Home » 国际竞赛 » 数学国际竞赛 » AMC美国数学竞赛 » Details

2002 AMC12B 真题及答案详细解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月5日上午10:19

2002 AMC 12B 真题

答案详细解析请参考文末

Problem 1

The arithmetic mean of the nine numbers in the set ${9, 99, 999, 9999, ldots, 999999999}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ does not contain the digit

$mathrm{(A)} 0 qquadmathrm{(B)} 2 qquadmathrm{(C)} 4 qquadmathrm{(D)} 6 qquadmathrm{(E)} 8$

Problem 2

What is the value of $[(3x - 2)(4x + 1) - (3x - 2)4x + 1]$

when $x=4$ ?

$mathrm{(A)} 0 qquadmathrm{(B)} 1 qquadmathrm{(C)} 10 qquadmathrm{(D)} 11 qquadmathrm{(E)} 12$

Problem 3

For how many positive integers $n$ is $n^2 - 3n + 2$ a prime number?

$mathrm{(A)} text{none} qquadmathrm{(B)} text{one} qquadmathrm{(C)} text{two} qquadmathrm{(D)} text{more than two, but finitely many} qquadmathrm{(E)} text{infinitely many}$

Problem 4

Let $n$ be a positive integer such that $frac 12 + frac 13 + frac 17 + frac 1n$ is an integer. Which of the following statements is not true:

$mathrm{(A)} 2 text{divides }n qquadmathrm{(B)} 3 text{divides }n qquadmathrm{(C)} 6 text{divides }n qquadmathrm{(D)} 7 text{divides }n qquadmathrm{(E)} {n > 84}$

Problem 5

Let $v, w, x, y,$ and $z$ be the degree measures of the five angles of a pentagon. Suppose that $v < w < x < y < z$ and $v, w, x, y,$ and $z$ form an arithmetic sequence. Find the value of $x$ .

$mathrm{(A)} 72 qquadmathrm{(B)} 84 qquadmathrm{(C)} 90 qquadmathrm{(D)} 108 qquadmathrm{(E)} 120$

Problem 6

Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2 + ax + b = 0$ has solutions $a$ and $b$ . Then the pair $(a,b)$ is

$mathrm{(A)} (-2,1) qquadmathrm{(B)} (-1,2) qquadmathrm{(C)} (1,-2) qquadmathrm{(D)} (2,-1) qquadmathrm{(E)} (4,4)$

Problem 7

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?

$mathrm{(A)} 50 qquadmathrm{(B)} 77 qquadmathrm{(C)} 110 qquadmathrm{(D)} 149 qquadmathrm{(E)} 194$

Problem 8

Suppose July of year $N$ has five Mondays. Which of the following must occur five times in August of year $N$ ? (Note: Both months have 31 days.)

$mathrm{(A)} text{Monday} qquadmathrm{(B)} text{Tuesday} qquadmathrm{(C)} text{Wednesday} qquadmathrm{(D)} text{Thursday} qquadmathrm{(E)} text{Friday}$

Problem 9

If $a,b,c,d$ are positive real numbers such that $a,b,c,d$ form an increasing arithmetic sequence and $a,b,d$ form a geometric sequence, then $frac ad$ is

$mathrm{(A)} frac 1{12} qquadmathrm{(B)} frac 16 qquadmathrm{(C)} frac 14 qquadmathrm{(D)} frac 13 qquadmathrm{(E)} frac 12$

Problem 10

How many different integers can be expressed as the sum of three distinct members of the set ${1,4,7,10,13,16,19}$ ?

$mathrm{(A)} 13 qquadmathrm{(B)} 16 qquadmathrm{(C)} 24 qquadmathrm{(D)} 30 qquadmathrm{(E)} 35$

Problem 11

The positive integers $A, B, A-B,$ and $A+B$ are all prime numbers. The sum of these four primes is

$mathrm{(A)} mathrm{even} qquadmathrm{(B)} mathrm{divisible by }3 qquadmathrm{(C)} mathrm{divisible by }5 qquadmathrm{(D)} mathrm{divisible by }7 qquadmathrm{(E)} mathrm{prime}$

Problem 12

For how many integers $n$ is $dfrac n{20-n}$ the square of an integer?

