, 2013

We make a table of the 36 possible combinations of rolls and the resulting sums:

2
3
5
7
11
13

2
4
5
7
9
13
15

3
5
6
8
10
14
16

5
7
8
10
12
16
18

7
9
10
12
14
18
20

11
13
14
16
18
22
24

13
15
16
18
20
24
26

Of the 36 entries in the table, 6 are prime numbers (two entries each of 5, 7 and 13).

Therefore, the probability that the sum is a prime number is $\frac{6}{36}$ or $\frac{1}{6}$.

(Note that each sum is at least 4 and so must be odd to be prime. Since odd plus odd equals even, then the only possibilities that really need to be checked are even plus odd and odd plus even (that is, the first row and first column of the table).)
First, we find the coordinates of $V$.

To do this, we use the given equation for the parabola and complete the square: $$y = -x^2+4x+1 = -(x^2-4x-1) = -(x^2-4x+2^2-2^2-1) = - ((x-2)^2-5) = -(x-2)^2+5$$ Therefore, the coordinates of the vertex $V$ are $(2,5)$.

Next, we find the coordinates of $A$ and $B$.

Note that $A$ and $B$ are the points of intersection of the line with equation $y=-x+1$ and the parabola with equation $y=-x^2+4x+1$.

We equate $y$-values to obtain $-x+1=-x^2+4x+1$ or $x^2-5x=0$ or $x(x-5)=0$.

Therefore, $x=0$ or $x=5$.

If $x=0$, then $y=-x+1=1$, and so $A$ (which is on the $y$-axis) has coordinates $(0,1)$.

If $x=5$, then $y=-x+1=-4$, and so $B$ has coordinates $(5,-4)$.

We now have the points $V(2,5)$, $A(0,1)$, $B(5,-4)$.

This gives $$\begin{aligned} AV^2 & = (0-2)^2+(1-5)^2 = 20 \\ BV^2 & = (5-2)^2+(-4-5)^2 = 90 \\ AB^2 & = (0-5)^2+(1-(-4))^2 = 50\end{aligned}$$ and so $AV^2+BV^2-AB^2 = 20+90-50=60$.