{y}}^{2})-B'_{y}({C'_{x}}^{2}+{C'_{y}}^{2})\right],\\[5pt]U'_{y}&={\frac {1}{D'}}\left[B'_{x}({C'_{x}}^{2}+{C'_{y}}^{2})-C'_{x}({B'_{x}}^{2}+{B'_{y}}^{2})\right]\end{aligned}}

with

D'=2(B'_{x}C'_{y}-B'_{y}C'_{x}).\,

Due to the translation of vertexA to the origin, the circumradiusr can be computed as

r=\|U'\|={\sqrt {{U'_{x}}^{2}+{U'_{y}}^{2}}}

and the actual circumcenter of△ABC follows as

U=U'+A

Trilinear coordinates

The circumcenter hastrilinear coordinates^[2]

\cos \alpha :\cos \beta :\cos \gamma

whereα, β, γ are the angles of the triangle.

In terms of the side lengthsa, b, c, the trilinears are^[3]

a\left(b^{2}+c^{2}-a^{2}\right):b\left(c^{2}+a^{2}-b^{2}\right):c\left(a^{2}+b^{2}-c^{2}\right).

Barycentric coordinates

The circumcenter hasbarycentric coordinates^[4]

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),\,

wherea, b, c are edge lengthsBC,CA,AB respectively) of the triangle.

In terms of the triangle's anglesα, β, γ, the barycentric coordinates of the circumcenter are^[3]

\sin 2\alpha :\sin 2\beta :\sin 2\gamma .

Circumcenter vector

Since the Cartesian coordinates of any point are a weighted average of those of the vertices, with the weights being the point's barycentric coordinates normalized to sum to unity, the circumcenter vector can be written as

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)}}.

HereU is the vector of the circumcenter andA, B, C are the vertex vectors. The divisor here equals16S² whereS is the area of the triangle. As stated previously

{\begin{aligned}\mathbf {a} &=\mathbf {A} -\mathbf {C} ,\\\mathbf {b} &=\mathbf {B} -\mathbf {C} .\end{aligned}}

Cartesian coordinates from cross- and dot-products

InEuclidean space, there is a unique circle passing through any given three non-collinear pointsP₁,P₂,P₃. UsingCartesian coordinates to represent these points asspatial vectors, it is possible to use thedot product andcross product to calculate the radius and center of the circle. Let

\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

\mathrm {r} ={\frac {\left|P_{1}-P_{2}\right|\left|P_{2}-P_{3}\right|\left|P_{3}-P_{1}\right|}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|}}

The center of the circle is given by thelinear combination

\mathrm {P_{c}} =\alpha \,P_{1}+\beta \,P_{2}+\gamma \,P_{3}

where

{\begin{aligned}\alpha ={\frac {\left|P_{2}-P_{3}\right|^{2}\left(P_{1}-P_{2}\right)\cdot \left(P_{1}-P_{3}\right)}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|^{2}}}\\\beta ={\frac {\left|P_{1}-P_{3}\right|^{2}\left(P_{2}-P_{1}\right)\cdot \left(P_{2}-P_{3}\right)}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|^{2}}}\\\gamma ={\frac {\left|P_{1}-P_{2}\right|^{2}\left(P_{3}-P_{1}\right)\cdot \left(P_{3}-P_{2}\right)}{2\left|\left(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:

For an acute triangle (all angles smaller than a right angle), the circumcenter always lies inside the triangle.
For a right triangle, the circumcenter always lies at the midpoint of thehypotenuse. This is one form ofThales' theorem.
For an obtuse triangle (a triangle with one angle bigger than a right angle), the circumcenter always lies outside the triangle.

The circumcenter of an acute triangle is inside the triangle

The circumcenter of a right triangle is at the midpoint 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 angleα meets the circle twice: once at each end; in each case at angleα (similarly for the other two angles). This is due to thealternate segment theorem, which states that the angle between the tangent and chord equals the angle in the alternate segment.

