Department of Mathematics Education

Dr. J. Wilson, EMAT 6690

by David Wise

**Problem and Solution**

**1. Given any line segment AB.**

- Construct segment AB and label it x.

**2. Construct a perpendicular segment BD usch that 2BD
= AB.**

- Construct a perpendicular line to segment AB through point B.
- Construct the midpoint of segment AB.
- Construct a circle by center and point with point B as the center and the midpoint of segment AB the point.
- Construct the intersection point of the circle and the perpendicular line to segment AB. Label one of the constructed intersection points D and construct the segment BD.
- Hide the midpoint of segment AB, the circle, the perpendicular line, and the other intersection point (not labelled D) of the circle and the perpendicular line to segment AB.
*Note: The described construction procedure is based upon Greek constructions. This construction could also be done using a rotation.*- Since AB is x, BD must equal (1/2)x.

**3. Draw AD.**

- Since segment BD is perpendicular to segment AB, triangle ABD is a right triangle.
- Using the Pythagorean theorem, the length of AD is calculated to be (sqrt (5)/2)x.

**4. Mark point E on AD such that DE = BD.**

- Construct a circle by center and point with point D as the center and point B as the point.
- Construct the point on intersection of the circle and segment AD and label point E.
- Since DE = BD, DE must equal (1/2)x. Since AE + ED = AD,
AE + (1/2)x = (sqrt (5)/2)x.

Therefore, AE = ((sqrt (5) - 1)/2)x.

**5. Mark point F on segment AB such that AE = AF.**

- Construct a circle by center and point with point A as the center and point E as the point.
- Construct the point on intersection of the circle and segment AB and label point F.
- Since AE = AF, AF must equal ((sqrt (5) - 1)/2)x. Since AF
+ FB = AB, ((sqrt (5) - 1)/2)x + FB = x.

Therefore, FB = x - ((sqrt (5) - 1)/2)x, or ((-sqrt (5) + 3)/2)x.

**6. Find AF/BF.**

- AF/BF = ((sqrt (5) - 1)/2)x / ((-sqrt (5) + 3)/2)x = (sqrt (5) - 1) / (-sqrt (5) + 3).
- Through rationalizing the denominator,
**AF/BF = (1 + sqrt (5))/2**, which is approximately equal to**1.618**. - Therefore, point F divides segment AB into the
**golden ratio**.

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