The distances from a certain point inside a regular hexagon to three of its consecutive vertices are equal to 1, 1 and 2, respectively. Find the length of this hexagon's side.
No computers needed, just a good plan.
Call the three vertices A, B, and C and the extra point P.
Then AB=BC=x, AP=BP=1 and CP=2. Angle ABC=120.
ABP is an isosceles triangle so add point M, the midpoint of AB. Then BM=x/2 and PM=sqrt[1-x^2/4].
Let angle MBP be T. Then by basic right angle trigonometry we have:
sin(T) = cos(90-T) = sqrt[1-x^2/4]
cos(T) = sin(90-T) = x/2
Angle PBC is 120-T. Applying the law of cosines to triangle PBC yields:
(3-x^2)/x = cos(120-T)
The trick here is to use the cosine sum identity cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) with a=30 and b=90-T.
Then the five expressions can all be plugged into the identity:
(3-x^2)/x = (sqrt[3]/2)*sqrt[1-x^2/4] - (1/2)*(x/2)
Multiplying through and isolating the radical cleans it up
12-3x^2 = x*sqrt[12-3x^2]
Then factor sqrt[12-3x^2] out and we can break the equation into
sqrt[12-3x^2] = 0 -OR- sqrt[12-3x^2] = x
Negative solutions are rejected as x is a positive length. Each has one positive solution, x=1/2 -OR- x=sqrt[3]
x=1/2 is rejected since the hexagon would be so small that P is outside of it, rather than inside as the problem states. So that leaves the single answer that the length of the hexagon's sides is sqrt[3].
Solution
