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考题15-16 2017 AMC 8

2018-08-06 重点归纳

AMC 8数学竞赛专为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 15

In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.

[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label(

$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$

Solution

Notice that the $A$ is adjacent to $4$ $M$ s, each $M$ is adjacent to $3$ $C$ s, and each $C$ is adjacent to $2$ $8$ s. Thus, the answer is $4\cdot 3\cdot 2 = \boxed{\textbf{(D)}\ 24}.$

Solution 2

There are three different kinds of paths that are on this diagram. The first kind is when you directly count A, M, C in a straight line. The second is when you count A, turn left or right to get M, then go straight to count M and C. The third is the one where you start with A, move forward to count M, turn left or right to count C, then move straight again to get 8.

There are 8 paths for each kind of path, making for $8 \cdot 3=\boxed{24}$ paths.

Problem 16

In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$ ?

$[asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label($

$\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}$

Solution 1

Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that $x$ is the length of the line from B to D. So, the perimeter of $\triangle{ABD}$ would be $\overline{AD} + 4 + x$ , while the perimeter of $\triangle{ACD}$ would be $\overline{AD} + 3 + (5 - x)$ . Notice that we can find out $x$ from these two equations. We can find out that $x = 2$ , so that means that the area of $\triangle{ABD} = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}$

Solution 2

We know that the perimeters of the two small triangles are $3+CD+AD$ and $4+BD+AD$ . Setting both equal and using $BD+CD = 5$ , we have $BD = 2$ and $CD = 3$ . Now, we simply have to find the area of $\triangle ABD$ . Since $\frac{BD}{CD} = \frac{2}{3}$ , we must have $\frac{[ABD]}{[ACD]} = 2/3$ . Combining this with the fact that $[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6$ , we get $[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) } \frac{12}{5}}$

Solution 3

Since point $D$ is on line $BC$ , it will split it into $CD$ and $DB$ . Let $CD = 5 - x$ and $DB = x$ . Triangle $CAD$ has side lengths $3, 5 - x, AD$ and triangle $DAB$ has side lengths $x, 4, AD$ . Since both perimeters are equal, we have the equation $3 + 5 - x + AD = 4 + x + AD$ . Eliminating $AD$ and solving the resulting linear equation gives $x = 2$ . Draw a perpendicular from point $D$ to $AB$ . Call the point of intersection $F$ . Because angle $ABC$ is common to both triangles $DBF$ and $ABC$ , and both are right triangles, both are similar. The hypotenuse of triangle $DBF$ is 2, so the altitude must be $6/5$ Because $DBF$ and $ABD$ share the same altitude, the height of $ABD$ therefore must be $6/5$ . The base of $ABD$ is 4, so $[ABD] = \frac{1}{2} * 4 * \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}$

Solution 4

Using any preferred method, realize $BD = 2$ . Since we are given a 3-4-5 right triangle, we know the value of $\sin(\angle ABC) = \frac{3}{5}$ . Since we are given $AB = 4$ , apply the Sine Area Formula to get $\frac{1}{2} * 4 * 2 * \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}$ .