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考题23-24 2015 AMC 12A

2018-08-22 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 23

Let $S$ be a square of side length 1. Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\frac12$ is $\frac{a-b\pi}{c}$ , where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a+b+c$ ?

$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63$

Solution

Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the 5 segments not bordering the bottom-left segment will be distance at least $\dfrac{1}{2}$ apart from $A$ . Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least 0.5 apart from $A$ is $\dfrac{0 + 1}{2} = \dfrac{1}{2}$ because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)

If the second point $B$ is on the left-bottom segment, then if $A$ is distance $x$ away from the left-bottom vertex, then $B$ must be at least $\dfrac{1}{2} - \sqrt{0.25 - x^2}$ away from that same vertex. Thus, using an averaging argument we find that the probability in this case is $\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left(\frac{1}{4} - \frac{\pi}{16}\right) = 1 - \frac{\pi}{4}.$

(Alternatively, one can equate the problem to finding all valid $(x, y)$ with $0 < x, y < \dfrac{1}{2}$ such that $x^2 + y^2 \ge \dfrac{1}{4}$ , i.e. (x, y) is outside the unit circle with radius 0.5.)

Thus, averaging the probabilities gives $P = \frac{1}{8} \left(5 + \frac{1}{2} + 1 - \frac{\pi}{4}\right) = \frac{1}{32} (26 - \pi).$

Our answer is $\textbf{(A)}$ .

Case Solution

Fix one of the points on a SIDE. There are three cases: the other point is on the same side, a peripheral side, or the opposite side, with probability $0.25, 0.5, 0.25$ , respectively.

Opposite side: Probability is obviously $1$ , no matter what.

Same side: Pretend the points are on a line with coordinates $x$ and $y$ . If $amc数学竞赛试题及答案$ , drawing a graph will give probability $\frac{1}{4}$ .

Peripheral side: superimpose a coordinate system over the points; call them $(x, 0)$ and $(0, y)$ . WLOG set $x, y >= 0$ and $x, y <= 1$ . We need $0.25x^2+0.25y^2>0.25$ , and drawing the coordinate system with bounds $(0, 0), (1, 0), (0, 1), (1, 1)$ gives probability $1-\frac{\pi}{16}$ that the distance between the points is $>0.25$ .

Adding these up and finding the fraction gives us $\frac{1}{32} (26 - \pi)$ for an answer of $\boxed{A}$ .

Problem 24

Rational numbers $a$ and $b$ are chosen at random among all rational numbers in the interval $[0,2)$ that can be written as fractions $\frac{n}{d}$ where $n$ and $d$ are integers with $1 \le d \le 5$ . What is the probability that $amc数学竞赛官网$ is a real number?

$amc数学竞赛试题及答案$

Solution

Let $\cos(a\pi) = x$ and $\sin(b\pi) = y$ . Consider the binomial expansion of the expression: $x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.$

We notice that the only terms with $i$ are the second and the fourth terms. Thus for the expression to be a real number, either $\cos(a\pi)$ or $\sin(b\pi)$ must be $0$ , or the second term and the fourth term cancel each other out (because in the fourth term, you have $i^2 = -1$ ).

$\text{Case~1:}$ Either $\cos(a\pi)$ or $\sin(b\pi)$ is $0$ .

The two $\text{a's}$ satisfying this are $\tfrac{1}{2}$ and $\tfrac{3}{2}$ , and the two $\text{b's}$ satisfying this are $0$ and $1$ . Because $a$ and $b$ can both be expressed as fractions with a denominator less than or equal to $5$ , there are a total of $20$ possible values for $a$ and $b$ :

$0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},$ $\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},$ $\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},$ $\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.$

Calculating the total number of sets of $(a,b)$ results in $20 \cdot 20 = 400$ sets. Calculating the total number of invalid sets (sets where $a$ doesn't equal $\tfrac{1}{2}$ or $\tfrac{3}{2}$ and $b$ doesn't equal $0$ or $1$ ), resulting in $(20-2) \cdot (20-2) = 324$ .

Thus the number of valid sets is $76$ .

$\text{Case~2}$ : The two terms cancel.

We then have:

$amc数学竞赛官网$

So:

$\cos^2(a\pi) = \sin^2(b\pi),$

which means for a given value of $\cos(a\pi)$ or $\sin(b\pi)$ , there are $4$ valid values(one in each quadrant).

When either $\cos(a\pi)$ or $\sin(b\pi)$ are equal to $1$ , however, there are only two corresponding values. We don't count the sets where either $\cos(a\pi)$ or $\sin(b\pi)$ equals $0$ , for we would get repeated sets. We also exclude values where the denominator is an odd number, for we cannot find any corresponding values(for example, if $a$ is $\tfrac{1}{5}$ , then $b$ must be $\tfrac{3}{10}$ , which we don't have). Thus the total number of sets for this case is $4 \cdot 4 + 2 \cdot 2 = 20$ .