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Area of a circle
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Part of a series of articles on the
mathematical constant π
3.1415926535897932384626433...
Uses
Area of a circle
Circumference
Use in other formulae
Properties
Irrationality
Transcendence
Value
Less than 22/7
Approximations
Memorization
People
Archimedes
Liu Hui
Zu Chongzhi
Aryabhata
Madhava
Ludolph van Ceulen
Seki Takakazu
Takebe Kenko
William Jones
John Machin
William Shanks
John Wrench
Chudnovsky brothers
Yasumasa Kanada
History
Chronology
Book
In culture
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Related topics
Squaring the circle
Basel problem
Six nines in π
Other topics related to π
v
t
e
In geometry, the area enclosed by a circle of radius r is π r2. Here the Greek letter π represents a constant, approximately equal to 3.14159, which is equal to the ratio of the circumference of any circle to its diameter.
One method of deriving this formula, which originated with Archimedes, involves viewing the circle as the limit of a sequence of regular polygons. The area of a regular polygon is half its perimeter multiplied by the distance from its center to its sides, and the corresponding formula (that the area is half the perimeter times the radius, i.e. 1⁄2 × 2πr × r) holds in the limit for a circle.
Although often referred to as the area of a circle in informal contexts, strictly speaking the term disk refers to the interior of the circle, while circle is reserved for the boundary only, which is a curve and covers no area itself. Therefore, the area of a disk is the more precise phrase for the area enclosed by a circle.
Contents
[hide]
1 History
2 Historical arguments
2.1 Using polygons
2.2 Archimedes's proof
2.2.1 Not greater
2.2.2 Not less
2.3 Rearrangement proof
3 Modern proofs
3.1 Onion proof
3.2 Triangle proof
3.3 Semicircle proof
4 Isoperimetric inequality
5 Fast approximation
5.1 Archimedes' doubling method
5.2 The Snell–Huygens refinement
5.3 Derivation of Archimedes' doubling formulae
6 Dart approximation
7 Finite rearrangement
8 Non-Euclidean circles
9 Generalizations
10 Bibliography
11 References
12 External links
History[edit]
Modern mathematics can obtain the area using the methods of integral calculus or its more sophisticated offspring, real analysis. However the area of a disk was studied by the Ancient Greeks. Eudoxus of Cnidus in the fifth century B.C. had found that the area of a disk is proportional to its radius squared.[1] Archimedes used the tools of Euclidean geometry to show that the area inside a circle is equal to that of a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius in his book Measurement of a Circle. The circumference is 2πr, and the area of a triangle is half the base times the height, yielding the area π r2 for the disk. Prior to Archimedes, Hippocrates of Chios was the first to show that the area of a disk is proportional to the square of its diameter, as part of his quadrature of the lune of Hippocrates,[2] but did not identify the constant of proportionality.
Historical arguments[edit]
A variety of arguments have been advanced historically to establish the equation A = π r 2 {\displaystyle A=\pi r^{2}} of varying degrees of mathematical rigor. The most famous of these is Archimedes' method of exhaustion, one of the earliest uses of the mathematical concept of a limit, as well as the origin of Archimedes' axiom which remains part of the standard analytical treatment of the real number system. The original proof of Archimedes is not quite rigorous by modern standards, because it assumes that we can compare the length of arc of a circle to the length of a secant and a tangent line, and similar statements about the area, as geometrically evident.
Using polygons[edit]
The area of a regular polygon is half its perimeter times the apothem. As the number of sides of the regular polygon increases, the polygon tends to a circle, and the apothem tends to the radius. This suggests that the area of a disk is half the circumference of its bounding circle times the radius.[3]
Archimedes's proof[edit]
Following Archimedes & c. 260 BCE, compare the area enclosed by a circle to a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius. If the area of the circle is not equal to that of the triangle, then it must be either greater or less. We eliminate each of these by contradiction, leaving equality as the only possibility. We use regular polygons in the same way.
Not greater[edit]
Circle with square and octagon inscribed, showing area gap
Suppose that the area C enclosed by the circle is greater than the area T = 1⁄2cr of the triangle. Let E denote the excess amount. Inscribe a square in the circle, so that its four corners lie on the circle. Between the square and the circle are four segments. If the total area of those gaps, G4, is greater than E, split each arc in half. This makes the inscribed square into an inscribed octagon, and produces eight segments with a smaller total gap, G8. Continue splitting until the total gap area, Gn, is less than E. Now the area of the inscribed polygon, Pn = C − Gn, must be greater than that of the triangle.
E = C − T > G n P n = C − G n > C − E P n > T {\displaystyle {\begin{aligned}E&{}=C-T\\&{}>G_{n}\\P_{n}&{}=C-G_{n}\\&{}>C-E\\P_{n}&{}>T\end{aligned}}}
But this forces a contradiction, as follows. Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, h, is less than the circle radius. Also, let each side of the polygon have length s; then the sum of the sides, ns, is less than the circle circumference. The polygon area consists of n equal triangles with height h and base s, thus equals 1⁄2nhs. But since h < r and ns < c, the polygon area must be less than the triangle area, 1⁄2cr, a contradiction. Therefore, our supposition that C might be greater than T must be wrong.
Not less[edit]
Circle with square and octagon circumscribed, showing area gap
Suppose that the area enclosed by the circle is less than the area T of the triangle. Let D denote the deficit amount. Circumscribe a square, so that the midpoint of each edge lies on the circle. If the total area gap between the square and the circle, G4, is greater than D, slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the gap area is less than D. The area of the polygon, Pn, must be less than T.
D = T − C > G n P n = C + G n < C + D P n < T {\displaystyle {\begin{aligned}D&{}=T-C\\&{}>G_{n}\\P_{n}&{}=C+G_{n}\\&{}<C+D\\P_{n}&{}<T\end{aligned}}}
This, too, forces a contradiction. For, a perpendicular to the midpoint of each polygon side is a radius, of length r. And since the total side length is greater than the circumference, the polygon consists of n identical triangles with total area greater than T. Again we have a contradiction, so our supposition that C might be less than T must be wrong as well.
Therefore, it must be the case that the area enclosed by the circle is precisely the same as the area of the triangle. This concludes the proof.
Rearrangement proof[edit]
Circle area by rearrangement
Following Satō Moshun (Smith & Mikami 1914, pp. 130–132) and Leonardo da Vinci (Beckmann 1976, p. 19), we can use inscribed regular polygons in a different way. Suppose we inscribe a hexagon. Cut the hexagon into six triangles by splitting it from the center. Two opposite triangles both touch two common diameters; slide them along one so the radial edges are adjacent. They now form a parallelogram, with the hexagon sides making two opposite edges, one of which is the base, s. Two radial edges form slanted sides, and the height is h (as in the Archimedes proof). In fact, we can assemble all the triangles into one big parallelogram by putting successive pairs next to each other. The same is true if we increase to eight sides and so on. For a polygon with 2n sides, the parallelogram will have a base of length ns, and a height h. As the number of sides increases, the length of the parallelogram base approaches half the circle circumference, and its height approaches the circle radius. In the limit, the parallelogram becomes a rectangle with width πr and height r.
