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

Continue reading at:

gogeometry.com/ArchBooLem15.htm

## Tuesday, December 16, 2008

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

Labels:
Archimedes,
book of lemmas,
circle,
diameter,
midpoint,
pentagon,
perpendicular

Subscribe to:
Post Comments (Atom)

http://img839.imageshack.us/img839/3698/problem655.png

ReplyDeleteLet H is the projection of C over AB

Observe that (CAD)=(DAB)=(DCB)=18

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

Sketch:

ReplyDeleteOG = 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)

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.

ReplyDeleteHence 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