6.5 Geometric Constructions
1.
a. Given the line through A and B and the point C, locate D on AB such that |AD| = |AC| . Then locate E such that |CE| = |DE|. The line through C and E is parallel to AB (ACED is a parallelogram).
2.
a. The line through (a, b) and (c, d) has equation 
c. Given ax + by = c and dx + ey = f in F[x], the solution (if any) involves only field operations in F, and so both x and y are in F (if they exist).
3. The question asks whether 
can be constructed. This is the result of bisecting 
. But the angle π/6 is constructible because 
is constructible (see diagram) so angle π/12 is constructible.
can be constructed. This is the result of bisecting . But the angle π/6 is constructible because is constructible (see diagram) so angle π/12 is constructible.
5. No. A sphere of radius 1 has volume 
and, if this is the volume of a cube with side a, then 
. But a is not constructible since it is not even algebraic over 
. For if a is a root of 
. Then π is a root of 
. This is impossible as π is transcendental over 
and, if this is the volume of a cube with side a, then . But a is not constructible since it is not even algebraic over . For if a is a root of . Then π is a root of . This is impossible as π is transcendental over
7.
a. If a = sin θ is constructible so is 
by the Lemma. The converse is similar.
by the Lemma. The converse is similar.
9. If a heptagon could be constructed so could a = cos (2π/7). Now De Moivre's Theorem gives cos 7θ + i · sin 7θ = (cos θ + i · sin θ)7. Writing c = cos θ and s = sin θ, the real parts are
Taking 
, this shows that a is a root of f = 64x7 − 112x5 + 56x3 − 7x − 1. By the hint f = pq, where p = 8x3 + 4x2 − 4x − 1, and
, this shows that a is a root of f = 64x7 − 112x5 + 56x3 − 7x − 1. By the hint f = pq, where p = 8x3 + 4x2 − 4x − 1, and
With a calculator, show that q(a) = − . 7 approximately. Hence p(a) = 0. But p is irreducible over 
(no roots by Theorem 9 §4.1) so it is the minimal polynomial of a. Since deg p = 3, a is not constructible.
(no roots by Theorem 9 §4.1) so it is the minimal polynomial of a. Since deg p = 3, a is not constructible.
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