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AMC数学竞赛真题2016年12A 13-14

2018-09-01 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 13

Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$ ?

$amc12$

Solution 1

Let $n = \frac{N}{5}$ . Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the red ball is inserted in the middle block of $n$ balls, and there are $n - 1$ positions where this happens. Thus, $P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}$ , so

$amc数学竞赛$

Multiplying both sides of the inequality by $400(5n+1)$ , we have

$400(4n+2)<321(5n+1),$

and by the distributive property,

$1600n+800<1605n+321.$

Subtracting $1600n+321$ on both sides of the inequality gives us

$479<5n.$

Therefore, $N=5n>479$ , so the least possible value of $N$ is $480$ . The sum of the digits of $480$ is $\boxed{\textbf{(A) } 12}$ .

Solution 2 (Pattern Solution)

Let $N$ $=$ $5$ , $P(N)$ $=$ 1 ( $Given$ )

Let $N$ $=$ $10$ , $P(N)$ $=$ $\frac{10}{11}$

Let $N$ $=$ $15$ , $P(N)$ $=$ $\frac{14}{16}$

Notice that the fraction can be written as $1$ $-$ $\frac{\frac{N}{5}-1}{N+1}$

Now it's quite simple to write the inequality as $1$ $-$ $\frac{\frac{N}{5}-1}{N+1}$ $<$ $\frac{321}{400}$

We can subtract $1$ on both sides to obtain $-$ $\frac{\frac{N}{5}-1}{N+1}$ $<$ $-$ $\frac{79}{400}$

Dividing both sides by $-1$ , we derive $\frac{\frac{N}{5}-1}{N+1}$ $>$ $\frac{79}{400}$ . (Switch the inequality sign when dividing by $-1$ )

We then cross multiply to get $80N - 400 > 79N + 79$

Finally we get $N > 479$

To achieve $N = 480$

So the sum of the digits of $N$ = $\boxed{\textbf{(A) } 12}$

Solution 3

We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$ : Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3}{5}N+1$ to $N+1$ . This is a total of $\frac{4}{5}N + 2$ spaces, over a total possible $N + 1$ places to put the ball. So:

$\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.$ And we know that the next value is what we are looking for, so $N+1 = 480$ , and the sum of it's digits is $\boxed{\textbf{(A) } 12}$ .

Problem 14

Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?

$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$

Solution 1

First of all, the adjacent faces have the same sum $(18$ , because $1+2+3+4+5+6+7+8=36$ , $36/2=18)$ , so now consider the $\text{opposite sides}$ (the two sides which are parallel but not on same face of the cube); they must have the same sum value too. Now think about the extreme condition 1 and 8, if they are not sharing the same side, which means they would become endpoints of $\text{opposite sides}$ , we should have $1+X=8+Y$ , but no solution for $[2,7]$ , contradiction.

Now we know $1$ and $8$ must share the same side, which sum is $9$ , the $\text{opposite side}$ also must have sum of $9$ , same thing for the other two parallel sides.

Now we have $4$ parallel sides $amc真题$ . thinking about $4$ endpoints number need to have a sum of $18$ . It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.

So if we fix one direction $1-8 ($ or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $amc数学竞赛$ or $(8-1,2-7,3-6,5-4)$

Now, the problem is same as the problem to arrange $4$ points in a two-dimensional square. which is $\frac{4!}{4}$ = $amc美国数学竞赛$

Solution 2

Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to one, then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x<y<z.$

$3,7,8:$ Does not work. $4,6,8:$ Works. $5,6,7:$ Does not work. $5,6,8:$ Works. $5,7,8:$ Does not work. $6,7,8:$ Works.

So our answer is $amc竞赛$

Solution 3

We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$ , it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same face, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same face, the only way to get the sum is with $1$ and $2$ . This means that $6$ and $7$ are not on the same edge as $8$ , or in other words they are diagonally across from it on the same face, or on the other end of the cube.

Now we look at three cases, each yielding two solutions which are reflections of each other:

1) $6$ and $7$ are diagonally opposite $8$ on the same face. 2) $6$ is diagonally across the cube from $8$ , while $7$ is diagonally across from $8$ on the same face. 3) $7$ is diagonally across the cube from $8$ , while $6$ is diagonally across from $8$ on the same face.

This means the answer is $amc数学竞赛$

以上就是小编对AMC12数学竞赛试题及答案的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！