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2001 AMC12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4日下午5:45

2001 AMC 12 真题

答案详细解析请参考文末

Problem 1

The sum of two numbers is $S$ . Suppose $3$ is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

$text{(A)} 2S + 3qquad text{(B)} 3S + 2qquad text{(C)} 3S + 6 qquadtext{(D)} 2S + 6 qquad text{(E)} 2S + 12$

Problem 2

Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$ . For example, $P(23) = 6$ and $S(23) = 5$ . Suppose $N$ is a two-digit number such that $N = P(N)+S(N)$ . What is the units digit of $N$ ?

$text{(A)} 2qquad text{(B)} 3qquad text{(C)} 6qquad text{(D)} 8qquad text{(E)} 9$

Problem 3

The state income tax where Kristin lives is levied at the rate of $p%$ of the first $textdollar 28000$ of annual income plus $(p + 2)%$ of any amount above $textdollar 28000$ . Kristin noticed that the state income tax she paid amounted to $(p + 0.25)%$ of her annual income. What was her annual income?

$text{(A)} $28000qquad text{(B)} $32000qquad text{(C)} $35000qquad text{(D)} $42000qquad text{(E)} $56000$

Problem 4

The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$ . What is their sum?

$text{(A)} 5qquad text{(B)} 20qquad text{(C)} 25qquad text{(D)} 30qquad text{(E)} 36$

Problem 5

What is the product of all positive odd integers less than 10000?

$text{(A)} dfrac{10000!}{(5000!)^2}qquad text{(B)} dfrac{10000!}{2^{5000}}qquad text{(C)} dfrac{9999!}{2^{5000}}qquad text{(D)} dfrac{10000!}{2^{5000} cdot 5000!}qquad text{(E)} dfrac{5000!}{2^{5000}}$

Problem 6

A telephone number has the form $text{ABC-DEF-GHIJ}$ , where each letter represents a different digit. The digits in each part of the number are in decreasing order; that is, $A > B > C$ , $D > E > F$ , and $G > H > I > J$ . Furthermore, $D$ , $E$ , and $F$ are consecutive even digits; $G$ , $H$ , $I$ , and $J$ are consecutive odd digits; and $A + B + C = 9$ . Find $A$ .

$text{(A)} 4qquad text{(B)} 5qquad text{(C)} 6qquad text{(D)} 7qquad text{(E)} 8$

Problem 7

A charity sells $140$ benefit tickets for a total of $textdollar 2001$ . Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?

$text{(A) }$782qquad text{(B) }$986qquad text{(C) }$1158qquad text{(D) }$1219qquad text{(E) }$1449$

Problem 8

Which of the cones listed below can be formed from a $252^circ$ sector of a circle of radius $10$ by aligning the two straight sides?

$[asy] import graph; unitsize(1.5cm); defaultpen(fontsize(8pt)); draw(Arc((0,0),1,-72,180),linewidth(.8pt)); draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt)); label("$10$",(-0.5,0),S); draw(Arc((0,0),0.1,-72,180)); label("$252^{circ}$",(0.05,0.05),NE); [/asy]$

$text{(A) A cone with slant height of } 10 text{ and radius } 6$

$text{(B) A cone with height of } 10 text{ and radius } 6$

$text{(C) A cone with slant height of } 10 text{ and radius } 7$

$text{(D) A cone with height of } 10 text{ and radius } 7$

$text{(E) A cone with slant height of } 10 text{ and radius } 8$

Problem 9

Let $f$ be a function satisfying $f(xy) = frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$ ?

$(mathrm{A}) 1 qquad (mathrm{B}) 2 qquad (mathrm{C}) frac52 qquad (mathrm{D}) 3 qquad (mathrm{E}) frac{18}5$

Problem 10

The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to

$text{(A) }50 qquad text{(B) }52 qquad text{(C) }54 qquad text{(D) }56 qquad text{(E) }58$

$[asy] unitsize(3mm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; for(int i=0; i<3; ++i) { for(int j=0; j<3; ++j) { draw(shift(3*i,3*j)*p); } } [/asy]$

Problem 11

A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?

$text{(A) }frac {3}{10} qquad text{(B) }frac {2}{5} qquad text{(C) }frac {1}{2} qquad text{(D) }frac {3}{5} qquad text{(E) }frac {7}{10}$

Problem 12

How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$ ?

$text{(A) }768 qquad text{(B) }801 qquad text{(C) }934 qquad text{(D) }1067 qquad text{(E) }1167$

Problem 13

The parabola with equation $y=ax^2+bx+c$ and vertex $(h,k)$ is reflected about the line $y=k$ . This results in the parabola with equation $y=dx^2+ex+f$ . Which of the following equals $a+b+c+d+e+f$ ?

$text{(A) }2b qquad text{(B) }2c qquad text{(C) }2a+2b qquad text{(D) }2h qquad text{(E) }2k$

Problem 14

Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$ , how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set ${A_1,A_2,dots,A_9}$ ?

