binder@dosadi.DEC (You are what you do when it counts.) (02/06/86)
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
hopp@nbs-amrf.UUCP (Ted Hopp) (02/10/86)
> 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
andrews@yale.ARPA (Thomas O. Andrews) (02/10/86)
In article <964@decwrl.DEC.COM> binder@dosadi.DEC (You are what you do when it counts.) writes: >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? > > (long detailed description of constructrion procedure which I have not > read.) Pentagon constructions are relatively easy, if you know the length of the side of a triangle. It is downright foolish, though, to claim that a construction of a pentagon is correct but not provably correct. People on this net seem to have Goedel on the brain. Specifically, in the coordinate plane, given the parameters for two lines, we can *compute* the coordinates of the point of intersection. Similarly, we can compute the the coordinates of the intersection point(s) of two circles or a line and a circle. Thus, given any construction with compass and straight edge, we can pick one point on the plane as an origin, choose arbitrary axis, conveniently label some length "1" and compute the coordinates of the point(s) constructed. This is difficult and unpleasant, but it can be done. The sort of "non-existance proofs" in geometric constructions are: 1) There is no construction of a regular heptagon with straight edge and compass. 2) Given only two points which are 1 unit apart, the distance (cube root of)(2) can't be constructed. 3) Given only two points which are 1 unit apart, the distance pi can't be constructed. 4) There is no way to construct a 10 degree angle (and hence no way to trisect an arbitrary angle.) In none of these statements is it suggested that constructions exist which are correect but not provably correct. They instead say that these constructions cannot be made at all, no matter how long you diddle with your compass and straight egde, as long as you follow two simple rules. 1) Lines can only be constructed through pairs of points already constructed. (A constructed point is either a point given in the premise of the problem, or the intersection of two figures.) 2) Circles can only be constructed if the center point has already been constructed, and at least one point on the circle has been constructed. (in other words, if a radius is constructed.) These two rules make it impossible for ramdomess to enter into constructions. Interesting point about constructions: 1) All constructions possible with a straight edge and compass are also possible with only a compass, and, indeed, this compass can be "rusty"; that is, stuck in one position. 2) If you draw a single circle, and then throw away your compass, the straight edge alone can be used to make all possible constructions. Courant and Robbins, *What is Mathematics* has a particularly good elemtary dicussion of these two statements. -- Thomas Andrews andrews-thomas@yale
binder@dosadi.DEC (You are what you do when it counts.) (02/12/86)
Many thanks to Lambert Meertens, =Ned=, Mark Brader, and David Moews, who provided proofs for my little pentagon problem. I've not gone through the proofs thoroughly yet, but at first glance they look reasonable. As a matter of clarification, Lambert, the instructor who supposed the problem to be unprovable wasn't a mathematician at all - he was an engineering drawing and descriptive geometry teacher who was retired from the US Army; neither of these subjects requires much math ability at all, relying as they do on the use of drafting instruments, and he'd probably forgotten what little math he knew. But he was a helluva fine draftsman! Cheers, Dick Binder (The Stainless Steel Rat)
ljdickey@water.UUCP (Lee Dickey) (02/12/86)
The question is this: Does this proceedure (paraphrased) give the vertices of a regular polygon? > Procedure: > > Let O:(0,0) be the center of the circle, radius 1. > Then A:(1,0) is on the circle and > B:(0,1) is on the circle. > > 1. Bisect OA to locate C. > 2. From C, with CB as radius, strike an arc to locate D (on x-axis, x <0). > 3. From B, with BD as radius, strike an arc to locate E (on circle). > 4. From E, still using BD as radius, locate F (on circle). > 5. Repeat step 4, moving around circumference, to locate G and H. > 6. BEFGHB is the required regular inscribed pentagon. ------------------------------------------------------------------------ Answer: Yes, the construction is correct, and provable. I think you will find that this proof is not too difficult. ------------------------------------------------------------------------ Proof: By simple calculations, the square of the distance from B to D is found to be dist(B,D) = sqrt ( ( 5 - sqrt(5) )/ 2) . Now, is that really the length of the side of a regular pentagon? I propose to show it is by finding the coordinates of the regular polygon whose vertices are the roots of the complex equation z^5=1. We will find the complex number x+iy which is the root of z^5 - 1 for which both x and y positive. Then (x+iy)^5 = 1. This equation, when expanded, can be separated into two parts, the real part and the imaginary part. The imaginary part has a factor of y. Since y>0, that factor can be removed. Call the two resulting equations (1), for the real part, and (2), for the imaginary part. There is a third equation (3), which says that (x,y) has to be on the circle: x^2 + y^2 = 1. Now, both (1) and (2) have only even powers of y, so (3) can be used to advantage. From (1) and (3), you can derive this: (1)': 16x^5 - 20x^3 + 5x - 1 = 0. But the value x=1 is a root of this equation (corresponding to y=0), but (1,0) is not the point we want (we want both x and y positive). Divide (1)' by the polynomial x-1 to get: (1)": 16x^4 + 16x^3 - 4x^2 - 4x + 1 = 0. Similarly, working with (2) and (3) gives: (2)': 16x^4 - 12x^2 + 1 = 0. Now, use (1)" and (2)' to get: 4x^2 + 2x - 1 = 0. Since x>0, this gives exactly the x-coordinate of the point (x,y), x = ( sqrt(5) - 1 ) / 4 Knowing this, the square of the distance from (x,y) to (1,0) is found to be (x-1)^2 + y^2 = 2 - 2x = ( 5 - sqrt (5) ) / 2. This is exactly what we wanted to know. Since the length of the side BD agrees with the distance from (x,y) to (1,0), the figure given by the construction is a regular pentagon.