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AMC数学竞赛真题2016年12A 9-10

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 9

The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What is $a+b$ ?

$amc真题$

$amc数学竞赛$

Solution

Let $s$ be the side length of the small squares.

The diagonal of the big square can be written in two ways: $\sqrt{2}$ and $s \sqrt{2} + s + s \sqrt{2}$ .

Solving for $s$ , we get $s = \frac{4 - \sqrt{2}}{7}$ , so our answer is $美国数学竞赛$

Problem 10

Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?

$amc12$

Solution 1

Bash: we see that the following configuration works.

Bea - Ada - Ceci - Dee - Edie

After moving, it becomes

Ada - Ceci - Bea - Edie - Dee.

Thus, Ada was in seat $\boxed{2}$ .

Solution 2

Process of elimination of possible configurations.

Let's say that Ada= $A$ , Bea= $B$ , Ceci= $C$ , Dee= $D$ , and Edie= $E$ . Since $B$ moved more to the right than $C$ did left, this implies that $B$ was in a LEFT end seat originally: $B,-,C\rightarrow -,C,B$

This is affirmed because $DE\rightarrow ED$ , which there is no new seats uncovered. So $A,B,C$ are restricted to the same $1,2,3$ seats. Thus, it must be $B,A,C\rightarrow A,C,B$ , and more specifically: $amc10$

So $A$ , Ada, was originally in seat $\boxed{\textbf{(B)}\text{ 2}}$ .

Solution 3

The seats are numbered 1 through 5, so let each letter ( $A,B,C,D,E$ ) correspond to a number. Let a move to the left be subtraction and a move to the right be addition.

We know that $1+2+3+4+5=A+B+C+D+E=15$ . After everyone moves around, however, our equation looks like $(A+x)+B+2+C-1+D+E=15$ because $D$ and $E$ switched seats, $B$ moved two to the right, and $C$ moved 1 to the left.

For this equation to be true, $x$ has to be -1, meaning $A$ moves 1 left from her original seat. Since $A$ is now sitting in a corner seat, the only possible option for the original placement of $A$ is in seat number $美国数学竞赛$ .