Can I get a hint?
A square has its vertices on the edges of a regular hexagon. Two of the edges of the square are parallel to two edges of the hexagon, as shown in the diagram. The sides of the hexagon have length 1.
What is the length of the sides of the square?
Let O be the center of the square
Let M be the top right vertex of the square
Let N be the rightmost vertex of the hexagon and P be the top right vertex of the hexagon
Triangle PON is equilateral so PN = ON = 1
OM = (1/2) diagonal of square = S /sqrt 2 where S is the side of the square
Angle OMN = 75° Angle MON = 45° Angle MNO = 60°
Using the Law of Sines
sin (OMN) / 1 = sin MNO / (S/sqrt 2)
sin (75) /1 = sin (60)/ (S/ sqrt 2)
S / sqrt (2) = sin 60 / sin 75
S = sqrt (2) sin (60) / sin (75)
S = sqrt (2) (sqrt (3) / 2) / ( sin (45 + 30)
S = sqrt (6) / (2 (sin 45cos 30 + cos 45 sin 30) )
S = sqrt (6) / ( 2 ( sqrt (2)/2 * sqrt (3)/2 + sqrt(2/2 )(1/2) )
S = sqrt (6) / (2 (sqrt (6) /4 + sqrt (2) / 4))
S = sqrt (6) / [ (sqrt (6) + sqrt (2)) / 2
S = 2 sqrt (6) / (sqrt 6 + sqrt 2)
S = 2 sqrt (6) ( sqrt 6 - sqrt 2) / ( 6-2)
S = (12 - 2sqrt (12)) /4
S = 3 - 2sqrt (4) sqrt (3) / 4
S = 3 - 2*2 * sqrt (3) / 4
S = 3 - 4sqrt (3) / 4
S = 3 - sqrt (3) ≈ 1.27
Here's a pic (AM = 1/2 the side of the square ≈ .63 )