_____ _I/ X \__ /  \ / M \ /  \ YKOY' \  / \ N / \_J  __/ \__X'_/ 

Hint: with this problem, there are two ways of drawing a regular pentagon within a circle.
Source: Reader Philippe Fondanaiche's personal wording from different books on how to draw a regular pentagon within a given circle.
I suppose there's a strictly geometric proof which would be more elegant, but: Without loss of generality, assume the radius of c is 2 so that we immediately have: Y=(2,0) O=(0,0) X=(0,2) M=(1,0), etc. Then XM=MN=sqr(5) => N=(0,1sqr(5)) Since YMN is isoceles, the altitude from M to YN bisects YN. Call the midpoint of YN, P. Then P=(1,(1sqr(5))/2) Then, the equation of line PM is given by: y = ((1+sqr(5))/2)x + 1. K is the pt on this line where y = 0 => x coordinate of K = 1/((1+sqr(5))/2) = 2/(1+sqr(5) = 2(1sqr(5))/4 = (1sqr(5))/2 So K = ((1sqr(5))/2, 0) On the other hand: Since Y=(2,0) and N=(0,1sqr(5)), YN^2 = 4 + (1  2sqr(5) + 5) = 10  2sqr(5) Since I and J are on the circle centered at Y with radius YN, they satisfy: (x+2)^2 + y^2 = 10  2sqr(5) => x^2 + 4x + y^2 = 6  2sqr(5) Since they are on c, they also satisfy x^2 + y^2 = 4 Subtracting: 4x = 2  2sqr(5) => x = (1  sqr(5))/2 Since this holds for both I and J, the line IJ must be the vertical line: x = (1  sqr(5))/2 and K is clearly on that line. QED