Problem:
Let A=(0,0) and B=(b,2) be points on the coordinate plane. Let ABCDEF be a convex equilateral hexagon such that ∠FAB=120∘,AB∥DE,BC∥EF, CD∥FA, and the y-coordinates of its vertices are distinct elements of the set {0,2,4,6,8,10}. The area of the hexagon can be written in the form mn, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.
Solution:
Let the x-coordinates of C,D,E, and F be c,d,e, and f, respectively. Note that the y-coordinate of C is not 4 , since, if it were, the fact that AB=BC would imply that A,B, and C are collinear or that c is 0 . Therefore F=(f,4). Since AF and CD are both parallel and congruent, C=(c,6) and D=(d,10), and then E=(e,8). Because the y-coordinates of B,C, and D are 2,6 , and 10 , respectively, and BC=CD, conclude that b=d. Since AB and DE are both parallel and congruent, e=0. Let a denote the side-length of the hexagon. Then f2+16=AF2=a2=AB2=b2+4. Apply the Law of Cosines in △ABF to obtain 3a2=BF2=(b−f)2+4. Without loss of generality, assume b>0. Then f<0 and b=a2−4,f=−a2−16, and b−f=3a2−4. Now
2a2−20+a2−4+a2−16=3a2−4, so 2(a2−4)(a2−16)=3a2−4, and 2(a2−4)(a2−16)=a2+16
Squaring again and simplifying yields a2=112/3, so b=10/3 and f=−8/3. Hence A=(0,0),B=(10/3,2),C=(63,6),D=(10/3,10), E=(0,8),F=(−8/3,4). Thus [ABCDEF]=[ABDE]+2[AEF]=b⋅AE+(−f)⋅AE=8(b−f)=483, so m+n=51.
OR
Let α denote the measure of the acute angle formed by AB and the x-axis. Then the measure of the acute angle formed by AF and the x-axis is 60∘−α. Note that asinα=2, so
4=asin(60∘−α)=a23cosα−a⋅21sinα=a23cosα−1
Thus a3cosα=10, and b=acosα=10/3. Then a2=b2+4=112/3, and f2=a2−16=64/3. Also, (c−b)2=a2−16=64/3, so c=b+8/3=63; c−d=0−f=8/3, so d=10/3; and e−f=c−b=8/3, so e=0. Proceed as above to obtain [ABCDEF]=483.