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AMC数学竞赛真题2016年12A 22-23

2018-09-03 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 22

How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600$ and $\text{lcm}(y,z)=900$ ?

$amc 真题$

Solution 1

We prime factorize $72,600,$ and $900$ . The prime factorizations are $2^3\times 3^2$ , $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$ , respectively. Let $x=2^a\times 3^b\times 5^c$ , $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$ . We know that $\max(a,d)=3$ $\max(b,e)=2$ $\max(a,g)=3$ $\max(b,h)=1$ $\max(c,i)=2$ $\max(d,g)=2$ $amc美国数学竞赛试题$ and $c=f=0$ since $\text{lcm}(x,y)$ isn't a multiple of 5. Since $\max(d,g)=2$ we know that $a=3$ . We also know that since $\max(b,h)=1$ that $e=2$ . So now some equations have become useless to us...let's take them out. $\max(b,h)=1$ $\max(d,g)=2$ are the only two important ones left. We do casework on each now. If $\max(b,h)=1$ then $(b,h)=(1,0),(0,1)$ or $(1,1)$ . Similarly if $\max(d,g)=2$ then $(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)$ . Thus our answer is $5\times 3=\boxed{\textbf{(A)15}}$ .

Solution 2

It is well known that if the $\text{lcm}(a,b)=c$ and $c$ can be written as $p_1^ap_2^bp_3^c\dots$ , then the highest power of all prime numbers $p_1,p_2,p_3\dots$ must divide into either $a$ and/or $b$ . Or else a lower $c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots$ is the $\text{lcm}$ .

Start from $x$ : $\text{lcm}(x,y)=72$ so $8\mid x$ or $9\mid x$ or both. But $9\nmid x$ because $\text{lcm}(x,z)=600$ and $9\nmid 600$ . So $x=8,24$ .

$y$ can be $9,18,36$ in both cases of $x$ but NOT $72$ because $\text{lcm}{y,z}=900$ and $72\nmid 900$ .

So there are six sets of $x,y$ and we will list all possible values of $z$ based on those.

$25\mid z$ because $z$ must source all powers of $5$ . $z\in\{25,50,75,100,150,300\}$ . $z\ne\{200,225\}$ because of $\text{lcm}$ restrictions.

By different sourcing of powers of $2$ and $3$ ,

$澳洲amc数学竞赛$ $(8,18):z=300$ $(8,36):z=75,150,300$ $(24,9):z=100,300$ $美国数学竞赛amc官方$ $(24,36):z=25,50,75,100,150,300$

$z=100$ is "enabled" by $x$ sourcing the power of $3$ . $z=75,150$ is uncovered by $y$ sourcing all powers of $2$ . And $z=25,50$ is uncovered by $x$ and $y$ both at full power capacity.

Counting the cases, $美国数学竞赛amc10题$

Solution 3 (Less Casework!)

As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: $72=2^3\cdot 3^2,$ $600=2^3\cdot 3\cdot 5^2,$ $\text{and } 900=2^2\cdot 3^2\cdot 5^2$ To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges. $[\textbf{\emph{insert diagram here}}]$ Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$ . $美国amc数学竞赛报名$ Analyzing for powers of $2$ , it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$ . Turning towards the vertices $y \text{and} z$ , we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$ . Powers of $3$ . $[\textbf{\emph{insert diagram here for powers of 3}}]$ Using the same logic as we did for powers of $2$ , it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$ , we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$ . Uh oh, where da diagram? The final and last case is the powers of $5$ . $[\textbf{\emph{insert diagram here for powers of 5}}]$ This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.

Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$ , we get the answer: $5\cdot3\cdot1=\boxed{\textbf{(A) }15}$ .

Problem 23

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?

$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$

Solution

Solution 1: Super WLOG

WLOG assume $a$ is the largest. Scale the triangle to $1,{b}/{a},{c}/{a}$ or $1,x,y$ Then the solution is $\boxed{\textbf{(C)}\;1/2}$ (Insert graph with square of side length 1 and the line $x+y=1$ that cuts it in half)

Solution 2: Conditional Probability

WLOG, let the largest of the three numbers drawn be $a>0$ . Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ . The probability that their sum is greater than $a$ is $amc12数学竞赛$

Solution 3: Calculus

When $a>b$ , consider two cases:

1) $0<a<\frac{1}{2}$ , then $\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}$

2) $\frac{1}{2}<a<1$ , then $美国数学竞赛 amc$

$a<b$ is the same. Thus the answer is $\frac{1}{2}$ .

Solution 4: Geometry

The probability of this occurring is the volume of the corresponding region within a $1 \times 1 \times 1$ cube, where each point $(x,y,z)$ corresponds to a choice of values for each of $x, y,$ and $z$ . The region where, WLOG, side $z$ is too long, $z\geq x+y$ , is a pyramid with a base of area $\frac{1}{2}$ and height $1$ , so its volume is $\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}$ . Accounting for the corresponding cases in $x$ and $y$ multiplies our answer by $3$ , so we have excluded a total volume of $\frac{1}{2}$ from the space of possible probabilities. Subtracting this from $1$ leaves us with a final answer of $\frac{1}{2}$ .

Solution 5: More Calculus

The probability of this occurring is the volume of the corresponding region within a $1 \times 1 \times 1$ cube, where each point $(x,y,z)$ corresponds to a choice of values for each of $x, y,$ and $z$ . We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when $x + y < z$ , which has area $\frac{z^2}{2}$ or when $x+z<y$ or $y+z<x$ , which have an area of $\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.$ Integrating this expression from 0 to 1 in the form

$amc数学竞赛证书$

Solution 6: Geometry in 2-D

WLOG assume that $z$ is the largest number and hence the largest side. Then $x,y \leq z$ . We can set up a square that is $z$ by $z$ in the $xy$ plane. We are wanting all the points within this square that satisfy $x+y > z$ . This happens to be a line dividing the square into 2 equal regions. Thus the answer is $\frac{1}{2}$ .

[][] diagram for this problem goes here (z by z square)

Solution 7: More WLOG, Complementary Probability

The triangle inequality simplifies to considering only one case: $\text{the smallest side+ the second smallest side} > \text{the largest side}$ . Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) $a$ is the largest, so on average $a=1/2$ (now equal to becomes a degenerate case with probability $0$ , so we no longer need to consider it). We now want $b+c<1/2$ , so imagine choosing $b+c$ at once rather than independently. But we know that $b+c$ is between $0$ and $2$ . The complement is thus: $(1/2-0)/2=1/4$ . But keep in mind that we choose each $b$ and $c$ randomly and independently, so if there are $k$ ways to choose $b+c$ together, there are $2k$ ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if $b+c=3$ , then we only count this once, but in reality: we have two cases $1+2$ , and $2+1$ ; similar reasoning also generalizes to non-integral values). The complement is then actually $2(1/4)=1/2$ . Therefore, our desired probability is given by $1-\text{complement}=1/2, C$