$mathrm{(A)} 1 qquadmathrm{(B)} 2 qquadmathrm{(C)} 3 qquadmathrm{(D)} 4 qquadmathrm{(E)} 10$

Problem 13

The sum of $18$ consecutive positive integers is a perfect square. The smallest possible value of this sum is

$mathrm{(A)} 169 qquadmathrm{(B)} 225 qquadmathrm{(C)} 289 qquadmathrm{(D)} 361 qquadmathrm{(E)} 441$

Problem 14

Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

$mathrm{(A)} 8 qquadmathrm{(B)} 9 qquadmathrm{(C)} 10 qquadmathrm{(D)} 12 qquadmathrm{(E)} 16$

Problem 15

How many four-digit numbers $N$ have the property that the three-digit number obtained by removing the leftmost digit is one ninth of $N$ ?

$mathrm{(A)} 4 qquadmathrm{(B)} 5 qquadmathrm{(C)} 6 qquadmathrm{(D)} 7 qquadmathrm{(E)} 8$

Problem 16

Juan rolls a fair regular octahedral die marked with the numbers $1$ through $8$ . Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of 3?

$mathrm{(A)} frac1{12} qquadmathrm{(B)} frac 13 qquadmathrm{(C)} frac 12 qquadmathrm{(D)} frac 7{12} qquadmathrm{(E)} frac 23$

Problem 17

Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?

$mathrm{(A)} text{Andy} qquadmathrm{(B)} text{Beth} qquadmathrm{(C)} text{Carlos} qquadmathrm{(D)} text{Andy and Carlos tie for first.} qquadmathrm{(E)} text{All three tie.}$

Problem 18

A point $P$ is randomly selected from the rectangular region with vertices $(0,0),(2,0),(2,1),(0,1)$ . What is the probability that $P$ is closer to the origin than it is to the point $(3,1)$ ?

$mathrm{(A)} frac 12 qquadmathrm{(B)} frac 23 qquadmathrm{(C)} frac 34 qquadmathrm{(D)} frac 45 qquadmathrm{(E)} 1$

Problem 19

If $a,b,$ and $c$ are positive real numbers such that $a(b+c) = 152, b(c+a) = 162,$ and $c(a+b) = 170$ , then $abc$ is

$mathrm{(A)} 672 qquadmathrm{(B)} 688 qquadmathrm{(C)} 704 qquadmathrm{(D)} 720 qquadmathrm{(E)} 750$

Problem 20

Let $triangle XOY$ be a right-angled triangle with $mangle XOY = 90^{circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$ .

$mathrm{(A)} 24 qquadmathrm{(B)} 26 qquadmathrm{(C)} 28 qquadmathrm{(D)} 30 qquadmathrm{(E)} 32$

Problem 21

For all positive integers $n$ less than $2002$ , let

$begin{eqnarray*} a_n =left{ begin{array}{lr} 11, & text{if }n text{is divisible by }13 text{and }14; 13, & text{if }n text{is divisible by }14 text{and }11; 14, & text{if }n text{is divisible by }11 text{and }13; 0, & text{otherwise}. end{array} right. end{eqnarray*}$

Calculate $sum_{n=1}^{2001} a_n$ .

$mathrm{(A)} 448 qquadmathrm{(B)} 486 qquadmathrm{(C)} 1560 qquadmathrm{(D)} 2001 qquadmathrm{(E)} 2002$

Problem 22

For all integers $n$ greater than $1$ , define $a_n = frac{1}{log_n 2002}$ . Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$ . Then $b- c$ equals

$mathrm{(A)} -2 qquadmathrm{(B)} -1 qquadmathrm{(C)} frac{1}{2002} qquadmathrm{(D)} frac{1}{1001} qquadmathrm{(E)} frac 12$

Problem 23

In $triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $overline{BC}$ and the median from $A$ to $overline{BC}$ have the same length. What is $BC$ ?

$mathrm{(A)} frac{1+sqrt{2}}{2} qquadmathrm{(B)} frac{1+sqrt{3}}2 qquadmathrm{(C)} sqrt{2} qquadmathrm{(D)} frac 32 qquadmathrm{(E)} sqrt{3}$

Problem 24

A convex quadrilateral $ABCD$ with area $2002$ contains a point $P$ in its interior such that $PA = 24, PB = 32, PC = 28, PD = 45$ . Find the perimeter of $ABCD$ .