Unit disk area by rearranging n polygons.
polygon
parallelogram
n
side
base
height
area
4 1.4142136 2.8284271 0.7071068 2.0000000
6 1.0000000 3.0000000 0.8660254 2.5980762
8 0.7653669 3.0614675 0.9238795 2.8284271
10 0.6180340 3.0901699 0.9510565 2.9389263
12 0.5176381 3.1058285 0.9659258 3.0000000
14 0.4450419 3.1152931 0.9749279 3.0371862
16 0.3901806 3.1214452 0.9807853 3.0614675
96 0.0654382 3.1410320 0.9994646 3.1393502
∞ 1/∞ π 1 π
Modern proofs[edit]
There are various equivalent definitions of the constant π. The conventional definition in pre-calculus geometry is the ratio of the circumference of a circle to its diameter:
π = C D . {\displaystyle \pi ={\frac {C}{D}}.}
However, because the circumference of a circle is not a primitive analytical concept, this definition is not suitable in modern rigorous treatments. A standard modern definition is that π is equal to twice the least positive root of the cosine function or, equivalently, the half-period of the sine (or cosine) function. The cosine function can be defined either as a power series, or as the solution of a certain differential equation. This avoids any reference to circles in the definition of π, so that statements about the relation of π to the circumference and area of circles are actually theorems, rather than definitions, that follow from the analytical definitions of concepts like "area" and "circumference".
The analytical definitions are seen to be equivalent, if it is agreed that the circumference of the circle is measured as a rectifiable curve by means of the integral
C = 2 ∫ − R R R d x R 2 − x 2 = 2 R ∫ − 1 1 d x 1 − x 2 . {\displaystyle C=2\int _{-R}^{R}{\frac {R\,dx}{\sqrt {R^{2}-x^{2}}}}=2R\int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}.}
The integral appearing on the right is an abelian integral whose value is a half-period of the sine function, equal to π. Thus C = 2 π R = π D {\displaystyle C=2\pi R=\pi D} is seen to be true as a theorem.
Several of the arguments that follow use only concepts from elementary calculus to reproduce the formula A = π r 2 {\displaystyle A=\pi r^{2}} , but in many cases to regard these as actual proofs, they rely implicitly on the fact that one can develop trigonometric functions and the fundamental constant π in a way that is totally independent of their relation to geometry. We have indicated where appropriate how each of these proofs can be made totally independent of all trigonometry, but in some cases that requires more sophisticated mathematical ideas than those afforded by elementary calculus.
Onion proof[edit]
Area of the disk via ring integration
Using calculus, we can sum the area incrementally, partitioning the disk into thin concentric rings like the layers of an onion. This is the method of shell integration in two dimensions. For an infinitesimally thin ring of the "onion" of radius t, the accumulated area is 2πt dt, the circumferential length of the ring times its infinitesimal width (one can approximate this ring by a rectangle with width=2πt and height=dt). This gives an elementary integral for a disk of radius r.
A r e a ( r ) = ∫ 0 r 2 π t d t = 2 π [ t 2 2 ] 0 r = π r 2 . {\displaystyle {\begin{aligned}\mathrm {Area} (r)&{}=\int _{0}^{r}2\pi t\,dt\\&{}=2\pi \left[{\frac {t^{2}}{2}}\right]_{0}^{r}\\&{}=\pi r^{2}.\end{aligned}}}
It is rigorously justified by the multivariate substitution rule in polar coordinates. Namely, the area is given by a double integral of the constant function 1 over the disk itself. If D denotes the disk, then the double integral can be computed in polar coordinates as follows:
A r e a ( r ) = ∬ D 1 d ( x , y ) = ∬ D t d t d θ = ∫ 0 r ∫ 0 2 π t d θ d t = ∫ 0 r [ t θ ] 0 2 π d t = ∫ 0 r 2 π t d t {\displaystyle {\begin{aligned}\mathrm {Area} (r)&{}=\iint _{D}1\ d(x,y)\\&{}=\iint _{D}t\ dt\ d\theta \\&{}=\int _{0}^{r}\int _{0}^{2\pi }t\ d\theta \ dt\\&{}=\int _{0}^{r}\left[t\theta \right]_{0}^{2\pi }dt\\&{}=\int _{0}^{r}2\pi t\,dt\\\end{aligned}}}
which is the same result as obtained above.
An equivalent rigorous justification, without relying on the special coordinates of trigonometry, uses the coarea formula. Define a function ρ : R 2 → R {\displaystyle \rho :\mathbb {R} ^{2}\to \mathbb {R} } by ρ ( x , y ) = x 2 + y 2 {\displaystyle \rho (x,y)={\sqrt {x^{2}+y^{2}}}} . Note ρ is a Lipschitz function whose gradient is a unit vector | ∇ ρ | = 1 {\displaystyle |\nabla \rho |=1} (almost everywhere). Let D be the disc ρ < 1 {\displaystyle \rho <1} in R 2 {\displaystyle \mathbb {R} ^{2}} . We will show that L 2 ( D ) = π {\displaystyle {\mathcal {L}}^{2}(D)=\pi } , where L 2 {\displaystyle {\mathcal {L}}^{2}} is the two-dimensional Lebesgue measure in R 2 {\displaystyle \mathbb {R} ^{2}} . We shall assume that the one-dimensional Hausdorff measure of the circle ρ = r {\displaystyle \rho =r} is 2 π r {\displaystyle 2\pi r} , the circumference of the circle of radius r. (This can be taken as the definition of circumference.) Then, by the coarea formula,
L 2 ( D ) = ∬ D | ∇ ρ | d L 2 = ∫ R H 1 ( ρ − 1 ( r ) ∩ D ) d r = ∫ 0 1 H 1 ( ρ − 1 ( r ) ) d r = ∫ 0 1 2 π r d r = π . {\displaystyle {\begin{aligned}{\mathcal {L}}^{2}(D)&=\iint _{D}|\nabla \rho |\,d{\mathcal {L}}^{2}\\&=\int _{\mathbb {R} }{\mathcal {H}}^{1}(\rho ^{-1}(r)\cap D)\,dr\\&=\int _{0}^{1}{\mathcal {H}}^{1}(\rho ^{-1}(r))\,dr\\&=\int _{0}^{1}2\pi r\,dr=\pi .\end{aligned}}}
Triangle proof[edit]
Circle unwrapped to form a triangle
The circle and the triangle are equal in area.
Similar to the onion proof outlined above, we could exploit calculus in a different way in order to arrive at the formula for the area of a disk. Consider unwrapping the concentric circles to straight strips. This will form a right angled triangle with r as its height and 2πr (being the outer slice of onion) as its base.
Finding the area of this triangle will give the area of the disk
Area = 1 2 ⋅ base ⋅ height = 1 2 ⋅ 2 π r ⋅ r = π r 2 {\displaystyle {\begin{aligned}{\text{Area}}&{}={\frac {1}{2}}\cdot {\text{base}}\cdot {\text{height}}\\&{}={\frac {1}{2}}\cdot 2\pi r\cdot r\\&{}=\pi r^{2}\end{aligned}}}
The opposite and adjacent angles for this triangle are respectively in degrees 9.0430611..., 80.956939... and in radians 0.1578311... A233527, 1.4129651... A233528.
