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AMC数学竞赛真题2016年12A 7-8

2018-08-31 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 7

Which of these describes the graph of $x^2(x+y+1)=y^2(x+y+1)$ ?

$\textbf{(A)}\ \text{two parallel lines}\\ \qquad\textbf{(B)}\ \text{two intersecting lines}\\ \qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\\ \qquad\textbf{(D)}\ \text{three lines that do not all pass through a common point}\\ \qquad\textbf{(E)}\ \text{a line and a parabola}$

Solution 1

The equation $x^2(x+y+1)=y^2(x+y+1)$ tells us $x^2=y^2$ or $x+y+1=0$ . $x^2=y^2$ generates two lines $y=x$ and $y=-x$ . $x+y+1=0$ is another straight line. The only intersection of $y=x$ and $y=-x$ is $(0,0)$ , which is not on $x+y+1=0$ . Therefore, the graph is three lines that do not have a common intersection, or $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$

Solution 2

If $x+y+1\neq0$ , then dividing both sides of the equation by $x+y+1$ gives us $x^2=y^2$ . Rearranging and factoring, we get $x^2-y^2=(x+y)(x-y)=0$ . If $x+y+1=0$ , then the equation is satisfied. Thus either $x+y=0$ , $x-y=0$ , or $x+y+1=0$ . These equations can be rearranged into the lines $y=-x$ , $y=x$ , and $y=-x-1$ , respectively. Since these three lines are distinct, the answer is $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$ .

Solution 3

Subtract $y^2(x+y+1)$ on both sides of the equation to get $x^2(x+y+1)-y^2(x+y+1)=0$ . Factoring $x+y+1$ gives us $(x+y+1)(x^2-y^2)=(x+y+1)(x+y)(x-y)=0$ , so either $x+y+1=0$ , $x+y=0$ , or $x-y=0$ . Continue on with the second half of solution 2.

Diagram:

$AB: y=x$

$CD: y=-x$

$EF: x+y+1=0$

$amc真题$

Problem 8

Find the area of the shaded region.

$amc数学竞赛$

$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$

Solution 1

$美国数学竞赛$

The bases of these triangles are all $1$ , and their heights are $4$ , $\frac{5}{2}$ , $4$ , and $\frac{5}{2}$ . Thus, their areas are $2$ , $\frac{5}{4}$ , $2$ , and $\frac{5}{4}$ , which add to the area of the shaded region, which is $\boxed{6\frac{1}{2}}$ .

Solution 2

Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. We can do this by splitting up the unshaded areas into various triangles and rectangles as shown.

$美国数学竞赛真题$

Notice that the two added lines bisect each of the $4$ sides of the large rectangle.

Subtracting the unshaded area from the total area gives us $40-33\frac{1}{2}=\boxed{6\frac{1}{2}}$ , so the correct answer is $\boxed{\textbf{(D)}}$ .