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AMC数学竞赛真题2016年12B 9-10

2018-09-06 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

Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?

$美国数学竞赛$

Solution

By Albert471

To start, use algebra to determine the number of posts on each side. You have (the long sides count for $2$ because there are twice as many) $6x = 20 + 4$ (each corner is double counted so you must add $4$ ) Making the shorter end have $4$ , and the longer end have $8$ . $((8-1)*4)*((4-1)*4) = 28*12 = 336$ . Therefore, the answer is $amc数学竞赛$

Problem 10

A quadrilateral has vertices $P(a,b)$ , $Q(b,a)$ , $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$ ?

$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$

Solution

By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$ . SImplifying we get $美国数学竞赛真题$ . Thus $a+b$ and $a-b$ have to be a factor of 8. The only way for them to be factors of $8$ and remain integers is if $a+b = 4$ and $a-b = 2$ . So the answer is $\boxed{\textbf{(A)}\ 4}$

Solution by I_Dont_Do_Math

Solution 2

Solution by e_power_pi_times_i

By the Shoelace Theorem, the area of the quadrilateral is $2a^2 - 2b^2$ , so $amc数学竞赛12$ . Since $a$ and $b$ are integers, $a = 3$ and $b = 1$ , so $a + b = \boxed{\textbf{(A)}\ 4}$ .