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

2006 AMC12A 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日下午6:12

2006 AMC 12 A 真题

答案详细解析请参考文末

Problem 1

Sandwiches at Joe's Fast Food cost $3$ dollars each and sodas cost $2$ dollars each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?

$mathrm{(A) } 31qquad mathrm{(B) } 32qquad mathrm{(C) } 33qquad mathrm{(D) } 34qquad mathrm{(E) } 35$

Problem 2

Define $xotimes y=x^3-y$ . What is $hotimes (hotimes h)$ ?

$mathrm{(A) } -hqquad mathrm{(B) } 0qquad mathrm{(C) } hqquad mathrm{(D) } 2hqquad mathrm{(E) } h^3$

Problem 3

The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?

$mathrm{(A) } 15qquad mathrm{(B) } 18qquad mathrm{(C) } 20qquad mathrm{(D) } 24qquad mathrm{(E) } 50$

Problem 4

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

$mathrm{(A) } 17qquad mathrm{(B) } 19qquad mathrm{(C) } 21qquad mathrm{(D) } 22qquad mathrm{(E) } 23$

Problem 5

Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $8$ dollars, and there was an additional cost of $2$ dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?

$mathrm{(A) } 1qquad mathrm{(B) } 2qquad mathrm{(C) } 3qquad mathrm{(D) } 4qquad mathrm{(E) } 5$

Problem 6

The $8times 18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$ ? $[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]$ $mathrm{(A) } 6qquad mathrm{(B) } 7qquad mathrm{(C) } 8qquad mathrm{(D) } 9qquad mathrm{(E) } 10$

Problem 7

Mary is $20%$ older than Sally, and Sally is $40%$ younger than Danielle. The sum of their ages is $23.2$ years. How old will Mary be on her next birthday?

$mathrm{(A) } 7qquad mathrm{(B) } 8qquad mathrm{(C) } 9qquad mathrm{(D) } 10qquad mathrm{(E) } 11$

Problem 8

How many sets of two or more consecutive positive integers have a sum of $15$ ?

$mathrm{(A) } 1qquad mathrm{(B) } 2qquad mathrm{(C) } 3qquad mathrm{(D) } 4qquad mathrm{(E) } 5$

Problem 9

Oscar buys $13$ pencils and $3$ erasers for $textdollar 1.00$ . A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser?

$mathrm{(A) } 10qquad mathrm{(B) } 12qquad mathrm{(C) } 15qquad mathrm{(D) } 18qquad mathrm{(E) } 20$

Problem 10

For how many real values of $x$ is $sqrt{120-sqrt{x}}$ an integer?

$mathrm{(A) } 3qquad mathrm{(B) } 6qquad mathrm{(C) } 9qquad mathrm{(D) } 10qquad mathrm{(E) } 11$

Problem 11

Which of the following describes the graph of the equation $(x+y)^2=x^2+y^2$ ?

$mathrm{(A)} text{the empty set}qquadmathrm{(B)} text{one point}qquadmathrm{(C)} text{two lines}qquadmathrm{(D)} text{a circle}qquadmathrm{(E)} text{the entire plane}$

Problem 12

A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring? $[asy]size(7cm); pointpen = black; pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E); [/asy]$ $mathrm{(A) } 171qquad mathrm{(B) } 173qquad mathrm{(C) } 182qquad mathrm{(D) } 188qquad mathrm{(E) } 210$

Problem 13

The vertices of a $3-4-5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles? $[asy] unitsize(5mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); pair B=(0,0), C=(5,0); pair A=intersectionpoints(Circle(B,3),Circle(C,4))[0]; draw(A--B--C--cycle); draw(Circle(C,3)); draw(Circle(A,1)); draw(Circle(B,2)); label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("3",midpoint(B--A),NW); label("4",midpoint(A--C),NE); label("5",midpoint(B--C),S);[/asy]$

$mathrm{(A) } 12piqquad mathrm{(B) } frac{25pi}{2}qquad mathrm{(C) } 13piqquad mathrm{(D) } frac{27pi}{2}qquad mathrm{(E) } 14pi$

Problem 14

Two farmers agree that pigs are worth $300$ dollars and that goats are worth $210$ dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a $390$ dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?

