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today we will be looking at alcuzzi's
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construction of a heptagon
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and show that the construction of a
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heptagon is the equivalent to the
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construction of two conic sections which
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intersect at a particular point
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we first start by supposing that we
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construct a regular heptagon
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we then inscribe this heptagon in a
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circle
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by using three vertices we are able to
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construct
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triangle abg with gb being one side of
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the heptagon
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now remember that according to euclidean
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geometry
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equal chords in a circle subtend arcs of
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equal length
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and if the angles on the circumference
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of the circle see arcs of equal length
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then the arcs are equal to each other
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with this information we can say that
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angle g sees arc
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a b which we can call unit one
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angle a sees two chords therefore it is
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two units and
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angle b sees four chords therefore it is
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four units
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thus we can construct triangle abg and
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label the angle g
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as a angle a as 2a and angle b
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as 4a we have now completed the first
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step
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which was converting the heptagon into a
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triangle
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now we will convert the triangle into a
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line segment
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first construct line g d so that g
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d is equal to a g
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then draw a d to create isosceles
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triangle dga
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we know that angle gad is equal to angle
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gda
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because triangle dga is isosceles
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also angle gad plus angle gda
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is equal to 2a because of the exterior
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angle property
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we thus know the angle gad equals angle
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gda which is equal to a
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we now construct line eb so that eb is
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equal to a b
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then we draw e a to create isosceles
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triangle e
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a b we know that angle b bea is equal to
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angle bae
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because triangle eba is isosceles
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and that angle bea plus angle bae
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equals 4a because of the exterior angle
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property
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thus we can say that angle bea equals
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angle bae
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which is equal to 2a
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we know that angle eba is the exterior
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angle of triangle abg
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thus using the exterior angle property
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we know that angle eba
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is equal to 2a plus a which is equal to
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3a
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we can now claim that triangle abg
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is similar to triangle dba because their
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angles are equal
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we then write db over a b is equal to a
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b
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over gb and cross multiplied to get
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a b squared equals b d times b g
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remember that a
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b is equal to e b thus we can replace a
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b
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with e b writing e b squared is equal to
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b d times b g
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we can also claim that triangle aeb is
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similar to triangle gea
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because their angles are also equal
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we then write g e over a e
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is equal to a e over e b and cross
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multiply to get
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ae squared is equal to ge times eb
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we know that ae is equal to a g because
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triangle ge a is isosceles and
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a g is equal to gd thus
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ae is equal to gd and we can write
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gd squared is equal to ge times eb
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we have now completed step two and move
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on to transforming the divided line
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segment into conic sections
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remember that we now have two formulas
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gd squared equals ge times eb
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and eb squared is equal to bd times bg
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first we draw a z so that a b is equal
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to b
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g and b z is equal to g b
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then we find point t where e b
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is equal to t z and e t is equal to b z
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since eb is equal to tz we can say that
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tz
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squared is equal to bd times bg
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this equation is the symptom of a
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parabola with a vertex at
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a axis a z and parameter bg
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we also know that g d equals e t since g
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d equals b z which is equal to e t
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we can now write e t squared is equal to
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g
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e times e b this equation is the symptom
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of a hyperbola
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with transverse site fbg and parameter
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t z we have now constructed a parabola
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and a hyperbola with a point of
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intersection
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if we reverse this process we can
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construct a heptagon
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thus the construction of a heptagon is
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the equivalent to the construction of
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two conic sections which intersect at a
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particular point