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考题11-12 2015 AMC 12A

2018-08-22 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

On a sheet of paper, Isabella draws a circle of radius $2$ , a circle of radius $3$ , and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge 0$ lines. How many different values of $k$ are possible?

$amc数学竞赛官网$

Solution

Isabella can get $0$ lines if the circles are concentric, $1$ if internally tangent, $2$ if overlapping, $3$ if externally tangent, and $4$ if non-overlapping and not externally tangent. There are $amc数学竞赛官网$ values of $k$ .

Problem

The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$ . What is $a+b$ ?

$AMC12数学竞赛试题及答案$

Solution 12

Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by $4 - (-2)$ (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are $(2, 0), (-2, 0).$ Then $0 = 4a - 2 \rightarrow a = 0.5$ , and $0 = 4 - 4b \rightarrow b = 1.$ Then $a + b = \boxed{\textbf{(B)}\ 1.5}$ .

Alternative Solution

The parabolas must intersect the x-axis at the same two points for the kite to form. We find the x values at which they intersect by equating them and solving for x as shown below. $y = ax^2-2$ and $AMC12数学竞赛试题及答案$ or $-\sqrt{\dfrac{6}{a+b}}$ . The x-values of the y-intercepts is 0, so we plug in zero in each of them and get $-2$ and $4$ . The area of a kite is $\dfrac{d_1*d_2}{2}$ . The $d_1$ is $2+4 = 6$ . The $d_2$ is $2\sqrt{\dfrac{6}{a+b}}$ . Solving for the area $AMC12数学竞赛试题及答案$ , therefore $a + b = \boxed{\textbf{(B)}\ 1.5}$ .