Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Path: utzoo!linus!decvax!genrad!panda!talcott!harvard!seismo!lll-crg!gymble!neurad!nbs-amrf!hopp From: hopp@nbs-amrf.UUCP (Ted Hopp) Newsgroups: net.math Subject: Re: A geometrical construction poser Message-ID: <140@nbs-amrf.UUCP> Date: Mon, 10-Feb-86 14:15:40 EST Article-I.D.: nbs-amrf.140 Posted: Mon Feb 10 14:15:40 1986 Date-Received: Thu, 13-Feb-86 17:38:13 EST References: <964@decwrl.DEC.COM> Organization: National Bureau of Standards Lines: 82 > I've got a real poser for the geometry wizards out there. Back in Engineering > Drawing class, oh, some 25 years ago, I learned a technique for the construc- > tion of a regular inscribed pentagon. The instructor, who was not a geometer, > stated that its validity was not proven and was probably unproveable. In the > following figure, I've faked it as best I can without a graphics tube, and > I've described the procedure. Is there anyone who can prove that this little > gem is valid, or that its validity is unproveable? > > | B > ____--+--____ | > _-'' _-/|\-_ ``-_|_-' > _-' _-" / | \ "-_ _-|-_ > .' _-" / | \ "-_ | `. > .' _-" / | \ "|_ `. > ' _-" / | \ | "-_ ` > -_ '_-" / | \ | "-_` > "+" / | \ | + H > E :\"-_ / | \ | /: > | "-_ / D | O \| C | A > --+-\-----+------------+----------+-------/-+-- > | | | | > ` \ | | / ' > | | | | > . \ | | / ' > `. | | .' > `\ | | /' > +_-------------|----------|--_+ > F "-_ | "-|-" G > "-__ | __-"|"-_ > """"--+--"""" | > | > > Procedure: > > 1. Bisect OA to locate C. > > 2. From C, with CB as radius, strike an arc to locate D. > > 3. From B, with BD as radius, strike an arc to locate E. > > 4. From E, still using BD as radius, locate F. > > 5. Repeat step 4, moving around circumference, to locate G and H. > > 6. BEFGHB is the required regular inscribed pentagon. > > Please reply by direct E-mail, as I don't subscribe to net.math. > > Cheers, > Dick Binder (The Stainless Steel Rat) > > UUCP: { decvax, allegra, ucbvax... }!decwrl!dec-rhea!dec-dosadi!binder > ARPA: binder%dosadi.DEC@decwrl.ARPA > The construction is valid, and it's not too hard to prove. Let the radius of the circle (which is also the length of OA and OB) be 2. By construction, OC=1. Since the angle BOC is pi/2, BC=sqrt(5) by the Pythagorean Theorem. Hence CD=sqrt(5) and OD=(sqrt(5)-1). Since angle BOD is pi/2, the length of BD is sqrt(10-2*sqrt(5)). The construction is valid, then, if the sides of an inscribed pentagon in a circle of radius 2 is sqrt(10-2*sqrt(5)). But this can be verified as follows. Assume there is a regular pentagon inscribed in the circle with one vertex at B. Let F be the pentagon vertex closest to E. We will show BF=sqrt(10-2*sqrt(5)), hence BF=BE, hence F and E are the same point. The angle BOF is 2*pi/5 (by definition of a regular pentagon). The lines OB and OF each are length 2, being radii of the circle. Thus, by the law of cosines, BF^2 = OB^2 + OF^2 - 2*OB*OF*cos(2*pi/5) = 8-8*cos(2*pi/5). If BF=BE, we must have 8-8*cos(2*pi/5) = 10-2*sqrt(5), or, solving for cos(2*pi/5), the condition for the construction to be valid is that: cos(2*pi/5) = (sqrt(5) - 1) / 4 This can be confirmed using the identity cos(5x) = 16*cos(x)^5 - 20*cos(x)^3 + 5*cos(x) setting x=2*pi/5 (the left hand side becomes 1), multiplying out the right hand side, and simplifying. -- Ted Hopp {seismo,umcp-cs}!nbs-amrf!hopp