$mathrm{(A) } textdollar5 qquad mathrm{(B) } textdollar 10 qquad mathrm{(C) } textdollar 30 qquad mathrm{(D) } textdollar 90 qquad mathrm{(E) } textdollar 210$

Problem 15

Suppose $cos x=0$ and $cos (x+z)=1/2$ . What is the smallest possible positive value of $z$ ?

$mathrm{(A) } frac{pi}{6}qquad mathrm{(B) } frac{pi}{3}qquad mathrm{(C) } frac{pi}{2}qquad mathrm{(D) } frac{5pi}{6}qquad mathrm{(E) } frac{7pi}{6}$

Problem 16

Circles with centers $A$ and $B$ have radii $3$ and $8$ , respectively. A common internal tangent intersects the circles at $C$ and $D$ , respectively. Lines $AB$ and $CD$ intersect at $E$ , and $AE=5$ . What is $CD$ ?
$[asy]unitsize(2.5mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair A=(0,0), Ep=(5,0), B=(5+40/3,0); pair M=midpoint(A--Ep); pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1]; pair D=B+8*dir(180+degrees(C)); dot(A); dot(C); dot(B); dot(D); draw(C--D); draw(A--B); draw(Circle(A,3)); draw(Circle(B,8)); label("$A$",A,W); label("$B$",B,E); label("$C$",C,SE); label("$E$",Ep,SSE); label("$D$",D,NW);[/asy]$

$mathrm{(A)} 13qquadmathrm{(B)} frac{44}{3}qquadmathrm{(C)} sqrt{221}qquadmathrm{(D)} sqrt{255}qquadmathrm{(E)} frac{55}{3}$

Problem 17

Square $ABCD$ has side length $s$ , a circle centered at $E$ has radius $r$ , and $r$ and $s$ are both rational. The circle passes through $D$ , and $D$ lies on $overline{BE}$ . Point $F$ lies on the circle, on the same side of $overline{BE}$ as $A$ . Segment $AF$ is tangent to the circle, and $AF=sqrt{9+5sqrt{2}}$ . What is $r/s$ ? $[asy]unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B=(0,0), C=(3,0), D=(3,3), A=(0,3); pair Ep=(3+5*sqrt(2)/6,3+5*sqrt(2)/6); pair F=intersectionpoints(Circle(A,sqrt(9+5*sqrt(2))),Circle(Ep,5/3))[0]; pair[] dots={A,B,C,D,Ep,F}; draw(A--F); draw(Circle(Ep,5/3)); draw(A--B--C--D--cycle); dot(dots); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,SW); label("$E$",Ep,E); label("$F$",F,NW); [/asy]$ $mathrm{(A) } frac{1}{2}qquad mathrm{(B) } frac{5}{9}qquad mathrm{(C) } frac{3}{5}qquad mathrm{(D) } frac{5}{3}qquad mathrm{(E) } frac{9}{5}$

Problem 18

The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and

$f(x)+fleft(frac{1}{x}right)=x$

What is the largest set of real numbers that can be in the domain of $f$ ?

$mathrm{(A) } {x|xne 0}qquad mathrm{(B) } {x|x<0}qquad mathrm{(C) } {x|x>0}qquad mathrm{(D) } {x|xne -1;$ $mathrm{and}; xne 0;mathrm{and}; xne 1}qquad mathrm{(E) } {-1,1}$

Problem 19

Circles with centers $(2,4)$ and $(14,9)$ have radii $4$ and $9$ , respectively. The equation of a common external tangent to the circles can be written in the form $y=mx+b$ with $m>0$ . What is $b$ ?