$mathrm{(A)} 4sqrt{2002} qquadmathrm{(B)} 2sqrt{8465} qquadmathrm{(C)} 2(48+$ $sqrt{2002}) qquadmathrm{(D)} 2sqrt{8633} qquadmathrm{(E)} 4(36 + sqrt{113})$

Problem 25

Let $f(x) = x^2 + 6x + 1$ , and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that $[f(x) + f(y) le 0 qquad text{and} qquad f(x)-f(y) le 0]$ The area of $R$ is closest to $mathrm{(A)} 21 qquadmathrm{(B)} 22 qquadmathrm{(C)} 23 qquadmathrm{(D)} 24 qquadmathrm{(E)} 25$

2002 AMC12 B 真题答案详细解析

Solution 1
We wish to find $frac{9+99+cdots +999999999}{9}$ , or $frac{9(1+11+111+cdots +111111111)}{9}=123456789$ . This does not have the digit 0, so the answer is $boxed{mathrm{(A)} 0}$ Solution 2
Notice that the final number is guaranteed to have the digits ${1, 3, 5, 7, 9}$ and that each of these digits can be paired with an even number adding up to 9. $boxed{mathrm{(A)} 0}$ can be taken out, with the other digits fulfilling divisibility by 9.
By the distributive property, $(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = boxed{mathrm{(D)} 11}$ .
Solution 1

Factoring, we get $n^2 - 3n + 2 = (n-2)(n-1)$ . Either $n-1$ or $n-2$ is odd, and the other is even. Their product must yield an even number. The only prime that is even is $2$ , which is when $n$ is $3$ . The answer is $boxed{mathrm{(B)} text{one}}$ .

Solution 2

Considering parity, we see that $n^2 - 3n + 2$ is always even. The only even prime is $2$ , and so $n^2-3n=0$ whence $n=3Rightarrowboxed{mathrm{(B)} text{one}}$ .
Since $frac 12 + frac 13 + frac 17 = frac {41}{42}$ , $[0 < lim_{n rightarrow infty} left(frac{41}{42} + frac{1}{n}right) < frac {41}{42} + frac 1n < frac{41}{42} + frac 11 < 2]$ From which it follows that $frac{41}{42} + frac 1n = 1$ and $n = 42$ . The only answer choice that is not true is $boxed{mathrm{(E)} n>84}$ .
Solution 1

The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into $5- 2 = 3$ triangles) is $3 cdot 180 = 540^{circ}$ . If we let $v = x - 2d, w = x - d, y = x + d, z = x+2d$ , it follows that

$[(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 Longrightarrow x = 108 mathrm{(D)}]$

Note that since $x$ is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence.

You can always assume the values are the same so $frac{540}{5}=108$

Solution 2

Let $v$ , $w$ , $x$ , $y$ , $z$ be $v$ , $v + d$ , $v+2d$ , $v+3d$ , $v+4d$ , respectively. Then we have $[v + w + x + y + z = 5v + 10d = 180^{circ} (5 - 2) = 540^{circ}]$ Dividing the equation by $5$ , we have $[v + 2d = x = 108^{circ} mathrm {(D)}]$
Solution 1

Since $(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0$ , it follows by comparing coefficients that $-a - b = a$ and that $ab = b$ . Since $b$ is nonzero, $a = 1$ , and $-1 - b = 1 Longrightarrow b = -2$ . Thus $(a,b) = boxed{mathrm{(C)} (1,-2)}$ .

Solution 2

Another method is to use Vieta's formulas. The sum of the solutions to this polynomial is equal to the opposite of the $x$ coefficient, since the leading coefficient is 1; in other words, $a + b = -a$ and the product of the solutions is equal to the constant term (i.e, $a*b = b$ ). Since $b$ is nonzero, it follows that $a = 1$ and therefore (from the first equation), $b = -2a = -2$ . Hence, $(a,b) = boxed{mathrm{(C)} (1,-2)}$

Solution 3 (Using the Answer Choices)

