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考题11-12 2015 AMC 12B

2018-08-06 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 11

The line $12x+5y=60$ forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?

$考题11-12 2015 AMC 12B$

Solution

Clearly the line and the coordinate axes form a right triangle. Since the x-intercept and y-intercept are 5 and 12 respectively, 5 and 12 are two sides of the triangle that are not the hypotenuse, and are thus two of the three heights. In order to find the third height, we can use different equations of the area of the triangle. Using the lengths we know, the area of the triangle is $\tfrac{1}{2} \times 5 \times 12= 30$ . We can use the hypotenuse as another base to find the third height. Using the distance formula, the length of the hypotenuse is $\sqrt{5^2+12^2}=13$ . Then $\frac{1}{2} \times 13 \times h=30$ , and so $h = \frac{60}{13}$ . Therefore the sum of all the heights is $考题11-12 2015 AMC 12B$ .

Problem 12

Let $a$ , $b$ , and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$ ?

$考题11-12 2015 AMC 12B$

Solution 1

The left-hand side of the equation can be factored as $(x-b)(x-a+x-c) = (x-b)(2x-(a+c))$ , from which it follows that the roots of the equation are $x=b$ , and $x=\tfrac{a+c}{2}$ . The sum of the roots is therefore $b + \tfrac{a+c}{2}$ , and the maximum is achieved by choosing $b=9$ , and $\{a,c\}=\{7,8\}$ . Therefore the answer is $9 + \tfrac{7+8}{2} = 9 + 7.5 = \boxed{\textbf{(D)}\; 16.5}.$

Solution 2

Expand the polynomial. We get $(x-a)(x-b)+(x-b)(x-c)=x^2-(a+b)x+ab+x^2-(b+c)+bc=2x^2-(a+2b+c)x+(ab+bc).$

Now, consider a general quadratic equation $ax^2+bx+c=0.$ The two solutions to this are

The sum of these roots is

$\dfrac{-b}{a}.$

Therefore, reconsidering the polynomial of the problem, the sum of the roots is

Now, to maximize this, it is clear that $b=9.$ Also, we must have $a=8, b=7$ (or vice versa). The reason $a,b$ have to equal these values instead of larger values is because each of $a,b,c$ is distinct.