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On a blank piece of paper (typing paper, no lines) fairly carefully draw the ideal shape that comes to mind when you hear the word “rectangle”. Now measure rectangle and compute the ratio of the longer side (numerator) to the shorter side (denominator). List the ratios from the class.
Starting with a square , find the midpoint of the side . Extend the side . Drop the arc of a circle with center and radius to find the point on the extension of . Now erect a perpendicular to at , and also extend to meet the perpendicular at . The resulting rectangle is a Golden Rectangle.
Some people say that the proportions of the Parthenon in Athens are the proportions of a Golden Rectangle. What do you think? Somme people say that the Golden Rectangle appears many other places in art, including the Mona Lisa, The Last Supper, and early twentieth-century art by Suerat and Mondrian. What do you think?
This section is adapted from: Instructor Resources and Adjunct Guide for the second edition of The Heart of Mathematics by E. Burger, M. Starbird, and D. Bergstrand.
Why does the equation have the same solution as the equation
As a solution, suppose that the number is the solution, and then multiply both sides of the first equation to obtain:
Then divide both sides of the equation by the value to see that
The result is that the value satisfies the second equation too!
Since we are only interested in proportions or ratios, we can set some convenient length to be 1 and measure all other lengths in terms of this standard. I will take the length of the vertical side of the original (blue in the figure in the book) rectangle. Then the length of the longer horizontal side is 2. The sidelength from the midpoint of the longer horizontal side is 1. Then the radius of the circle which swings to extend the rectangle (by the Pythagorean Theorem) satisfies r2 = 12 + 12 = 2. The radius is and the length of the extended side is . The ratio of the long side to the short side is then 1 + and this is not a Golden Rectangle.
Again, since we are concerned with ratios or proportions, we can choose a convenient side length to be our standard length. Let the sidelength of the vertical side of the rectangle be 1, that is the height of the rectangle is 1. Let the sidelength of the base be b and assume that as shown, so that the largest square is removed from the base as shown. Then removing the square, we are left with a rectangle of dimensions and 1. Then the desired ratio is
But this is an impossible equation for lengths, since it implies b = -1. So that means b < 1 so the initial rectangle s tall and thin. (Now draw your own diagram to help you visualize the problem solution.) Then removing the square, we obtain
and so solving for b, b = 1/2. This is not a Golden Rectangle, it is a rectangle twice as high as it is wide.
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