textot

nice answer, just note that the question asks for the entire solution set, which is 360n±51.13, where n is an integer.

It's indeed an interesting problem.

The proof here is based on the brick wall problem in the picture below (the proof is quite simple and it involves some simple induction) 

Rephrasing this problem, it basically asks for the proof that the number of ways to have n-1 digits, each one of them either 1 or 0, without having consecutive 1's, is equal to the n+1th Fibonacci number. (2^n will always be a fibbinary number, so we can add 1) 

Other than the rightmost digit, a 1 must be followed by a 0, so we can group them together as '10', or a block of length 2, and the other zeroes other than the rightmost digit can be treated as a block of length 1.

If the last digit is a 1, then the number of possibilities is equal to that of a brick wall of length n-2. If the last digit is a zero, then the number of possibilities is equal to that of a brick wall of length n-1.

As mentioned before, this proof is based on the brick wall problem, so they are equal to the n-1th and nth Fibonacci numbers, respectively.

I think you can finish it

Hint:

use the Euclidean algorithm to find $\gcd(4\cdot\frac{n(n+1)}{2}, n-2)$

Also, note that $ax^2+bx+c$ has the reciprocal of the roots of $cx^2+bx+a$. 

thanks melody! this problem was pretty hard

$2(x^2+y^2)=x+y+8\\ 2(x^2-\frac{1}{2}x+y^2-\frac{1}{2}y)=8\\ x^2-\frac{1}{2}x+y^2-\frac{1}{2}y=4\\ x^2-\frac{1}{2}x+\frac{1}{16}+y^2-\frac{1}{2}x+\frac{1}{16}=4+\frac{1}{8}\\ (x-\frac{1}{4})^2+(y-\frac{1}{4})^2=\frac{33}{8}$

Notice that for the equation $x^2+y^2=1$, we can substitute $x=\cos(\theta)$ and $y=\sin(\theta)$.

Similarly, here we can substitute $x=\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta))+\frac{1}{4}, y=\frac{\sqrt{33}}{2\sqrt{2}}(\sin(\theta))+\frac{1}{4}$, but with slight adjustments due to the translation and the dilation of the circle.

So, we want to find the maximum value of:

$(\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta))+\frac{1}{4})-(\frac{\sqrt{33}}{2\sqrt{2}}(\sin(\theta))+\frac{1}{4})\\ =\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta)-\sin(\theta))$

I know there is a way to find a simpler form of $\cos(\theta)-\sin(\theta)$, but it's hard to see and I didn't see it, so what I ended up doing is just taking the derivative with respect to theta, which is:

$-\sin(\theta)-\cos(\theta)$

The extreme points are when:

$-\sin(\theta)=\cos(\theta)$, which only happens when $\theta = 135+180n$, where n is an integer. From now, It shouldn't be hard to figure out that the maximum value of $\cos(\theta)-\sin(\theta)$ is $\sqrt{2}$.

This means that the maximum value of our original expression is $\frac{\sqrt{33}}{2\sqrt{2}} \cdot \sqrt{2}=\boxed{\frac{\sqrt{33}}{2}}$

For completeness, try to find which values of x and y give that maximum value.

Edit: I noticed that my previous answer contained a huge mistake lol

Thanks, Melody.

To fix my answer:

From the reduced problem, I noticed that the triples with repeated numbers are all in the form of (k, k, l), (k, l, k), and (l, k, k), where k is some number between 0 and 11 (inclusive) and l is a number that is forced by that particular value of k, meaning that there are 3*12=36 triples that are repeated.

We now have 276-36=240 triples that are not repeated. But now we can divide by 3! to account for the overcounted triples. That gives a final answer of 40, which matches your answer :)

hmmm

let $x=2a+1, y = 2b+1, z=2c+1$, so

$2a+1+ 2b+1+2c+1=47\\ 2a+2b+2c=44\\ a+b+c=22$

The problem reduces to:

"How many nonnegative ordered triples (a, b, c) satisfy $a+b+c=22$?"

which is a direct application of stars and bars, having an answer of ${22+3-1 \choose 3-1} = \frac{24!}{2!22!}=\boxed{276}$

I'm getting the same answer as guest

quadratic in terms of cos(x)?