Triangle centers on the circumcircle

In this section, the vertex angles are labeledA, B, C and all coordinates aretrilinear coordinates:

Steiner point: the non-vertex point of intersection of the circumcircle with the Steiner ellipse.

{\frac {bc}{b^{2}-c^{2}}}:{\frac {ca}{c^{2}-a^{2}}}:{\frac {ab}{a^{2}-b^{2}}}

(TheSteiner ellipse, with center = centroid (ABC), is the ellipse of least area that passes throughA, B, C. An equation for this ellipse is

{\tfrac {1}{ax}}+{\tfrac {1}{by}}+{\tfrac {1}{cz}}=0

.)

Tarry point: antipode of the Steiner point

\sec(A+\omega ):\sec(B+\omega ):\sec(C+\omega )

Focus of theKiepert parabola:

\csc(B-C):\csc(C-A):\csc(A-B).

Other properties

Thediameter of the circumcircle, called thecircumdiameter and equal to twice thecircumradius, can be computed as the length of any side of the triangle divided by thesine of the oppositeangle:

{\text{diameter}}={\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}.

As a consequence of thelaw 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

{\begin{aligned}{\text{diameter}}&{}={\frac {abc}{2\cdot {\text{area}}}}={\frac {|AB||BC||CA|}{2|\Delta ABC|}}\\[5pt]&{}={\frac {abc}{2{\sqrt {s(s-a)(s-b)(s-c)}}}}\\[5pt]&{}={\frac {2abc}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}}\end{aligned}}

wherea, b, c are the lengths of the sides of the triangle and $s={\tfrac {a+b+c}{2}}$ is the semiperimeter. The expression $\scriptstyle {\sqrt {s(s-a)(s-b)(s-c)}}$ above is the area of the triangle, byHeron's formula.^[5] Trigonometric expressions for the diameter of the circumcircle include^[6]

{\text{diameter}}={\sqrt {\frac {2\cdot {\text{area}}}{\sin A\sin B\sin C}}}.

The triangle'snine-point circle has half the diameter of the circumcircle.

In any given triangle, the circumcenter is always collinear with thecentroid andorthocenter. The line that passes through all of them is known as theEuler line.

Theisogonal conjugate of the circumcenter is theorthocenter.

The usefulminimum 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 threecollinear points is the line on which the three points lie, often referred to as acircle of infinite radius. Nearly collinear points often lead tonumerical instability in computation of the circumcircle.

Circumcircles of triangles have an intimate relationship with theDelaunay triangulation of aset of points.

ByEuler's theorem in geometry, the distance between the circumcenterO and theincenterI is

{\overline {OI}}={\sqrt {R(R-2r)}},

wherer is the incircle radius andR is the circumcircle radius; hence the circumradius is at least twice the inradius (Euler's triangle inequality), with equality only in theequilateral case.^[7]^[8]

The distance betweenO and theorthocenterH is^[9]^[10]

{\overline {OH}}={\sqrt {R^{2}-8R^{2}\cos A\cos B\cos C}}={\sqrt {9R^{2}-(a^{2}+b^{2}+c^{2})}}.

ForcentroidG andnine-point centerN we have

{\begin{aligned}{\overline {IG}}&<{\overline {IO}},\\2{\overline {IN}}&<{\overline {IO}},\\{\overline {OI}}^{2}&=2R\cdot {\overline {IN}}.\end{aligned}}

The product of the incircle radius and the circumcircle radius of a triangle with sidesa, b, c is^[11]

rR={\frac {abc}{2(a+b+c)}}.

