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AMC数学竞赛真题2016年12A 24-25

2018-09-04 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 24

There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$ ?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution

Solution 1

The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value. Derive $f(x)$ so that the acceleration $f''(x)=0$ : $x^3-ax^2+bx-a\rightarrow 3x^2-2ax+b\rightarrow 6x-2a\rightarrow x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).

The function with the minimum $a$ :

$amc数学竞赛$ $x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}$ Since this is equal to the original equation $x^3-ax^2+bx-a$ ,

$amc竞赛$ $b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}$

The actual function: $f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}$

$f(x)=0\rightarrow x=\sqrt{3}$ triple root. "Complete the cube."

Solution 2

Note that since both $a$ and $b$ are positive, all 3 roots of the polynomial are positive as well.

Let the roots of the polynomial be $r, s, t$ . By Vieta's $美国数学竞赛$ and $a=rst$ .

Since $r, s, t$ are positive we can apply AM-GM to get $\frac{r+s+t}{3} \ge \sqrt[3]{rst} \rightarrow \frac{a}{3} \ge \sqrt[3]{a}$ . Cubing both sides and then dividing by $a$ (since $a$ is positive we can divide by $a$ and not change the sign of the inequality) yields $\frac{a^2}{27} \ge 1 \rightarrow a \ge 3\sqrt{3}$ .

Thus, the smallest possible value of $a$ is $3\sqrt{3}$ which is achieved when all the roots are equal to $\sqrt{3}$ . For this value of $a$ , we can use Vieta's to get $b=\boxed{\textbf{(B) }9}$ .

Solution 3

All three roots are identical. Therefore, comparing coefficients, the root of this cubic function is $\sqrt{3}$ . Using Vieta's, the coefficient we desire is the sum of the pairwise products of the roots. Because our root is unique, the answer is simply $b=\boxed{\textbf{(B) }9}$ .

Problem 25

Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$ , then the numbers that Bernardo writes are $16, 25, 36, 49, 64$ , and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6$ , and thus $f(1) = 5$ . What is the sum of the digits of $f(2) + f(4)+ f(6) + \dots + f(2016)$ ?

$amc真题$

Solution

Consider $f(2)$ . The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$ . Calculus $amc 真题数学竞赛$ and a bit of thinking says this first happens at $x\ge 100/2 = 50$ *. The perfect squares from here go: $2500, 2601, 2704, 2809\dots$ . Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$ . After the ones and tens make $100$ , the hundreds place will go up by $2$ , thus reaching our goal. Since $10^2=100$ , the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$ . The missing number is one less than the number of hundreds $(k=2)$ of $3600$ , or $35$ .

Now consider f(4). Instead of the difference between two squares needing to be $100$ , the difference must now be $10000$ . This first happens at $x\ge 5000$ . After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$ . This will take place at $\left(5000+100\right)^2=5100^2= 26010000$ . Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value.

In general, each new value of $f(k+2)$ will add two digits to the " $5$ " and one digit to the " $1$ ". This means that the last number Bernardo writes for $k=6$ is $\left(500000+1000\right)^2$ , the last for $k = 8$ will be $\left(50000000+10000\right)^2$ , and so on until $k=2016$ . Removing the last $k$ digits as Silvia does will be the same as removing $k/2$ trailing zeroes on the number to be squared. This means that the last number on the board for $k=6$ is $5001^2$ , $k=8$ is $50001^2$ , and so on. So the first missing number is $5001^2-1,50001^2-1\text{ etc.}$ The squaring will make a " $25$ " with two more digits than the last number, a " $10$ " with one more digit, and a " $1$ ". The missing number is one less than that, so the "1" will be subtracted from $f(k)$ . In other words, $f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}$ .

Therefore:

$f(2) =35 =25 +10$ $f(4) =2600 =2500 +100$ $f(6) =251000 =250000 +1000$ $amc12数学竞赛考试题$

And so on. The sum $f(2) + f(4) + f(6) +\dots + f(2016)$ is:

$2.52525252525\dots 2525\cdot 10^{2015}$ + $1.11111\dots 110\cdot 10^{1008}$ , with $2016$ repetitions each of " $25$ " and " $1$ ". There is no carrying in this addition. Therefore each $f(k)$ adds $2 + 5 + 1 = 8$ to the sum of the digits. Since $2n = 2016$ , $n = 1008$ , and $8n = 8064$ , or $\boxed{\textbf{(E)}\text{ 8064}}$ .

Addendum: *You could also use the fact that $(x+1)^2 = x^2 +2x+1$ In other words, the difference between $x^2$ and $(x+1)^2$ is equal to $2x+1$ . We can set the inequality $2x+1 \geq 100$ . Obviously, the first integer $x$ that satisfies this is 50. This way, while being longer, is IMO more motivated and doesn't use calculus.