$text{(A) }30 qquad text{(B) }36 qquad text{(C) }63 qquad text{(D) }66 qquad text{(E) }72$

Problem 15

An insect lives on the surface of a regular tetrahedron with edges of length 1. It wishes to travel on the surface of the tetrahedron from the midpoint of one edge to the midpoint of the opposite edge. What is the length of the shortest such trip? (Note: Two edges of a tetrahedron are opposite if they have no common endpoint.)

$text{(A) }frac {1}{2} sqrt {3} qquad text{(B) }1 qquad text{(C) }sqrt {2} qquad text{(D) }frac {3}{2} qquad text{(E) }2$

Problem 16

A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

$text{(A) }8! qquad text{(B) }2^8 cdot 8! qquad text{(C) }(8!)^2 qquad text{(D) }frac {16!}{2^8} qquad text{(E) }16!$

Problem 17

A point $P$ is selected at random from the interior of the pentagon with vertices $A = (0,2)$ , $B = (4,0)$ , $C = (2 pi + 1, 0)$ , $D = (2 pi + 1,4)$ , and $E=(0,4)$ . What is the probability that $angle APB$ is obtuse?

$text{(A) }frac {1}{5} qquad text{(B) }frac {1}{4} qquad text{(C) }frac {5}{16} qquad text{(D) }frac {3}{8} qquad text{(E) }frac {1}{2}$

Problem 18

A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is

$[asy] unitsize(0.75cm); pair A=(0,1), B=(4,4); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( A -- (A+(-1,0)) ); label("$1$", A -- (A+(-1,0)), N ); draw( B -- (B+(4,0)) ); label("$4$", B -- (B+(4,0)), N ); label("$A$",A,E); label("$B$",B,W); filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); [/asy]$

$text{(A) }frac {1}{3} qquad text{(B) }frac {2}{5} qquad text{(C) }frac {5}{12} qquad text{(D) }frac {4}{9} qquad text{(E) }frac {1}{2}$

Problem 19

The polynomial $P(x)=x^3+ax^2+bx+c$ has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the $y$ -intercept of the graph of $y=P(x)$ is 2, what is $b$ ?

$text{(A) }-11 qquad text{(B) }-10 qquad text{(C) }-9 qquad text{(D) }1 qquad text{(E) }5$

Problem 20

Points $A = (3,9)$ , $B = (1,1)$ , $C = (5,3)$ , and $D=(a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ABCD$ . The quadrilateral formed by joining the midpoints of $overline{AB}$ , $overline{BC}$ , $overline{CD}$ , and $overline{DA}$ is a square. What is the sum of the coordinates of point $D$ ?

$text{(A) }7 qquad text{(B) }9 qquad text{(C) }10 qquad text{(D) }12 qquad text{(E) }16$

Problem 21

Four positive integers $a$ , $b$ , $c$ , and $d$ have a product of $8!$ and satisfy:

$begin{align*} ab + a + b & = 524 \ bc + b + c & = 146 \ cd + c + d & = 104 end{align*}$

What is $a-d$ ?

$text{(A) }4 qquad text{(B) }6 qquad text{(C) }8 qquad text{(D) }10 qquad text{(E) }12$

Problem 22

In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $overline{DC}$ . Also, $overline{AC}$ intersects $overline{EF}$ at $H$ and $overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$ .

$text{(A) }frac {5}{2} qquad text{(B) }frac {35}{12} qquad text{(C) }3 qquad text{(D) }frac {7}{2} qquad text{(E) }frac {35}{8}$

Problem 23

A polynomial of degree four with leading coefficient 1 and integer coefficients has two real zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

$text{(A) }frac {1 + i sqrt {11}}{2} qquad text{(B) }frac {1 + i}{2} qquad text{(C) }frac {1}{2} + i qquad text{(D) }1 + frac {i}{2} qquad text{(E) }frac {1 + i sqrt {13}}{2}$

Problem 24

In $triangle ABC$ , $angle ABC=45^circ$ . Point $D$ is on $overline{BC}$ so that $2cdot BD=CD$ and $angle DAB=15^circ$ . Find $angle ACB$ .

$text{(A) }54^circ qquad text{(B) }60^circ qquad text{(C) }72^circ qquad text{(D) }75^circ qquad text{(E) }90^circ$

Problem 25

Consider sequences of positive real numbers of the form $x, 2000, y, dots$ in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of $x$ does the term 2001 appear somewhere in the sequence?

$text{(A) }1 qquad text{(B) }2 qquad text{(C) }3 qquad text{(D) }4 qquad text{(E) more than }4$

2001 AMC12 真题答案详细解析

Suppose the two numbers are $a$ and $b$ , with $a+b=S$ . Then the desired sum is $2(a+3)+2(b+3)=2(a+b)+12=2S +12$ , which is answer $boxed{text{(E)}}$ .
Denote $a$ and $b$ as the tens and units digit of $N$ , respectively. Then $N = 10a+b$ . It follows that $10a+b=ab+a+b$ , which implies that $9a=ab$ . Since $aneq0$ , $b=9$ . So the units digit of $N$ is $boxed{(text{E})9}$ .
Solution 1

Let the income amount be denoted by $A$ .