$[asy]size(150); defaultpen(linewidth(0.7)+fontsize(8)); draw(circle((2,4),4));draw(circle((14,9),9)); draw((0,-2)--(0,20));draw((-6,0)--(25,0)); draw((2,4)--(2,4)+4*expi(pi*4.5/11)); draw((14,9)--(14,9)+9*expi(pi*6/7)); label("4",(2,4)+2*expi(pi*4.5/11),(-1,0)); label("9",(14,9)+4.5*expi(pi*6/7),(1,1)); label("(2,4)",(2,4),(0.5,-1.5));label("(14,9)",(14,9),(1,-1)); draw((-4,120*-4/119+912/119)--(11,120*11/119+912/119)); dot((2,4)^^(14,9));[/asy]$

$mathrm{(A) } frac{908}{119}qquad mathrm{(B) } frac{909}{119}qquad mathrm{(C) } frac{130}{17}qquad mathrm{(D) } frac{911}{119}qquad mathrm{(E) } frac{912}{119}$

Problem 20

A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?

$mathrm{(A) } frac{1}{2187}qquad mathrm{(B) } frac{1}{729}qquad mathrm{(C) } frac{2}{243}qquad mathrm{(D) } frac{1}{81}qquad mathrm{(E) } frac{5}{243}$

Problem 21

Let

$S_1={(x,y)|log_{10}(1+x^2+y^2)le 1+log_{10}(x+y)}$

and

$S_2={(x,y)|log_{10}(2+x^2+y^2)le 2+log_{10}(x+y)}$ .

What is the ratio of the area of $S_2$ to the area of $S_1$ ?

$mathrm{(A) } 98qquad mathrm{(B) } 99qquad mathrm{(C) } 100qquad mathrm{(D) } 101qquad mathrm{(E) } 102$

Problem 22

A circle of radius $r$ is concentric with and outside a regular hexagon of side length $2$ . The probability that three entire sides of hexagon are visible from a randomly chosen point on the circle is $1/2$ . What is $r$ ?

$mathrm{(A) } 2sqrt{2}+2sqrt{3}qquad mathrm{(B) } 3sqrt{3}+sqrt{2}qquad mathrm{(C) } 2sqrt{6}+sqrt{3} qquad mathrm{(D) } 3sqrt{2}+sqrt{6}qquad mathrm{(E) } 6sqrt{2}-sqrt{3}$

Problem 23

Given a finite sequence $S=(a_1,a_2,ldots ,a_n)$ of $n$ real numbers, let $A(S)$ be the sequence

$left(frac{a_1+a_2}{2},frac{a_2+a_3}{2},ldots ,frac{a_{n-1}+a_n}{2}right)$

of $n-1$ real numbers. Define $A^1(S)=A(S)$ and, for each integer $m$ , $2le mle n-1$ , define $A^m(S)=A(A^{m-1}(S))$ . Suppose $x>0$ , and let $S=(1,x,x^2,ldots ,x^{100})$ . If $A^{100}(S)=(1/2^{50})$ , then what is $x$ ?

$mathrm{(A) } 1-frac{sqrt{2}}{2}qquad mathrm{(B) } sqrt{2}-1qquad mathrm{(C) } frac{1}{2}qquad mathrm{(D) } 2-sqrt{2}qquad mathrm{(E) } frac{sqrt{2}}{2}$

Problem 24

The expression

$(x+y+z)^{2006}+(x-y-z)^{2006}$

is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

$mathrm{(A) } 6018qquad mathrm{(B) } 671,676qquad mathrm{(C) } 1,007,514qquad mathrm{(D) } 1,008,016qquad mathrm{(E) } 2,015,028$

Problem 25

How many non-empty subsets $S$ of $lbrace 1,2,3,ldots ,15rbrace$ have the following two properties?

$(1)$ No two consecutive integers belong to $S$ .

