Download Construct a regular hexagon in as few steps as possible. Construction

Constructing a Regular Hexagon
Jeremy Cue
Problem 72: Construct a regular hexagon in as few steps as possible.
Construction:
E
D
O
F
A
C
B
Construction Steps:
1) Construct a line segment AB.
2) Construct a circle with center A and radius AB.
3) Construct a circle with center B and radius BA. Get the point O as one of the two
points where these two circles intersect.
4) Construct a circle with center O and radius OB. Get the point C where this circle
intersects the circle with center B and radius BA and get the point F where this circle
intersects the circle with center A and radius AB.
5) Construct a circle with center C and radius CO. Get the point D where this circle
intersects the circle with center O and radius OB.
6) Construct a circle with center F and radius FO. Get the point E where this circle
intersects the circle with center O and radius OB.
7 – 11) Construct line segments BC, CD, DE, EF, FA.
Claim: Hexagon ABCDEF is both equilateral and equiangular, meaning it is a regular
hexagon.
Proof:
Construct line segments AO, BO, CO, DO, EO, and FO to obtain the following diagram:
E
D
O
F
A
C
B
First, note that all the circles used in this construction are congruent because they all have
radii congruent to line segment AB. Therefore, line segments AB, BC, CD, EF, FA, AO,
BO, CO, DO, EO, and FO are all congruent to one another because they are radii of
congruent circles. Therefore, triangles AOB, BOC, COD, EOF, and FOA are all
congruent to one another by I.8 (SSS), and they are also all equilateral. Because we know
that equilateral triangles are equiangular by Theorem 26, the three angles in each
equilateral triangle are congruent, and therefore all the interior angles of these triangles
are congruent because corresponding parts of congruent triangles are congruent. In short,
we have constructed 5 congruent regular triangles.
Now, consider triangle DOE. If we show that DE is congruent to AB, BC, CD, EF, and
FA, we know that the hexagon is equilateral. If we consider one of the regular triangles,
we know that the three interior angles are congruent, and since they must add up to 2
right angles by I.32, each interior angle must be congruent to 2/3 of a right angle.
Therefore, each of the interior angles of the 5 congruent regular triangles is congruent to
2/3 of a right angle.
Consider the sum of angles AOB, FOA, and EOF. Each of these angles is congruent to
2/3 of a right angle, so the sum of these 3 angles is 2 right angles. Therefore, line
segments BO and OE are in a straight line by I.14. Similarly (using angles AOB, BOC,
and COD), line segments AO and OD are in a straight line by I.14. Therefore, because
angles AOB and DOE are vertical angles, they are congruent by I.15. Therefore, triangle
AOB is congruent to triangle DOE by I.4 (SAS). Thus, all the corresponding sides and
angles are congruent, so line segment DE is congruent to AB, BC, CD, EF, and FA, and
all the interior angles are congruent to the interior angles of triangle AOB and the
remaining 4 congruent regular triangles. Therefore, all the sides of hexagon ABCDEF are
congruent, so it is equilateral. Also, all the angles of the hexagon are congruent, so it is
equiangular. Hence, hexagon ABCDEF is regular.