We know that $frac{A(p+.25)}{100}=frac{28000p}{100}+frac{(p+2)(A-28000)}{100}$ .

We can now try to solve for $A$ :

$(p+.25)A=28000p+Ap+2A-28000p-56000$

$.25A=2A-56000$

$A=32000$

So the answer is $boxed{B}$

Solution 2

Let $A$ , $T$ be Kristin's annual income and the income tax total, respectively. Notice that $begin{align*} T &= p%cdot28000 + (p + 2)%cdot(A - 28000) \ &= [p%cdot28000 + p%cdot(A - 28000)] + 2%cdot(A - 28000) \ &= p%cdot A + 2%cdot(A - 28000) end{align*}$ We are also given that $[T = (p + 0.25)%cdot A = p%cdot A + 0.25%cdot A]$ Thus, $[p%cdot A + 2%cdot(A - 28000) = p%cdot A + 0.25%cdot A]$ $[2%cdot(A - 28000) = 0.25%cdot A]$ Solve for $A$ to obtain $A = 32000$ . $boxed{B}$
Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, 5. So $dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which implies that $m=10$ . Hence, the sum of the three numbers is $3(10) = boxed{(text{D})30}$ .
$1 cdot 3 cdot 5 cdots 9999 = dfrac{1 cdot 2 cdot 3 cdot 4 cdots 10000}{2 cdot 4 cdot 6 cdots 10000}= dfrac{10000!}{2^{5000} cdot 1 cdot 2 cdot 3 cdots 5000}= dfrac{10000!}{2^{5000}cdot5000!}$ $text{(D)}$
The last four digits $text{GHIJ}$ are either $9753$ or $7531$ , and the other odd digit ( $1$ or $9$ ) must be $A$ , $B$ , or $C$ . Since $A + B + C = 9$ , that digit must be $1$ . Thus the sum of the two even digits in $text{ABC}$ is $8$ . $text{DEF}$ must be $864$ , $642$ , or $420$ , which respectively leave the pairs $2$ and $0$ , $8$ and $0$ , or $8$ and $6$ , as the two even digits in $text{ABC}$ . Only $8$ and $0$ has sum $8$ , so $text{ABC}$ is $810$ , and the required first digit is $boxed{(text{E})8}$ .
Solution 1

Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.

Let $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.

Hence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single half price ticket costs $frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$ . Keeping in mind that $0leq hleq 140$ , we are looking for a divisor between $140$ and $280$ , inclusive.

The prime factorization of $4002$ is $4002=2cdot 3cdot 23cdot 29$ . We can easily find out that the only divisor of $4002$ within the given range is $2cdot 3cdot 29 = 174$ .

This gives us $280-h=174$ , hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.

In our modified setting (with prices multiplied by $2$ ) the price of a half price ticket is $frac{4002}{174} = 23$ . In the original setting this is the price of a full price ticket. Hence $23cdot 34 = boxed{(text{A})782}$ dollars are raised by the full price tickets.

Solution 2

Let the cost of the full price ticket be $x$ , let the number of full price tickets be $A$ and half price tickets be $B$

Multiplying everything by two first to make cancel out fractions.

We have

$2Ax+Bx=4002$

And we have $A+B=140implies B=140-A$

Plugging in, we get $implies 2Ax+(140-A)(x)=4002$

Simplifying, we get $Ax+140x=4002$

Factoring out the $x$ , we get $x(A+140)=4002implies x=frac{4002}{A+140}$

Obviously, we see that the fraction has to simplify to an integer.

Hence, $A+140$ has to be a factor of 4002.

By inspection, we see that the prime factorization of $4002=2times3times23times29$

We see that $A=34$ through inspection. We also find that $x=23$

Hence, the price of full tickets out of $2001$ is $23times34=boxed{782}implies boxed{A}$ .
$[asy] import graph; unitsize(1.5cm); defaultpen(fontsize(8pt)); draw(Arc((0,0),1,-72,180),linewidth(.8pt) + red); draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt) + blue); label("$10$",(-0.5,0),S); draw(Arc((0,0),0.1,-72,180)); label("$252^{circ}$",(0.05,0.05),NE); [/asy]$ The blue lines will be joined together to form a single blue line on the surface of the cone, hence $boxed{10}$ will be the $boxed{text{slant height}}$ of the cone.The red line will form the circumference of the base. We can compute its length and use it to determine the radius.The length of the red line is $dfrac{252}{360}cdot 2pi cdot 10 = 14pi$ . This is the circumference of a circle with radius $boxed{7}$ .Therefore the correct answer is $boxed{text{C}}$ .
Solution 1

Letting $x = 500$ and $y = dfrac65$ in the given equation, we get $f(500cdotfrac65) = frac3{frac65} = frac52$ , or $f(600) = boxed{textbf{C } frac52}$ .

