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AMC数学竞赛8真题2016年 21-22

2018-08-28 重点归纳

AMC数学竞赛8专为8年级及以下的初中学生设计,但近年来的数据显示,越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中,而当这些学生能在成绩中取得“A”类标签,则是对孩子数学天赋的优势证明,不管是对美高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AMC 8的官方真题以及官方解答吧:

Problem 21

A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

amc数学竞赛

Solution 1

We put five chips randomly in order, and then pick the chips from the left to the right. To find the number of ways to rearrange the three red chips and two green chips, we solve for $\binom{5}{2} = 10$. However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. This means that the last chip must be green in all the situations that work. This means we are left with finding the number of ways to rearrange three red chips and one green chip, which is amc真题. Because a green chip will be last $4$ out of the $10$ situations, our answer is $\boxed{\textbf{(B) } \frac{2}{5}}$.

Solution 2

There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out $3$red chips, $3$ red chips and $1$ green chip, $2$ green chips, $2$ green chips and $1$ red chip, and $2$ green chips and $2$ red chips. Because order is important in this problem, there are $1+4+1+3+6=15$ ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which $15-5=10$. Out of the 10 ways to end the game, 4 of them ends with a red chip. The answer is $\frac{4}{10} = \frac{2}{5}$, or $\boxed{\textbf{(B) } \frac{2}{5}}$.

Problem 22

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA$. What is the area of the "bat wings" (shaded area)?amc竞赛

$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$

Solution

The area of trapezoid $CBFE$ is $\frac{1+3}2\cdot 4=8$. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one is $3,$ while the height of the smaller one is $1.$ Thus, their areas are $\frac12$ and $\frac92$. Subtracting these areas from the trapezoid, we get amc数学竞赛. Therefore, the answer to this problem is $\boxed{\textbf{(C) }3}$

Solution 2

Setting coordinates!

Let $E=(0,0)$$F=(3,0)$

amc竞赛

Now, we easily discover that line $CF$ has lattice coordinates at $(1,4)$ and $(3,0)$. Hence, the slope of line $CF=-2$

Plugging in the rest of the coordinate points, we find that line $CF=-2x+6$

Doing the same process to line $BE$, we find that line $BE=2x$.

Hence, setting them equal to find the intersection point...

$y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3$.

Hence, we find that the intersection point is $(\frac{3}{2},3)$. Call it Z.

Now, we can see that

$E=(0,0)$

amc真题

$C=(1,4)$.

Shoelace!

Using the well known Shoelace Formula(https://en.m.wikipedia.org/wiki/Shoelace_formula), we find that the area of one of those small shaded triangles is $\frac{3}{2}$.

Now because there are two of them, we multiple that area by $2$ to get $\boxed{\textbf{(C) }3}$

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