Explicitly, we imagine dividing up a circle into triangles, each with a height equal to the circle's radius and a base that is infinitesimally small. The area of each of these triangles is equal to 1 / 2 ⋅ r ⋅ d u {\displaystyle 1/2\cdot r\cdot du} . By summing up (integrating) all of the areas of these triangles, we arrive at the formula for the circle's area:
A r e a ( r ) = ∫ 0 2 π r 1 2 r d u = [ 1 2 r u ] 0 2 π r = π r 2 . {\displaystyle {\begin{aligned}\mathrm {Area} (r)&{}=\int _{0}^{2\pi r}{\frac {1}{2}}r\,du\\&{}=\left[{\frac {1}{2}}ru\right]_{0}^{2\pi r}\\&{}=\pi r^{2}.\end{aligned}}}
It too can be justified by a double integral of the constant function 1 over the disk by reversing the order of integration and using a change of variables in the above iterated integral:
A r e a ( r ) = ∬ D 1 d ( x , y ) = ∬ D t d t d θ = ∫ 0 2 π ∫ 0 r t d t d θ = ∫ 0 2 π 1 2 r 2 d θ {\displaystyle {\begin{aligned}\mathrm {Area} (r)&{}=\iint _{D}1\ d(x,y)\\&{}=\iint _{D}t\ dt\ d\theta \\&{}=\int _{0}^{2\pi }\int _{0}^{r}t\ dt\ d\theta \\&{}=\int _{0}^{2\pi }{\frac {1}{2}}r^{2}\ d\theta \\\end{aligned}}}
Making the substitution u = r θ , d u = r d θ {\displaystyle u=r\theta ,\ du=r\ d\theta } converts the integral to
∫ 0 2 π r 1 2 r 2 r d u = ∫ 0 2 π r 1 2 r d u {\displaystyle {\begin{aligned}\int _{0}^{2\pi r}{\frac {1}{2}}{\frac {r^{2}}{r}}du=\int _{0}^{2\pi r}{\frac {1}{2}}r\ du\end{aligned}}}
which is the same as the above result.
The triangle proof can be reformulated as an application of Green's theorem in flux-divergence form (i.e. a two-dimensional version of the divergence theorem), in a way that avoids all mention of trigonometry and the constant π. Consider the vector field r = x i + y j {\displaystyle \mathbf {r} =x\mathbf {i} +y\mathbf {j} } in the plane. So the divergence of r is equal to two, and hence the area of a disc D is equal to
A = 1 2 ∬ D div  r d A . {\displaystyle A={\frac {1}{2}}\iint _{D}\operatorname {div} \mathbf {r} \,dA.}
By Green's theorem, this is the same as the outward flux of r across the circle bounding D:
A = 1 2 ∮ ∂ D  r ⋅ n d s {\displaystyle A={\frac {1}{2}}\oint _{\partial D}\mathbf {r} \cdot \mathbf {n} \,ds}
where n is the unit normal and ds is the arc length measure. For a circle of radius R centered at the origin, we have | r | = R {\displaystyle |\mathbf {r} |=R} and n = r / R {\displaystyle \mathbf {n} =\mathbf {r} /R} , so the above equality is
A = 1 2 ∮ ∂ D  r ⋅ r R d s = R 2 ∮ ∂ D d s . {\displaystyle A={\frac {1}{2}}\oint _{\partial D}\mathbf {r} \cdot {\frac {\mathbf {r} }{R}}\,ds={\frac {R}{2}}\oint _{\partial D}\,ds.}
The integral of ds over the whole circle ∂ D {\displaystyle \partial D} is just the arc length, which is its circumference, so this shows that the area A enclosed by the circle is equal to R / 2 {\displaystyle R/2} times the circumference of the circle.
Semicircle proof[edit]
Note that the area of a semicircle of radius r can be computed by the integral ∫ − r r r 2 − x 2 d x {\displaystyle \int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}\,dx} .
A semicircle of radius r
By trigonometric substitution, we substitute x = r sin  θ {\displaystyle x=r\sin \theta } , hence d x = r cos  θ d θ . {\displaystyle dx=r\cos \theta \,d\theta .}
∫ − r r r 2 − x 2 d x {\displaystyle \int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}\,dx}
= ∫ − π 2 π 2 r 2 ( 1 − sin 2  θ ) ⋅ r cos  θ d θ {\displaystyle =\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\sqrt {r^{2}(1-\sin ^{2}\theta )}}\cdot r\cos \theta \,d\theta }
= 2 r 2 ∫ 0 π 2 cos 2  θ d θ {\displaystyle =2r^{2}\int _{0}^{\frac {\pi }{2}}\cos ^{2}\theta \,d\theta }
= π r 2 2 . {\displaystyle ={\frac {\pi r^{2}}{2}}.}
The last step follows since the trigonometric identity cos  ( θ ) = sin  ( π / 2 − θ ) {\displaystyle \cos(\theta )=\sin(\pi /2-\theta )} implies that cos 2  θ {\displaystyle \cos ^{2}\theta } and sin 2  θ {\displaystyle \sin ^{2}\theta } have equal integrals over the interval [ 0 , π / 2 ] {\displaystyle [0,\pi /2]} , using integration by substitution. But on the other hand, since cos 2  θ + sin 2  θ = 1 {\displaystyle \cos ^{2}\theta +\sin ^{2}\theta =1} , the sum of the two integrals is the length of that interval, which is π / 2 {\displaystyle \pi /2} . Consequently, the integral of cos 2  θ {\displaystyle \cos ^{2}\theta } is equal to half the length of that interval, which is π / 4 {\displaystyle \pi /4} .
Therefore, the area of a circle of radius r, which is twice the area of the semi-circle, is equal to 2 ⋅ π r 2 2 = π r 2 {\displaystyle 2\cdot {\frac {\pi r^{2}}{2}}=\pi r^{2}} .
This particular proof may appear to beg the question, if the sine and cosine functions involved in the trigonometric substitution are regarded as being defined in relation to circles. However, as noted earlier, it is possible to define sine, cosine, and π in a way that is totally independent of trigonometry, in which case the proof is valid by the change of variables formula and Fubini's theorem, assuming the basic properties of sine and cosine (which can also be proved without assuming anything about their relation to circles).
Isoperimetric inequality[edit]
The circle is the closed curve of least perimeter that encloses the maximum area. This is known as the isoperimetric inequality, which states that if a rectifiable Jordan curve in the Euclidean plane has perimeter C and encloses an area A (by the Jordan curve theorem) then
4 π A ≤ C 2 . {\displaystyle 4\pi A\leq C^{2}.}
Moreover, equality holds in this inequality if and only if the curve is a circle, in which case A = π r 2 {\displaystyle A=\pi r^{2}} and C = 2 π r {\displaystyle C=2\pi r} .
Fast approximation[edit]
The calculations Archimedes used to approximate the area numerically were laborious, and he stopped with a polygon of 96 sides. A faster method uses ideas of Willebrord Snell (Cyclometricus, 1621), further developed by Christiaan Huygens (De Circuli Magnitudine Inventa, 1654), described in Gerretsen & Verdenduin (1983, pp. 243–250).