Solution 2

The only function that satisfies the given condition is $y = frac{k}{x}$ , for some constant $k$ . Thus, the answer is $frac{500 cdot 3}{600} = frac52$ .
Consider any single tile: $[asy] unitsize(1cm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; draw(p); [/asy]$ If the side of the small square is $a$ , then the area of the tile is $9a^2$ , with $4a^2$ covered by squares and $5a^2$ by pentagons. Hence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $55.overline{5}%$ , and the closest integer to this value is $boxed{(mathrm{D})56}$ .
Solution 1

Imagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. To draw out all the white chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip of the five is red (and so also the probability that the last chip drawn is white) is $boxed{(text{D}) frac {3}{5}}$ .

Solution 2

Let's assume we don't stop picking until all of the balls are picked. To satisfy this condition, we have to arrange the letters: W, W, R, R, R such that both R's appear in the first 4. We find the number of ways to arrange the red balls in the first 4 and divide that by the total ways to choose all the balls. The probability of this occurring is $dfrac{dbinom{4}{2}}{dbinom{5}{2}} = boxed{text{(D)}dfrac35}$
Solution 1

Out of the numbers $1$ to $12$ four are divisible by $3$ and three by $4$ , counting $12$ twice. Hence $6$ out of these $12$ numbers are multiples of $3$ or $4$ .

The same is obviously true for the numbers $12k+1$ to $12k+12$ for any positive integer $k$ .

Hence out of the numbers $1$ to $60=5cdot 12$ there are $5cdot 6=30$ numbers that are divisible by $3$ or $4$ . Out of these $30$ , the numbers $15$ , $20$ , $30$ , $40$ , $45$ and $60$ are divisible by $5$ . Therefore in the set ${1,dots,60}$ there are precisely $30-6=24$ numbers that satisfy all criteria from the problem statement.

Again, the same is obviously true for the set ${60k+1,dots,60k+60}$ for any positive integer $k$ .

We have $1980/60 = 33$ , hence there are $24cdot 33 = 792$ good numbers among the numbers $1$ to $1980$ . At this point we already know that the only answer that is still possible is $boxed{text{(B)}}$ , as we only have $20$ numbers left.

By examining the remaining $20$ by hand we can easily find out that exactly $9$ of them match all the criteria, giving us $792+9=boxed{801}$ good numbers. This is correct.

Solution 2

We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $leftlfloor frac{2001}{3} rightrfloor$ , $leftlfloor frac{2001}{4} rightrfloor$ , and $leftlfloor frac{2001}{12} rightrfloor$ . The first two floor functions were for calculating the number of individual cases for $3$ and $4$ . The third case was to find any overlapping numbers. The numbers were $667$ , $500$ , and $166$ , respectively. We add the first two terms and subtract the third to get $1001$ . The first case is finished.

The second case is more or less the same, except we are applying $3$ and $4$ to $5$ . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $leftlfloor frac{2001}{3*5} rightrfloor$ , $leftlfloor frac{2001}{4*5} rightrfloor$ , and $leftlfloor frac{2001}{3*4*5} rightrfloor$ yields the numbers $133$ , $100$ , and $33$ . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$ . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$ . Subtracting this number from the original $1001$ numbers procures $boxed{textbf{(B)} 801}$ .

Solution 3

First find the number of such numbers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of 3.

There are $frac45*2000=1600$ numbers that are not multiples of $5$ . $frac23*frac34*1600=800$ are not multiples of $3$ or $4$ , so $800$ numbers are. $800+1=boxed{textbf{(B)} 801}$

Solution 4

Take a good-sized sample of consecutive integers; for example, the first 25 positive integers. Determine that the numbers 3, 4, 6, 8, 9, 12, 16, 18, 21, and 24 exhibit the properties given in the question. 25 is a divisor of 2000, so there are $frac{10}{25}*2000=800$ numbers satisfying the given conditions between 1 and 2000. Since 2001 is a multiple of 3, add 1 to 800 to get $800+1=boxed{textbf{(B)} 801}$ .
We write $p(x)$ as $a(x-h)^2+k$ (this is possible for any parabola). Then the reflection of $p(x)$ is $q(x) = -a(x-h)^2+k$ . Then we find $p(x) + q(x) = 2k$ . Since $p(1) = a+b+c$ and $q(1) = d+e+f$ , we have $a+b+c+d+e+f = 2k$ , so the answer is $fbox{E}$ .
Each of the ${9choose 2}=36$ pairs of vertices determines two equilateral triangles, one on each side of the segment. This would give us $72$ triangles. However, note that there are three equilateral triangles that have all three vertices among the vertices of the polygon. These are the triangles $A_1A_4A_7$ , $A_2A_5A_8$ , and $A_3A_6A_9$ . We counted each of these three times (once for each side). Hence we overcounted by $2$ for each of these triangles for a total of $6$ overcounted, and the correct number of equilateral triangles is $72-6=boxed{66}$ .
Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. Its length is the same as the length of the tetrahedron's edge, i.e. $boxed{1}$ . $[asy] unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(1,0), C=rotate(60)*B, D1=B+C-A, D2=A+C-B, D3=A+B-C; draw(A--B--C--cycle); draw(D1--B--C--cycle); draw(D2--A--C--cycle); draw(D3--B--A--cycle); dot( A+0.5*(C-A) ); dot( B+0.5*(C-A) ); draw( ( A+0.5*(C-A) ) -- ( B+0.5*(C-A) ), dashed ); [/asy]$
Solution 1

