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PUZZLE # 016 SEPTEMBER 19, 2016 PUZZLE # 016 THE FIVE-POINTED STAR You draw a five-pointed star, like a pentagram, with a single pen stroke of five lines. Each of the five lines have a length of exactly 1 unit (Fig. 1). The Question: What is the distance between two adjacent points of the pentagram? x Fig 1: The five-pointed star METHOD Triangle A A Triangle B x C x D Triangle C x Triangle D x PROCESS How We Can Look At The Problem: The goal is to find the distance between two adjacent points of the star (x). x B 1 1 unit We can approach the problem by dividing up the star, and pentagon that is formed by connecting the points, into more manageable geometry. What We Know: x Triangle A We know that the length of the lines making up the star are 1 unit long. 0.5 units By dividing the pentagram into smaller geometry, we can use a right angle triangle that has the line connecting two adjacent points of the star (x) as the hypotenuse (Fig. 2). Fig 2: Triangle A with base 0.5 units, and hypotenuse x What We Can Find: 360º/5 = 72º 72º 2 In order to solve for x, we need this height, or the angles of Triangle A (Fig. 3). Fig 3: The angle of a pentagon at the center 180º total in triangle 54º Triangle B 3 54º 72º The base of Triangle A is 0.5 units, but we are missing the height. = (180º - 72º)/2 = 108º/2 = 54º 108º From the center, the angle between two lines connected to the points of the star is 72º. Solving Triangle B: Knowing the angle at the center, we can solve Triangle B. Here we find that the angle between the edges of the pentagon is 108º This means the top angle of Triangle A, is 54º (Fig. 4). Fig 4: Angles at edge of the pentagon SOLUTION Triangle A x = 0.618 54º units Sin(54º) = 0.5/x x = 0.5/[Sin(54º)] x = 0.618 units 36º 0.5 units Using Trigonometry: Now that we have the angles of Triangle A, we can use sine to solve for x (Fig. 5). The distance between two adjacent points of this five-pointed star is 0.618 units. Fig 5: Use sine to solve for x: x = 0.618 units