Archimedes' doubling method[edit]
Given a circle, let un be the perimeter of an inscribed regular n-gon, and let Un be the perimeter of a circumscribed regular n-gon. Then un and Un are lower and upper bounds for the circumference of the circle that become sharper and sharper as n increases, and their average (un + Un)/2 is an especially good approximation to the circumference. To compute un and Un for large n, Archimedes derived the following doubling formulae:
u 2 n = U 2 n u n {\displaystyle u_{2n}={\sqrt {U_{2n}u_{n}}}} (geometric mean), and
U 2 n = 2 U n u n U n + u n {\displaystyle U_{2n}={\frac {2U_{n}u_{n}}{U_{n}+u_{n}}}} (harmonic mean).
Starting from a hexagon, Archimedes doubled n four times to get a 96-gon, which gave him a good approximation to the circumference of the circle.
In modern notation, we can reproduce his computation (and go further) as follows. For a unit circle, an inscribed hexagon has u6 = 6, and a circumscribed hexagon has U6 = 4√3. Doubling seven times yields
Archimedes doubling seven times; n = 6×2k.
k
n
un
Un
(un + Un)/4
0 6 6.0000000 6.9282032 3.2320508
1 12 6.2116571 6.4307806 3.1606094
2 24 6.2652572 6.3193199 3.1461443
3 48 6.2787004 6.2921724 3.1427182
4 96 6.2820639 6.2854292 3.1418733
5 192 6.2829049 6.2837461 3.1416628
6 384 6.2831152 6.2833255 3.1416102
7 768 6.2831678 6.2832204 3.1415970
(Here (un + Un)/2 approximates the circumference of the unit circle, which is 2π, so (un + Un)/4 approximates π.)
The last entry of the table has 355⁄113 as one of its best rational approximations; i.e., there is no better approximation among rational numbers with denominator up to 113. The number 355⁄113 is also an excellent approximation to π, better than any other rational number with denominator less than 16604.[4]
The Snell–Huygens refinement[edit]
Snell proposed (and Huygens proved) a tighter bound than Archimedes':
n 3 sin  π n 2 + cos  π n < π < n ( 2 sin  π 3 n + tan  π 3 n ) . {\displaystyle n{\frac {3\sin {\frac {\pi }{n}}}{2+\cos {\frac {\pi }{n}}}}<\pi <n\left(2\sin {\frac {\pi }{3n}}+\tan {\frac {\pi }{3n}}\right).}
This for n = 48 gives a better approximation (about 3.14159292) than Archimedes' method for n = 768.
Derivation of Archimedes' doubling formulae[edit]
Circle with similar triangles: circumscribed side, inscribed side and complement, inscribed split side and complement
Let one side of an inscribed regular n-gon have length sn and touch the circle at points A and B. Let A′ be the point opposite A on the circle, so that A′A is a diameter, and A′AB is an inscribed triangle on a diameter. By Thales' theorem, this is a right triangle with right angle at B. Let the length of A′B be cn, which we call the complement of sn; thus cn2+sn2 = (2r)2. Let C bisect the arc from A to B, and let C′ be the point opposite C on the circle. Thus the length of CA is s2n, the length of C′A is c2n, and C′CA is itself a right triangle on diameter C′C. Because C bisects the arc from A to B, C′C perpendicularly bisects the chord from A to B, say at P. Triangle C′AP is thus a right triangle, and is similar to C′CA since they share the angle at C′. Thus all three corresponding sides are in the same proportion; in particular, we have C′A : C′C = C′P : C′A and AP : C′A = CA : C′C. The center of the circle, O, bisects A′A, so we also have triangle OAP similar to A′AB, with OP half the length of A′B. In terms of side lengths, this gives us
c 2 n 2 = ( r + 1 2 c n ) 2 r c 2 n = s n s 2 n . {\displaystyle {\begin{aligned}c_{2n}^{2}&{}=\left(r+{\frac {1}{2}}c_{n}\right)2r\\c_{2n}&{}={\frac {s_{n}}{s_{2n}}}.\end{aligned}}}
In the first equation C′P is C′O+OP, length r+1⁄2cn, and C′C is the diameter, 2r. For a unit circle we have the famous doubling equation of Ludolph van Ceulen,
c 2 n = 2 + c n . {\displaystyle c_{2n}={\sqrt {2+c_{n}}}.}
If we now circumscribe a regular n-gon, with side A″B″ parallel to AB, then OAB and OA″B″ are similar triangles, with A″B″ : AB = OC : OP. Call the circumscribed side Sn; then this is Sn : sn = 1 : 1⁄2cn. (We have again used that OP is half the length of A′B.) Thus we obtain
c n = 2 s n S n . {\displaystyle c_{n}=2{\frac {s_{n}}{S_{n}}}.}
Call the inscribed perimeter un = nsn, and the circumscribed perimenter Un = nSn. Then combining equations, we have
c 2 n = s n s 2 n = 2 s 2 n S 2 n , {\displaystyle c_{2n}={\frac {s_{n}}{s_{2n}}}=2{\frac {s_{2n}}{S_{2n}}},}
so that
u 2 n 2 = u n U 2 n . {\displaystyle u_{2n}^{2}=u_{n}U_{2n}.}
This gives a geometric mean equation.
We can also deduce
2 s 2 n S 2 n s n s 2 n = 2 + 2 s n S n , {\displaystyle 2{\frac {s_{2n}}{S_{2n}}}{\frac {s_{n}}{s_{2n}}}=2+2{\frac {s_{n}}{S_{n}}},}
or
2 U 2 n = 1 u n + 1 U n . {\displaystyle {\frac {2}{U_{2n}}}={\frac {1}{u_{n}}}+{\frac {1}{U_{n}}}.}
This gives a harmonic mean equation.
Dart approximation[edit]
Unit circle area Monte Carlo integration. Estimate by these 900 samples is 4×709⁄900 = 3.15111...
When more efficient methods of finding areas are not available, we can resort to “throwing darts”. This Monte Carlo method uses the fact that if random samples are taken uniformly scattered across the surface of a square in which a disk resides, the proportion of samples that hit the disk approximates the ratio of the area of the disk to the area of the square. This should be considered a method of last resort for computing the area of a disk (or any shape), as it requires an enormous number of samples to get useful accuracy; an estimate good to 10−n requires about 100n random samples (Thijsse 2006, p. 273).
Finite rearrangement[edit]
We have seen that by partitioning the disk into an infinite number of pieces we can reassemble the pieces into a rectangle. A remarkable fact discovered relatively recently (Laczkovich 1990) is that we can dissect the disk into a large but finite number of pieces and then reassemble the pieces into a square of equal area. This is called Tarski's circle-squaring problem. The nature of Laczkovich's proof is such that it proves the existence of such a partition (in fact, of many such partitions) but does not exhibit any particular partition.
Non-Euclidean circles[edit]
Circles can be defined in non-Euclidean geometry, and in particular in the hyperbolic and elliptic planes.
For example, the unit sphere S 2 ( 1 ) {\displaystyle S^{2}(1)} is a model for the two-dimensional elliptic plane. It carries an intrinsic metric that arises by measuring geodesic length. The geodesic circles are the parallels in a geodesic coordinate system.