Let the spider try to put on all $16$ things in a random order. Each of the $16!$ permutations is equally probable. For any fixed leg, the probability that he will first put on the sock and only then the shoe is clearly $frac{1}{2}$ . Then the probability that he will correctly put things on all legs is $frac{1}{2^{8}}$ . Therefore the number of correct permutations must be $boxed{frac {16!}{2^8}}$ .

Solution 2

Each dressing sequence can be uniquely described by a sequence containing two $1$ s, two $2$ s, ..., and two $8$ s -- the first occurrence of number $x$ means that the spider puts the sock onto leg $x$ , the second occurrence of $x$ means he puts the shoe onto leg $x$ . If the numbers were all unique, the answer would be $16!$ . However, since 8 terms appear twice, the answer is $frac{16!}{(2!)^8} = boxed{frac {16!}{2^8}}$ .

Solution 3

You can put all $8$ socks on first for $8!$ ways and then all $8$ shoes on next for $8!$ more ways. This is not the only possibility, so the lower bound is $(8!)^2$ . You can choose all $16$ in a random fashion, but some combinations would violate the rules, so the upper bound is $16!$ . $text{(C)}$ & $text{(E)}$ are the lower and upper bounds, so the answer is in between them, $boxed{frac {16!}{2^8}}$ .
The angle $APB$ is obtuse if and only if $P$ lies inside the circle with diameter $AB$ . (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.) $[asy] defaultpen(0.8); real pi=3.14159265359; pair A=(0,2), B=(4,0), C=(2*pi+1, 0), D=(2*pi+1,4), E=(0,4), F=(0,0); draw(A--B--C--D--E--cycle); draw(circle((A+B)/2,length(B-A)/2)); label("$A$",A,W); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,SW); draw(A--F--B,dashed); [/asy]$ The area of $AFB$ is $[AFB] = frac {AFcdot FB}2 = 4$ , and the area of $ABCDE$ is $CDcdot DE - [AFB] = 4cdot (2pi+1) - 4 = 8pi$ .From the Pythagorean theorem the length of $AB$ is $sqrt{2^2 + 4^2} = 2sqrt{5}$ , thus the radius of the circle is $sqrt{5}$ , and the area of the half-circle that is inside $ABCDE$ is $frac{ 5pi }2$ .Therefore the probability that $APB$ is obtuse is $frac{ frac{ 5pi }2 }{ 8pi } = boxed{text{(C) } frac 5{16}}$ .
Solution 1

$[asy] unitsize(1cm); pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( (A+(4,0)) -- A -- (A+(0,-1)) ); draw( A -- B -- (B+(0,-4)) ); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$S$",S,N); filldraw( circle(S,4/9), lightgray, black ); dot(S); draw( rightanglemark(A,C,B) ); draw( S -- A ); draw( S -- B ); [/asy]$

In the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$ , thus by the Pythagorean theorem we have $AC=4$ .

We can now pick a coordinate system where the common tangent is the $x$ axis and $A$ lies on the $y$ axis. In this coordinate system we have $A=(0,1)$ and $B=(4,4)$ .

Let $r$ be the radius of the small circle, and let $s$ be the $x$ -coordinate of its center $S$ . We then know that $S=(s,r)$ , as the circle is tangent to the $x$ axis. Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$ .

We have $SA = sqrt{s^2 + (1-r)^2}$ and $SB=sqrt{(4-s)^2 + (4-r)^2}$ . Hence we get the following two equations:

$begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \ (4-s)^2 + (4-r)^2 & = (4+r)^2 end{align*}$

Simplifying both, we get

$begin{align*} s^2 & = 4r \ (4-s)^2 & = 16r end{align*}$

As in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $left( frac{4-s}s right)^2 = 4$ .

Now there are two possibilities: either $frac{4-s}s=-2$ , or $frac{4-s}s=2$ . In the first case clearly $s<0$ , hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the $x$ axis - a large circle whose center is somewhere to the left of $A$ .) The second case solves to $s=frac 43$ . We then have $4r = s^2 = frac {16}9$ , hence $r = boxed{frac 49}$ .

