Xenosmilus

Obviously, \(k=0 \) is an answer. Let's try to find some more (or prove that k=0 is the only answer). Let the base of the triangle be the segment from (1, 0) to (0, 1). The length of this segment will then be \(\sqrt{2}\). We can see that the height of the triangle be \(\frac{\sqrt{2}}{2}\). Since the third point is (k, k), the point must lie on the line x=y which means that k can be 0 or 1.

First, we can simplify f(x): \(f(x)=-3x^2+6x+27\). Then, we factor out -3 for \(f(x)=-3(x^2-2x-9)=-3(x-1)^2+30\).

 From this equation, we get that the vertex of f(x) is (x, y) = (1, 30).

Ok, I found the answer.

We can use polynomial long division to simplify: 

\(\frac{5x^2+20x+33}{x^2+4x+7}=5-\frac{2}{x^2+4x+7}\)

Then, h(x)<5, and we can find the minimum by completing the square on the quadratic.

\(x^2+4x+7=(x+2)^2+3\),

so the minimum is 3. Hence, the least possible value of h(x) is \(\frac{13}{3}\), and the range is \(\boxed{h(x)\in[\frac{13}{3},5)}\)

Completing the square in \(x\) and \(y \), we get

\(\begin{align*} -4 (x + 1)^2 + (y - 1)^2 = 6. \end{align*}\)

We can re-write this as

\(\begin{align*} \frac{(y - 1)^2}{6} - \frac{(x + 1)^2}{3/2} = 1. \end{align*}\)

The center of the hyperbola is at \(\boxed{(-1,1)}\)

Couldn't you just substitute a=-14-b into the second equation and then factor/solve the cubic?

Start by moving all the terms to one side and combining like terms:

\(4x^2-8x+4y^2+14y+12=0\)

Then, divide both sides by \(4 \):

\(x^2-2x+y^2+\frac{7}{2}y+3=0\)

Next, complete the square for \(x \) and \(y \):

\((x-1)^2+(y+\frac{7}{4})^2-1-\frac{49}{16}+3=0\)

Simplify:

\((x-1)^2+(y+\frac{7}{4})^2=\frac{17}{16}\)

This is the equation (in standard form) for the circle. Now take the square root of \(\frac{17}{16} \):

\(\sqrt{\frac{17}{16}}=\frac{\sqrt{17}}{\sqrt{16}}=\boxed{\frac{\sqrt{17}}{4}}\)  Which is our answer.

The answer on that page (by CPhill) is wrong however: 

Ball splits             Ways to distribute the balls

7  1                                  2

6  2                                  2

5  3                                  2

4  4                                  1

2 * 2 * 2 * 1  =  8 ways

but you cannot multiply as these are separate cases; you add them to get 7.

Then the answer would be 75.

There is also another way (stars and bars):

Choose 3 boxes to be empty: \({5 \choose 3}=10\)

Use stars and bars allowing empty boxes to calculate (or just count like CPhill)

CPhills method if he was here:

Ball splits

0 8

1 7

2 6

3 5

4 4

5 3

6 2

7 1

8 0

to get 9 different cases (stars and bars would be \({n+k-1 \choose k-1}={8+2-1\choose 2-1}={9\choose 1}=9\) ways)

Multiply by 10 to get 90 ways. However, this overcounts (0 0 0 8 0 is counted four times, and so are all other permutations) so we subtract 5*3=15 from 90 to get our final answer, 75.