More precisely, fix a point z ∈ S 2 ( 1 ) {\displaystyle \mathbf {z} \in S^{2}(1)} that we place at the zenith. Associated to that zenith is a geodesic polar coordinate system ( ϕ , θ ) {\displaystyle (\phi ,\theta )} , 0 ≤ ϕ ≤ π {\displaystyle 0\leq \phi \leq \pi } , 0 ≤ θ < 2 π {\displaystyle 0\leq \theta <2\pi } , where z is the point ϕ = 0 {\displaystyle \phi =0} . In these coordinates, the geodesic distance from z to any other point x ∈ S 2 ( 1 ) {\displaystyle \mathbf {x} \in S^{2}(1)} having coordinates ( ϕ , θ ) {\displaystyle (\phi ,\theta )} is the value of ϕ {\displaystyle \phi } at x. A spherical circle is the set of points a geodesic distance R from the zenith point z. Equivalently, with a fixed embedding into R 3 {\displaystyle \mathbb {R} ^{3}} , the spherical circle of radius R ≤ π {\displaystyle R\leq \pi } centered at z is the set of x in S 2 ( 1 ) {\displaystyle S^{2}(1)} such that x ⋅ z = cos  R {\displaystyle \mathbf {x} \cdot \mathbf {z} =\cos R} .
We can also measure the area of the spherical disk enclosed within a spherical circle, using the intrinsic surface area measure on the sphere. The area of the disk of radius R is then given by
A = ∫ 0 2 π ∫ 0 R sin  ( ϕ ) d ϕ d θ = 2 π ( 1 − cos  R ) . {\displaystyle A=\int _{0}^{2\pi }\int _{0}^{R}\sin(\phi )d\phi \,d\theta =2\pi (1-\cos R).}
More generally, if a sphere S 2 ( ρ ) {\displaystyle S^{2}(\rho )} has radius of curvature ρ {\displaystyle \rho } , then the area of the disk of radius R is given by
A = 2 π ρ 2 ( 1 − cos  ( R / ρ ) ) . {\displaystyle A=2\pi \rho ^{2}(1-\cos(R/\rho )).}
Observe that, as an application of L'Hôpital's rule, this tends to the Euclidean area π R 2 {\displaystyle \pi R^{2}} in the flat limit ρ → ∞ {\displaystyle \rho \to \infty } .
The hyperbolic case is similar, with the area of a disk of intrinsic radius R in the (constant curvature − 1 {\displaystyle -1} ) hyperbolic plane given by
A = 2 π ( 1 − cosh  R ) {\displaystyle A=2\pi (1-\cosh R)}
where cosh is the hyperbolic cosine. More generally, for the constant curvature − k {\displaystyle -k} hyperbolic plane, the answer is
A = 2 π k − 2 ( 1 − cosh  ( k R ) ) . {\displaystyle A=2\pi k^{-2}(1-\cosh(kR)).}
These identities are important for comparison inequalities in geometry. For example, the area enclosed by a circle of radius R in a flat space is always greater than the area of a spherical circle and smaller than a hyperbolic circle, provided all three circles have the same (intrinsic) radius. That is,
2 π ( 1 − cos  R ) < π R 2 < 2 π ( 1 − cosh  R ) {\displaystyle 2\pi (1-\cos R)<\pi R^{2}<2\pi (1-\cosh R)}
for all R > 0 {\displaystyle R>0} . Intuitively, this is because the sphere tends to curve back on itself, yielding circles of smaller area than those in the plane, whilst the hyperbolic plane, when immersed into space, develops fringes that produce additional area. It is more generally true that the area of the circle of a fixed radius R is a strictly decreasing function of the curvature.
In all cases, if k {\displaystyle k} is the curvature (constant, positive or negative), then the isoperimetric inequality for a domain with area A and perimeter L is
L 2 ≥ 4 π A − k A 2 {\displaystyle L^{2}\geq 4\pi A-kA^{2}}
where equality is achieved precisely for the circle.[5]
Generalizations[edit]
We can stretch a disk to form an ellipse. Because this stretch is a linear transformation of the plane, it has a distortion factor which will change the area but preserve ratios of areas. This observation can be used to compute the area of an arbitrary ellipse from the area of a unit circle.
Consider the unit circle circumscribed by a square of side length 2. The transformation sends the circle to an ellipse by stretching or shrinking the horizontal and vertical diameters to the major and minor axes of the ellipse. The square gets sent to a rectangle circumscribing the ellipse. The ratio of the area of the circle to the square is π/4, which means the ratio of the ellipse to the rectangle is also π/4. Suppose a and b are the lengths of the major and minor axes of the ellipse. Since the area of the rectangle is ab, the area of the ellipse is πab/4.
We can also consider analogous measurements in higher dimensions. For example, we may wish to find the volume inside a sphere. When we have a formula for the surface area, we can use the same kind of “onion” approach we used for the disk.
Bibliography[edit]
Archimedes (1897), "Measurement of a circle", in Heath, T. L., The Works of Archimedes, Cambridge University Press
(Originally published by Cambridge University Press, 1897, based on J. L. Heiberg's Greek version.)
Beckmann, Petr (1976), A History of Pi, St. Martin's Griffin, ISBN 978-0-312-38185-1
Gerretsen, J.; Verdenduin, P. (1983), "Chapter 8: Polygons and Polyhedra", in H. Behnke; F. Bachmann; K. Fladt; H. Kunle, Fundamentals of Mathematics, Volume II: Geometry, translated by S. H. Gould, MIT Press, pp. 243–250, ISBN 978-0-262-52094-2
(Originally Grundzüge der Mathematik, Vandenhoeck & Ruprecht, Göttingen, 1971.)
Laczkovich, Miklós (1990), "Equidecomposability and discrepancy: A solution to Tarski's circle squaring problem", Journal für die reine und angewandte Mathematik, 404: 77–117, MR 1037431 [permanent dead link]
Lange, Serge (1985), "The length of the circle", Math! : Encounters with High School Students, Springer-Verlag, ISBN 978-0-387-96129-3
Smith, David Eugene; Mikami, Yoshio (1914), A history of Japanese mathematics, Chicago: Open Court Publishing, pp. 130–132, ISBN 978-0-87548-170-8
Thijsse, J. M. (2006), Computational Physics, Cambridge University Press, p. 273, ISBN 978-0-521-57588-1
References[edit]
Jump up ^ Stewart, James (2003). Single variable calculus early transcendentals (5th. ed.). Toronto ON: Brook/Cole. p. 3. ISBN 0-534-39330-6. However, by indirect reasoning, Eudoxus (fifth century B.C.) used exhaustion to prove the familiar formula for the area of a disk: A = π r 2 . {\displaystyle A=\pi r^{2}.}
Jump up ^ Heath, Thomas L. (2003), A Manual of Greek Mathematics, Courier Dover Publications, pp. 121–132, ISBN 0-486-43231-9 .
Jump up ^ Hill, George. Lessons in Geometry: For the Use of Beginners, page 124 (1894).
Jump up ^ Not all best rational approximations are the convergents of the continued fraction!
