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the exact ratio of the diagonal to the side of a square has now been proved; but an approximation may be made to it, as near as we please, by means of the continued fraction which is equal to that ratio. The first operation gave us 1 for a quotient; the second, and all the others to infinity, give two: thus the fraction in question is
If that fraction, for example, is computed to the fourth term inclusively, its value is found to be 18, or ; so that the approximate ratio of the diagonal to the side of a square, is :: 41: 29. A closer approximation to the ratio might be found by computing a greater number of terms.
REGULAR POLYGONS, AND THE MEASUREMENT OF THE CIRCLE.
A POLYGON, which is at once equilateral and equiangular, is called a regular polygon.
Regular polygons may have any number of sides: the equilateral triangle is one of three sides; the square is one of four.
PROPOSITION I. THEOREM.
Two regular polygons of the same number of sides are similar figures.
Suppose, for example, that ABCDEF, abcdef, are two regular hexagons. The sum of all the angles is the same in both figures, being in each equal to eight right angles (28. I.). The angle
A is the sixth part of that sum; so is the angle a: hence the angles A and a are equal; and for the same reason, the angles B and b, the angles C and c, and, so on.
Again, since from the nature of the polygons, the sides AB, BC, CD, &c. are equal, and likewise the sides ab, bc, cd, &c.; it is plain that AB: ab:: BC: bc:: CD; cd, &c.; hence the
two figures in question have their angles equal, and their homologous sides proportional; hence (Def. 2. III.) they are similar.
Cor. The perimeters of two regular polygons of the same number of sides, are to each other as the homologous sides, and their surfaces as the squares of those sides (27. III.).
Scholium. The angle of a regular polygon is determined by the number of sides, as that of an equiangular polygon (28. I.)
PROPOSITION II. THEOREM.
Any regular polygon may be inscribed in a circle, and circumscribed about one.
Let ABCDE, &c. be a regular polygon: describe a circle through the three points A, B, C, the centre being O, and OP the perpendicular let fall from it to the middle point of BC: join AO and OD. H If the quadrilateral OPCD be placed above the quadrilateral OPBA, they will coincide; for the side OP is common: the angle OPC=OPB, being right; hence the side PC will apply to its equal
PB, and the point C will fall on B: besides, from the nature of the polygon, the angle PCD=PBA; hence CD will take the direction BA; and since CD=BA, the point D will fall on A, and the two quadrilaterals will entirely coincide. The distance OD is therefore equal to AO; and consequently the circle which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it might be shewn, that the circle which passes through the three points B, C, D, will also pass through the point E; and so of all the rest: hence the circle which passes through the points A, B, C, passes through the vertices of all the angles in the polygon, which is therefore inscribed in this circle.
Again, in reference to this circle, all the sides AB, BC, CD, &c. are equal chords; they are therefore (8. II.) equally distant from the centre: hence, if from the point O with the distance OP, a circle be described, it will touch the side BC, and all the other sides of the polygon, each in its middle point, and the circle will be inscribed in the polygon, or the polygon
described about the circle.
Scholium 1. The point O, the common centre of the inscribed and circumscribed circles, may also be regarded as the centre of the polygon; and upon this principle the angle AOB is called the angle at the centre, being formed by two radii drawn to the extremities of the same side AB.
Since all the chords AB, BC, CD, &c. are equal, all the angles at the centre must evidently be equal likewise; and therefore the value of each will be found by dividing four right angles by the number of the polygon's sides.
Scholium 2. To inscribe a regular polygon of a certain number of sides in a given circle, we have only to divide the circumference into as many equal parts as the polygon has sides: for the arcs being equal (see the diagram of Prop. 4.) the chords AB, BC, CD, &c. will also be equal; hence likewise the triangles ABO, BOC, COD must be equal, because they are equiangular; hence all the angles ABC, BCD, CDE, &c. will be equal; hence the figure ABCDE &c. will be a regular polygon.
PROPOSITION III. PROBLEM.
To inscribe a square in a given circle.
Draw two diameters AC, BD, cutting each other at right angles; join their extremities A, B, C, D: the figure ABCD will be a square. For the angles AOB, BOC, &c. being equal, A the chords AB, BC, &c. are also equal; and the angles ABC, BCD, &c. being in semicircles, are right.
