AMC数学竞赛真题2016年12A 13-14

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  be a positive multiple of . One red ball and  green balls are arranged in a line in random order. Let  be the probability that at least  of the green balls are on the same side of the red ball. Observe that  and that  approaches  as  grows large. What is the sum of the digits of the least value of  such that ?

Solution 1

Let . Then, consider  blocks of  green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the  positions between the green balls to insert the red ball. Less than  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  balls, and there are  positions where this happens. Thus, , so

Multiplying both sides of the inequality by , we have

and by the distributive property,

Subtracting  on both sides of the inequality gives us

Therefore, , so the least possible value of  is . The sum of the digits of  is .

Solution 2 (Pattern Solution)

Let   ,   1 ()

Let   ,   

Let   ,   

Notice that the fraction can be written as   

Now it's quite simple to write the inequality as     

We can subtract  on both sides to obtain   

Dividing both sides by , we derive   . (Switch the inequality sign when dividing by )

We then cross multiply to get 

Finally we get 

To achieve 

So the sum of the digits of  = 

Solution 3

We are trying to find the number of places to put the red ball, such that  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 : Trying a few values, we see that the ball "works" in places  to and spaces  to . This is a total of  spaces, over a total possible  places to put the ball. So:

 And we know that the next value is what we are looking for, so , and the sum of it's digits is .

Problem 14

Each vertex of a cube is to be labeled with an integer  through , 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?

Solution 1

First of all, the adjacent faces have the same sum , because , , so now consider the  (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 , we should have , but no solution for , contradiction.

Now we know  and  must share the same side, which sum is , the  also must have sum of , same thing for the other two parallel sides.

Now we have  parallel sides . thinking about  endpoints number need to have a sum of . It is easy to notice only  and  would work.

So if we fix one direction or  all other  parallel sides must lay in one particular direction.  or 

Now, the problem is same as the problem to arrange  points in a two-dimensional square. which is =

Solution 2

Again, all faces sum to  If  are the vertices next to one, then the remaining vertices are  Now it remains to test possibilities. Note that we must have Without loss of generality, let 

 Does not work.  Works.  Does not work.  Works.  Does not work.  Works.

So our answer is 

Solution 3

We know the sum of each face is  If we look at an edge of the cube whose numbers sum to , it must be possible to achieve the sum  in two distinct ways, looking at the two faces which contain the edge. If  and  were on the same face, it is possible to achieve the desired sum only with the numbers  and  since the values must be distinct. Similarly, if  and  were on the same face, the only way to get the sum is with  and . This means that  and  are not on the same edge as , 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)  and  are diagonally opposite  on the same face. 2)  is diagonally across the cube from , while  is diagonally across from  on the same face. 3)  is diagonally across the cube from , while  is diagonally across from  on the same face.

This means the answer is 

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！