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Geometry Difficulty 3.2 AMC 10/12 Prove it Canada

IMG0 Two fair dice, called $D_1$ and $D_2$ , each have six faces. $D_1$ has the numbers $1$ , $2$ , $3$ , $4$ , $5$ , $6$ on its faces. $D_2$ has a $1$ on some of its faces and a $2$ on its remaining faces. When $D_1$ and $D_2$ are rolled, the probability that the sum of the numbers on the top faces is a prime number is $\dfrac{23}{36}$ . How many faces on $D_2$ have the number $1$ on them? In the diagram, square $ABCD$ has $A$ and $B$ on the $x$ -axis and $C$ and $D$ below the $x$ -axis on the parabola with equation $y = x^2 - 4$ .Determine the area of $ABCD$ , writing your answer in the form $r -
$\sqrt{t}$ for some positive integers

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Figure for this problem r $and$ t$.

Solution

Suppose that $n$ faces on $D_2$ have a $1$ on them and $6-n$ faces have a $2$ on them. We make a chart to enumerate the possible totals when the two dice are rolled. The number rolled on $D_1$ is shown in the left column and the number rolled on $D_2$ is shown across the top row. Inside the chart, we track the sum and the number of times this sum could occur. $\boldsymbol{1}$ ( $\boldsymbol{n}$ times) $\boldsymbol{2}$ ( $\boldsymbol{6-n}$ times) $\boldsymbol{1}$ $2$ ( $n$ times) $3$ ( $6-n$ times) $\boldsymbol{2}$ $3$ ( $n$ times) $4$ ( $6-n$ times) $\boldsymbol{3}$ $4$ ( $n$ times) $5$ ( $6-n$ times) $\boldsymbol{4}$ $5$ ( $n$ times) $6$ ( $6-n$ times) $\boldsymbol{5}$ $6$ ( $n$ times) $7$ ( $6-n$ times) $\boldsymbol{6}$ $7$ ( $n$ times) $8$ ( $6-n$ times) Of these sums, $2$ , $3$ , $5$ , and $7$ are prime. These occur a total of $n + (6-n) + n + (6-n)
+ n + (6-n) + n = 18 + n times out of the possible

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Figure for this problem 6 $\times$ 6 = 36 outcomes from rolling the two dice together. Since the probability of having a prime sum is

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Figure for this problem $\dfrac{23}{36}$ $, then$ 23 $of the$ 36 outcomes give a prime sum, and so

Figure for this problem

Figure for this problem 18 + n = 23 $or$ n = 5 $. Since$ ABCD $is a square and$ AB $is horizontal, then$ CD $is parallel to$ AB $and so is also horizontal. Since$ C $and$ D $are on a parabola and$ CD $is horizontal, then$ C $and$ D are equidistant from the axis of symmetry. Since the parabola has equation

Figure for this problem

Figure for this problem y = x^2 -
4 $, its$ x $-intercepts are$ 2 $and$ -2 and so its axis of symmetry has equation

Figure for this problem

Figure for this problem x = 0 $. Thus, we can say that$ C $and$ D $have$ x $-coordinates$ s $and$ -s $, respectively, for some$ s > 0 $. This means that$ A $has coordinates$ (-s, 0) $and$ B $has coordinates$ (s, 0). This means that the side length of square

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Figure for this problem ABCD $is$ s - (-s) = 2s$.

Since the height and width of $ABCD$ are equal, then $C$ has coordinates $(s, -2s)$ and $D$ has coordinates $(-s, -2s)$ . Since $C$ lies on the parabola with equation $y = x^2 - 4$ , then $-2s = s^2 - 4$ and so $s^2 + 2s - 4 = 0$ . By the quadratic formula, $s = \dfrac{-2 \pm \sqrt{2^2 - 4(1)(-4)}}{2} = \dfrac{-2 \pm \sqrt{20}}{2} = \dfrac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5}$ Since $s > 0$ , then $s = -1 + \sqrt{5}$ . This means that the area of square $ABCD$ is equal to $(2s)^2$ which equals $(-2+2\sqrt{5})^2$ . Expanding and simplifying, we obtain $4 + 20
- 8 $\sqrt{5}$ = 24 - $8\sqrt{5}$ = 24 - $\sqrt{320}\$.$

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