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AMC数学竞赛真题2016年12B 20-21

2018-09-11 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：

AMC真题20

A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ , $B$ beat $C$ , and $C$ beat $A?$

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AMC真题解答

We use complementary counting. Firstly, because each team played $20$ other teams, there are $21$ teams total. All sets that do not have $A$ beat $B$ , $B$ beat $C$ , and $C$ beat $A$ have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.

There are $21$ ways to choose the team that beat the two other teams, and $美国数学竞赛AMC真题$ to choose two teams that the first team both beat. This is $21 * 45 = 945$ sets. There are $\binom{21}{3} = 1330$ sets of three teams total. Subtracting, we obtain $1330 - 945 = \boxed{385}$ , $A$ as our answer.

AMC真题 21

Let $ABCD$ be a unit square. Let $Q_1$ be the midpoint of $\overline{CD}$ . For $i=1,2,\dots,$ let $P_i$ be the intersection of $\overline{AQ_i}$ and $\overline{BD}$ , and let $Q_{i+1}$ be the foot of the perpendicular from $P_i$ to $\overline{CD}$ . What is $amc数学竞赛$

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AMC真题解答

(By Qwertazertl)

We are tasked with finding the sum of the areas of every $\triangle DQ_i^{}P_i^{}$ where $i$ is a positive integer. We can start by finding the area of the first triangle, $\triangle DQ_1^{}P_1^{}$ . This is equal to $\frac{1}{2}$ ⋅ $DQ_1^{}$ ⋅ $P_1^{}Q_2^{}$ . Notice that since triangle $\triangle DQ_1^{}P_1^{}$ is similar to triangle $\triangle ABP_1^{}$ in a 1 : 2 ratio, $P_1^{}Q_2^{}$ must equal $\frac{1}{3}$ (since we are dealing with a unit square whose side lengths are 1). $DQ_1^{}$ is of course equal to $\frac{1}{2}$ as it is the mid-point of CD. Thus, the area of the first triangle is $\frac{1}{2}$ ⋅ $\frac{1}{2}$ ⋅ $\frac{1}{3}$ .

The second triangle has a base $DQ_2^{}$ equal to that of $P_1^{}Q_2^{}$ (see that $\triangle DQ_2^{}P_1^{}$ ~ $\triangle DCB$ ) and using the same similar triangle logic as with the first triangle, we find the area to be $\frac{1}{2}$ ⋅ $\frac{1}{3}$ ⋅ $\frac{1}{4}$ . If we continue and test the next few triangles, we will find that the sum of all $\triangle DQ_i^{}P_i^{}$ is equal to $\frac{1}{2} \sum\limits_{n=2}^\infty \frac{1}{n(n+1)}$ or $\frac{1}{2} \sum\limits_{n=2}^\infty \frac{1}{n} - \frac{1}{n+1}$

This is known as a telescoping series because we can see that every term after the first $\frac{1}{n}$ is going to cancel out. Thus, the the summation is equal to $\frac{1}{2}$ and after multiplying by the half out in front, we find that the answer is $\boxed{\textbf{(B) }\frac{1}{4}}$ .

Solution 2

(By mastermind.hk16)

Note that $AD \|\ P_iQ_{i+1}\ \forall i \in \mathbb{N}$ . So $amc真题美国数学竞赛$

Hence $\frac{Q_iQ_{i+1}}{DQ_{i}}=\frac{P_{i}Q_{i+1}}{AD} \ \ \Longrightarrow DQ_i \cdot P_iQ_{i+1}=Q_iQ_{i+1}$

We compute $\frac{1}{2} \sum_{i=1}^{\infty}DQ_i \cdot P_iQ_{i+1}= \frac{1}{2} \sum_{i=1}^{\infty}Q_iQ_{i+1}=\frac{1}{2} \cdot DQ_1 =\frac{1}{4}$ because $Q_i \rightarrow D$ as $i \rightarrow \infty$ .