﻿1 00:00:05,600 --> 00:00:09,440 today we will be looking at alcuzzi's 2 00:00:07,600 --> 00:00:11,040 construction of a heptagon 3 00:00:09,440 --> 00:00:12,960 and show that the construction of a 4 00:00:11,040 --> 00:00:15,519 heptagon is the equivalent to the 5 00:00:12,960 --> 00:00:19,199 construction of two conic sections which 6 00:00:15,519 --> 00:00:21,119 intersect at a particular point 7 00:00:19,199 --> 00:00:23,840 we first start by supposing that we 8 00:00:21,119 --> 00:00:26,080 construct a regular heptagon 9 00:00:23,840 --> 00:00:28,160 we then inscribe this heptagon in a 10 00:00:26,080 --> 00:00:30,560 circle 11 00:00:28,160 --> 00:00:31,199 by using three vertices we are able to 12 00:00:30,560 --> 00:00:34,800 construct 13 00:00:31,199 --> 00:00:37,200 triangle abg with gb being one side of 14 00:00:34,800 --> 00:00:39,520 the heptagon 15 00:00:37,200 --> 00:00:40,480 now remember that according to euclidean 16 00:00:39,520 --> 00:00:43,520 geometry 17 00:00:40,480 --> 00:00:44,879 equal chords in a circle subtend arcs of 18 00:00:43,520 --> 00:00:47,440 equal length 19 00:00:44,879 --> 00:00:49,840 and if the angles on the circumference 20 00:00:47,440 --> 00:00:52,960 of the circle see arcs of equal length 21 00:00:49,840 --> 00:00:55,120 then the arcs are equal to each other 22 00:00:52,960 --> 00:00:56,719 with this information we can say that 23 00:00:55,120 --> 00:01:00,800 angle g sees arc 24 00:00:56,719 --> 00:01:03,760 a b which we can call unit one 25 00:01:00,800 --> 00:01:04,879 angle a sees two chords therefore it is 26 00:01:03,760 --> 00:01:07,760 two units and 27 00:01:04,879 --> 00:01:09,840 angle b sees four chords therefore it is 28 00:01:07,760 --> 00:01:12,960 four units 29 00:01:09,840 --> 00:01:14,159 thus we can construct triangle abg and 30 00:01:12,960 --> 00:01:17,759 label the angle g 31 00:01:14,159 --> 00:01:21,840 as a angle a as 2a and angle b 32 00:01:17,759 --> 00:01:22,400 as 4a we have now completed the first 33 00:01:21,840 --> 00:01:24,479 step 34 00:01:22,400 --> 00:01:25,600 which was converting the heptagon into a 35 00:01:24,479 --> 00:01:27,920 triangle 36 00:01:25,600 --> 00:01:30,000 now we will convert the triangle into a 37 00:01:27,920 --> 00:01:33,119 line segment 38 00:01:30,000 --> 00:01:36,159 first construct line g d so that g 39 00:01:33,119 --> 00:01:39,280 d is equal to a g 40 00:01:36,159 --> 00:01:42,799 then draw a d to create isosceles 41 00:01:39,280 --> 00:01:42,799 triangle dga 42 00:01:42,960 --> 00:01:47,040 we know that angle gad is equal to angle 43 00:01:45,920 --> 00:01:50,720 gda 44 00:01:47,040 --> 00:01:54,640 because triangle dga is isosceles 45 00:01:50,720 --> 00:01:57,360 also angle gad plus angle gda 46 00:01:54,640 --> 00:01:59,680 is equal to 2a because of the exterior 47 00:01:57,360 --> 00:02:03,280 angle property 48 00:01:59,680 --> 00:02:08,319 we thus know the angle gad equals angle 49 00:02:03,280 --> 00:02:11,440 gda which is equal to a 50 00:02:08,319 --> 00:02:13,760 we now construct line eb so that eb is 51 00:02:11,440 --> 00:02:16,560 equal to a b 52 00:02:13,760 --> 00:02:17,360 then we draw e a to create isosceles 53 00:02:16,560 --> 00:02:22,239 triangle e 54 00:02:17,360 --> 00:02:23,599 a b we know that angle b bea is equal to 55 00:02:22,239 --> 00:02:27,120 angle bae 56 00:02:23,599 --> 00:02:31,440 because triangle eba is isosceles 57 00:02:27,120 --> 00:02:34,080 and that angle bea plus angle bae 58 00:02:31,440 --> 00:02:35,680 equals 4a because of the exterior angle 59 00:02:34,080 --> 00:02:38,640 property 60 00:02:35,680 --> 00:02:40,000 thus we can say that angle bea equals 61 00:02:38,640 --> 00:02:43,120 angle bae 62 00:02:40,000 --> 00:02:46,080 which is equal to 2a 63 00:02:43,120 --> 00:02:48,720 we know that angle eba is the exterior 64 00:02:46,080 --> 00:02:51,360 