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考题21-22 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 21

Suppose $a$ , $b$ , and $c$ are nonzero real numbers, and $a+b+c=0$ . What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$ ?

$考题21-22 2017 AMC 8$

Solution 1

There are $2$ cases to consider:

Case $1$ : $2$ of $a$ , $b$ , and $c$ are positive and the other is negative. WLOG assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that

Case $2$ : $2$ of $a$ , $b$ , and $c$ are negative and the other is positive. WLOG assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that

In both cases, we get that the given expression equals $\boxed{\textbf{(A)}\ 0}$ .

Solution 2

Assuming numbers:

WLOG $a=1, b=2,$ and $c=-3$ (Other numbers can apply for $a, b,$ and $c$ as long as their sum is $0$ .) . Then plug $a, b,$ and $c$ into the given equation. The result is always $\boxed{\textbf{(A)}\ 0}$ .

Problem 22

In the right triangle $ABC$ , $AC=12$ , $BC=5$ , and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?

$考题21-22 2017 AMC 8$

Solution

We can reflect triangle $ABC$ over line $AC.$ This forms the triangle $AB'C$ and a circle out of the semicircle. Let us call the center of the circle $O.$

We can see that Circle $O$ is the incircle of $AB'C.$ We can use the formula for finding the radius of the incircle to solve this problem. The area of $AB'C$ is $12\times5 = 60.$ The semiperimeter is $5+13 = 18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$

Solution 2

We immediately see that $AB=13$ , and we label the center of the semicircle $O$ . Drawing radius $OD$ with length $x$ such that $OD$ is perpendicular to $AB$ , we immediately see that $ODB\cong OCB$ because of $\operatorname{HL}$ congruence, so $BD=5$ and $DA=8$ . By similar triangles $ODA$ and $BCA$ , we see that $\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{(D)}\ \frac{10}{3}}$ .

Solution 3

Let the center of the semicircle be $O$ . Let the point of tangency between line $AB$ and the semicircle be $F$ . Angle $BAC$ is common to triangles $ABC$ and $AFO$ . By tangent properties, angle $AFO$ must be $90$ degrees. Since both triangles $ABC$ and $AFO$ are right and share an angle, $AFO$ is similar to $ABC$ . The hypotenuse of $AFO$ is $12 - r$ , where $r$ is the radius of the circle. (See for yourself) The short leg of $AFO$ is $r$ . Because $AFO$ ~ $ABC$ , we have $r/(12 - r) = 5/13$ and solving gives $r = \boxed{\textbf{(D)}\ \frac{10}{3}}$