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考题21-22 2015 AMC 12A

2018-08-22 重点归纳

AMC12是针对高中学生的数学测验，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 21

A circle of radius $r$ passes through both foci of, and exactly four points on, the ellipse with equation $x^2+16y^2=16.$ The set of all possible values of $r$ is an interval $[a,b).$ What is $a+b?$

$\textbf{(A)}\ 5\sqrt{2}+4\qquad\textbf{(B)}\ \sqrt{17}+7\qquad\textbf{(C)}\ 6\sqrt{2}+3\qquad\textbf{(D)}\ \sqrt{15}+8\qquad\textbf{(E)}\ 12$

Solution

We can graph the ellipse by seeing that the center is $(0, 0)$ and finding the ellipse's intercepts. The points where the ellipse intersects the coordinate axes are $(0, 1), (0, -1), (4, 0)$ , and $(-4, 0)$ . Recall that the two foci lie on the major axis of the ellipse and are a distance of $c$ away from the center of the ellipse, where $c^2 = a^2 - b^2$ , with $a$ being half the length of the major (longer) axis and $b$ being half the minor (shorter) axis of the ellipse. We have that $c^2 = 4^2 - 1^2 \implies$ $c^2 = 15 \implies c = \pm \sqrt{15}$ . Hence, the coordinates of both of our foci are $(\sqrt{15}, 0)$ and $(-\sqrt{15}, 0)$ . In order for a circle to pass through both of these foci, we must have that the center of this circle lies on the y-axis.

The minimum possible value of $r$ belongs to the circle whose diameter's endpoints are the foci of this ellipse, so $a = \sqrt{15}$ . The value for $b$ is achieved when the circle passes through the foci and only three points on the ellipse, which is possible when the circle touches $(0, 1)$ or $(0, -1)$ . Which point we use does not change what value of $b$ is attained, so we use $(0, -1)$ . Here, we must find the point $(0, y)$ such that the distance from $(0, y)$ to both foci and $(0, -1)$ is the same. Now, we have the two following equations. $(\sqrt{15})^2 + (y)^2 = b^2$ $y + 1 = b \implies y = b - 1$ Substituting for $y$ , we have that $15 + (b - 1)^2 = b^2 \implies -2b + 16 = 0.$

Solving the above simply yields that $b = 8$ , so our answer is $a + b = \sqrt{15} + 8 \textbf{ (D)}$ .

Solution 2

As above, we can show that the foci of the ellipse are $(\sqrt{15},0), (-\sqrt{15},0).$

To obtain the lower bound, note that the smallest circle is when the diameter is on the line segment formed by the two foci. We can check that this indeed passes through four points on the ellipse since $\sqrt{15}>1,$ so $a=\sqrt{15}.$

To get the upper bound, note that the circle must go through either $(0,1)$ or $(0,-1).$ WLOG, let let the circle go through $(0,1).$ We know that the circle must go through the foci of the ellipse $(\sqrt{15},0), (-\sqrt{15},0),$ So we can apply power of a point to find the diameter. Let $x$ denote the length of the line segment from the origin to the lower point on the circle. Note that $x$ lies on the diameter. Then by POP, we have $x * 1 = \sqrt{15} * \sqrt{15},$ yielding $x=15$ , and so the radius of the circle is $(15+1)/2=8,$ so $b=8.$ Thus $amc数学竞赛试题及答案$ .

Problem 22

For each positive integer $n$ , let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$ , with no more than three $A$ s in a row and no more than three $B$ s in a row. What is the remainder when $S(2015)$ is divided by $12$ ?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$

Solution

One method of approach is to find a recurrence for $S(n)$ .

Let us define $A(n)$ as the number of sequences of length $n$ ending with an $A$ , and $B(n)$ as the number of sequences of length $n$ ending in $B$ . Note that $A(n) = B(n)$ and $S(n) = A(n) + B(n)$ , so $S(n) = 2A(n)$ .

For a sequence of length $n$ ending in $A$ , it must be a string of $A$ s appended onto a sequence ending in $B$ of length $n-1, n-2, \text{or}\ n-3$ . So we have the recurrence: $amc数学竞赛试题及答案$

We can thus begin calculating values of $A(n)$ . We see that the sequence goes (starting from $A(0) = 1$ ): $1,1,2,4,7,13,24...$

A problem arises though: the values of $A(n)$ increase at an exponential rate. Notice however, that we need only find $S(2015)\ \text{mod}\ 12$ . In fact, we can use the fact that $S(n) = 2A(n)$ to only need to find $A(2015)\ \text{mod}\ 6$ . Going one step further, we need only find $A(2015)\ \text{mod}\ 2$ and $A(2015)\ \text{mod}\ 3$ to find $A(2015)\ \text{mod}\ 6$ .

Here are the values of $A(n)\ \text{mod}\ 2$ , starting with $A(0)$ : $1,1,0,0,1,1,0,0...$

Since the period is $4$ and $2015 \equiv 3\ \text{mod}\ 4$ , $A(2015) \equiv 0\ \text{mod}\ 2$ .

Similarly, here are the values of $A(n)\ \text{mod}\ 3$ , starting with $A(0)$ : $amc数学竞赛官网$

Since the period is $13$ and $2015 \equiv 0\ \text{mod}\ 13$ , $A(2015) \equiv 1\ \text{mod}\ 3$ .

Knowing that $A(2015) \equiv 0\ \text{mod}\ 2$ and $A(2015) \equiv 1\ \text{mod}\ 3$ , we see that $A(2015) \equiv 4\ \text{mod}\ 6$ , and $S(2015) \equiv 8\ \text{mod}\ 12$ . Hence, the answer is $\textbf{(D)}$ .

Note that instead of introducing $A(n)$ and $B(n)$ , we can simply write the relation $S(n)=S(n-1)+S(n-2)+S(n-3),$ and proceed as above.

Recursion Solution II

The huge $n$ value in place, as well as the "no more than... in a row" are key phrases that indicate recursion is the right way to go. Let's go with finding the case of $S(n)$ from previous cases. So how can we make the words of $S(n)$ ? Do we choose 3-in-a-row of one letter, $A$ or $B$ , or do we want 2 consecutive ones or 1? Note that this covers all possible cases of ending with $A$ and $B$ with a certain number of consecutive letters. And obviously they are all distinct.

[Convince yourself that each case for $S(n)$ is considered exactly once by using these cues: does it end in 3, 2, or 1 consecutive letter(s) (1 consecutive means a string like ...BA, ...AB, as in the letter switches) and does it WLOG consider both A and B?]

From there we realize that $amc数学竞赛官网$ because 3 in a row requires $S(n-3)$ , and so on. Let's start using all $S(n)$ values $(\mod12)$ . We also know that $S(1)=2, S(2)=4, S(3)=8$ , and so on. These residues are: $2, 4, 8, 2, 2, 0, 4, 6, 10, 8, 0, 8, 8, 4, 8, 8, 8, 0...$ , upon which the cycle repeats. Note that the cycle length is 7, and $2015-(286*7=2002)=13$ , so the residue of $S(2015) (\mod{12})$ is the residue of $S(13)=\boxed{008}$ .