Scholium. Since the triangle is rightangled and isosceles, we have (11. III.)
BC: BO:: √2: 1; hence the side of the inscribed
the radius, as the square root of 2 is to unity.
PROPOSITION IV. PROBLEM.
In a given circle, to inscribe a regular hexagon and an equi
Suppose the problem solved, and that AB is a side of the inscribed hexagon; the radii AO, OB being drawn, the triangle AOB will be equilateral.
For the angle AOB is the sixth part of four right angles; therefore, taking the right angle for unity, we shall have AOB-=: and the two other angles ABO, BAO, of the same triangle, are together equal to 2-1; and being mutually equal, each of them must be
equal to ; therefore the triangle ABO is equilateral; therefore the side of the inscribed hexagon is equal to the radius.
Hence to inscribe a regular hexagon in a given circle, the radius must be applied six times to the circumference; which will bring us round to the point we set out from.
And the hexagon ABCDEF being inscribed, the equilateral triangle ACE may be formed by joining the vertices of the alternate angles.
Scholium. The figure ABCO is a parallelogram, and even a rhombus, since AB=BC-CO-AO; hence (14. III.) the sum of the squares of the diagonals AC+BO is equal to the sum of the squares of the sides, that is, to 4AB2, or 4B02: and taking away BO from both, there will remain AC2—3 BO2; hence AC2: BO2:: 3: 1, or AC: BO:: √3: 1; hence the side of the inscribed equilateral triangle is to the radius, as the square root of three is to unity.
PROPOSITION V. PROBLEM.
In a given circle, to inscribe a regular decagon; then a pentagon, and a pentedecagon.
Divide the radius AO in extreme and mean ratio (Prob. 4 III.) at the point M; take the chord AB equal to OM the greater segment; AB will be the side of the regular decagon, and will require to be applied ten times to the circumference.
For, joining MB, we have by construction, AO: OM:: OM: AM; or, since AB=OM, AO: AB::AB: AM; hence the triangles ABO, AMB have
a common angle A, included between proportional sides; hence (20. III.) they are similar. Now the triangle OAB being isosceles, AMB must be isosceles also, and AB=BM; besides AB=OM; hence also MB=OM; hence the triangle BMO is isosceles.
Again, the angle AMB being exterior to the isosceles triangle BMO, is double of the interior angle O (27. I.): but the angle AMB=MAB; hence the triangle OAB is such, that each of the angles at its base, OAB or OBA, is double of O the angle at its vertex; hence the three angles of the triangle are together equal to five times the angle O, which consequently is the fifth part of the two right angles, or the tenth part of four
hence the arc AB is the tenth part of the circumference, and the chord AB is the side of the regular decagon.
Cor. 1. By joining the alternate corners of the regular decagon, the pentagon ACEGI will be formed, also regular.
Cor. 2. AB being still the side of the decagon, let AL be the side of the hexagon; the arc BL will then, with reference to the whole circumference, be fō, or; hence the chord BL will be the side of the pentedecagon or regular polygon of fifteen sides. It is evident, also, that the arc CL is the third of CB.
Scholium. Any regular polygon being inscribed, if the arcs subtended by its sides be severally bisected, the chords of those semi-arcs will form a new regular polygon of double the number of sides: thus it is plain, the square may enable us successively to inscribe regular polygons of 8, 16, 32, &c. sides. And in like manner, by means of the hexagon, regular polygons of 12, 24, 48, &c. sides may be inscribed; by means of the decagon, polygons of 20, 40, 80, &c. sides; by means of the pentedecagon, polygons of 30, 60, 120, &c. sides.*
PROPOSITION VI. PROBLEM.
A regular inscribed polygon ABCD &c. being given, to circumscribe a similar polygon about the same circle.
It is evident, in the
first place, that the three
points O, B, H, lie in the same straight line; for the right-angled
It was long supposed, that, besides the polygons here mentioned, no other could be inscribed by the operations of elementary geometry, or what amounts to the same, by the resolution of equations of the first and second degree. But M. Gauss of Göttingen at length proved, in a work entitled Disquisitiones Arithmeticae, Lipsiae, 1801, that by the method in question, a regular polygon of 17 sides might be inscribed, and generally a regular polygon of +1 sides, provided 2"+1 be a prime number.