Problem:
Let A=(0,0) and B=(b,2) be points on the coordinate plane. Let ABCDEF be a convex equilateral hexagon such that β FAB=120β,ABβ₯DE,BCβ₯EF, CDβ₯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 mnβ, 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 β³ABF to obtain 3a2=BF2=(bβf)2+4. Without loss of generality, assume b>0. Then f<0 and b=a2β4β,f=βa2β16β, and bβf=3a2β4β. Now
2a2β20+βa2β4β+a2β16β=3a2β4β, so 2(a2β4)(a2β16)β=3a2β4, and 2(a2β4)(a2β16)β=a2+16β
Squaring again and simplifying yields a2=112/3, so b=10/3β and f=β8/3β. Hence A=(0,0),B=(10/3β,2),C=(63β,6),D=(10/3β,10), E=(0,8),F=(β8/3β,4). Thus [ABCDEF]=[ABDE]+2[AEF]=bβ AE+(βf)β AE=8(bβf)=483β, so m+n=51β.
OR
Let Ξ± 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 60ββΞ±. Note that asinΞ±=2, so
Thus a3βcosΞ±=10, and b=acosΞ±=10/3β. Then a2=b2+4=112/3, and f2=a2β16=64/3. Also, (cβb)2=a2β16=64/3, so c=b+8/3β=63β; cβd=0βf=8/3β, so d=10/3β; and eβf=cβb=8/3β, so e=0. Proceed as above to obtain [ABCDEF]=483β.