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AMC数学竞赛真题2016年10A 11-12

2018-10-27 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 11

Find the area of the shaded region.

[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(

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

Solution 1

[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(

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.

[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(

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 $\boxed{\textbf{(D)}}$ .

Solution 3

Notice that we can graph this on the coordinate plane.

The top-left shaded figure has coordinates of $AMC数学竞赛$ .

Notice that we can apply the shoelace method to find the area of this polygon.

We find that the area of the polygon is $\frac{13}{4}$ .

However, notice that the two shaded regions are two congruent polygons.

Hence, the total area is $AMC真题$ or $\boxed{D}$ .

Problem 12

Three distinct integers are selected at random between $1$ and $2016$ , inclusive. Which of the following is a correct statement about the probability $p$ that the product of the three integers is odd?

$\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}$

Solution 1

For the product to be odd, all three factors have to be odd. The probability of this is $\frac{1008}{2016} \cdot \frac{1007}{2015} \cdot \frac{1006}{2014}$ .

$\frac{1008}{2016} = \frac{1}{2}$ , but $\frac{1007}{2015}$ and $\frac{1006}{2014}$ are slightly less than $\frac{1}{2}$ . Thus, the whole product is slightly less than $美国数学竞赛$ , so $\boxed{\textbf{(A) }p<\dfrac{1}{8}}$ .

Solution 2

For the product to be odd, all three factors have to be odd. There are a total of $\binom{2016}{3}$ ways to choose 3 numbers at random, and there are $\binom{1008}{3}$ to choose 3 odd numbers. Therefore, the probability of choosing 3 odd numbers is $\frac{\binom{1008}{3}}{\binom{2016}{3}}$ . Simplifying this, we obtain $\frac{1008*1007*1006}{2016*2015*2014}$ , which is slightly less than $\frac{1}{8}$ , so our answer is $\boxed{\textbf{(A) }p<\dfrac{1}{8}}$ .