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AMC数学竞赛真题2016年B 22-23

2018-12-07 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 22

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 $A?$

$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

Solution

There are $21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:

$AMC数学竞赛$

But we know that every team beat exactly $10$ other teams, so for each possible $X$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $Y$ and $Z$ . Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles.

Thus the answer is $1330-945=385$ which is $\boxed{\textbf{(A)}}$ .

Problem 23

In regular hexagon $ABCDEF$ , points $W$ , $X$ , $Y$ , and $Z$ are chosen on sides $\overline{BC}$ , $\overline{CD}$ , $\overline{EF}$ , and $\overline{FA}$ respectively, so lines $AB$ , $ZW$ , $YX$ , and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$ ?

$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$

Solution 1

We draw a diagram to make our work easier:

$AMC竞赛辅导$

Assume that $AB$ is of length $1$ . Therefore, the area of $ABCDEF$ is $\frac{3\sqrt 3}2$ . To find the area of $WCXYFZ$ , we draw $\overline{CF}$ , and find the area of the trapezoids $WCFZ$ and $CXYF$ .

$[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label($

From this, we know that $CF=2$ . We also know that the combined heights of the trapezoids is $\frac{\sqrt 3}3$ , since $\overline{ZW}$ and $\overline{YX}$ are equally spaced, and the height of each of the trapezoids is $\frac{\sqrt 3}6$ . From this, we know $\overline{ZW}$ and $\overline{YX}$ are each $\frac 13$ of the way from $\overline{CF}$ to $\overline{DE}$ and $\overline{AB}$ , respectively. We know that these are both equal to $\frac 53$ .

We find the area of each of the trapezoids, which both happen to be $\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}$ , and the combined area is $\frac{11\sqrt 3}{18}^{*}$ .

We find that $\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}$ is equal to $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$ .

$^*$ At this point, you can answer $\textbf{(C)}$ and move on with your test.

Solution 2 (a lot faster than Solution 1)

$美国AMC8数学竞赛$

First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$ , and $9 \cdot 6 = 54$ small triangles in the whole hexagon.