Note that for roots $a$ and $b$ , $ab = b$ . This implies that $a$ is $1$ , and there is only one answer choice with $1$ in the position for $a$ , hence, $(a,b) = boxed{mathrm{(C)} (1,-2)}$
Let the three consecutive positive integers be $a-1$ , $a$ , and $a+1$ . So, $a(a-1)(a+1)=24aRightarrow(a-1)(a+1)=24$ . $24=4times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=boxed{mathrm{ (B)} 77}$ .
If there are five Mondays, there are only three possibilities for their dates: $(1,8,15,22,29)$ , $(2,9,16,23,30)$ , and $(3,10,17,24,31)$ .In the first case August starts on a Thursday, and there are five Thursdays, Fridays, and Saturdays in August.In the second case August starts on a Wednesday, and there are five Wednesdays, Thursdays, and Fridays in August.In the third case August starts on a Tuesday, and there are five Tuesdays, Wednesdays, and Thursdays in August.The only day of the week that is guaranteed to appear five times is therefore $boxed{textrm{(D)} text{Thursday}}$ .
Solution 1

We can let $a=1$ , $b=2$ , $c=3$ , and $d=4$ . $frac{a}{d}=boxed{frac{1}{4}} Longrightarrow mathrm{(C)}$

Solution 2

As $a, b, d$ is a geometric sequence, let $b=ka$ and $d=k^2a$ for some $k>0$ .

Now, $a, b, c, d$ is an arithmetic sequence. Its difference is $b-a=(k-1)a$ . Thus $d=a + 3(k-1)a = (3k-2)a$ .

Comparing the two expressions for $d$ we get $k^2=3k-2$ . The positive solution is $k=2$ , and $frac{a}{d}=frac{a}{k^2a}=frac{1}{k^2}=boxed{frac{1}{4}} Longrightarrow mathrm{(C)}$ .

Solution 3

Letting $n$ be the common difference of the arithmetic progression, we have $b = a + n$ , $c = a + 2n$ , $d = a + 3n$ . We are given that $b / a$ = $d / b$ , or $[frac{a + n}{a} = frac{a + 3n}{a + n}.]$ Cross-multiplying, we get $[a^2 + 2an + n^2 = a^2 + 3an]$ $[n^2 = an]$ $[n = a]$ So $frac{a}{d} = frac{a}{a + 3n} = frac{a}{4a} = boxed{frac{1}{4}} Longrightarrow mathrm{(C)}$ .
Solution 1

Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set ${-3, -2, -1, 0, 1, 2, 3}$ . It is easy to see that we can get any integer between $-6$ and $6$ inclusive as the sum of three elements from this set, for the total of $boxed{mathrm{(A) } 13}$ integers.

Solution 2

The set is an arithmetic sequence of numbers each $1$ more than a multiple of $3$ . Thus the sum of any three numbers will be a multiple of $3$ . All the multiples of $3$ from $1+4+7=12$ to $13+16+19=48$ are possible, totaling to $boxed{mathrm{(A) } 13}$ integers.
Solution1

Since $A-B$ and $A+B$ must have the same parity, and since there is only one even prime number, it follows that $2002 AMC12B 真题及答案详细解析$ and $A+B$ are both odd. Thus one of $A, B$ is odd and the other even. Since $A+B > A > A-B > 2$ , it follows that $A$ (as a prime greater than $2$ ) is odd. Thus $B = 2$ , and $A-2, A, A+2$ are consecutive odd primes. At least one of $A-2, A, A+2$ is divisible by $3$ , from which it follows that $A-2 = 3$ and $A = 5$ . The sum of these numbers is thus $17$ , a prime, so the answer is $boxed{mathrm{(E)} text{prime}}$ .

Solution 2

In order for both $A - B$ and $A + B$ to be prime, one of $A, B$ must be 2, or else both $A - B$ , $A + B$ would be even numbers.

If $A = 2$ , then $A < B$ and $A - B < 0$ , which is not possible. Thus $B = 2$ .

Since $A$ is prime and $A > A - B > 2$ , we can infer that $A > 3$ and thus $A$ can be expressed as $6n pm 1$ for some natural number $n$ .