With circumradiusR, sidesa, b, c, andmediansm_a, m_b, m_c, we have^[12]

{\begin{aligned}3{\sqrt {3}}R&\geq a+b+c\\[5pt]9R^{2}&\geq a^{2}+b^{2}+c^{2}\\[5pt]{\frac {27}{4}}R^{2}&\geq m_{a}^{2}+m_{b}^{2}+m_{c}^{2}.\end{aligned}}

If medianm, altitudeh, and internal bisectort all emanate from the same vertex of a triangle with circumradiusR, then^[13]

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 theinradius.^[14] Here a segment's length is considered to be negativeif and only if the segment lies entirely outside the triangle.

If a triangle has two particular circles as its circumcircle andincircle, 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 then = 3 case ofPoncelet's porism). A necessary and sufficient condition for such triangles to exist is the above equality ${\overline {OI}}={\sqrt {R(R-2r)}}.$ ^[15]

Cyclic polygons

Cyclic quadrilaterals

Main articles:Cyclic quadrilateral andConcyclic points

A set of points lying on the same circle are calledconcyclic, and a polygon whose vertices are concyclic is called acyclic polygon. Every triangle is concyclic, but polygons with more than three sides are not in general.

Cyclic polygons, especially four-sidedcyclic quadrilaterals, have various special properties. In particular, the opposite angles of a cyclic quadrilateral aresupplementary angles (adding up to 180° or π radians).

See also

References

^Whitworth, William Allen (1866).Trilinear Coordinates and Other Methods of Modern Analytical Geometry of Two Dimensions. Deighton, Bell, and Co. p. 199.
^Whitworth (1866),p. 19.
^^a ^bKimberling, Clark."Part I: Introduction and Centers X(1) – X(1000)".Encyclopedia of Triangle Centers. The circumcenter is listed under X(3).
^Weisstein, Eric W."Barycentric Coordinates".MathWorld.
^Coxeter, H.S.M. (1969). "Chapter 1".Introduction to geometry. Wiley. pp. 12–13.ISBN 0-471-50458-0.
^Dörrie, Heinrich (1965).100 Great Problems of Elementary Mathematics. Dover. p. 379.
^Nelson, Roger, "Euler's triangle inequality via proof without words,"Mathematics Magazine 81(1), February 2008, 58-61.
^Svrtan, Dragutin; Veljan, Darko (2012)."Non-Euclidean versions of some classical triangle inequalities".Forum Geometricorum.12:197–209. Archived fromthe original on 2019-10-28. Retrieved2015-01-18. See in particular p. 198.
^Gras, Marie-Nicole (2014)."Distances between the circumcenter of the extouch triangle and the classical centers".Forum Geometricorum.14:51–61.
^Smith, G. C.; Leversha, Gerry (November 2007). "Euler and triangle geometry".The Mathematical Gazette.91 (522):436–452.doi:10.1017/S0025557200182087.JSTOR 40378417.S2CID 125341434. See in particular p. 449.
^Johnson, Roger A. (1929).Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Houghton Mifflin Co. p. 189, #298(d).hdl:2027/wu.89043163211. Republished by Dover Publications asAdvanced Euclidean Geometry, 1960 and 2007.
^Posamentier, Alfred S.; Lehmann, Ingmar (2012).The Secrets of Triangles. Prometheus Books. pp. 289–290.
^Altshiller Court, Nathan (1952).College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle (2nd ed.). Barnes & Noble. p. 122, #96. Reprinted by Dover Publications, 2007.
^Altshiller Court (1952), p. 83.
^Johnson (1929), p. 188.

External links

Derivation of formula for radius of circumcircle of triangle at Mathalino.com
Semi-regular angle-gons and side-gons: respective generalizations of rectangles and rhombi atDynamic Geometry Sketches, interactive dynamic geometry sketch.
Weisstein, Eric W."Circumcircle","Cyclic Polygon".MathWorld.
Triangle circumcircle andcircumcenter With interactive animation
An interactive Java applet for the circumcenter

Retrieved from "https://en.wikipedia.org/w/index.php?title=Circumcircle&oldid=1278217412#Circumcenter_coordinates"

Hidden categories:

[8]ページ先頭

©2009-2025 Movatter.jp