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AMC数学竞赛真题2016年12B 13-14

2018-09-07 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 13

Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?

$美国数学竞赛$

Solution

Let's set the altitude = z, distance from Alice to airplane's ground position (point right below airplane)=y and distance from Bob to airplane's ground position=x

From Alice's point of view, $\tan(\theta)=\frac{z}{y}$ . $\tan{30}=\frac{\sin{30}}{\cos{30}}=\frac{1}{\sqrt{3}}$ . So, $amc数学竞赛$

From Bob's point of view, $\tan(\theta)=\frac{z}{x}$ . $amc 真题$ . So, $x = \frac{z}{\sqrt{3}}$

We know that $x^2$ + $y^2$ = $10^2$

Solving the equation (by plugging in x and y), we get z= $\sqrt{30}$ = about 5.5.

So, answer is $E) 5.5$

solution by sudeepnarala

Solution 2

Non-trig solution by e_power_pi_times_i

Set the distance from Alice's and Bob's position to the point directly below the airplane to be $x$ and $y$ , respectively. From the Pythagorean Theorem, $x^2 + y^2 = 100$ . As both are $30-60-90$ triangles, the altitude of the airplane can be expressed as $\dfrac{x\sqrt{3}}{3}$ or $y\sqrt{3}$ . Solving the equation $\dfrac{x\sqrt{3}}{3} = y\sqrt{3}$ , we get $x = 3y$ . Plugging this into the equation $x^2 + y^2 = 100$ , we get $10y^2 = 100$ , or $y = \sqrt{10}$ ( $y$ cannot be negative), so the altitude is $\sqrt{3*10} = \sqrt{30}$ , which is closest to $\boxed{\textbf{E)}\ 5.5}$

Problem 14

The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$

$amc真题$

Solution

The second term in a geometric series is $a_2 = a \cdot r$ , where $r$ is the common ratio for the series and $a$ is the first term of the series. So we know that $a\cdot r = 1$ and we wish to find the minimum value of the infinite sum of the series. We know that: $S_\infty = \frac{a}{1-r}$ and substituting in $a=\frac{1}{r}$ , we get that $S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}$ . From here, you can either use calculus or AM-GM.

$\textbf{Calculus}$

Let $f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}$ , then $f'(x) = -(x-x^2)^{-2}\cdot (1-2x)$ . Since $f(0)$ and $f(1)$ are undefined $x \neq 0,1$ . This means that we only need to find where the derivative equals $0$ , meaning $amc竞赛$ . So $r = \frac{1}{2}$ , meaning that $S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4$

$\textbf{AM-GM}$

For 2 positive real numbers $a$ and $b$ , $\frac{a+b}{2} \geq \sqrt{ab}$ . Let $a = \frac{1}{r}$ and $b = \frac{1}{1-r}$ . Then: $\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}$ . This implies that $\frac{S_\infty}{2} \geq \sqrt{S_\infty}$ . or $S_\infty^2 \geq 4 \cdot S_\infty$ . Rearranging : $美国数学竞赛$ . Thus, the smallest value is $S_\infty = 4$ .

Solution 2

A geometric sequence always looks like

$a,ar,ar^2,ar^3,\dots$

and they say that the second term $ar=1$ . You should know that the sum of an infinite geometric series (denoted by $S$ here) is $\frac{a}{1-r}$ . We now have a system of equations which allows us to find $S$ in one variable.

$\begin{align*} ar&=1 \\ S&=\frac{a}{1-r} \end{align*}$

$\textbf{Solving in terms of \textit{a} then graphing}$

$S=\frac{a^2}{a-1}$

We seek the smallest positive value of $S$ . We proceed by graphing in the $aS$ plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is $\boxed{\textbf{(E)}\ 4}.$

$\textbf{Solving in terms of \textit{r} then graphing}$

$美国数学竞赛$

We seek the smallest positive value of $S$ . We proceed by graphing in the $rS$ plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is $\boxed{\textbf{(E)}\ 4}.$

$\textbf{Solving in terms of \textit{a} then doing some calculus}$

$S=\frac{a^2}{a-1}$

We seek the smallest positive value of $S$ . $\frac{(a-2)a}{(a-1)^2}=S'$ and $\frac{(a-2)a}{(a-1)^2}=0$ at $a=0$ and $a=2$ . $\frac{2}{(a-1)^3}=S''$ and $\frac{2}{(0-1)^3}$ is negative (implying a relative maximum occurs at $a=0$ ) and $\frac{2}{(2-1)^3}$ is positive (implying a relative minimum occurs at $a=2$ ). At $a=2$ , $S=4$ . Since this a competition math problem with an answer, we know this relative minimum must also be the absolute minimum among the "positive parts" of $S$ and that our answer is indeed $\boxed{\textbf{(E)}\ 4}.$ However, to be sure of this outside of this cop-out, one can analyze the end behavior of $S$ , how $S$ behaves at its asymptotes, and the locations of its maxima and minima relative to the asymptotes to be sure that 4 is the absolute minimum among the "positive parts" of $S$ .

$\textbf{Solving in terms of \textit{r} then being clever}$

$S=\frac{1}{-r^2+r}$

We seek the smallest positive value of $S$ . We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at $x=\frac{1}{2}$ and $\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}$ .

Solution 3

$\textbf{Completing the Square and Quadratics}$ Let $r$ be the common ratio. If the second term is $1$ , the first must be $\frac{1}{r}$ . By the infinite geometric series formula, the sum must be $S=\frac{\frac{1}{r}}{1-r}$ This equals $\frac{1}{r(1-r)}$ . To find the minimum value of S, we must find the maximum value of the denominator, $r(1-r)$ , which is $\frac{1}{4}$ , completing the square. Thus, the minimum value of $S$ is $\boxed{\textbf{(E)}\ 4}$ .