## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #15

Problem 655: Inscribed Regular Pentagon, Arc, Midpoint, Perpendicular, Radius
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #15 (high school level) and lift up your geometry skills.

Click the figure below to see the complete problem 655

gogeometry.com/ArchBooLem15.htm

1. http://img839.imageshack.us/img839/3698/problem655.png
Let H is the projection of C over AB
And (OBC)=(OCB)=54
(CEB)=36
AF is angle bisector of (CAB)
So we have CF/FB=AC/AB=HG/GB=HC/BC
So CG is angle bisector of (HCB) and (HCG)=(GCB)=18=(OCH)
And OCG and ACE are isosceles triangles and OH=HG, HA=HE
So GE=OA= radius of the circle
Peter Tran

2. Sketch:
OG = AG - R = AC - R = 2R cos 36 - R
BE = BD = 2R sin 18
OG - BE = 2R (cos 36 - sin 18) - R
=2R(1/2)- R = 0
So OG = BE, OG + GB = GB + BE,
EG = R
(Center is O & radius is R)

3. We first note that chord BC subtends an angle of 36 at the circumference and thus 72 at the centre. CD and DB subtend 18 at the circumference and AC subtends 54.

Hence Tr.s ACF and AGF are congruent ASA. Hence AC = AG and thus < AGC = 72 = < COG. Hence CO = CG = R the radius.

Now < CBO = 54 and < BCD = 18 so < CEB = 36 and so CG = GE = R

Sumith Peiris
Moratuwa
Sri Lanka