$(2)$ If $S$ contains $k$ elements, then $S$ contains no number less than $k$ .

$mathrm{(A) } 277qquad mathrm{(B) } 311qquad mathrm{(C) } 376qquad mathrm{(D) } 377qquad mathrm{(E) } 405$

2006 AMC12 A 真题答案详细解析

请扫描下方二维码咨询

The $5$ sandwiches cost $5cdot 3=15$ dollars. The $8$ sodas cost $8cdot 2=16$ dollars. In total, the purchase costs $15+16=31$ dollars. The answer is $mathrm{(A)}$ .
By the definition of $otimes$ , we have $hotimes h=h^{3}-h$ . Then $hotimes (hotimes h)=hotimes (h^{3}-h)=h^{3}-(h^{3}-h)=h$ . The answer is $mathrm{(C)}$ .
Solution 1

Let $m$ be Mary's age. Then $frac{m}{30}=frac{3}{5}$ . Solving for $m$ , we obtain $m=18$ . The answer is $mathrm{(B)}$ .

Solution 2

We can see this is a combined ratio of $8$ , $(5+3)$ . We can equalize by doing $30div5=6$ , and $6cdot3=18$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6cdot8=30+18$ therefore we can see our answer is correct.
Solution 1

From the greedy algorithm, we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=23Rightarrowmathrm{(E)}$

Solution 2

With a matrix we can see $begin{bmatrix} 1+2&9&6&3\ 1+11&8&5&2\ 1+0&7&4&2 end{bmatrix}$ The largest digit sum we can see is $9$ . For the minutes digits, we can combine the largest $2$ digits, which are $9,5=9+5=14$ which we can then do $14+9=23$
Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $frac{3}{8}cdot 8=3$ . Dave paid $10-3=7$ dollars. Dave paid $7-3=4$ more dollars than Doug. The answer is $mathrm{(D)}$ .
Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below. $[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$A$",A,W); label("$B$",B,NE); label("$C$",(12.6,4)); label("$D$",D,SW); label("$12$",E--B,N); label("$12$",D--F,S); label("$4$",E--A,W); label("$4$",(12.4,-1.75),E); label("$8$",A--D,W); label("$8$",(12.4,4),E); label("$y$",A--H,S); label("$y$",G--C,N); [/asy]$ As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = frac{12}{2} = 6 Longrightarrow mathrm{(A)}$ .
Let $m$ be Mary's age, let $s$ be Sally's age, and let $d$ be Danielle's age. We have $s=.6d$ , and $m=1.2s=1.2(.6d)=.72d$ . The sum of their ages is $m+s+d=.72d+.6d+d=2.32d$ . Therefore, $2.32d=23.2$ , and $d=10$ . Then $m=.72(10)=7.2$ . Mary will be $8$ on her next birthday. The answer is $mathrm{(B)}$ .
Notice that if the consecutive positive integers have a sum of 15, then their average (which could be a fraction) must be a divisor of 15. If the number of integers in the list is odd, then the average must be either 1, 3, or 5, and 1 is clearly not possible. The other two possibilities both work:
- $1 + 2 + 3 + 4 + 5 = 15$
- $4 + 5 + 6 = 15$
If the number of integers in the list is even, then the average will have a $frac{1}{2}$ . The only possibility is $frac{15}{2}$ , from which we get:
- $15 = 7 + 8$
Thus, the correct answer is 3, answer choice $mathrm{(C) }$ .
Let the price of a pencil be $p$ and an eraser $e$ . Then $13p + 3e = 100$ with $p > e > 0$ . Since $p$ and $e$ are positive integers, we must have $e geq 1$ and $p geq 2$ .Considering the equation $13p + 3e = 100$ modulo 3 (that is, comparing the remainders when both sides are divided by 3) we have $p + 0e equiv 1 pmod 3$ so $p$ leaves a remainder of 1 on division by 3.Since $p geq 2$ , possible values for $p$ are 4, 7, 10 ....Since 13 pencils cost less than 100 cents, $13p < 100$ . $13 times 10 = 130$ is too high, so $p$ must be 4 or 7.If $p = 4$ then $13p = 52$ and so $3e = 48$ giving $e = 16$ . This contradicts the pencil being more expensive. The only remaining value for $p$ is 7; then the 13 pencils cost $7 times 13= 91$ cents and so the 3 erasers together cost 9 cents and each eraser costs $frac{9}{3} = 3$ cents.Thus one pencil plus one eraser cost $7 + 3 = 10$ cents, which is answer choice $mathrm{(A) }$ .
For $sqrt{120-sqrt{x}}$ to be an integer, $120-sqrt{x}$ must be a perfect square.Since $sqrt{x}$ can't be negative, $120-sqrt{x} leq 120$ .The perfect squares that are less than or equal to $120$ are ${0,1,4,9,16,25,36,49,64,81,100}$ , so there are $11$ values for $120-sqrt{x}$ .Since every value of $120-sqrt{x}$ gives one and only one possible value for $x$ , the number of values of $x$ is $11 Rightarrow boxed{E}$ .
$begin{align*}(x+y)^2&=x^2+y^2\ x^2 + 2xy + y^2 &= x^2 + y^2\ 2xy &= 0end{align*}$
Either $x = 0$ or $y = 0$ . The union of them is 2 lines, and thus the answer is $mathrm{(C)}$ . $[asy] draw((0,-50)--(0,50));draw((-50,0)--(50,0));[/asy]$
Solution 1