Jump up ^ Isaac Chavel (2001), Isoperimetric inequalities, Cambridge University Press
External links[edit]
Area of a Circle Calculator
Area enclosed by a circle (with interactive animation)
Science News on Tarski problem
Calculate disk area on fxSolver
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Algonquin people
From Wikipedia, the free encyclopedia
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Algonquin
Total population
(16,900)
Regions with significant populations
Quebec, Ontario
Languages
French, English, Algonquin
Religion
Midewiwin
Related ethnic groups
Abnaki, Innu, Anicinàpek (Nipissing, Ojibwa, Mississaugas, Saulteaux, Odawa and Potawatomi)
Range of Algonquins around year 1800 in green
The Algonquins are indigenous inhabitants of North America who speak the Algonquin language, a divergent dialect of the Ojibwe language, which is part of the Algonquian language family.[1] Culturally and linguistically, they are closely related to the Odawa and Ojibwe, with whom they form the larger Anicinàpe (Anishinaabe) grouping. The Algonquin people call themselves Omàmiwinini (plural: Omàmiwininiwak) or the more generalised name of Anicinàpe.
Though known by several names in the past, the most common term "Algonquin" has been suggested to derive from the Maliseet word elakómkwik (IPA: [ɛlæˈɡomoɡwik]): "they are our relatives/allies".[2][3] The much larger heterogeneous group of Algonquian-speaking peoples, who, according to Brian Conwell, stretch from Virginia to the Rocky Mountains and north to Hudson Bay, was named after the tribe.
Most Algonquins live in Quebec. The nine Algonquin bands in that province and one in Ontario have a combined population of about 11,000. The Algonquin are original natives of southern Quebec and eastern Ontario in Canada. Today they live in nine communities in Quebec and one in Ontario. The Algonquin were a small tribe that also lives in northern Michigan and southern Quebec and eastern Ontario. (Popular usage reflects some confusion on the point. The term "Algonquin" is sometimes used, such as in the Catholic Encyclopedia, to refer to all Algonquian-speaking societies, although this is not correct.)
Many Algonquins still speak the Algonquin language, called generally Anicinàpemowin or specifically Omàmiwininìmowin. The language is considered one of several divergent dialects of the Anishinaabe languages. Among younger speakers, the Algonquin language has experienced strong word borrowings from the Cree language.[4] Traditionally, the Algonquins lived in either a birch bark wìkiwàms or in wooden mìkiwàms.[5] Today Algonquins live in housings like those of the general public.
Traditionally, the Algonquins were practitioners of Midewiwin (the right path). They believed they were surrounded by many manitòk or spirits in the natural world. French missionaries converted many Algonquins to Catholicism in the 17th and 18th centuries. Today, many of the people practice traditional Midewiwin or a syncretic merging of Christianity and Midewiwin.
In the earliest oral history, the Algonquins say they migrated from the Atlantic coast. Together with other Anicinàpek, they arrived at the "First Stopping Place" near Montreal. While the other Anicinàpe peoples continued their journey up the St. Lawrence River, the Algonquins settled along the Kitcisìpi (Ottawa River), a long-important highway for commerce, cultural exchange and transportation. Algonquin identity, though, was not fully realized until after the dividing of the Anicinàpek at the "Third Stopping Place". Scholars have used the oral histories, archeology, and linguistics to estimate this took place about 2000 years ago, near present-day Detroit.
After contact with the Europeans, especially French and Dutch, the Algonquin nations became active in the fur trade. This led them to fight against the powerful Iroquois, whose confederacy was based in present-day New York. In 1570, the Algonquins formed an alliance with the Montagnais to the east, whose territory extended to the ocean.
Contents
[hide]
1 French contact
2 French-Indian War/Seven Years' War
3 Settlement in Quebec
4 Economy
5 Ethnobotany
6 Modern events
7 Communities
7.1 Historic
7.2 Contemporary
8 See also
9 References
10 Further reading
11 External links
French contact[edit]
Algonquin couple, 18th-century watercolor. The first Algonquian encountered by the French were the Kitcisìpiriniwak ("Ottawa River Men"; singular: Kitcisìpirini), whose village was located on an island in the Ottawa River; the French called this group La Nation de l'Isle.
The Algonquin first met Europeans when Samuel de Champlain came upon a party led by the Kitcisìpirini Chief Tessouat at Tadoussac, in eastern present-day Quebec, in the summer of 1603. They were celebrating a recent victory over the Iroquois, with the allied Montagnais and Etechemins (Malecite). Champlain did not understand that the Algonquins were socially united by a strong totem/clan system rather than the European-styled political concept of nationhood. The several Algonquin bands each had its own chief. Within each band, the chief depended on political approval from each of the band's clan leaders. Champlain needed to cultivate relationships with numerous chiefs and clan leaders. From 1603, some of the Algonquin allied with the French under Champlain. This alliance proved useful to the Algonquin, who previously had little to no access to European firearms.
Champlain made his first exploration of the Ottawa River during May 1613 and reached the fortified Kitcisìpirini village at Morrison Island. Unlike the other Algonquin communities, the Kitcisìpiriniwak did not change location with the seasons. They had chosen a strategic point astride the trade route between the Great Lakes and the St. Lawrence River. They prospered through the collection of beaver pelts from native traders passing through their territory. They also were proud of their corn fields.
At first, the French used the term "Algonquin" only for a second group, the Wàwàckeciriniwak. However, by 1615, they applied the name to all of the Algonquin bands living along the Ottawa River. Because of keen interest by tribes to gain control of the lower Ottawa River, the Kitcisìpiriniwak and the Wàwàckeciriniwak came under fierce opposition. These two large groups allied together, under the leadership of Sachem Charles Parcharini, to maintain the Omàmiwinini identity and territory.
French-Indian War/Seven Years' War[edit]
Main article: French and Indian War
The Iroquois Confederacy drove the Algonquins from their lands. They were aided by having been traded arms by the Dutch, and later by the English. The Iroquois and the English defeated the French and Algonquins.
In 1623, after Sir David Kirke's occupation of New France demonstrated French colonial vulnerability, the French began to trade muskets to the Algonquins and their allies. French Jesuits began to seek Algonquin conversions to Roman Catholicism.
Through all of these years, the Iroquois never attacked the Kitcisìpirinik fortress. But, in 1642, they made a surprise winter raid, attacking the Algonquin while most of the warriors were absent, and causing severe casualties. On March 6, 1647 (Ash Wednesday), a large Mohawk war party attacked the Kitcisìpiriniwak living near Trois-Rivières and almost exterminated them. The Kitcisìpiriniwak were still at Morrison Island in 1650 and inspired respect with their 400 warriors. When the French retreated from the Huron country that year, Tessouat is reported to have had the superior of the Jesuit mission suspended by his armpits because he refused to offer him the customary presents for being allowed to travel through Algonquin territory.
Some joined the mission at Sillery, where they were mostly destroyed by an infectious disease epidemic by 1676. Encouraged by the French, others remained at Trois-Rivières. Their settlement at nearby Pointe-du-Lac continued until about 1830. That year the last 14 families, numbering about 50, moved to Kanesatake near Oka. (The families who stayed in Trois Rivieres can be found in the Algonquin census of Trois Rivieres in the mid-19th century).
The Sulpician Mission of the Mountain was founded at Montreal in 1677, and some Algonquins settled there together with Iroquois converts. The mostly Mohawk community became known as Kahnawake. But many Algonquin maintained their attachment to the traditional territory and fur trading. Those who agreed to move to established reserves or joined other historical bands were federally "recognized". Many others who did not re-locate were later called "stragglers" in the Ottawa and Pontiac counties.
