Alan

36° is pi/5 so put n = 1 in (2n-1)pi/5

108° is 3pi/5 so put n = 2 in (2n-1)pi/5

Here's my solution:

Correct.  I forgot about b having to be less than or equal to 1 at the end!

I solved the quadratic in terms of a and b (well, ok, I used a piece of software to do so!). This resulted in an expression under a square root sign that meant if b>=2-2a, two of the four roots would be real.

As long as b >= 2 - 2a the equation has at least one real root.  the line b = 2 - 2a is a straight line going from b = 2 when a = 0, to b = 0 when a = 1.  Hence the area of b >= 2 - 2a is a triangle of base 1 and height 2; i.e. the area is (1/2)*1*2 = 1.

Hyperbola C2:  \(\frac{(x-a)^2}{9}-\frac{(y-0)^2}{4}=1\)

On the x-axis y = 0, and the values of x are: \(x = a\pm3\) so \(x=a\) is the value half way between the two parts of the curve.

The two asymptotes cross at this half-way point, so from \(y=\frac{2}{3}x-2\) we have \(x = 3\) when y = 0.  

So \(a = 3\)

As follows:

"equation of the line tangent to the circle with center at C(3,1) at point (6,-9)"

It seems to me that the question as stated is incomplete - see my analysis below.

The only way I can see that 16 is the smallest value for abc+d-2 is if a,b,c and d must all be positive integers.

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I guess 27 is the answer being sought here. However, it seems to me that if m and n are positive integers, then the largest integer that mn is necessarily divisible by is mn.  !!!