Solution 2

The horizontal line is the equivalent of a circle of curvature $0$ , thus we can apply Descartes' Circle Formula.

The four circles have curvatures $0, 1, frac 14$ , and $frac 1r$ .

We have $2left(0^2+1^2+frac {1}{4^2}+frac{1}{r^2}right)=left(0+1+frac 14+frac 1rright)^2$

Simplifying, we get $frac{34}{16}+frac{2}{r^2}=frac{25}{16}+frac{5}{2r}+frac{1}{r^2}$

$[frac{1}{r^2}-frac{5}{2r}+frac{9}{16}=0]$ $[frac{16}{r^2}-frac{40}{r}+9=0]$ $[left(frac{4}{r}-9right)left(frac{4}{r}-1right)=0]$

Obviously $r$ cannot equal $4$ , therefore $r = boxed{frac 49}$ .
We are given $c=2$ . So the product of the roots is $-c = -2$ by Vieta's formulas. These also tell us that $frac{-a}{3}$ is the average of the zeros, so $frac{-a}3=-2 implies a = 6$ . We are also given that the sum of the coefficients is $-2$ , so $1+6+b+2 = -2 implies b=-11$ . So the answer is $fbox{A}$ .
$[asy] pair A=(3,9), B=(1,1), C=(5,3), D=(7,3); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,E); pair AB = (A + B)/2, BC = (B + C)/2, CD = (C + D)/2, DA = (D + A)/2; draw(AB--BC--CD--DA--cycle); [/asy]$ We already know two vertices of the square: $(A+B)/2 = (2,5)$ and $(B+C)/2 = (3,2)$ .There are only two possibilities for the other vertices of the square: either they are $(6,3)$ and $(5,6)$ , or they are $(0,1)$ and $(-1,4)$ . The second case would give us $D$ outside the first quadrant, hence the first case is the correct one. As $(6,3)$ is the midpoint of $CD$ , we can compute $D=(7,3)$ , and $7+3=boxed{10}$ .
Solution 1

Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:

$begin{align*} (a+1)(b+1) & = 525 \ (b+1)(c+1) & = 147 \ (c+1)(d+1) & = 105 end{align*}$

Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$ . We get:

$begin{align*} ef & = 3cdot 5cdot 5cdot 7 \ fg & = 3cdot 7cdot 7 \ gh & = 3cdot 5cdot 7 end{align*}$

Clearly $7^2$ divides $fg$ . On the other hand, $7^2$ can not divide $f$ , as it then would divide $ef$ . Similarly, $7^2$ can not divide $g$ . Hence $7$ divides both $f$ and $g$ . This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$ .

The first case solves to $(e,f,g,h)=(75,7,21,5)$ , which gives us $(a,b,c,d)=(74,6,20,4)$ , but then $abcd not= 8!$ . We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$ . (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.)

The second case solves to $(e,f,g,h)=(25,21,7,15)$ , which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$ , and we have $a-d=24-14 =boxed{10}$ .
Solution 1

Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $frac {AH}{HC} = frac{AF}{EC} = frac 23$ .

Also, triangles $AGJ$ and $CEJ$ are similar, hence $frac {AJ}{JC} = frac {AG}{EC} = frac 43$ .

We can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$ . We have:
- $[ACD]=frac{[ABCD]}2 = 35$ .
- $[AHD]$ is $2/5$ of $[ACD]$ , as these two triangles have the same base $AD$ , and $AH$ is $2/5$ of $AC$ , therefore also the height from $H$ onto $AD$ is $2/5$ of the height from $C$ . Hence $[AHD]=14$ .
- $[HED]$ is $3/10$ of $[ACD]$ , as the base $ED$ is $1/2$ of the base $CD$ , and the height from $H$ is $3/5$ of the height from $A$ . Hence $[HED]=frac {21}2$ .
- $[JEC]$ is $3/14$ of $[ACD]$ for similar reasons, hence $[JEC]=frac{15}2$ .
Therefore $[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-frac {21}2-frac{15}2 = boxed{3}$ .

Solution 2

As in the previous solution, we note the similar triangles and prove that $H$ is in $2/5$ and $J$ in $4/7$ of $AC$ .

We can then compute that $HJ = AC cdot left( frac 47 - frac 25 right) = AC cdot frac{6}{35}$ .