The inside diameters of the rings are the positive integers from 1 to 18. The total distance needed is the sum of these values plus 2 for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series, the answer is $frac{18 cdot 19}{2} + 2 = 173 Rightarrow mathrm{(B)}$ .

Solution 2

Alternatively, the sum of the consecutive integers from 3 to 20 is $frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = 173$ .
Let the radius of the smallest circle be $r_A$ , the radius of the second largest circle be $r_B$ , and the radius of the largest circle be $r_C$ . $[r_A + r_B = 3]$ $[r_A + r_C = 4]$ $[r_ B + r_C = 5]$ Adding up all these equations and then dividing both sides by 2, we get, $[r_A + r_B + r_C = 6]$ Then, we get $r_A = 1$ , $r_B = 2$ , and $r_C = 3$ Then we get $1^2 pi + 2^2 pi + 3^2 pi = 14 pi iffmathrm{(E)}$
Solution 1

The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest x possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation. Bezout’s Identity tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the greatest common divisor of $a$ and $b$ . Therefore, the answer is $gcd(300,210)$ , which is $30$ , $mathrm{(C) }$

Solution 2

Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by 30 no matter what $p$ and $g$ are, so our answer must be divisible by 30. In addition, three goats minus two pigs gives us $630 - 600 = 30$ exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is $mathrm{(C) }$ . debt that can be resolved.

Solution 3

Let us simplify this problem. Dividing by $30$ , we get a pig to be: $frac{300}{30} = 10$ , and a goat to be $frac{210}{30}= 7$ . It becomes evident that if you exchange $5$ pigs for $7$ goats, we get the smallest positive difference - $5cdot 10 - 7cdot 7 = 50-49 = 1$ . Since we originally divided by $30$ , we need to multiply again, thus getting the answer: $1cdot 30 = mathrm{(C) 30}$
- For $cos x = 0$ , x must be in the form of $frac{pi}{2} + pi n$ , where $n$ denotes any integer.
- For $cos (x+z) = 1 / 2$ , $x + z = frac{pi}{3} +2pi n, frac{5pi}{3} + 2pi n$ .
The smallest possible value of $z$ will be that of $frac{5pi}{3} - frac{3pi}{2} = frac{pi}{6} Rightarrow mathrm{(A)}$ .
$angle AED$ and $angle BEC$ are vertical angles so they are congruent, as are angles $angle ADE$ and $angle BCE$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $triangle ACE sim triangle BDE$ . By the Pythagorean Theorem, line segment $DE = 4$ . The sides are proportional, so $frac{DE}{AD} = frac{CE}{BC} Rightarrow frac{4}{3} = frac{CE}{8}$ . This makes $CE = frac{32}{3}$ and $CD = CE + DE = 4 + frac{32}{3} = frac{44}{3} Longrightarrow mathrm{B}$ .
Solution 1

