A point Q lies inside regular hexagon ABCDEF such that [QAB]=10, [QCD]=17 and [QEF]=5, where the square brackets indicate the area of the enclosed triangle. Find the side of the hexagon.
Extend sides AB, CD, and EF into lines. Let AB and CD intersect at X, AB and EF intersect at Y, and CD and EF intersect at Z.
Triangle XYZ is equilateral, it will have an area equal to 1.5 times the area of the hexagon and the sides of XYZ will be three times the length of the sides of the hexagon.
Triangles QXY and QAB have the same altitude and bases in a 3:1 ratio. Then QAB having an area of 10 implies the area of QXY is 30. Similarly, QXZ and QYZ can be determined to have areas of 51 and 15, respectively.
The union of triangles QXY, QXZ, and QYZ is triangle XYZ. Then XYZ has a area of 96. The formula area of an equilateral triangle with side s is Area=s^2*sqrt(3)/4. Then the sides of XYZ are 8*sqrt(2)*3^(1/4), which then means the sides of the hexagon are 1/3 of that or 8*sqrt(2)/3^(3/4).
Solution