As $E$ is the midpoint of $CD$ , the height from $E$ onto $AC$ is $1/2$ of the height from $D$ onto $AC$ . Therefore we have $[EHJ] = frac{6}{35} cdot frac 12 cdot [ACD] = frac 3{35} cdot 35 = boxed{3}$ .
Let the polynomial be $P$ and let the two integer zeros be $z_1$ and $z_2$ . We can then write $P(x)=(x-z_1)(x-z_2)(x^2+ax+b)$ for some integers $a$ and $b$ .If a complex number $p+qi$ with $qnot=0$ is a root of $P$ , it must be the root of $x^2+ax+b$ , and the other root of $x^2+ax+b$ must be $p-qi$ .We can then write $x^2+ax+b = (x-p-qi)(x-p+qi) = (x-p)^2 - (qi)^2 = x^2 - 2px + p^2 + q^2$ .We can now examine each of the five given complex numbers, and find the one for which the values $-2p$ and $p^2+q^2$ are integers. This is $boxed{frac {1 + i sqrt {11}}{2}}$ , for which we have $-2p = -2cdotfrac 12 = -1$ and $p^2+q^2 = left( frac 12 right)^2 + left( frac {sqrt{11}}2 right)^2 = frac 14 + frac {11}4 = frac {12}4 = 3$ .(As an example, the polynomial $x^4 - 2x^3 + 4x^2 - 3x$ has zeroes $0$ , $1$ , and $frac {1 pm i sqrt {11}}{2}$ .)

Solution 2

By Vieta, we know that the product of all four zeros of the polynomial equals the constant at the end of the polynomial. We also know that the two imaginary roots are a conjugate pair (I.E if one is a+bi, the other is a-bi). So the two imaginary roots must multiply to give you an integer. Taking the 5 answers into hand, we find that $boxed{frac {1 + i sqrt {11}}{2}}$ is our only integer giving solution.

Solution 3

After dividing the polynomial out by $(x-p)$ and $(x-q)$ , where p and q are the real roots of the polynomial, we will obtain a quadratic with two complex roots. We can then use the quadratic formula to solve for these complex roots.

Let's start by using synthetic division to divide $x^4+ax^3+bx^2+cx+d$ by $(x-p)$ . Using this method, the quotient becomes $1x^3+(a-p)x^2+(b-pa+p^2)x+(c-bp+ap^2-p^3)+frac{d-pc+bp^2-ap^3+p^4}{x-q}$ . However, we know that there should be no remainder because $(x-p)$ is a factor of the polynomial, so $frac{d-pc+bp^2-ap^3+p^4}{x-q}$ must equal 0, so $d=-pc+bp^2-ap^3+p^4$ . When we divide the expression on the left by -p, we get $c-bp+ap^2-p^3$ , so we can replace it in our original synthetic division equation with $frac {d}{-k}$ .

We then want to synthetically divide $x^3+(a-p)x^2+(b-pa+p^2)x+frac {d}{-k}$ by the next factor, $(x-q)$ . Using the same method as before, we can simplify the quotient to $x^2+(a-p-q)x+frac{d}{pq}$ . Now for the easy part!

Use the quadratic formula to determine the form of the complex roots.

$frac{p+q-apmsqrt{(a-p-q)^2-frac{4d}{pq}}}{2}$

Now this is starting to look a lot like answers A and E. Noticing that the real part in each answer choice is $frac{1}{2}$ , $(k+n-a)=1$ and $(a-k-n)^2=1$ , and the imaginary part is positive. Furthermore, by Vieta's Formulas, we know that d must be a multiple of p and q, so $frac{4d}{pq}$ is a multiple of 4. Rearranging the expression, we get:

$frac{1+isqrt{frac{4d}{pq}-1}}{2}$

The radicand therefore must be one less than a multiple of four, which is only the case in $frac {1 + i sqrt {11}}{2}$ or $boxed{A}$ .
Solution 1

$[asy] unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); pair ortho=rotate(-90)*(D-A); pair E=extension(C, ortho, A, B); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(E--D); draw(A--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,S); [/asy]$

We start with the observation that $angle ADB = 180^circ - 15^circ - 45^circ = 120^circ$ , and $angle ADC = 15^circ + 45^circ = 60^circ$ .

We can draw the height $CE$ from $C$ onto $AD$ . In the triangle $CED$ , we have $frac {ED}{CD} = cos EDC = cos 60^circ = frac 12$ . Hence $ED = CD/2$ .

By the definition of $D$ , we also have $BD=CD/2$ , therefore $BD=DE$ . This means that the triangle $BDE$ is isosceles, and as $angle BDE=120^circ$ , we must have $angle BED = angle EBD = 30^circ$ .

Then we compute $angle ABE = 45^circ - 30^circ = 15^circ$ , thus $angle ABE = angle BAE$ and the triangle $ABE$ is isosceles as well. Hence $AE=BE$ .

Now we can note that $angle DCE = 180^circ - 90^circ - 60^circ = 30^circ$ , hence also the triangle $EBC$ is isosceles and we have $BE=CE$ .

Combining the previous two observations we get that $AE=EC$ , and as $angle AEC=90^circ$ , this means that $angle CAE = angle ACE = 45^circ$ .

Finally, we get $angle ACB = angle ACE + angle ECD = 45^circ + 30^circ = boxed{75^circ}$ .