However in either case, one of $A - B$ and $A + B$ can be expressed as $6n pm 3 = 3(2n pm 1)$ which is a multiple of 3. Therefore the only possibility that works is when $A - B = 3$ and $[A + B + (A - B) + (A + B) = 5 + 2 + 3 + 7 = 17]$

Which is a prime number. $boxed{(E)}$
Solution 1

Let $x^2 = frac{n}{20-n}$ , with $x ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$ ). Then rewriting, $n = frac{20x^2}{x^2 + 1}$ . Since $text{gcd}(x^2, x^2 + 1) = 1$ , it follows that $x^2 + 1$ divides $20$ . Listing the factors of $20$ , we find that $x = 0, 1, 2 , 3$ are the only $boxed{mathrm{(D)} 4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).

Solution 2

For $n<0$ and $n>20$ the fraction is negative, for $n=20$ it is not defined, and for $nin{1,dots,9}$ it is between 0 and 1.

Thus we only need to examine $n=0$ and $nin{10,dots,19}$ .

For $n=0$ and $n=10$ we obviously get the squares $0$ and $1$ respectively.

For prime $n$ the fraction will not be an integer, as the denominator will not contain the prime in the numerator.

This leaves $nin{12,14,15,16,18}$ , and a quick substitution shows that out of these only $n=16$ and $n=18$ yield a square. Therefore, there are only $boxed{mathrm{(D)} 4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).

Solution 3

If $frac{n}{20-n} = k^2 ge 0$ , then $n ge 0$ and $20-n > 0$ , otherwise $frac{n}{20-n}$ will be negative. Thus $0 le n le 19$ and $[0 = frac{0}{20-(0)} le frac{n}{20-n} le frac{19}{20-(19)} = 19]$ Checking all $k$ for which $0 le k^2 le 19$ , we have $0$ , $1$ , $2$ , $3$ as the possibilities. $boxed{(D)}$
Solution 1

Let $a, a+1, ldots, a + 17$ be the consecutive positive integers. Their sum, $18a + frac{17(18)}{2} = 9(2a+17)$ , is a perfect square. Since $9$ is a perfect square, it follows that $2a + 17$ is a perfect square. The smallest possible such perfect square is $25$ when $a = 4$ , and the sum is $225 Rightarrow mathrm{(B)}$ .

Solution 2

Notice that all five choices given are perfect squares.

Let $a$ be the smallest number, we have $[a+(a+1)+(a+2)+...+(a+17)=18a+sum_{k=1}^{17}k=18a+153]$

Subtract $153$ from each of the choices and then check its divisibility by $18$ , we have $225$ as the smallest possible sum. $mathrm {(B)}$
Solution 1

For any given pair of circles, they can intersect at most $2$ times. Since there are ${4choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 cdot 2 = 12$ . We can construct such a situation as below, so the answer is $boxed{mathrm{(D)} 12}$ .

Solution 2

Because a pair or circles can intersect at most $2$ times, the first circle can intersect the second at $2$ points, the third can intersect the first two at $4$ points, and the fourth can intersect the first three at $6$ points. This means that our answer is $2+4+6=boxed{mathrm{(D)} 12}.$

Solution 3

Pick a circle any circle- $4$ ways. Then, pick any other circle- $3$ ways. For each of these circles, there will be $2$ intersections for a total of $4*3*2$ = $24$ intersections. However, we have counted each intersection twice, so we divide for overcounting. Therefore, we reach a total of $frac{24}{2}=boxed{12}$ , which corresponds to $text{(D)}$ .
Solution

Let $N = overline{abcd} = 1000a + overline{bcd}$ , such that $frac{N}{9} = overline{bcd}$ . Then $1000a + overline{bcd} = 9overline{bcd} Longrightarrow 125a = overline{bcd}$ . Since $100 le overline{bcd} < 1000$ , from $a = 1, ldots, 7$ we have $7$ three-digit solutions, and the answer is $mathrm{(D)}$ .

Solution 2

Since N is a four digit number, assume WLOG that $N = 1000a + 100b + 10c + d$ , where a is the thousands digit, b is the hundreds digit, c is the tens digit, and d is the ones digit. Then, $frac{1}{9}N = 100b + 10c + d$ , so $N = 900b + 90c + 9d$ Set these equal to each other: $[1000a + 100b + 10c + d = 900b + 90c + 9d]$ $[1000a = 800b + 80c + 8d]$ $[1000a = 8(100b + 10c + d)]$ Notice that $100b + 10c + d = N - 1000a$ , thus: $[1000a = 8(N - 1000a)]$ $[1000a = 8N - 8000a]$ $[9000a = 8N]$ $[N = 1125a]$

