Tiggsy

Easiest is to use the parametric equations of the ellipse.

For the ellipse

 $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \\\text{ the parametric equations are }\\x = a\cos \theta,\qquad y = b\sin \theta.$

Have you thought of taking up a different subject ?

(a - 1/a)^2 = a^2 +1/a^2 - 2 = 25/36,

a^2 + 1/a^2 + 2 = 25/36 + 4 = 169/36,

(a + 1/a)^2 = 169/36,

a + 1/a  = 13/6.

I arrived at the now accepted result, but via a different method.

Think of a qualifying sequence as any containing exactly two ones,two twos, and two "others".

Every qualifying sequence has the same probability of occurring, (1/6).(1/6).(1/6).(1/6).(4/6).(4/6)=16/(6^6).

The order doesn't matter.

The number of qualifying sequences will be 6!/(2^3), the three twos to remove the duplication of the 1's, the 2's, and the "others".

The total probability will be the product of these. 

That works out to be 5/162, approximately 0.030864.

Hi Melody

Consider it from the point of view of the equation

$\displaystyle x^{n}-1=(x-1)k(x).$

All that you are looking for is that $\displaystyle (x-1) \text{ is a factor of }x^{n}-1.$  

It's a perfectly reasonable question.

The actual number of twin primes is not relevant.

(Has it been proven that they are infinite in number, or is it still conjecture ?)

All twin primes are of the form $\displaystyle 6n\pm 1,$

(the integer between them has to be divisible by both 2 and 3, so has to be divisible by 6).

Their sum will be of the form 12n, meaning that their sum will be divisible by 12.

Could there be a larger common divisor ?

Look at the first few cases.

5   +  7  = 12 = 12*1,

11 + 13 = 24 = 12*2,

17 + 19 = 36 = 12*3.

Clearly, 12 is going to be the largest.

Let

 $\displaystyle f(x)=x^{3}+x^{6}+x^{9}+x^{27}\quad \text{and}\quad \frac{f(x)}{x^{2}-1}=Q(x)+\frac{R(x)}{x^{2}-1}\\ \text{where R(x) is linear, i.e.}\; R(x)=ax+ b \quad \text{for some}\; a\; \text{and}\; b.\\ \text{Then}\\ f(x)=Q(x)(x^{2}-1)+R(x)\\ \text{so}\\ f(-1)=-2=R(-1)=-a+b\dots\dots(1)\\ f(1)= 4=R(1)=a+b\dots\dots\dots(2)$

and from (1) and (2), a = 3 and b = 1.

See diagram from previous post.

Triangles BEO and CGO are congruent.

Let BE = x , the angle at E is a right angle so  x = (R + h)cos(theta).

EA = 5 - x, and if AD is to be a tangent to the semi circle, then the line from A to the point of tangency will have length 5 - x also.

Similarly on the RHS, the line from D to the point of tangency will have length 4 - x.

So, if AD is to be a tangent to the semi circle, it will have a length 9 - 2x = 9 - 2(R + h)cos(theta).

Now move into co-ordinate geometry mode.

Take BC to be the horizontal axis, with O as the origin.

Point A will have an x co-ordinate   -(R + h) + 5cos(theta) and a y co-ordinate 5sin(theta).

Point D will have an x co-ordinate    (R + h) - 4cos(theta) and a y co-ordinate 4sin(theta).

So,

$\displaystyle AD^{2}= \{9-2(R+h)\cos\theta\}^{2}=\{(R+h)-4\cos\theta-\{-(R+h)+5\cos\theta\}\}^{2}+(4\sin\theta-5\sin\theta)^{2},\\ 81-36(R+h)\cos\theta+4(R+h)^{2}\cos^{2}\theta=4(R+h)^{2}-36(R+h)\cos\theta+81\cos^{2}\theta+\sin^{2}\theta,\\ 81+4(R+h)^{2}\cos^{2}\theta-4(R+h)^{2}-81\cos^{2}\theta-\sin^{2}\theta = 0, \\ 4(R+h)^{2}(\cos^{2}\theta-1)+81\sin^{2}\theta -\sin^{2}\theta=0,\\ 4(R+h)^{2}(-\sin^{2}\theta)+80\sin^{2}\theta=0,\\ (R+h)^{2}=20,\\ R+h =\sqrt{20}=2\sqrt{5} .$

$\displaystyle \text{BC}=2(R+h)=4\sqrt{5}.$

I get an answer of 4sqrt(5).

I need to check my algebra and, if correct, I'll try to a post a solution tomorrow, (no time at the moment).