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AMC数学竞赛真题2016年12B 11-12

2018-09-07 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 11

How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$

Solution

Solution by e_power_pi_times_i Revised by Kinglogic

$amc数学竞赛$ (red shows lattice points within the triangle)

If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1*\pi)$ , and the limit for the $x$ -value is $5$ squares. First we count the $1*1$ squares. In the back row, there are $12$ squares with length $1$ because $y=4*\pi$ generates squares from $(4,0)$ to $(4,4\pi)$ , and continuing on we have $9$ , $6$ , and $3$ for $x$ -values for $1$ , $2$ , and $3$ in the equation $y=\pi x$ . So there are $12+9+6+3 = 30$ squares with length $1$ in the figure. For $2*2$ squares, each square takes up $2$ units left and $2$ units up. Squares can also overlap. For $2*2$ squares, the back row stretches from $(3,0)$ to $(3,3\pi)$ , so there are $8$ squares with length $2$ in a $2$ by $9$ box. Repeating the process, the next row stretches from $(2,0)$ to $(2,2\pi)$ , so there are $5$ squares. Continuing and adding up in the end, there are $8+5+2=15$ squares with length $2$ in the figure. Squares with length $3$ in the back row start at $(2,0)$ and end at $(2,2\pi)$ , so there are $4$ such squares in the back row. As the front row starts at $(1,0)$ and ends at $(1,\pi)$ there are $4+1=5$ squares with length $3$ . As squares with length $4$ would not fit in the triangle, the answer is $30+15+5$ which is $\boxed{\textbf{D)}\ 50}$ .

Problem 12

All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?

$美国数学竞赛$

Solution

Solution by Mlux: Draw a $3\times3$ matrix. Notice that no adjacent numbers could be in the corners since two consecutive numbers must share an edge. Now find 4 nonconsecutive numbers that add up to $18$ . Trying $1+3+5+9 = 18$ works. Place each odd number in the corner in a clockwise order. Then fill in the spaces. There has to be a $2$ in between the $1$ and $3$ . There is a $4$ between $3$ and $5$ . The final grid should similar to this.

$amc真题$

$\textbf{(C)}7$ is in the middle.

Solution 2

If we color the square like a chessboard, since the numbers alternate between even and odd, and there are five odd numbers and four even numbers, the odd numbers must be in the corners/center, while the even numbers must be on the edges. Since the odd numbers add up to 25, and the numbers in the corners add up to 18, the number in the center must be $25 - 18 = 7$ .