1062 The area of the inscribed regular hexagon is the mean proportional between the areas of the inscribed and circumscribed equilateral triangles. 1063 Two diagonals of a regular hexagon, not drawn from the same vertex, are parallel or perpendicular; or, they bisect or trisect each other. 1064 The intersections of the diagonals joining the alternate vertices of a regular hexagon form a regular hexagon whose area is one third that of the given hexagon. 1065 The radius of an inscribed regular polygon is the mean proportional between its apothem and the radius of the similar circumscribed polygon. 1066 The side of a circumscribed regular hexagon is two thirds of the side of an inscribed equilateral triangle. 1067 The perimeter of the inscribed equilateral triangle is three fourths of the perimeter of the circumscribed regular hexagon. 1068 The sum of the squares of the three lines drawn from any point on the circumference to the vertices of an inscribed equilateral triangle is constant. 1069 The difference of the squares of the diagonal and the side of a regular pentagon is equal to the product of the diagonal and the side. 1070 The inscribed regular octagon is equivalent to a rectangle whose dimensions are the sides of the inscribed and the circumscribed squares. 1071 If from any point within a regular polygon of n sides perpendiculars are drawn to the sides, the sum of these perpendiculars is equal to n times the apothem. 1072 An angle of a regular polygon of n sides is to an angle of a regular polygon of n + 2 sides as n2 - 4 is to n2. 1073 In a regular polygon of n sides the diagonals drawn from the vertex of any angle to the opposite vertices divide the angle into n - 2 equal parts. 1074 If the diameter of a circle is divided into two parts, and on opposite sides of these parts as diameters semicircles are described, the circle is divided by the semicircumferences into two parts which are proportional to the segments of the diameter. 1075 Arcs of equal length on two circumferences subtend angles at the center which are inversely proportional to the radii. PROBLEMS OF CONSTRUCTION 1076 To circumscribe an equilateral triangle about a given circle. 1077 To circumscribe a square about a given circle. 1078 To circumscribe a regular pentagon about a given circle. 1079 To circumscribe a regular hexagon about a given circle. 1080 To circumscribe a regular octagon about a given circle. 1081 To construct a circle equivalent to twice a given circle. SOLUTION. Let R be the radius of the given circle. Construct an isosceles right ▲ with legs equal to R. Call the hypotenuse R'. R' is the radius of the required circle. PROOF. R/2≈2 R2 (§ 418). ... πR12 = 2 TR2. But #R/2 is the area of *.TR/2 the required O, and 2 πR2 is twice the area of the given O (§ 464). 1082 To construct a circle equivalent to the sum of two given circles. 1083 To construct a circle equivalent to the difference of two given circles. 1084 To bisect a circle by a concentric circumference. 1085 To trisect a circle by concentric circumferences. 1086 To construct an equilateral triangle on a given line. 1087 To construct a square on a given line. 1088 To construct a regular pentagon on a given line. 1094 To divide a right angle into five equal parts. 1095 To cut off the corners of a square so as to form a regular octagon. 1096 To cut off the corners of a regular pentagon so as to form a regular decagon. 1097 To construct a circle equivalent to five times a given circle. 1098 To construct a circle equivalent to two thirds of a given circle. 1099 To divide a circle into n equal parts by concentric circumferences. 1100 To construct a circle which shall be to a given circle as m is to n. SUPPLEMENTARY MAXIMA AND MINIMA DEFINITIONS 486 A maximum magnitude is the greatest of its class. 487 A minimum magnitude is the least of its class. Thus, the diameter of a circle is the maximum chord, and the perpendicular is the minimum line that can be drawn from a given point to a given straight line. 488 Isoperimetric figures are figures which have equal perimeters. PROPOSITION I. THEOREM 489 Of all triangles having two given sides, that in which these two sides include a right angle is the maximum. HYPOTHESIS. In the ABC and DEF, AB = DE, BC = EF, AB is 1 to BC, PROPOSITION II. THEOREM 490 Of all isoperimetric triangles on the same base, the isosceles is the maximum. HYPOTHESIS. In the A ABC and DBC, the perimeters are equal, AB = AC, and DB > DC. CONCLUSION. ▲ ABC > ▲ DBC. PROOF On BA produced take AE = AB, join EC, ED, and draw AH and DK L to EC. Then the BCE is a rt. 4, for it may be inscribed in a semicircle whose center is A and radius AC. The ▲ AEC is isosceles by const. But BD + DE > BE. § 277 .. EH = HC. § 178 BD + DC. That is, K falls between H and C. ... HC > KC. .:. Δ ABC > Δ DBC. Ax. 15 Iden. Ax. 5 § 191 Ax. 12 § 408 Q. E. D. Of all isoperimetric triangles the equi 491 COROLLARY. lateral is the maximum. PROPOSITION III. THEOREM 492 Of all polygons having all the sides given but one, the maximum can be inscribed in a semicircle whose diameter is the undetermined side. HYPOTHESIS. ABCDEF is the maximum of all polygons whose given sides are AB, BC, CD, DE, EF. CONCLUSION. is AF. ABCDEF can be inscribed in a semicircle whose diameter PROOF Join any vertex C to A and F. Then the ACF is a rt. ; for if not, without changing the length of AC and CF, the area of the triangle ACF can be increased by making the ▲ ACF a rt. ≤, while the area of the remainder of the polygon remains unchanged. § 489 But this would increase the area of the polygon, which is contrary to the hypothesis that the polygon is a maximum. .. the ACF is a rt. ▲, and the vertex C lies in the semicircumference whose diameter is AF. § 277 Likewise every vertex of the polygon lies in the semicircumference. .. the polygon can be inscribed in a semicircle whose diameter is AF. Q. E. D. |