Settlement in Quebec[edit]
Starting in 1721, many Christian Algonquins began to settle for the summer at Kahnesatake, near Oka. The Mohawk Nation was then considered one of the Seven Nations of Canada. Algonquin warriors continued to fight in alliance with France until the British conquest of Quebec in 1760 during the Seven Years' War. After the British took over colonial rule of Canada, their officials sought to make allies of the First Nations. Fighting on behalf of the British Crown, the Algonquins took part in the Barry St Leger campaign during the American Revolutionary War.
Loyalist settlers began encroaching on Algonquin lands shortly after the Revolution. Later in the 19th century, the lumber industry began to move up the Ottawa valley, and a lot of Algonquins were relegated to a string of small reserves.
Economy[edit]
Although the historical Algonquin society was largely hunting- and fishing-based, some Algonquin practiced agriculture and cultivated corn, beans and squash, the famous "Three Sisters" of indigenous horticulture. Being primarily a hunting nation, the people emphasized mobility. They used materials that were light and easy to transport. Canoes were made of birch bark, sewed with spruce roots and rendered waterproof by the application of heated spruce resin and bear grease. During winter, toboggans were used to transport material, and people used snowshoes to get around. The women used tikinaagan (cradleboards) to carry their babies. It was built with wood and covered with an envelope made of leather or material. The baby was standing up with his feet resting on a small board. The mother would then put the tikinàgan on her back. This allowed the infant to look around and observe his surroundings. The child was kept close to the mother but also had much stimulation.
Algonquian-speaking people also practiced agriculture, particularly south of the Great Lakes, where the climate allows for a longer growing season. Notable indigenous crops historically farmed by Algonquins are the sunflower and tobacco. Around 800CE, Eastern Algonquians adopted maize agriculture from their neighbors in the interior.[6] Even among groups who mainly hunted, agricultural products were an important source of food. They obtained what was needed by trading with or raiding societies that practiced more agriculture. Eastern Algonquians created pots that could withstand not only thermal stress but, mechanical stress of rough use.[6]
Archaeological sites on Morrison Island near Pembroke, within the territory of the later Kitcisìpiriniwak, reveal a 1,000-year-old culture that manufactured copper tools and weapons. Copper ore was extracted north of Lake Superior and distributed down to today's northern New York. Local pottery artifacts from this period show widespread similarities that indicate the continuing use of the river for cultural exchange throughout the Canadian Shield and beyond.
Some centuries later, the Algonquin tribe moved in and inhabited the islands and shores along the Ottawa. By the 17th century the first Europeans found them well-established as a hunter-gatherer society in control of the river. The Kitcisìpiriniwak showed entrepreneurial spirit. On Morrison Island, at the location where 5,000-year-old copper artifacts were discovered, the Kitcisìpirini band levied a toll on canoe flotillas descending the river.
Ethnobotany[edit]
The Algonquin of Quebec use an infusion of leaves of Epigaea repens for kidney disorders.[7]
Modern events[edit]
In 1981, members of the Algonquin tribe successfully blockaded a commercial rice-harvesting venture which the Government of Canada had granted permission to harvest wild rice. The tribe had traditionally gathered this grain by hand for centuries.[8] Hundreds of people blockaded roads, and despite police helicopters, paddy wagons, and "a lot of hostility and pushing and shoving," according to Harold Perry, honorary chief of the Ardoch Algonquins, the tribe and its supporters held their ground for 27 days—long enough for the federal government to reverse its decision and revoke the commercial permit.
In recent years, tensions with the lumber industry have flared up again among Algonquin communities, in response to the practice of clear-cutting.
In Ontario, an Algonquin land claim has been ongoing since 1983, encompassing much of the southeastern part of the province, stretching from near North Bay to near Hawkesbury and including Ottawa, Pembroke, and most of Algonquin Provincial Park.[9] The Algonquins never relinquished title to this area. An agreement-in-principle between the Algonquins of Ontario, the Government of Canada and the Government of Ontario was reached in 2015.[10]
In 2000, Algonquins from Timiskaming First Nation played a significant part in the local popular opposition to the plan to convert Adams Mine into a garbage dump.
Members of the Algonquin tribe began a peaceful blockade of an operation on their sacred lands north of Kingston, Ontario on June 29, 2007.[11] Oakville-based Frontenac Ventures Corporation, the prospecting company, sought a court order to force the protesters from the area. A court injunction was obtained on August 27, 2007, and a series of arrests followed, including that of the Ardoch Algonquin First Nation co-Chiefs Robert Lovelace and Paula Sherman. Chief Lovelace served a six-month sentence for contempt of court for violating the injunction, which required all protesters to remain at least 200 metres (660 ft) from the mining site. Chief Sherman also received a six-month sentence, but it was suspended as she agreed to respect the injunction. Tens of thousands of dollars in fines were levied against them.[12]
In addition to the charges of contempt, Frontenac Ventures is suing the Algonquins for $77 million in civil damages. On March 18, 2008, contempt charges were dropped "without costs" against three non-native activists: Frank Morrison and Christian Peacemakers David Milne and Reverend John Hudson. They had been charged with violating the same injunction as Lovelace and Sherman, but Frontenac Ventures declined to prosecute. During the same proceedings, however, warrants were obtained for the arrest of five other non-native activists who allegedly violated the injuncton.[13]
Communities[edit]
At the time of their first meeting with the French in 1603, the various Algonquin bands probably had a combined population somewhere in the neighborhood of 6000. The British estimate in 1768 was 1500. As of 2000, there are close to 8000 Algonquins in Canada, organized into ten separate First Nations: nine in Quebec and one in Ontario.
Historic[edit]
Algonquian Nations documented as early as 1630:
Quebec
Kichesipirini ("people of the great river") — They were the largest and most powerful group of Algonquins. Known variously as: Kitcisìpirini, Kitcisìpiriniwak, Algoumequins de l'Isle-aux-Allumettes, Big River People, Gens d l'Isle, Honkeronon (Wyandot language), Island Algonquian, Island Indians, Island Nation, People from the Island, Kichesippiriniwek, Nation de l'Isle, Nation of the Isle, and Savages de l'Isle. Their main village was on Morrison Island.
Kinounchepirini ("people of the Pickerel-waters") — Also known as Keinouche, Kinouchebiiriniwek, Kinònjepìriniwak, Kinonche, Pickerel, Pike and Quenongebin. Sometimes they were listed as an Algonquian band, but after 1650 they were associated with the Ottawa and were originally found along the lower Ottawa River below Allumette Island.
"Otaguottaouemin" — Also known as Kotakoutouemi or Outaoukotwemiwek. They were located along the Upper Ottawa River above Allumette Island.
Sàgaiganininiwak ("people of the lake") — Also known as Saghiganirini.
"Saginitaouigama" — Also known as Sagachiganiriniwek.
Weskarini ("people of the deer[-clan]") — Also known as the Wàwàckeciriniwak, La Petite Nation, Little Nation, Ouaouechkairini, Ouassouarini, Ouescharini, Ouionontateronon (Wyandot language), or Petite Nation. Their traditional home land is located on the north side of the Ottawa River along the Lievre River and the Rouge River in Quebec.
