Unveil
the mystery of <<Trisecting Angle>> problem
Wang
Changyuan
Summary :
As it is known to all, the <<Trisecting Angle>> problem has long
been a mystery. Yet, for any given angle E(360 °
≥
E>0 °),
the trisection angle of angle E is already decided. This thesis introduces a
<<Three free points natural convergence method>> 1, which is a
method to trisect any angle E using <<Ruler and Compasses
Construction>> according to the <<Construction Public Law>>.
Key
Words: trisect angle; three
free points natural convergence method; ruler and compasses construction.
This thesis
consists of two parts: 1, Graphic Construction; 2, Calculation and Demonstration
1. Graphic Construction
Give a random angle E. Specify Angle E' s position in the plane rectangular
coordination system.
a
. Suppose: Angle E' s vertex angle is Point O. The vertex of one side is Point X
and the vertex of another side is Point y'. Namely: ∠
XOy'= E. (According to chart 1)
b. Prolong Side
XO with a ruler; take Point O as the midpoint; the prolonged side is in the
opposite direction to Point X. Suppose:
the direction of Point O to Point X is the positive direction and the prolonged
side is along the negative direction. Take
Side OX as the axis X of the plane rectangular coordination system. Take Point O
as the center point and r as the radius and draw a circle. The arc and axis X
intersect at Point A in the positive direction and at Point A' in the negative
direction.
c. Take Point A and A' respectively as the center points and take R as the radius( R> r), draw circles. There are two intersection points on the arcs of the two circles. Take the intersection point above axis X as Point F; the intersection point below axis X as Point F'. Draw a straight line passing Point F and F' and as a result, the line definitely passes Point O( the diagonals of rhombus are orthogonal). That is: FF' ⊥ AA'. Take the direction of the Line FF' above Point O as the positive direction and below as the negative. Make one point on the positive direction of Line FF' as Point Y, which make: OY=OX. Thus, the new plane rectangular coordination system YOX is established.
d. Now divide
Angle E' s position in the YOX( plane rectangular coordination system) into
three areas. 1.
90°
≥
E>0 °;
2. 180°
≥
E>90 °;
3. 360°
≥
E>180 °.
e. Take r as the
radius (r is the random given value) in the following graphic construction. The
detailed construction is omitted.
f. Suppose 90 °
≥
E>0 °
in the following construction. Take Point Q(i), a free point on axis X, as the
circle center point and use a pair of compasses to draw a circle; the arc
intersects the straight line L2. Take Point Q'( i), a free point on Straight
line Oy' as the circle center point to draw a circle; the arc intersects the
straight line L1. Normally, each has two intersection points. (When E=90 °
there' s only one intersection point). Name the two intersection points on the
straight line L2 as Point R(i) and Point r(i), make |BR(i)|>|Br(i)|, Point
r(i) is a point on Line segment BR(i). In a similar way, the two intersection
points on the straight line L1 are Point R'( i) and Point r'( i), make| AR'( i)|>|
Ar'( i)|; Point r'( i) is a point on Line segment AR'( i). But the method that
this thesis introduces only uses Point R( i) on the straight line L2 and Point
R'( i) on the straight line L1 and Point r( i) and Point r'( i) are ignored.
(According to chart 3). (Mainly in the construction 1.1.5-1.1.8 below). The
detailed construction is omitted.
g. In the
constructions below, a random Angle E is given and E=e1+e2+e3.
1.1.1 While 90 °
≥
E>0 °:
Take Point O as the center point to draw a circle; the arc intersects with axis
X at Point A and intersects the straight line Oy' at Point B.
1.1.2 Take Point
A and Point B as the circle center points to draw circles; the two arcs
intersect at Point Z. (There is another intersection Point O, which is
neglected). Connect Point O and Z, then Line segment OZ is the angular bisector
of Angle E..
1.1.3
Draw straight line L1 that passes Point A and Z and
straight line L2 that passes Point B and Z, respectively. Point
Z is
the intersection point of the straight lines L1 and L2. In the quadrilateral OAZB,
because OA=AZ=BZ=OB=r, so OAZB is a parallelogram (rhombus); OA//L2,OB//L1.