angle of triangle abg 65 00:02:48,720 --> 00:02:53,200 thus using the exterior angle property 66 00:02:51,360 --> 00:02:55,840 we know that angle eba 67 00:02:53,200 --> 00:02:57,599 is equal to 2a plus a which is equal to 68 00:02:55,840 --> 00:03:00,720 3a 69 00:02:57,599 --> 00:03:03,680 we can now claim that triangle abg 70 00:03:00,720 --> 00:03:05,760 is similar to triangle dba because their 71 00:03:03,680 --> 00:03:09,760 angles are equal 72 00:03:05,760 --> 00:03:10,080 we then write db over a b is equal to a 73 00:03:09,760 --> 00:03:13,280 b 74 00:03:10,080 --> 00:03:17,920 over gb and cross multiplied to get 75 00:03:13,280 --> 00:03:18,879 a b squared equals b d times b g 76 00:03:17,920 --> 00:03:22,480 remember that a 77 00:03:18,879 --> 00:03:22,720 b is equal to e b thus we can replace a 78 00:03:22,480 --> 00:03:26,480 b 79 00:03:22,720 --> 00:03:30,000 with e b writing e b squared is equal to 80 00:03:26,480 --> 00:03:33,040 b d times b g 81 00:03:30,000 --> 00:03:35,200 we can also claim that triangle aeb is 82 00:03:33,040 --> 00:03:38,400 similar to triangle gea 83 00:03:35,200 --> 00:03:41,440 because their angles are also equal 84 00:03:38,400 --> 00:03:44,319 we then write g e over a e 85 00:03:41,440 --> 00:03:45,360 is equal to a e over e b and cross 86 00:03:44,319 --> 00:03:50,480 multiply to get 87 00:03:45,360 --> 00:03:54,080 ae squared is equal to ge times eb 88 00:03:50,480 --> 00:03:57,280 we know that ae is equal to a g because 89 00:03:54,080 --> 00:04:00,720 triangle ge a is isosceles and 90 00:03:57,280 --> 00:04:04,720 a g is equal to gd thus 91 00:04:00,720 --> 00:04:09,840 ae is equal to gd and we can write 92 00:04:04,720 --> 00:04:12,159 gd squared is equal to ge times eb 93 00:04:09,840 --> 00:04:14,159 we have now completed step two and move 94 00:04:12,159 --> 00:04:16,880 on to transforming the divided line 95 00:04:14,159 --> 00:04:19,280 segment into conic sections 96 00:04:16,880 --> 00:04:22,479 remember that we now have two formulas 97 00:04:19,280 --> 00:04:26,800 gd squared equals ge times eb 98 00:04:22,479 --> 00:04:29,759 and eb squared is equal to bd times bg 99 00:04:26,800 --> 00:04:30,240 first we draw a z so that a b is equal 100 00:04:29,759 --> 00:04:33,840 to b 101 00:04:30,240 --> 00:04:36,880 g and b z is equal to g b 102 00:04:33,840 --> 00:04:41,520 then we find point t where e b 103 00:04:36,880 --> 00:04:44,960 is equal to t z and e t is equal to b z 104 00:04:41,520 --> 00:04:45,600 since eb is equal to tz we can say that 105 00:04:44,960 --> 00:04:49,440 tz 106 00:04:45,600 --> 00:04:51,440 squared is equal to bd times bg 107 00:04:49,440 --> 00:04:53,120 this equation is the symptom of a 108 00:04:51,440 --> 00:04:57,520 parabola with a vertex at 109 00:04:53,120 --> 00:05:01,039 a axis a z and parameter bg 110 00:04:57,520 --> 00:05:05,039 we also know that g d equals e t since g 111 00:05:01,039 --> 00:05:07,759 d equals b z which is equal to e t 112 00:05:05,039 --> 00:05:08,080 we can now write e t squared is equal to 113 00:05:07,759 --> 00:05:11,680 g 114 00:05:08,080 --> 00:05:12,960 e times e b this equation is the symptom 115 00:05:11,680 --> 00:05:16,000 of a hyperbola 116 00:05:12,960 --> 00:05:20,000 with transverse site fbg and parameter 117 00:05:16,000 --> 00:05:21,759 t z we have now constructed a parabola 118 00:05:20,000 --> 00:05:23,039 and a hyperbola with a point of 119 00:05:21,759 --> 00:05:24,800 intersection 120 00:05:23,039 --> 00:05:26,720 if we reverse this process we can 121 00:05:24,800 --> 00:05:29,039 construct a heptagon 122 00:05:26,720 --> 00:05:30,800 thus the construction of a heptagon is 123 00:05:29,039 --> 00:05:33,120 the equivalent to the construction of 124 00:05:30,800 --> 00:05:36,560 two conic sections which intersect at a 125 00:05:33,120 --> 00:05:36,560 particular point