Ontario
"Iroquet" — They were known as Hiroquet, Hirocay, Iroquay, Yroquetto, and to the Huron as the Atonontrataronon or Ononchataronon; they lived along Ontario's South Nation River.
Matàwackariniwak "people of the bulrushed-shore" — Also known as Madawaska, Madwaska, Matouchkarine, Matouashita, Mataouchkarini, Matouechkariniwek and Matouescarini; the Madawaska River in the Upper Ottawa Valley is named after this band.
"Nibachis" — Located at Muskrat Lake near present-day Cobden, Ontario.
Contemporary[edit]
Status nations in Quebec
Kitigan Zibi Anishinabeg, Kitigan Zibi near Maniwaki (population 3237)
Timiskaming First Nation, Notre-Dame-du-Nord, Quebec (population 2129)
Nation Anishinabe du Lac Simon, Lac-Simon, Quebec (population 2149)
Conseil de la Première Nation Abitibiwinn, Pikogan, Quebec (population 1059)
Eagle Village First Nation, Témiscaming, Quebec (population 991)
Long Point First Nation, Winneway River, Quebec (population 866)
Algonquins of Barrière Lake, Lac Rapide (population 791)
Communauté anicinape de Kitcisakik, Val-d'Or, Quebec (population 494)
Wolf Lake First Nation, Témiscaming, Quebec (population 232) — formed from three historical bands:
Dumoine Lake Band of Algonquin, (historical)
Grassy Lake Band of Algonquin, (historical)
Lac des Quinze Band of Algonquin, (historical)
Status nations in Ontario
Matachewan First Nation, Matachewan, Ontario (population 787) — formed in part by Algonquins
Temagami First Nation, Temagami, Ontario (population 841) — formed in part by Algonquins
Wahgoshig First Nation, Black River-Matheson, Ontario (population 332) — formed in part by Algonquins
Algonquins of Pikwàkanagàn First Nation, Golden Lake, Ontario (population 2635)
Non-status nations
Ardoch Algonquin First Nation, comprising 11 smaller bands:
Big Rideau Algonquin First Nation
Bob’s Lake Algonquin First Nation
Calabogie Algonquin First Nation
Eagle Lake Algonquin First Nation
Eel Lake Algonquin First Nation
Fall River Algonquin First Nation
Golden Lake Algonquin First Nation, (historical): divided into two groups—
Algonquins of Pikwàkanagàn First Nation, (status - see above)
Kijicho Manito Madaouskarini Algonquin Nation (non-status)
>>"NOT" OF ALGONQUIN LINEAGE. Kijicho Manito, a "NIPISSING CHIEF, of NIPISSING DESCENT from Lower Canada, now Quebec. Ancestry.ca. NO valid Baptiste Chiefs left, traditional
Bonnechere Algonquin Community
Mattawachen Algonquin First Nation
Ottawa Algonquin First Nation
Shabot Obaadjiwan First Nation
Tay River Algonquin First Nation
Beaverhouse Algonquin First Nation
Kichesipirini Algonquin First Nation
Weskarini Algonquin First Nation
Temagami Lake Band of Algonquin, (historical)
Washagami Lake Band of Algonquin, (historical)
These population figures are from Canada's Department of Indian and Northern Affairs.[14]
The Nipissing First Nation of North Bay, Ontario is also sometimes considered to belong to the Algonquin group of Anishinaabeg.
See also[edit]
Algonquian Bible
Algonquin mythology
Algonquin Round Table
Great Trail
Kingdom of Saguenay
List of Algonquin Chiefs
References[edit]
Jump up ^ "Assembly of First Nations - The Story". The Assembly of First Nations. Archived from the original on 2009-08-02.
Jump up ^ Campbell (1997:401 n. 133, 136)
Jump up ^ Bright, William (2004). Native American Place Names of the United States. Norman: University of Oklahoma Press, pg. 32
Jump up ^ Artuso, Christian. 1998. noogom gaa-izhi-anishinaabemonaaniwag: Generational Difference in Algonquin. Winnipeg: The University of Manitoba Press.
Jump up ^ Cuoq, Jean André. 1886. Lexique de la Langue Algonquine, Montréal: J. Chapleau & Fils.
^ Jump up to: a b Snow, Dean R. (2010). Archaeology of Native North America. New York: Prentice Hall. p. 223.
Jump up ^ Black, Meredith Jean 1980 Algonquin Ethnobotany: An Interpretation of Aboriginal Adaptation in South Western Quebec. Ottawa. National Museums of Canada. Mercury Series Number 65 (p. 216)
Jump up ^ "Ardoch Algonquin First Nation". Aafna.ca. Archived from the original on 2012-08-13. Retrieved 2012-05-18.
Jump up ^ Ministry of Aboriginal Affairs of Ottawa. Claim Map. Accessed 2010-11-25.
Jump up ^ "Algonquins of Ontario Proposed Agreement-in-Principle". Government of Ontario. Retrieved May 4, 2016.
Jump up ^ "Ardoch Algonquin First Nation". Aafna.ca. 2007-08-23. Archived from the original on 2012-02-11. Retrieved 2012-05-18.
Jump up ^ The Whig Standard - Ontario, CA[dead link]
Jump up ^ The Whig Standard - Ontario, CA[dead link]
Jump up ^ http://sdiprod2.inac.gc.ca/FNProfiles/FNProfiles_home.htm Archived November 6, 2004, at the Wayback Machine.
Further reading[edit]
Daniel Clément, The Algonquins. Hull, Quebec: Canadian Museum of Civilization, 1996.
Yvon H. Couture, Les Algonquins. Val d’Or, Quebec: Éditions Hyperborée, 1983.
Robert Michael Morrissey, Empire by Collaboration: Indians, Colonists, and Governments in Colonial Illinois Country. Philadelphia, PA: University of Pennsylvania Press, 2015.
Evan T. Pritchard, No Word for Time: The Way of the Algonquin People. Tulsa, OK: Council Oak Books, 1997.
Richard White, The Middle Ground: Indians, Empires, and Republics in the Great Lakes Region, 1650-1815. Cambridge, England: Cambridge University Press, 1991.
External links[edit]
Wikimedia Commons has media related to Algonquin people.
Algonquin Nation
Ardoch Algonguin First Nation's website
Ottawa Algonquin First Nation's website
The Bonnechere Algonquin Community's website
Kichesipirini Algonquin First Nation
Tanakiwin - Algonquin Nation in present-day Ontario, Canada
History of the Algonkin
Algonquin Language sample
Watch the documentary Ojigkwanong - Encounter with an Algonquin Sage
Authority control
GND: 4079700-4
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values: Алгонкинал (миллат) – Avaric

values: Амамівініні – Belarusian

values: Алгонкин – Bulgarian

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Algonkin – Bavarian
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values:
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Algonkin – German
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Algonkin – Basque
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eu
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https://ga.wikipedia.org/wiki/Algancach
Algancach – Irish
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Algonquin – Croatian
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hr
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Algonkini – Latvian
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values: Алгонкины (народ) – Russian

values: Алгонквини (народ) – Serbian

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Algonquin – Serbo-Croatian
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values: Алгонкіни (плем'я) – Ukrainian

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