are generally considered as especially harmonious. Golden rectangle, golden ellipse, pentagram, golden spiral. The golden angle above corresponds to the division of the circle into two arcs with lengths proportional to 1 and φ: a/1 = b/φ = (a+b)/(1+φ) = 2π/φ² rad or 360°/φ². These objects are then laid-out in arcs of spirals oriented in the two directions the numbers of arcs (in the two directions) are two consecutive numbers of the Fibonacci's sequence. on a spiral by creating an object every 137,5.°. Plant species implant scales (pine cone), seeds (sunflower). Miss Nature uses them to provide harmonious growths (flowers, fruits, shells, horns. On the spiral on the right this coefficient is around 5, while it is around 3 for the nautilus' shell! On the left drawing the center of similitude is the intersection of the rectangles' diagonals, the ratio 1/φ, and the angle -π/2 the radius is thus multiplied by φ 4≅6,9 at each turn. On Bernoulli's tombstone is engraved "eadem mutata resurgo" which can be translated by "moved, I rise again the same". These drawings are close to logarithmic spirals, also known as equiangle spirals (constant tangent angle) or Bernoulli's spirals they are invariant by similitude. With sequences of embedded golden rectangles or triangles, we get easily nice " golden spirals". Its eccentricity e is equal to the distance focus-directrix d (if AB=2 then e = d = 1/√φ). Its foci F' and F are easy to construct (M being a point on the ellipse, if AB=1 then MF'+MF = F'F² = φ). These two triangles are at the origin of the Penrose tilings.Īn ellipse inscribed in a golden rectangle is a golden ellipse (ratio of the axis equal to φ). The pentagram shows several golden sections and several examples of the two types of golden triangles: isosceles triangles with ratio of the sides equal to φ, their angles measure 72°-36°-72° and 36°-108°-36° (remark: cos π /5 = φ /2). The vertices of three golden rectangles two by two orthogonal are the vertices of a regular icosahedron more generally two opposite edges of a regular icosahedron define a golden rectangle (thus there are 15).Ī rhombus whose diagonal's ratio is the golden ratio is a golden rhombus (its vertices are the midpoints of the sides of a golden rectangle). Its construction is simple (with AB=2AU, ABCD and ABC'D' are two golden rectangles, and we get the first by adding a square to the second). The ratio of the sides of this two regular pentagons is the golden ratio φ.Ī rectangle whose length/width ratio is the golden ratio is a golden rectangle. MO=NU=PG (N midpoint of and P of )Ī simple knot made with a strip of paper, and then carefully flatted is a " golden knot" just fold over one of the strip's ends and you get a complete pentagram (convex regular pentagon with its five diagonals which are the sides of a regular star pentagon). Starting from three segments with same length Starting from an equilateral triangle and a square Starting from a rectangular triangle OIV with OV=2×OI We construct 0, U and G such as OG/OU=φ, thus OG=φ if we choose OU as unit. Remark: Any non-zero natural integer is written uniquely as sum of non-consecutive numbers of the Fibonacci sequence (Zeckendorf's theorem).Įxample: 33=21+8+3+1 by successive subtractions of numbers from the Fibonacci sequence (the largest possible): 33- 21=12 12- 8=4 4- 3= 1 three constructions of the golden ratio But the golden ratio is also the irrational number the worst approximated by rational numbers because all the integer parts in its continuous fraction are all equal to 1! E.It is interesting to point out that every sequence defined as the Fibonacci sequence by f(n+1)=f(n)+f(n-1) leads to the golden ratio, no matter what the two initial values f(0) and f(1) are: f(n+1)/f(n) -> φ. (Incidently, the circle is tangent to the hypotenuse AB.) References By the Power of a Point Theorem,įrom which, BO = 3 √ 5/2. Indeed, BO being an angle bisector, O divides AC in the ratio of the sides AB : BC:įrom here, AO = 5/2 and CO = 3/2. Extend BO to meet the circle at Q and let P be the other point of intersection of BO with the circle. Draw a circle with center O and radius CO. Let O be the foot of the angle bisector at B. Let ABC be such a triangle with BC = 3, AC = 4 and AB = 5.
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