Go back to our first equation, in which we set $N = 1000a + 100b + 10c + d$ , Then: $[1125a = 1000a + 100b + 10c + d]$ $[125a = 100b + 10c + d]$ The upper limit for the right hand side (RHS) is $999$ (when $b = 9$ , $c = 9$ , and $d = 9$ ). It's easy to prove that for an $a$ there is only one combination of $b, c,$ and $d$ that can make the equation equal. Just think about the RHS as a three digit number $bcd$ . There's one and only one way to create every three digit number with a certain combination of digits. Thus, we test for how many as are in the domain set by the RHS. Since $125cdot7 = 975$ which is the largest $a$ value, then $a$ can be $1$ through $7$ , giving us the answer of $boxed {D) 7}$
Solution 1

On both dice, only the faces with the numbers $3,6$ are divisible by $3$ . Let $P(a) = frac{2}{8} = frac{1}{4}$ be the probability that Juan rolls a $3$ or a $6$ , and $P(b) = frac{2}{6} = frac 13$ that Amal does. By the Principle of Inclusion-Exclusion,

$[P(a cup b) = P(a) + P(b) - P(a cap b) = frac{1}{4} + frac{1}{3} - frac{1}{4} cdot frac{1}{3} = frac{1}{2} Rightarrow mathrm{(C)}]$

Alternatively, the probability that Juan rolls a multiple of $3$ is $frac{1}{4}$ , and the probability that Juan does not roll a multiple of $3$ but Amal does is $left(1 - frac{1}{4}right) cdot frac{1}{3} = frac{1}{4}$ . Thus the total probability is $frac 14 + frac 14 = frac 12$ .

Solution 2

The probability that neither Juan nor Amal rolls a multiple of $3$ is $frac{6}{8} cdot frac{4}{6} = frac{1}{2}$ ; using complementary counting, the probability that at least one does is $1 - frac 12 = frac 12 Rightarrow mathrm{(C)}$ .
We say Andy's lawn has an area of $x$ . Beth's lawn thus has an area of $frac{x}{2}$ , and Carlos's lawn has an area of $frac{x}{3}$ .We say Andy's lawn mower cuts at a speed of $y$ . Carlos's cuts at a speed of $frac{y}{3}$ , and Beth's cuts at a speed $frac{2y}{3}$ .Each person's lawn is cut at a speed of $frac{text{area}}{text{rate}}$ , so Andy's is cut in $frac{x}{y}$ time, as is Carlos's. Beth's is cut in $frac{3}{4}timesfrac{x}{y}$ , so the first one to finish is $boxed{mathrm{(B)} text{Beth}}$ .
Solution 1

The region containing the points closer to $(0,0)$ than to $(3,1)$ is bounded by the perpendicular bisector of the segment with endpoints $(0,0),(3,1)$ . The perpendicular bisector passes through midpoint of $(0,0),(3,1)$ , which is $left(frac 32, frac 12right)$ , the center of the unit square with coordinates $(1,0),(2,0),(2,1),(1,1)$ . Thus, it cuts the unit square into two equal halves of area $1/2$ . The total area of the rectangle is $2$ , so the area closer to the origin than to $(3,1)$ and in the rectangle is $2 - frac 12 = frac 32$ . The probability is $frac{3/2}{2} = frac 34 Rightarrow mathrm{(C)}$ .

Solution 2

Assume that the point $P$ is randomly chosen within the rectangle with vertices $(0,0)$ , $(3,0)$ , $(3,1)$ , $(0,1)$ . In this case, the region for $P$ to be closer to the origin than to point $(3,1)$ occupies exactly $frac{1}{2}$ of the area of the rectangle, or $1.5$ square units.

If $P$ is chosen within the square with vertices $(2,0)$ , $(3,0)$ , $(3,1)$ , $(2,1)$ which has area $1$ square unit, it is for sure closer to $(3,1)$ .

Now if $P$ can only be chosen within the rectangle with vertices $(0,0)$ , $(2,0)$ , $(2,1)$ , $(0,1)$ , then the square region is removed and the area for $P$ to be closer to $(3,1)$ is then decreased by $1$ square unit, left with only $0.5$ square unit.