(according to chart 4)
1.1.4 Make
Point z' the intersection point of the straight lines OZ and Arc AB. The
distribution of the three free points( P( i). P'( i), Q( i). Q'( i), R( i). R'(
i)) are: a , Point P( i) and Point P'( i) are free points respectively on
Arc Az' and Arc Bz'( in which P(1) and P'(1) are any points respectively on Arc
Az' and Arc Bz'), b, Point Q( i) and Point Q'( i) are free points respectively
on axis X and straight line Oy'; c, Point R( i) and R'( i) are free points
respectively on straight line L2 and straight line L1. (i=1,2,3 …).
1.1.5 Take Point
P( i), a free point on Arc Az' as the center to draw a circle( in which the
initial point P(1) is a random point on arc Az'), the arc intersects with axis X
at Point Q( i)( The other intersection Point O on axis X is ignored). Take Point
Q(i) as the center point to draw a circle; the arc intersects with straight line
L2 at Point R(i) (The other intersection Point r(i) is ignored).
Connect Points O and R( i) and the straight line OR( i) and Arc Az'
intersect at Point P( i+1).
1.1.6 Take
P(i+1) as the center point to draw a circle.
a.
While lim| P( i+1) –
P( i)| ≠
0( P( i+1) takes the place of P( i) in process 1.1.5. ), repeat process1.1.5.
b.
While lim| P( i+1) –
P( i)|=0, lim| Q( i+1) –
Q( i)|=0, lim| R( i+1) –
R( i)|=0, then P( i+1) coincides with P( i) ,P(i+1)=P(i).then Q(
i+1) coincides with Q( i),Q(i+1)=Q(i).then
R( i+1) coincides with R( i),R(i+1)=R(i).
the P( i) is a fixed point on the straight line OR( i+1), make ∠
XOR( i)= e1.
1.1.7 Take a
free point, Point P'( i) on Arc Bz' as the center point to draw a circle;( in
which the initial point P'(1) is a random point on arc Bz'), the arc intersects
with the straight line Oy' at Point Q'( i)( The other intersection Point O on
Arc Oy' is ignored). Take Q'( i) as the center point to draw a circle, the arc
intersects with the straight line L1 at Point R'( i)( The other intersection
Point r'( i) is ignored). Connect Point O and R'( i); Straight line OR'( i) and
Arc Bz' intersect at Point P'( i+1).
1.1.8 Take Point
P'( i+1) as the center point to draw a circle.
a.
While lim| P'-( i+1) P'( i)| ≠
0( Point P'( i+1) replaces Point P'( i) in process 1.1.7. ), repeat
process1.1.7.
b.
While lim| P'( i+1) –
P'( i)|=0, lim| Q'( i+1) –
Q'( i)|=0, lim| R'( i+1) –
R'( i)|=0, then P'( i+1) coincides with P'( i) ,P'(i+1)=P'(i).then
Q'( i+1) coincides with Q'(
i),Q'(i+1)=Q'(i).then R'(
i+1) coincides with R'( i),R'(i+1)=R'(i).
the P'( i) is a fixed point on the
straight line OR'( i+1), make ∠
y' OR'( i)= e3, ∠
R( i) OR'( i)= e2. (According to chart 5).
1.1.9 (a) Prove
the relationship of Angle e1, e2 and e3.
In the same way, ∠
R'( i) Q'( i) y'= E=3* e3, e3= E/3. Because:
E=e1+e2+e3, so: e2=E-e1-e3=E/3, then e1=e2=e3=E/3.
(b) As Point
P(1) is a random point on Arc Az', will there be: P( i)-> Q( i)-> R( i)->
P( i+1)-> Q( i+1)-> R( i+1)-> ……, yet P(i+1) and P(i) will never
coincide? The answer is negative.
While 90 °>= E>0 °, use the graphical construction according to
the<< three free points natural convergence method>> mentioned
above, in finite construction steps, Point P( i+1) and P( i) will naturally
coincide, e(i) converges E/3. This is an inevitable rule .
1.2.1 While 180 °
≥
E>90 °,
draw the angular bisector Oy'' of Angle E, make ∠
XOy''= ∠
y'' Oy'= E/2 , 90 °>=
∠
XOy''>0 °.
Trisect the Angle XOy'', make ∠
XOy''= q1+ q2+ q3 and q1= q2= q3. Make:
e1= q1+ q2, one side of Angle e1 coincides with axis X, the other side
intersects the arc, which takes Point O as the circle center point, at Point P'(
i). Take Arc AP'( i) as a fixed value in order to trisect Angle E by the
corresponding arcs of e1, e2 and e3 on the arc with Point O as the circle center
point. So, while 180 °>= E>90 °, just bisect Angle E and trisect the
first subangle to( q1.q2.q3), take the arc q1+ q2 as a fixed value, Angle E can
be trisected on the arc that takes the Point O as circle center point. Connect
the intersection point on the trisected arc with the Point O, then the Angle E
can be trisected.
