高斯 正十七边形到底怎么做?网上的不同做法到底哪个有权威?

在网上搜索高斯的正十七边形做法.居然有不同的方法?到底哪个是对的哪个是错的?简单的问题为什么搞那么复杂?只需要做法步骤.
还有高斯正十七边形的历史?到底是他研究半年得出做法.还是他一晚上突发奇想?
philip_zg.你有没有找到过另外一种说法.第5步是∠DCE=π/4.

高斯做出来的是所有正N边形,哪些是可以做的,哪些是不可以做的
以及可以做的画法
在这里赞一下天才的高斯!!!
而且一个图形是可以有很多种画法的,高斯也没有证明那种是最简单的……
十七边形的画法是高斯的得意之作,之前他的教授教他不要再学数学了,他自己也在犹豫是数学还是拉丁文,而这做出来之后,他就决定这辈子奉献给数学,在他死前,他还要求在他的墓碑上不用刻其他的东西,只要刻一个正十七边形就好了……

给一圆O,作两垂直的直径OA、OB,
作C点使OC=1/4OB,
作D点使∠OCD=1/4∠OCA
作AO延长线上E点使得∠DCE=45度
作AE中点M,并以M为圆心作一圆过A点,
此圆交OB于F点,再以D为圆心,作一圆
过F点,此圆交直线OA于G4和G6两点。
过G4作OA垂直线交圆O于P4,
过G6作OA垂直线交圆O于P6,
则以圆O为基准圆,A为正十七边形之第一顶点P4为第四顶点,P6为第六顶点。
以1/2弧P4P6为半径,即可在此圆上截出正十七边形的所有顶点。

π/4和45度是一回事……
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第1个回答  2007-08-23
正十七边形
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在几何学中,正十七边形是有17边的多边形。正十七边形的每个内角约为158.823529411765°。

1796年,高斯成功利用尺规作图作出正十七边形,同时发现了可作图多边形的条件,并定下他要成为数学家的决心。

可作图性亦同时显示2π/17的三角函数可以只用基本算术和平方根来表示。高斯的书Disquisitiones包含了这条等式:

NOTE: I did NOT write this, it was taken with permission at http://www.seanet.com/~ksbrown/kmath487.htm

Constructing the Heptadecagon
The ancient Greek geometers devoted considerable thought to the
question of which regular n-gons could be constructed by straightedge
and compass. They knew how to construct an equilateral triangle
(3-gon), a square (4-gon), and a regular pentagon (5-gon), and of
course they could double the number of sides of any polygon simply
by bisecting the angles, and they could construct the 15-gon by
combining a triangle and a pentagon. For over 2000 years no other
constructible n-gons were known.

Then, on 30 March 1796, the 19 year old Gauss discovered that it
was possible to construct the regular heptadecagon (17-gon). (This
discovery apparently convinced him to pursue a career in mathematics
rather than philology.) The result was announced in the "New
Discoveries" column of the journal "Intellegenzblatt der allgemeinen
Litteraturzeitung" on 1 June 1796 by A. W. Zimmermann, a professor
at the Collegium Carolinum and an early mentor of the young Gauss.

Subsequently Gauss presented this result at the end of Disquistiones
Arithmeticae, in which he proves the constructibility of the n-gon
for any n that is a prime of the form 2^(2^k) + 1, also known as
Fermat primes. Gauss's Disquisitiones gives only the algebraic
expression for the cosine of 2pi/17 in terms of nested square
roots, i.e.,

cos(2pi/17) = -1/16 + 1/16 sqrt(17) + 1/16 sqrt[34 - 2sqrt(17)]

+ 1/8 sqrt[17 + 3sqrt(17) - sqrt(34-2sqrt(17)) - 2sqrt(34+2sqrt(17)]

which is just the solution of three nested quadratic equations.
Interestingly, although Gauss states in the strongest terms (all
caps) that his criteria for constructibility (based on Fermat
primes) is necessary as well as sufficient, he never published a
proof of the necessity, nor has any evidence of one ever been
found in his papers (according to Buhler's biography).

One of the nicest actual constructions of the 17-gon is Richmond's
(1893), as reproduced in Stewart's "Galois Theory". Draw a circle
centered at O, and choose one vertex V on the circle. Then locate
the point A on the circle such that OA is perpindicular to OV, and
locate point B on OA such that OB is 1/4 of OA. Then locate the
point C on OV such that angle OBC is 1/4 the angle OBV. Then find
the point D on OV (extended) such that DBC is half of a right angle.

Let E denote the point where the circle on DV cuts OA. Now draw a
circle centered at C through the point E, and let F and G denote
the two points where this circle strikes OV. Then, if perpindiculars
to OV are drawn at F and G they strike the main circle (the one
centered at O through V) at points V3 and V5, as shown below:

The points V, V3, and V5 are the zeroth, third, and fifth verticies
of a regular heptadecagon, from which the remaining verticies are
easily found (i.e., bisect angle V3 O V5 to locate V4, etc.).

Gauss was clearly fond of this discovery, and there's a story that
he asked to have a heptadecagon carved on his tombstone, like the
sphere incribed in a cylinder on Archimedes' tombstone. The story
is probably apochryphal, because if Gauss had seriously wanted such
a monument located in the proximity of his actual remains, it would
have to be placed, not at his grave site, but above the jar in the
anatomical collection of the University of Gottingen where his brain
has been preserved (rather goulishly, in my opinion). On the other
hand, if proximity to the actual remains is not important, then the
heptadecagon on the monument to Gauss in his native town of Brunswick,
or even the figure above, may suffice.

Back to Formulae

参考资料:http://www.geocities.com/RainForest/Vines/2977/gauss/formulae/heptadecagon.html

第2个回答  2007-08-23
一篇文章上说“2000年与一个晚上”,说的就是这个事儿!
高斯用的是Z~n=1的所谓单位圆将这个问题给解决了!
17=2~(2~2)+1,即所谓的费马数。
Z~17=1有17个根,其中有一个为1,令其他的根为a1,a2,a3,.....,a16.
令Z1=a1,Zn=(Zn-1)~3.
根据韦达定理:令A=Z1+Z3+。。。+Z15。
B=Z2+Z4+。。。+Z16。
A+B=-1.AB=1。
得B=。。。
再将B拆开。。。
一直到最后,即得解的模。
第3个回答  2007-08-30
已知一边AB,用此边作正三角形,另一点为C,将BC平分六等份,得到B,1,2,3,4,5,6各点.每等份为a,在AB的中垂线上从C点开始向上以a截取各等份段,得7,8,9,10,11,12,13,14,15,16,17点,以17点为圆心17A为半径作圆,以AB长依次截取圆得各点,连接各点即为所求正17边行.
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