One possibility is to use the coordinate plane, setting $B$ at the origin. Point $A$ will be $(0,s)$ and $E$ will be $left(s + frac{r}{sqrt{2}}, s + frac{r}{sqrt{2}}right)$ since $B, D$ , and $E$ are collinear and contain a diagonal of $ABCD$ . The Pythagorean theorem results in

$[AF^2 + EF^2 = AE^2]$

$[r^2 + left(sqrt{9 + 5sqrt{2}}right)^2 = left(left(s + frac{r}{sqrt{2}}right) - 0right)^2 + left(left(s + frac{r}{sqrt{2}}right) - sright)^2]$

$[r^2 + 9 + 5sqrt{2} = s^2 + rssqrt{2} + frac{r^2}{2} + frac{r^2}{2}]$

$[9 + 5sqrt{2} = s^2 + rssqrt{2}]$

This implies that $rs = 5$ and $s^2 = 9$ ; dividing gives us $frac{r}{s} = frac{5}{9} Rightarrow B$ .

Solution 2

First note that angle $angle AFE$ is right since $overline{AF}$ is tangent to the circle. Using the Pythagorean Theorem on $triangle AFE$ , then, we see $[AE^2 = 9 + 5sqrt{2} + r^2.]$

But it can also be seen that $angle BDA = 45^circ$ . Therefore, since $D$ lies on $overline{BE}$ , $angle ADE = 135^circ$ . Using the Law of Cosines on $triangle ADE$ , we see

$[AE^2 = s^2 + r^2 - 2srcos(135^circ)]$ $[AE^2 = s^2 + r^2 - 2srleft(-frac{1}{sqrt{2}}right)]$ $[AE^2 = s^2 + r^2 + sqrt{2}sr]$ $[9 + 5sqrt{2} + r^2 = s^2 + r^2 + sqrt{2}sr]$ $[9 + 5sqrt{2} = s^2 + sqrt{2}sr.]$

Thus, since $r$ and $s$ are rational, $s^2 = 9$ and $sr = 5$ . So $s = 3$ , $r = frac{5}{3}$ , and $frac{r}{s} = frac{5}{9}$ .

Solution 3

(Similar to Solution 1) First, draw line AE and mark a point Z that is equidistant from E and D so that $angle{DZE} = 90^circ$ and that line $overline{AZ}$ includes point D. Since DE is equal to the radius $r$ , $DZ=EZ=frac{r}{sqrt2}=frac{rsqrt2}{2}.$

Note that triangles $AFE$ and $AZE$ share the same hypotenuse $(AE)$ , meaning that $[AZ^2+EZ^2=AF^2+EF^2]$ Plugging in our values we have: $[(s+frac{rsqrt{2}}{2})^2+(frac{rsqrt{2}}{2})^2=(sqrt{9+5sqrt{2}})^2+r^2]$ $[s^2+srsqrt{2}+frac{r^2}{2}+frac{r^2}{2}=9+5sqrt{2}+r^2]$ $[s^2+srsqrt{2}=9+5sqrt{2}]$ By logic $s=3$ and $sr=5 implies r=5/3.$

