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考题9-10 2015 AMC 12B

2018-08-06 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

Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is $\tfrac{1}{2}$ , independently of what has happened before. What is the probability that Larry wins the game?

$考题9-10 2015 AMC 12B$

Solution

Solution 1

If Larry wins, he either wins on the first move, or the third move, or the fifth move, etc. Let $W$ represent "player wins", and $L$ represent "player loses". Then the events corresponding to Larry winning are $W, LLW, LLLLW, LLLLLLW, \ldots$

Thus the probability of Larry winning is

$考题9-10 2015 AMC 12B$

This is a geometric series with ratio $\frac{1}{2^2}=\frac{1}{4}$ , hence the answer is $考题9-10 2015 AMC 12B$ .

Solution 2

Break the problem up into two separate cases: (a) Larry wins on the first throw or (b) Larry wins after the first throw.

a: The probability that Larry wins on the first throw is $\frac{1}{2}$ .

b: The probability that Larry wins after the first throw is half the probability that Julius wins because it only occurs half the time. This probability is $\frac{1}{2}(1-x)$ , where $x$ is the probability that Larry wins.

Therefore, $x = \frac{1}{2} + \frac{1}{2}(1 - x)$ . This equation can be solved for $x$ to find that the probability that Larry wins is $\boxed{\textbf{(C)}\; \frac{2}{3}}$ .

Problem 10

How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?

$考题9-10 2015 AMC 12B$

Solution

Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths $(a,b,c)$ with $a<b<c$ . Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$ .

There are no triangles when $a = 1$ because then $c$ must be less than $b+1$ , implying that $b \geq c$ , contrary to $b < c$ .

When $a=2$ , similar to above, $c$ must be less than $b+2$ , so this leaves the only possibility $c = b+1$ . This gives 3 triangles $(2,3,4), (2,4,5), (2,5,6)$ within our perimeter constraint.

When $a=3$ , $c$ can be $b+1$ or $b+2$ , which gives triangles $(3,4,5), (3,4,6), (3,5,6)$ . Note that $(3,4,5)$ is a right triangle, so we get rid of it and we get only 2 triangles.