| First, I'll shift the problem slightly, to an equivalent one.
Find a regular pentagon in the complex plane all of whose vertices
have rational real parts. Or show that no such pentagon can exist.
It can't exist. Proof follows:
Let the vertices be p_0, p_1, p_2, p_3, and p_4, (the underscore is
used to indicate subscripts), in order counterclockwise. Without loss
of generality, let p_0 = 0. Let p_1 = V = V_r + i V_i. And also let
W = cos(2pi/5) + i sin(2pi/5), so that multiplying by W represents a
2pi/5 rotation counterclockwise (2pi/5 = 72�).
let p_1 = V = V_r + i V_i. Then:
p_0 = 0
p_1 = V
p_2 = V + V*W
p_3 = V + V*W + V*W�
p_4 = V + V*W + V*W� + V*W�
Now, we expand all of the above. In so doing, the following table
is useful:
k sin(k pi/5) cos(k pi/5)
- ----------- -----------
0 0 1
1 sqrt((5-sqrt(5))/8) (sqrt(5)+1)/4
2 sqrt((5+sqrt(5))/8) (sqrt(5)-1)/4
3 sqrt((5+sqrt(5))/8) -(sqrt(5)-1)/4
4 sqrt((5-sqrt(5))/8) -(sqrt(5)+1)/4
5 0 -1
We get:
p_0 = 0
p_1 = V_r + i V_i
p_2 = {V_r (3+sqrt(5))/4 - V_i sqrt((5+sqrt(5))/8)}
+ i {V_i (3+sqrt(5))/4 + V_r sqrt((5+sqrt(5))/8)}
p_3 = {V_r/2 - V_i sqrt(5+2 sqrt(5))/2}
+ i {V_i/2 + V_r sqrt(5+2 sqrt(5))/2}
p_4 = {V_r (1-sqrt(5))/4 - V_i sqrt((5+sqrt(5))/8)}
+ i {V_i (1-sqrt(5))/4 - V_r sqrt((5+sqrt(5))/8))
And so we take just the real parts of these, to see whether they
might all be rational. Let R_i = Re(p_i), the real part of each p_i;
we also want each of these R_i to be rational.
R_1 = V_r
R_3 = R_1/2 - V_i sqrt(5+2 sqrt(5))/2, so
V_i = (R_1 - 2 R_3) sqrt((5-2 sqrt(5))/5)
R_2 = R_1 (3+sqrt(5))/4 - (R_1 - 2 R_3) sqrt((3-sqrt(5))/8)
= R_1 (3+sqrt(5))/4 - (R_1 - 2 R_3) (sqrt(5)-1)/4, so
R_2 = R_1 + R_3 (sqrt(5)-1)/2
sqrt(5) = 2 (R_2 - R_1)/R_3 + 1
Thus, if R_1, R_2, and R_3 are rational, we've discovered that sqrt(5)
is really a rational number! The alternative is that no four vertices
(we never got to R_4) of a regular pentagon in the plane may have
rational x coordinates.
|