Therefore, $frac{frac{5}{3}}{3}=frac{5}{9}=boxed{B}$
Quickly verifying by plugging in values verifies that $-1$ and $1$ are in the domain. $f(x)+fleft(frac{1}{x}right)=x$ Plugging in $frac{1}{x}$ into the function: $fleft(frac{1}{x}right)+fleft(frac{1}{frac{1}{x}}right)=frac{1}{x}$ $fleft(frac{1}{x}right)+ f(x)= frac{1}{x}$ Since $f(x) + fleft(frac{1}{x}right)$ cannot have two values: $x = frac{1}{x}$ $x^2 = 1$
$x=pm 1$

Therefore, the largest set of real numbers that can be in the domain of $f$ is ${-1,1} Rightarrow E$
Let $L_1$ be the line that goes through $(2,4)$ and $(14,9)$ , and let $L_2$ be the line $y=mx+b$ . If we let $theta$ be the measure of the acute angle formed by $L_1$ and the x-axis, then $tantheta=frac{5}{12}$ . $L_1$ clearly bisects the angle formed by $L_2$ and the x-axis, so $m=tan{2theta}=frac{2tantheta}{1-tan^2{theta}}=frac{120}{119}$ . We also know that $L_1$ and $L_2$ intersect at a point on the x-axis. The equation of $L_1$ is $y=frac{5}{12}x+frac{19}{6}$ , so the coordinate of this point is $left(-frac{38}{5},0right)$ . Hence the equation of $L_2$ is $y=frac{120}{119}x+frac{912}{119}$ , so $b=frac{912}{119}$ , and our answer choice is $boxed{mathrm{E}}$ .
Solution 1

$[asy] import three; unitsize(1cm); size(50); currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); [/asy]$

Let us count the good paths. The bug starts at an arbitrary vertex, moves to a neighboring vertex (3 ways), and then to a new neighbor (2 more ways). So, without loss of generality, let the cube have vertices $ABCDEFGH$ such that $ABCD$ and $EFGH$ are two opposite faces with $A$ above $E$ and $B$ above $F$ . The bug starts at $A$ and moves first to $B$ , then to $C$ .

From this point, there are two cases.

Case 1: the bug moves to $D$ . From $D$ , there is only one good move available, to $H$ . From $H$ , there are two ways to finish the trip, either by going $H to G to F to E$ or $H to E to F to G$ . So there are 2 good paths in this case.

Case 2: the bug moves to $G$ . Case 2a: the bug moves $G to H$ . In this case, there are 0 good paths because it will not be possible to visit both $D$ and $E$ without double-visiting some vertex. Case 2b: the bug moves $G to F$ . There is a unique good path in this case, $F to E to H to D$ .

Thus, all told we have 3 good paths after the first two mo, for a total of $3cdot 3 cdot 2 = 18$ good paths. There were $3^7 = 2187$ possible paths the bug could have taken, so the probability a random path is good is the ratio of good paths to total paths, $frac{18}{2187} = frac2{243}Rightarrow boxed{mathrm (C)}$ .

Solution 2 (using the answer choices)

As in Solution 1, the bug can move from its arbitrary starting vertex to a neighboring vertex in 3 ways. After this, the bug can move to a new neighbor in 2 ways (it cannot return to the first vertex). The total number of paths (as stated above) is $3^7$ or $2187$ . Therefore, the probability of the bug following a good path is equal to $frac{6x}{2187}$ for some positive integer $x$ . The only answer choice which can be expressed in this form is $frac2{243}Rightarrowboxed{mathrm (C)}$ .
Looking at the constraints of $S_1$ : $log_{10}(1+x^2+y^2)le 1+log_{10}(x+y)$ $log_{10}(1+x^2+y^2)le log_{10} 10 +log_{10}(x+y)$ $log_{10}(1+x^2+y^2)le log_{10}(10x+10y)$ $1+x^2+y^2 le 10x+10y$ $x^2-10x+y^2-10y le -1$ $x^2-10x+25+y^2-10y+25 le 49$ $(x-5)^2 + (y-5)^2 le (7)^2$
$S_1$ is a circle with a radius of $7$ . So, the area of $S_1$ is $49pi$ .

