GEOMETRIC CONSTRUCTION OF THE REGULAR DECAGON AND PENTAGON INSCRIBED IN A CIRCLE BY HENRY H. LUDLOW MAJOR IN THE ARTILLERY CORPS, U. S. ARMY PRINATED FOR ITHE A U'HOR THE OPEN CHICAGO COUR;T PUBLISHING CO. 1904 COPYRIGHT, 1904 H-ENRY H. LUDLOW STARKVILLE, MISS. CONTENTS. PAGE Theorem I................. 7 Theorem II................ 8 Theorem III............... 9 Problem I................ 9 Problem II................ 11 Problem III............... 11 PREFACE. In the Spring of 1873, at West Point, N. Y., the writer's attention was called to the construction in the last paragraph of this paper by the late Albert E. Church, then Professor of Mathematics at the U. S. Military Academy, West Point, N. Y. He stated to the writer's class, then studying Geometry as Cadets, that this was an ancient construction, probably of Chinese origin; that he had tested it by trigonometric computation with but slight discrepancy which might be attributed to tabular inaccuracy. He expressed doubt as to its geometric truth and proposed to the class as a problem, " To prove or disprove its truth."* Two solutions resulted, one by the late Edward S. Farrow, the other by the writer. Incidental to the writer's solution came the construction in paragraph 4, which is really a part of the ancient construction. It is so much simpler than the construction usually given in elementary geometry that the writer taught it to Cadets at West Point, N. Y., in 1880 and subsequently elsewhere with this proof. * The construction without proof may be found in Davies and Peck's Mathematical Dictionary. 6 PREFACE. The question having been recently raised whether geometric proofs of these constructions have been published, the writer offers them to the public, believing them worthy of introduction in text-books on elementary geometry. The introductory paragraphs were put in to lead up as simply as possible to these constructions. H. H. L. AGRICULTURAL COLLEGE MISSISSIPPI, December, 1903. GEOMETRIC CONSTRUCTION OF THE REGULAR DECAGON AND PENTAGON INSCRIBED IN A CIRCLE. 1. THEOREM I. The side of a regular decagon inscribed in a circle is equal to the mean segment of the radius divided in extreme and mean ratio. A 21 C Let AB represent the side of a regular decagon inscribed in the circle whose center is C. Draw the radii CA, CB, and bisect the angle ABC by BD, meeting AC at D. Since AB is the side of the regular inscribed decagon, it is the chord of one tenth the circumference, and the angle ACB is one tenth of four right angles, or two fifths of a right angle. The sum of the interior angles of the isosceles triangle ABC being two right angles, the angles A and B are each equal to - (2 - -) - of a right angle, and half of B is equal to C. Hence the triangle BCD is isosceles, giving BD = CD. The triangles CBA and BDA, having the angle 8 CONSTRUCTION OF DECAGON AND PENTAGON. A in common and the angle BCA equal to the angle DBA, are mutually equiangular and similar. But the triangle CBA is isosceles, hence BDA is also isosceles, and AB = BD - CD. Since the triangles CBA and BDA are similar, we have CA: AB:: AB: DA, or CA: CD::CD:CA, in which AB = CD. Q. E. D. 2. THEOREM II. The height of the arc subtended by the side of a regular pentagon inscribed in a circle is equal to half the extreme segment of the radius divided in extreme and mean ratio. A t4 Resume the construction of ~1. From B draw BE perpendicular to CA, meeting it at E and the arc produced at F. Since BF is perpendicular to the radius CA, the chord BF and its arc BAF are both bisected by CA. Hence BF is the side of the regular inscribed pentagon. Since BE is a perpendicular from the vertex B to the base DA of the isosceles triangle BDA, it bisects the base at E, and EA - - DA. Q. E. D. CONSTRUCTION OF DECAGON AND PENTAGON. 9 3. THEOREM III. In any circle, the square of the side of the regular inscribed pentagon exceeds the square of the side of regular inscribed decagon by the square of the radius. Resume the construction of ~2. Since FB- 2 A E 2 C EB, we have FB2 4 EB2. The right triangle AEB gives EB2 = AB2 - EA2. Hence, FB2 =4 AB - 4EA2. Since AB = CD = CA- DA, we have FB2 3 AB2 + (CA2 - 2 CA x DA + DA2) - 4 EA2. But CA is divided in extreme and mean ratio atD, ~ 1, giving CA x DA = CD2 = AB2, and by ~ 2, DA2 4 EA2 which reduces the above to FB2 = AB2 + CA2. Q. E. D. 4. PROBLEM I. To divide a given straight line in extreme and mean ratio. Let AB represent the given line. At B erect the perpendicular BC = AB and produce AB to D, making BD = I AB. Join CD. From D as 10 CONSTRUCTION OF DECAGON AND PENTAGON. a center, with DC as a radius, draw an arc cutting AB at E. Then will AB: EB:: EB: AE. For, produce the arc EC until it cuts AB proA E P duced at G. Since BC is a perpendicular from a point of a semi-circumference to the diameter, it is a mean proportional between the segments of the diameter, giving EB: BC:: BC: BG. By division we have BC-EB: EB:: BG- BC: BC, but BC - 4B EBAB EB AE, BG - BC BGAB = BG - 2BD = EB, and BC = AB, giving AE: EB:: EB: AB. Q. E. D. 5. If an exterior point of division be required, it will lie at the second point of intersection G. For, as above, BG: BC:: BC: EB. By composition BG + BC BG:: BC: + EB: BC. CONSTRUCTION OF DECAGON AND PENTAGON. 11 But BG +BC BG+AB =AG, BC+EB =AB + EB = 2 BD EB = BG, and BC = AB, giving AG: BG:: BG: AB. Q. E. D. 6. PROBLEM II. To inscribe a regular decagon in a circle given. A Let C be the center and CA any radius. At A draw the tangent line AB and lay off AB A AC. Join CB. From B as a center with BC as a radius describe an arc cutting BA produced at D. Then, ~~ 4, 1, DA is the required side. From A lay off successive chords, each equal to AD. 7. PROBLEM III. To inscribe a regular pentagon in a given circle. First method. Let C be the center of the given circle, AC any radius. Draw the radius 12 CONSTRUCTION OF DECAGON AND PENTAGON. CB perpendicular to AC. Produce AC to D, making CD -i AC. Join BD. From D as a A C JP center, with the radius DB, draw an arc cutting AC at E. Draw the chord BE. It will be the side of the required pentagon. For, by ~ 4, AC is divided in extreme and mean ratio at E, and EC is equal to the side of the regular inscribed decagon, ~~ 1, 6. But in the right triangle BCE, BE2 = EC2 + CB2. Therefore, ~ 3, BE is equal to the side of the regular inscribed pentagon. From B lay off successive chords, each equal to BE. Second method. Draw the perpendicular bisector of AE. It will be the required side. For, its height of arc from its chord is the same as that of the side of the regular inscribed pentagon, ~ 2, the two chords are therefore equally distant from the center, and are equal. Third method. Inscribe a regular decagon, ~ 6, and join the alternate vertices.