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AMC数学竞赛真题2017年10A 11-12

2018-08-27 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，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。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

Problem 11

The region consisting of all points in three-dimensional space within 3 units of line segment $\overline{AB}$ has volume 216 $\pi$ . What is the length $\textit{AB}$ ?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution

In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$ . However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$ ):

$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$ , where $x$ is equal to the length of our line segment.

Solving, we find that $amc数学竞赛真题$ .

Problem 12

Let $S$ be a set of points $(x,y)$ in the coordinate plane such that two of the three quantities $3,~x+2,$ and $y-4$ are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description for $S?$

$amc10$

Solution

If the two equal values are $3$ and $x+2$ , then $x=1$ . Also, $y-4\le 3$ because 3 is the common value. Solving for $y$ , we get $y \le 7$ . Therefore the portion of the line $x=1$ where $y \le 7$ is part of $S$ . This is a ray with an endpoint of $(1, 7)$ .

Similar to the process above, we assume that the two equal values are $3$ and $y-4$ . Solving the equation $3=y-4$ then $y=7$ . Also, $x+2\le 3$ because 3 is the common value. Solving for $x$ , we get $x\le1$ . Therefore the portion of the line $y=7$ where $x\le 1$ is also part of $S$ . This is another ray with the same endpoint as the above ray: $(1, 7)$ .

If $x+2$ and $y-4$ are the two equal values, then $x+2=y-4$ . Solving the equation for $y$ , we get $y=x+6$ . Also $3\le y-4$ because $y-4$ is one way to express the common value. Solving for $y$ , we get $y\ge 7$ . Therefore the portion of the line $y=x+6$ where $y\ge 7$ is part of $S$ like the other two rays. The lowest possible value that can be achieved is also $(1, 7)$ .