Voldemort

Hmm  the thing is, it may be talking about the major arc instead of the minor arc. 

Edit: Nevermind, I thought nerdiest was saying that 8pi was wrong. Sorry! :(

The first letter we have to choose is M which is our only option. This gives us a probabililty of 1/10.

Next, the only letter where we can get our A is from the second letter. Again, this gives a probability of 1/10. 

When choosing the letter T, however, we have to consider the fact that there are two T's in MATHCOUNTS. This gives a probability of 1/5 of choosing it next.

Again, there is only one H so there is 1/10 probability

Multiplying all of these gives us: 1/5000.

Which I think is somehow wrong,  someone please correct me. 

This problem avoids direct vieta's formulas! Smart!

We know that a^2 + b^2 = (a + bi)(a - bi), which are just the product of the roots!

x^2 + 5x + 22 = 0

By vieta's,  the product of the roots is -22.

What about the case where we get 4, 6, 6 :)

Use Simon's Favorite Factoring trick!

ab + 2a + 2b = -16 becomes a(b + 2) + 2b = - 16.

Adding 4 on each side gives us:

a(b + 2) + 2(b + 2) = -12, simplifying to (a + 2)(b + 2) = -12

Doing this factoring method for other equations gives us:

(a + 2)(b + 2) = -12

(a + 2)(c + 2) = 30

(b + 2)(c + 2) = -40.

Multiplying those equations and square rooting gives  us:

(a + 2)(b + 2)(c + 2) = 120

Applying thiss gives us

120/(c + 2) = - 12
120/(b + 2) = 30
120/(a + 2) = - 40

Can you solve for a, b, c from here?

By the British Flag theorem,

DX^2 + 81 = 121 + 100

DX^2 = 140

DX = sqrt(140) = 2sqrt(35)

240 0= 24 * 10 = 2^3 * 3 * 10 = 2^4 * 3 * 5

We need 2 2s, 2 3s, and 2 5s. This gives us:

2^2 * 3^2 * 5^2 = 900. 

a^3 + b^3 = (a + b)((a + b)^2 - 3ab)

Use vieta's formulas!

x^2 - 2x + 6 = 0 

a + b = 2

ab = 6

Plugging this in gives us

(2)(4 - 18) =  -28.