Straightedge and Compass Only Construction: Regular Pentagon Inscribed in a Unit Circle

1. Fill in the missing angle measures in the isosceles triangle below. Given: 7ÐnEGFÑœ$'‰, EGœFG , and EH is the bisector of nGEF . (Be sure to find 7ÐnEFGÑ, 7ÐnGEHÑ , 7ÐnHEFÑ , and 7ÐnEHGÑ .)

2a. In class, you showed ˜EFH µ ˜GEF . Use this result to find the missing side lengths in ˜EFG , assuming EGœFGœ" . (Hint: Let BœEF and find BÞÑ

" b. Find B , where B is the value you found in part (a). What do you notice? (Hint: It's "È & customary to write the golden ratio as 9 œ # .) 3a. For this problem you are only allowed to use GeoGebra's compass tool, point tool, line tool, segment tool, and circle tool (and perpendicular bisector tool that YOU created using the straightedge and compass only). Starting with a segment of length 1 and only the tools already mentioned: i. construct a circle with center E and radius with length EFœ 1. ii. construct the circle's diameters, FG and HI , iii. construct the midpoint, J , of radius EFÞ iv. draw segment HJ Þ v. draw a circle with center J and radius of length HJÞ vi. let point K be the intersection of the circle from part (v) and EGÞ b. Now you are going to prove the claim: HK is the side length of a regular pentagon inscribed in a circle of radius 1. First convince yourself by using GeoGebra to construct such a pentagon in the circle from part (a) (use straightedge and compass only). In the parts below, you will prove this by (I) computing the length of HK and (II) computing the length of a regular pentagon inscribed in a circle of radius 1.

I) i) Your work from part (a) should look like the following figure. In particular, the circle shown with center E has radius EFœ" and point J is the midpoint of EF . Compute the lengths EJ and DF.

ii) In the figure above, compute the length EKÞ

iii) In the figure above, compute the length HK . II) i) The figure below shows a regular pentagon inscribed in a circle of radius EF œ 1. Compute the length MNœ= (in terms of the constant HMœ a and the golden ratio, "È & 9 œ # ). Hint: Use your work from problem 2b above. (You may use the fact that ˜MHN is isosceles.)

"È & ii) Compute the length INœ> . Your answer should be in terms of 9 œ # . (Hint: The circle's radius is 1.)

iii) Observe that nHN I is inscribed in a semicircle. Since an inscribed angle's measure is half the central angle, QÐnHNIÑ œ *! ‰ . Use this fact together with the Pythagorean theorem to find a relationship between the lengths HNœ+ and > . (Hint: The circle's radius is 1.)

iii) Parts (i)-(iii) yield 3 equations and 3 unknowns: +ß=ß and > . Put these together to "È & compute the length = . (Hint: Write # as 9 and use substitution. Once you have = in "È & terms of 9 only, plug in # and simplify.) How does your answer compare to the length HK , which you found in part (I, iii)?