## Points on a Circle

 ``` _____ _I/ X \__ / | \ / M \ / | \ Y----K---O--------Y' \ | / \ N / \_J | __/ \__X'_/ ``` Given a circle (c) with center O and diameters X'OX and Y'OY perpendicular to each other. M is the middle of OX. The circle of center M and radius MY cuts OX' at N. The circle of center Y and radius YN cuts (c) at I and J. The perpendicular from M to YN cuts OY at K. Demonstrate that I, J, K are on the same straight line.

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.

Larry Baum sent the following solution:
```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=(0,1-sqr(5))
Since YMN is isoceles, the altitude from M to YN bisects YN.  Call the
midpoint of YN, P.  Then P=(-1,(1-sqr(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(1-sqr(5))/-4 = (1-sqr(5))/2
So K = ((1-sqr(5))/2, 0)

On the other hand:
Since Y=(-2,0) and N=(0,1-sqr(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
```

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