Thus the probability that $P$ is closer to $(3.1)$ is $frac{0.5}{2}=frac{1}{4}$ and that of $P$ is closer to the origin is $1-frac{1}{4}=frac{3}{4}$ . $mathrm{(C)}$
Adding up the three equations gives $2(ab + bc + ca) = 152 + 162 + 170 = 484 Longrightarrow ab + bc + ca = 242$ . Subtracting each of the above equations from this yields, respectively, $bc = 90, ca = 80, ab = 72$ . Taking their product, $ab cdot bc cdot ca = a^2b^2c^2 = 90 cdot 80 cdot 72 = 720^2 Longrightarrow abc = boxed{720} Rightarrow mathrm{(D)}$ .
Let $OM = x$ , $ON = y$ . By the Pythagorean Theorem on $triangle XON, MOY$ respectively, $begin{align*} (2x)^2 + y^2 &= 19^2 x^2 + (2y)^2 &= 22^2end{align*}$ Summing these gives $5x^2 + 5y^2 = 845 Longrightarrow x^2 + y^2 = 169$ .By the Pythagorean Theorem again, we have $[(2x)^2 + (2y)^2 = XY^2 Longrightarrow XY = sqrt{4(x^2 + y^2)} = sqrt{4(169)} = sqrt{676} = boxed{mathrm{(B)} 26}]$ Alternatively, we could note that since we found $x^2 + y^2 = 169$ , segment $MN=13$ . Right triangles $triangle MON$ and $triangle XOY$ are similar by Leg-Leg with a ratio of $frac{1}{2}$ , so $XY=2(MN)=boxed{mathrm{(B)} 26}$
Since $2002 = 11 cdot 13 cdot 14$ , it follows that $begin{eqnarray*} a_n =left{ begin{array}{lr} 11, & text{if }n=13 cdot 14 cdot k, quad k = 1,2,cdots 10; 13, & text{if }n=14 cdot 11 cdot k, quad k = 1,2,cdots 12; 14, & text{if }n=11 cdot 13 cdot k, quad k = 1,2,cdots 13; end{array} right. end{eqnarray*}$ Thus $sum_{n=1}^{2001} a_n = 11 cdot 10 + 13 cdot 12 + 14 cdot 13 = 448 Rightarrow mathrm{(A)}$ .
By the change of base formula, $a_n = frac{1}{frac{log 2002}{log n}} = left(frac{1}{log 2002}right) log n$ . Thus $begin{align*}b- c &= left(frac{1}{log 2002}right)(log 2 + log 3 + log 4 + log 5 - log 10 - log 11 - log 12 - log 13 - log 14) &= left(frac{1}{log 2002}right)left(log frac{2 cdot 3 cdot 4 cdot 5}{10 cdot 11 cdot 12 cdot 13 cdot 14}right) &= left(frac{1}{log 2002}right) log 2002^{-1} = -left(frac{log 2002}{log 2002}right) = -1 Rightarrow mathrm{(B)}end{align*}$
Solution 1

$[asy] unitsize(4cm); pair A, B, C, D, M; A = (1.768,0.935); B = (1.414,0); C = (0,0); D = (1.768,0); M = (0.707,0); draw(A--B--C--cycle); draw(A--D); draw(D--B); draw(A--M); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$M$",M,S); label("$x$",(A+D)/2,E); label("$y$",(B+D)/2,S); label("$a$",(C+M)/2,S); label("$a$",(M+B)/2,S); label("$2a$",(A+M)/2,SE); label("$1$",(A+B)/2,SE); label("$2$",(A+C)/2,NW); draw(rightanglemark(B,D,A,3)); [/asy]$

Let $D$ be the foot of the altitude from $A$ to $overline{BC}$ extended past $B$ . Let $AD = x$ and $BD = y$ . Using the Pythagorean Theorem, we obtain the equations

$begin{align*} x^2 + y^2 = 1 hspace{0.5cm}(1) x^2 + y^2 + 2ya + a^2 = 4a^2 hspace{0.5cm}(2) x^2 + y^2 + 4ya + 4a^2 = 4 hspace{0.5cm}(3) end{align*}$

Subtracting $(1)$ equation from $(2)$ and $(3)$ , we get

$begin{align*} 2ya + a^2 = 4a^2 - 1 hspace{0.5cm}(4) 4ya + 4a^2 = 3 hspace{0.5cm}(5) end{align*}$

