Equilateral △ABC with side length 14 is rotated about its center by angle θ, where 0<θ<60∘, to form △DEF. See the figure. The area of hexagon ADBECF is 913. What is tanθ?
Let G be the center of △ABC.Since △ABC is equilateral, M is the midpoint of BC, so BM=MC=7. Hence, AM=73. Since G is the centroid, GMAG=2. Therefore, AG=32AM=32(73)=3143.
Since △DEF is a rotation of △ABC about G, AG=BG=CG=DG=EG=FG=3143. Using the areas of the six triangles, 913=[ADBECF]=[GBE]+[GEC]+[GCF]+[GFA]+[GAD]+[GDB]=3[GBE]+3[GEC]=321(3143)2sinθ+321(3143)2sin(120∘−θ)=23(3143)2(sinθ+sin(120∘−θ))=98(sinθ+sin(120∘−θ)). Thus, sinθ+sin(120∘−θ)=98913=14133.
Using the sum-to-product identity, sinθ+sin(120∘−θ)=2sin60∘cos(60∘−θ)=3cos(60∘−θ)=14133. Therefore, cos(60∘−θ)=1413.
Using sin2x+cos2x=1, sin(60∘−θ)=±1433. Since 0<θ<60∘⟹0<60∘−θ<60∘, we have sin(60∘−θ)=1433. Hence, tan(60∘−θ)=cos(60∘−θ)sin(60∘−θ)=14131433=1333.
Using the tangent subtraction identity, 1333=tan(60∘−θ)=1+tan60∘tanθtan60∘−tanθ. Thus, 1333=1+3tanθ3−tanθ. Cross-multiplying, 33+9tanθ=133−13tanθ. Therefore, 22tanθ=103, so tanθ=1153.
