Proposition XVI 406. Problem. To inscribe a regular pentadecagon in a Solution. Draw the chord AB equal to a side of the regular inscribed hexagon, and from a draw the chord AC equal to a side of the regular inscribed decagon. Draw CB. From B as a center with CB as a radius, describe an arc intersecting the circumference as at D. In like manner determine the points E, F, G, H, etc. Proof. and ... arc BC arc AB arc AC = 1, or of the circum., Q.E.F. and the chord CB, which subtends the arc CB, is a side of the regular inscribed pentadecagon CBDEF etc. 407. Cor. I. By joining the alternate vertices of any regular inscribed polygon of an even number of sides, a regular polygon of half the number of sides is inscribed. 408. Cor. II. By joining to the vertices of any regular inscribed polygon the middle points of the arcs subtended by its sides, a regular polygon of double the number of sides is inscribed. Ex. 672. To inscribe a regular octagon in a circle. Ex. 673. To inscribe a regular dodecagon in a circle. Ex. 674. To circumscribe a regular hexagon about a circle. Ex. 676. To inscribe a regular hexagon in an equilateral triangle. Ex. 677. To divide an angle of an equilateral triangle into five equal parts. Ex. 678. The segment of a circle is equal to of a similar segment. What is the ratio of their radii? Ex. 679. How many degrees are there in an arc 18 in. long on a circumference whose radius is 5 ft.? Ex. 680. The radii of two similar segments are as 3: 5. What is the ratio of their areas? Ex. 681. In a circle 3 ft. in diameter an equilateral triangle is inscribed. What is the area of a segment without the triangle? Ex. 682. Two chords drawn from the same point in a circumference to the extremities of a diameter of a circle are 6 in. and 8 in. respectively. What is the area of the circle ? MAXIMA AND MINIMA 409. Of any number of magnitudes of the same kind the greatest is called the Maximum, and the least is called the Minimum. Of all chords of any circle the diameter is the maximum; and of all lines from any point to a given line the perpendicular is the minimum. 410. Figures which have equal perimeters are called Isoperi metric. Proposition XVII 411. Theorem. Of all triangles having two given sides, that in which these sides are perpendicular to each other is the maximum. Proposition XVIII 412. Theorem. Of all isoperimetric triangles which have the same base, the isosceles triangle is the maximum. Data: Any two isoperimetric triangles upon the same base AB, as ABC and ABD, of which ABC is isosceles. To prove ▲ ABC the maximum. Proof. Produce AC to E, making CE equal to AC. Since C is equidistant from A, B, and E, ▲ ABE may be inscribed in a semicircumference; LABE is a right angle. Draw DF equal to DB meeting EB produced in F; CG and DH parallel to AB; CJ and DK perpendicular to AB; and draw AF. 413. Cor. Of all isoperimetric triangles, the equilateral triangle is the maximum. Ex. 683. Of all equivalent parallelograms having equal bases, the rectangle has the least perimeter. ▲ ABC is the maximum. Q.E.D. Proposition XIX 414. Theorem. Of isoperimetric polygons which have the same number of sides, the maximum is equilateral. Data: The maximum of isoperimetric poly gons of a given number of sides, as ABCDEF. Then, ▲ AEF must be the maximum of all the that can be formed upon AE with a B F G perimeter equal to that of ▲ AEF, for if not, a greater ▲, as AEG, could be substituted for ▲ AEF without changing the perimeter of ABCDEF. But it would be impossible to enlarge ABCDEF, for, data, it is the maximum. Hence, § 412, and AAEF is isosceles, AE EF. Similarly any two consecutive sides may be shown equal. Hence, ABCDEF is equilateral. Proposition XX Q.E.D. 415. Theorem. Of isoperimetric regular polygons, that which has the greatest number of sides is the maximum. Data: Any two isoperimetric regular polygons, as ABC and D, of which D has one side more than ABC. To prove Proof. D the maximum. To E, any point in AB, draw CE and construct the ▲ CEF equal to the ▲ ACE. 416. Cor. The area of a circle is greater than the area of any isoperimetric polygon. 417. Theorem. Proposition XXI Of regular polygons which have equal areas, that which has the greatest number of sides has the least perimeter. Data: Any regular polygons which have equal areas, as 4 and B, of which 4 has a greater number of sides than B. To prove the perimeter of 4 less than the perimeter of B. Proof. Construct the regular polygon C, having its perimeter equal to that of 4 and having the same number of sides as B. and, § 346, the perimeter of B is greater than that of C. But the perimeter of c is equal to that of 4; the perimeter of B is greater than that of 4; the perimeter of 4 is less than the perimeter of B. that is, Therefore, etc. Q.E.D. 418. Cor. The circumference of a circle is less than the perimeter of any polygon which has an equal area. Ex. 684. Of all rectangles of a given area, the square has the least perimeter. |