Then, subtracting $2 times (4)$ from $(5)$ and rearranging, we get $10a^2 = 5$ , so $BC = 2a = sqrt{2}Rightarrow boxed{mathrm{(C)}}$

Solution 2

Let $D$ be the foot of the median from $A$ to $overline{BC}$ , and we let $AD = BC = 2a$ . Then by the Law of Cosines on $triangle ABD, triangle ACD$ , we have $begin{align*} 1^2 &= a^2 + (2a)^2 - 2(a)(2a)cos ADB 2^2 &= a^2 + (2a)^2 - 2(a)(2a)cos ADC end{align*}$

Since $cos ADC = cos (180 - ADB) = -cos ADB$ , we can add these two equations and get

$[5 = 10a^2]$

Hence $a = frac{1}{sqrt{2}}$ and $BC = 2a = sqrt{2} Rightarrow mathrm{(C)}$ .

Solution 3

From Stewart's Theorem, we have $(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.$ Simplifying, we get $(5/4)a^3 = (5/2)a implies (5/4)a^2 = 5/2 implies a^2 = 2 implies a = boxed{sqrt{2}}.$
We have $[[ABCD] = 2002 le frac 12 (AC cdot BD)]$ (This is true for any convex quadrilateral: split the quadrilateral along $AC$ and then using the triangle area formula to evaluate $[ACB]$ and $[ACD]$ ), with equality only if $overline{AC} perp overline{BD}$ . By the triangle inequality, $begin{align*}AC &le PA + PC = 52 BD &le PB + PD = 77end{align*}$ with equality if $P$ lies on $overline{AC}$ and $overline{BD}$ respectively. Thus $[2002 le frac{1}{2} AC cdot BD le frac 12 cdot 52 cdot 77 = 2002]$ Since we have the equality case, $overline{AC} perp overline{BD}$ at point $P$ , as shown below. $[asy] size(200); defaultpen(0.6); pair A = (0,0), B = (40,0), C = (25.6 * 52 / 24, 19.2 * 52 / 24), D = (40 - (40-25.6)*77/32,19.2*77/32), P = (25.6,19.2), Q = (25.6, 18.5); pair E=(A+P)/2, F=(B+P)/2, G=(C+P)/2, H=(D+P)/2; draw(A--B--C--D--cycle); draw(A--P--B--P--C--P--D); label("(A)",A,WSW); label("(B)",B,ESE); label("(C)",C,ESE); label("(D)",D,NW); label("(P)",Q,SSW); label("24",E,WNW); label("32",F,WSW); label("28",G,ESE); label("45",H,ENE); draw(rightanglemark(C,P,D,50)); [/asy]$ By the Pythagorean Theorem, $begin{align*} AB = sqrt{PA^2 + PB^2} & = sqrt{24^2 + 32^2} = 40 BC = sqrt{PB^2 + PC^2} & = sqrt{32^2 + 28^2} = 4sqrt{113} CD = sqrt{PC^2 + PD^2} & = sqrt{28^2 + 45^2} = 53 DA = sqrt{PD^2 + PA^2} & = sqrt{45^2 + 24^2} = 51 end{align*}$ The perimeter of $ABCD$ is $AB + BC + CD + DA = 4(36 + sqrt{113}) Rightarrow mathrm{(E)}$ .
The first condition gives us that $[x^2 + 6x + 1 + y^2 + 6y + 1 le 0 Longrightarrow (x+3)^2 + (y+3)^2 le 16]$ which is a circle centered at $(-3,-3)$ with radius $4$ . The second condition gives us that $[x^2 + 6x + 1 - y^2 - 6y - 1 le 0 Longrightarrow (x^2 - y^2) + 6(x-y) le 0 Longrightarrow (x-y)(x+y+6) le 0]$ Thus either $[x - y ge 0,quad x+y+6 le 0]$ or $[x - y le 0,quad x+y+6 ge 0]$

Each of those lines passes through $(-3,-3)$ and has slope $pm 1$ , as shown above. Therefore, the area of $R$ is half of the area of the circle, which is $frac{1}{2} (pi cdot 4^2) = 8pi approx 25 Rightarrow mathrm{(E)}$ .