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考题13-14 2015 AMC 10B

2018-08-06 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的官方真题以及官方解答吧：

Problem 13

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?

$考题13-14 2015 AMC 10B$

Solution

We find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If $x=0$ , then $y=12$ . If $y$ is $0$ , then $x=5$ . Our three vertices are $(0,0)$ , $(5,0)$ , and $(0,12)$ . Two of our altitudes are $5$ and $12$ , and since it is a 5-12-13 right triangle, the hypotenuse is $13$ . Since the area of the triangle is $30$ , so our final altitude is $\frac{30(2)}{13}=\frac{60}{13}$ . The sum of our altitudes is $\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}$ . Note that there is no need to calculate the final answer after we know that the third altitude has length $\frac{60}{13}$ since $E$ is the only choice with a denominator of $13$ .

Problem 14

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 $(x-a)(x-b)+(x-b)(x-c)=0$ ?

$考题13-14 2015 AMC 10B$

Solution 1

Expanding the equation and combining like terms results in $2x^2-(a+2b+c)x+(ab+bc)=0$ . By Vieta's formula the sum of the roots is $考题13-14 2015 AMC 10B$ . To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$ . So let $b=9$ , $a=8$ , and $c=7$ . Then the answer is $\dfrac{8+18+7}{2}=\boxed{\textbf{(D)} 16.5}$ .

Solution 2

Factoring out $(x-b)$ from the equation yields $考题13-14 2015 AMC 10B$ . Therefore the roots are $b$ and $\frac{a+c}{2}$ . Because $b$ must be the larger root to maximize the sum of the roots, letting $a,b,$ and $c$ be $8,9,$ and $7$ respectively yields the sum $9+\frac{8+7}{2} = 9+7.5 = \boxed{\textbf{(D)}~16.5}$ .

Solution 3

There are 2 cases. Case 1 is that $(x-a)(x-b)=0$ and $(x-b)(x-c)=0$ . Lets test that 1st. If $x-b=0$ , the maximum value for $x$ and $b$ is $9$ . Then $b=9$ and $x=9.$ The next highest values are $7$ and $8$ so $a=8$ and $c=9$ . Therefore, $\frac{18+8+7}{2}= \boxed{\textbf{(D)}~16.5}$ .