Transcript screams
B divides AC into the golden ratio if AC AB AB BC AC AB AB BC AC BC AB 2 Are these definitions equivalent? AC BC AB 2 This definition screams AREA! (Area of the square equals area of the rectangle) Canonical construction of the golden ratio is proven in Euclid (Construct a square, bisect the base, construct “semidiagonal”, swing this length to base) Euclid starts with the following lemma. Extend a line segment AB out to a point C, then find the midpoint of AB. (C is an arbitrary distance from B). Then: MC2=MB2+AC∙BC One website says it must have taken a lot of brain power to prove this. But again, think area: MC2=MB2+AC∙BC Here we’ve drawn the square on MC, and the rectangle on AC and BC. Let’s draw a vertical line through B: 5 1 4 2 3 MC2=MB2+AC∙BC MC2 is the sum of 1+2+3+4 MB2 is area 1 The rectangle is 2+3+5=2+3+4 Now Euclid proves the golden ratio construction: MC2=MB2+AC∙BC (lemma) MC2=MD2=MB2+BD2=MB2+AB2 MB2+AB2=MB2+AC∙BC AB2=AC∙BC The lemma is also used in the next proof: B C D A AD●AC=AB2 That last proof is used to prove a theorem in solid geometry. Neither the lemma or golden ratio construction is used again. Do you suppose someone trying to prove that solid geometry theorem realized he needed that third theorem, then realized he needed the lemma? And maybe later, he or someone else realized the lemma would be useful in proving the golden ratio construction. It’s also interesting that Euclid does not use this construction to construct a regular pentagon, even though constructing a regular pentagon automatically involves constructing the golden ratio. Another construction of the golden ratio: Draw a circle, then using the same radius on the compass, strike off six equal arcs. If we connected consecutive arcs, we’d get a regular hexagon, but we’re only going to connect every other arc. This gives a regular triangle. Now connect the center of the circle with two of the unused tic marks. This is only to find the midpoint of two sides of the triangle. Now draw a line through those two midpoints and extend it to the circle on one side: Then B divides AC into the golden ratio This construction was discovered by George Odom around 1979. George was an amateur mathematician and also a sculptor. He spent quite a bit of his life in an institution for the mentally insane(!). He was acquainted with H.S.M. Coxeter, the premier geometer of the 20th century. Coxeter had not seen this proof, so submitted it in Odom’s name to the Mathematical Association of America (in Odom’s name), and it was published in their monthly in 1980. Here are a couple ways to prove that it works: Scale figure so BD=2. Then MB=1, MD= 3 Now find OD (which = OC), OM, MC, then AC A second proof follows: EDF ECF DEC DFC EB BD EB AB BF BC AB EA Interior angles cutting off same arc http://www.cut-the-knot.org/do_you_know/GoldenRatio.shtml 1 1 Proof Without Words http://www.cut-the-knot.org/do_you_know/GoldenRatio.shtml Here’s another construction. Draw a vertical line segment: Draw a circle centered at each end with radius equal to the length of the line segment: Extend the line segment to intersect the upper circle: Draw a circle with center at the center of the lower circle and passing through that upper point of the line segment: Draw a line through the intersection of the two smaller circles and extend it to the larger circle: To see why this works, draw DA and DB to intersect the outer circle: D A B O Now angle ODB is inscribed in a semicircle, so is a right angle. Also OD is twice OB, so angle ODB is 30o so the triangle is equilateral, and this is equivalent to Odom’s construction. Incidentally, the fact that an angle inscribed in a semicircle is a right angle is attributed to Thales, and is one of the first theorems ever proved in Greek geometry. I’ve often wondered what axioms and postulates Thales used, and what logical constructs he was familiar with (Aristotelian logic was not formalized by Aristotle until over a hundred years later), but there are no records to answer that question. One additional construction of the golden ratio is the “rusty compass” construction. A “rusty compass” is one that has a frozen hinge. It will still draw circles, but only of one radius. Constructions of this type were studied by Persian mathematicians in the 12th century. It is possible to do everything with a rusty compass and straightedge that can be done with a standard compass and straightedge except draw circles of a given radius. The following construction is from a neat website called “Cutthe-knot.” I copied their construction, but it did not reproduce very well. RUSTY COMPASS CONSTRUCTION OF GOLDEN RATIO http://www.cut-the-knot.org/do_you_know/GoldenRatio.shtml • http://web.aurora.edu/~bdillon/golden.ppt • http://web.aurora.edu/~bdillon/pentagon.ppt • http://web.aurora.edu/~bdillon/math.htm