Solution 2

Draw a good diagram! Now, let's call $BD=t$ , so $DC=2t$ . Given the rather nice angles of $angle ABD = 45^circ$ and $angle ADC = 60^circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$ ; call this point $H$ . We realize that there is no specific factor of $t$ we can call this just yet, so let $AH=kt$ . Notice that in $triangle{ABH}$ we get $BH=kt$ . Using the 60-degree angle in $triangle{ADH}$ , we obtain $DH=frac{sqrt{3}}{3}kt$ . The comparable ratio is that $BH-DH=t$ . If we involve our $k$ , we get:

$kt(frac{3}{3}-frac{sqrt{3}}{3})=t$ . Eliminating $t$ and removing radicals from the denominator, we get $k=frac{3+sqrt{3}}{2}$ . From there, one can easily obtain $HC=3t-kt=frac{3-sqrt{3}}{2}t$ . Now we finally have a desired ratio. Since $tanangle ACH = 2+sqrt{3}$ upon calculation, we know that $angle ACH$ can be simplified. Indeed, if you know that $tan(75)=2+sqrt{3}$ or even take a minute or two to work out the sine and cosine using $sin(x)^2+cos(x)^2=1$ , and perhaps the half- or double-angle formulas, you get $boxed{75^circ}$ .

Solution 3

Without loss of generality, we can assume that $BD = 1$ and $CD = 2$ . As above, we are able to find that $angle ADC = 60^circ$ and $angle ADB = 120^circ$ .

Using Law of Sines on triangle $ADB$ , we find that $frac{1}{sin15^circ} = frac{AD}{sin 45^circ} = frac{AB}{sin 120^circ}$ . Since we know that $sin 15^circ = frac{sqrt{6}-sqrt{2}}{4}$ , $sin 45^circ = frac{sqrt{2}}{2}$ , and $sin 120^circ = frac{sqrt{3}}{2}$ , we can compute $AD$ to equal $1+sqrt{3}$ and $AB$ to be $frac{3sqrt{2}+sqrt{6}}{2}$ .

Next, we apply Law of Cosines to triangle $ADC$ to see that $AC^2 = (1+sqrt{3})^2 + 2^2 - (2)(1+sqrt{3})(2)(cos 60^circ)$ . Simplifying the right side, we get $AC^2 = 6$ , so $AC = sqrt{6}$ .

Now, we apply Law of Sines to triangle $ABC$ to see that $frac{sqrt{6}}{sin 45^circ} = frac{frac{3sqrt{2}+sqrt{6}}{2}}{sin ACB}$ . After rearranging and noting that $sin 45^circ = frac{sqrt{2}}{2}$ , we get $sin ACB = frac{sqrt{6}+3sqrt{2}}{4sqrt{3}}$ .

Dividing the right side by $sqrt{3}$ , we see that $sin ACB = frac{sqrt{6}+sqrt{2}}{4}$ , so $angle ACB$ is either $75^circ$ or $105^circ$ . Since $105^circ$ is not a choice, we know $angle ACB = boxed{75^circ}$ .

Note that we can also confirm that $angle ACB neq 105^circ$ by computing $angle CAB$ with Law of Sines.
It never hurts to compute a few terms of the sequence in order to get a feel how it looks like. In our case, the definition is that $forall n>1:~ a_n = a_{n-1}a_{n+1} - 1$ . This can be rewritten as $a_{n+1} = frac{a_n +1}{a_{n-1}}$ . We have $a_1=x$ and $a_2=2000$ , and we compute: $begin{align*} a_3 & = frac{a_2+1}{a_1} = frac{2001}x \ a_4 & = frac{a_3+1}{a_2} = frac{ dfrac{2001}x + 1 }{ 2000 } = frac{2001 + x}{2000x} \ a_5 & = frac{a_4+1}{a_3} = frac{ frac{2001 + x}{2000x} + 1 }{ frac{2001}x } = frac{ frac{2001 + 2001x}{2000x} }{ frac{2001}x } = frac{1+x}{2000} \ a_6 & = frac{a_5+1}{a_4} = frac{ frac{1+x}{2000} + 1 }{ frac{2001 + x}{2000x} } = frac{ frac{2001+x}{2000} }{ frac{2001 + x}{2000x} } = x \ a_7 & = frac{a_6+1}{a_5} = frac{ x+1 }{ frac{1+x}{2000} } = 2000 end{align*}$ At this point we see that the sequence will become periodic: we have $a_6=a_1$ , $a_7=a_2$ , and each subsequent term is uniquely determined by the previous two.Hence if $2001$ appears, it has to be one of $a_1$ to $a_5$ . As $a_2=2000$ , we only have four possibilities left. Clearly $a_1=2001$ for $x=2001$ , and $a_3=2001$ for $x=1$ . The equation $a_4=2001$ solves to $x = frac{2001}{2000cdot 2001 - 1}$ , and the equation $a_5=2001$ to $x=2000cdot 2001 - 1$ .No two values of $x$ we just computed are equal, and therefore there are $boxed{4}$ different values of $x$ for which the sequence contains the value $2001$