1.3.1 While 360 °
≥
E>180 °
, firstly quarter the Angle E, make ∠
XOy'''= E/4 , 90 °>=
∠
XOy'''>0 °.
Trisect the Angle ∠
XOy''', make ∠
XOy'''= q1+ q2+ q3 and q1= q2= q3. Cut on the arc with Point O as the circle
center point by the arc corresponding to q1+q2, then the arc that corresponds to
the Angle E can be divided into six equal arcs. Take every two arcs as an e(i),
then: the six arcs can be divided into e1, e2 and e3. So, when 360 °>=
E>180 °, just quarter the Angle E and trisect the first quarter of the Angle
E into q1, q2 and q3. Take the arc that corresponds with q1+q2 as a fixed value,
then Angle E can be divided into six equal arcs by the arc with Point O as the
circle center point. Take Point A, the intersection point of axis X and the arc
as the initial point, connect the intersection point of every two equal arcs
with Point O, then: the angle E can be trisected into (e1, e2 and e3).
1. Give an
example for calculation. Trisect the angle E, E=0.123456789 °
2. List below
the e( i, j) trisection computation result while E=0.1 ° . 0.2°. …….
89.9° .90 °.
a. In a common equation ; aX^2+bX+c=0, because: X1={-b+sqr(b^2-4*a*c)}/(2*a), X2={-b-sqr(b^2-4*a*c)}/(2*a),
only X1 is extracted in the calculation and the root of X2 is ignored.
b. In (Xr-Xq)^2+(Yr-Yq)^2=r^2,
only (Xr-Xq)=sqr(r^2-(Yr-Yq)^2) is calculated and (Xr-Xq)=-sqr(r^2-(Yr-Yq)^2) is
ignored.
c.
In (Xq-Xp)^2+(Yq-Yp)^2=r^2, only (Xq-Xp)=sqr(r^2-(Yq-Yp)^2)
is calculated and (Xq-Xp)=-sqr(r^2-(Yq-Yp)^2) is ignored.
(Because 90°≥
E>0°, take the intersection Point R'( i) on straight line L1 and the
intersection Point R( i) on straight line L2, the intersection Point r'( i) and
r( i) respectively on straight line L1 and L2 are ignored and the intersection
point with the Point O is also ignored). There will be no repeated explanation
below anymore.
2.1.1 Give
a random Angle E(90°≥E＞0°).
Firstly, specify Angle E’s position in the rectangular coordination
system (According to chart 7). Their respective coordination points and linear
equations are A( r ,0 ) , z’( Xz’ , Yz’) , B( Xb , Yb )，L2=Yb
and linear equation is
2.1.2 Because
the initial point P(1) and P’(1) are random points on Arc Az’ and Bz’,
in order to prove that Angle e( i ) finally converges the fixed value E/3,
suppose P(1) and P’(1) are respectively the vertex points of Arc Az’ and Bz’.
Make: a,
P1=A ( r , 0 ), then the calculated subangle of Angle E is e(i,1);
b,
P1=z’( Xz’ ,Yz’), then the calculated subangle of Angle E is e(i,2).
c,
P’1=z’( Xz’ ,Yz’), then the calculated subangle of Angle E is e(i,3).
d,
P’1=B( Xb ,Yb), then the calculated subangle of Angle E is e(i,4). (i=1,2,3,…….).
(According to chart 8).
2.1.3 Make: Angle E=0.123456789°, radius r=9876543210, then straight slope k=tg( 0.123456789°)=0.00215473078668851.
A( 9876543210 , 0 ) .
L1=0.00215473078668851*(X-9876543210). L2=21281242.3177386. y'=0.00215473078668851*X ,
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While
a random Angle E=0.123456789° is given, take initial Point P(1) at A(9876543210,0)，after
three recurrent calculations, Subangle e(3,1)
naturally converges at E/3.
e(3,1)=E/3=0.041152263° |
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While
a random Angle E=0.123456789° is given, take initial Point P(1) at
z'(9876537478.08649 ,10640627.3342238)，after
three recurrent calculations, Subangle e(3,2) naturally converges at E/3.
e(3,2)=E/3=0.041152263° |
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