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Asked May 6, 2019

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I am stuck on a question from my history of math textbook. Here is the question:

The division of a line segment into two unequal parts

so that the whole segment will have the same ratio to

its larger part that its larger part has to its smaller

part is called the golden section. A classical

ruler-and-compass construction for the golden section

of a segment AB is as follows. At B erect BC equal

and perpendicular to AB. Let M be the midpoint of

AB, and with MC as a radius, draw a semicircle

cutting AB extended in D and E. Then the segment

B E laid off on AB gives P, the golden section.

(a) Show that 4DBC is similar to 4CBE, whence

DB=BC D BC=B E.

(b) Subtract 1 from both sides of the equality in

part (a) and substitute equals to conclude that

AB=AP D AP=P B.

(c) Prove that the value of the common ratio in part

(b) is (p5 C 1)=2, which is the “golden ratio.”

[Hint: Replace P B by AB AP to see that

AB2 AB Ð AP AP2 D 0. Divide this

equation by AP2 to get a quadratic equation in

the ratio AB=AP.]

(d) A golden rectangle is a rectangle whose sides

are in the ratio (p5 C 1)=2. (The golden

rectangle has dimensions pleasing to the eye

and was used for the measurements of the

facade of the Parthenon and other Greektemples.) Verify that both the rectangles AEFG and BEFC are golden rectangles

1 Rating

Step 1

As per norms , the first three parts a,b and c are answered. The problem pertains to the construction of the point P between given points A and B , which divides AB in the golden ratio.

Step 2

Proof of a) the triangles are similar as the angles are the same. (recall that DCE is a right angle, as it is the angle in a semicircle, by construction)

Step 3

b) Using BC=AB, AP=BE, we obtain...

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