Looking at the constraints of $S_2$ :

$log_{10}(2+x^2+y^2)le 2+log_{10}(x+y)$

$log_{10}(2+x^2+y^2)le log_{10} 100 +log_{10}(x+y)$

$log_{10}(2+x^2+y^2)le log_{10}(100x+100y)$

$2+x^2+y^2 le 100x+100y$

$x^2-100x+y^2-100y le -2$

$x^2-100x+2500+y^2-100y+2500 le 4998$

$(x-50)^2 + (y-50)^2 le (7sqrt{102})^2$

$S_2$ is a circle with a radius of $7sqrt{102}$ . So, the area of $S_2$ is $4998pi$ .

So the desired ratio is $frac{4998pi}{49pi} = 102 Rightarrow boxed{E}$ .
Project any two non-adjacent and non-opposite sides of the hexagon to the circle; the arc between the two points formed is the location where all three sides of the hexagon can be fully viewed. Since there are six such pairs of sides, there are six arcs. The probability of choosing a point is $1 / 2$ , or if the total arc degree measures add up to $frac{1}{2} cdot 360^{circ} = 180^{circ}$ . Each arc must equal $frac{180^{circ}}{6} = 30^mathrm{circ}$ .Call the center $O$ , and the two endpoints of the arc $A$ and $B$ , so $angle AOB = 30^{circ}$ . Let $P$ be the intersections of the projections of the sides of the hexagon corresponding to $overline{AB}$ . Notice that $triangle APO$ is an isosceles triangle: $angle AOP = 15^{circ}$ and $angle OAP = OAB - 60^{circ} = frac{180^{circ}-30^{circ}}{2} - 60^{circ} = 15^{circ}$ . Since $OA$ is a radius and $OP$ can be found in terms of a side of the hexagon, we are almost done.If we draw the altitude of $APO$ from $P$ , then we get a right triangle. Using simple trigonometry, $cos 15^{circ} = frac{frac{r}{2}}{2sqrt{3}} = frac{r}{4sqrt{3}}$ .Since $cos 15^{circ} = cos (45^{circ} - 30^{circ}) = frac{sqrt{6} + sqrt{2}}{4}$ , we get $r = left(frac{sqrt{6} + sqrt{2}}{4}right) cdot 4sqrt{3} = 3sqrt{2} + sqrt{6} Rightarrow mathrm{(D)}$ .
$A^1(S)=left(frac{1+x}{2},frac{x+x^2}{2},...,frac{x^{99}+x^{100}}{2}right)$ $A^2(S)=left(frac{1+2x+x^2}{2^2},frac{x+2x^2+x^3}{2^2},...,frac{x^{98}+2x^{99}+x^{100}}{2^2}right)$ $Rightarrow A^2(S)=left(frac{(x+1)^2}{2^2},frac{x(x+1)^2}{2^2},...,frac{x^{98}(x+1)^2}{2^2}right)$ In general, $A^n(S)=left(frac{(x+1)^n}{2^n},frac{x(x+1)^n}{2^n},...,frac{x^{100-n}(x+1)^n}{2^n}right)$ such that $A^n(s)$ has $101-n$ terms. Specifically, $A^{100}(S)=frac{(x+1)^{100}}{2^{100}}$ To find x, we need only solve the equation $frac{(x+1)^{100}}{2^{100}}=frac{1}{2^{50}}$ . Algebra yields $x=sqrt{2}-1$ .
Solution 1

By the Multinomial Theorem, the summands can be written as

$[sum_{a+b+c=2006}{frac{2006!}{a!b!c!}x^ay^bz^c}]$

and

$[sum_{a+b+c=2006}{frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},]$

respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:

$[{2006+2choose 2} = 2015028]$

terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:

if the exponent of $y$ is 1, the exponent of $z$ can be all even integers up to 2004, so there are 1003 terms.

if the exponent of $y$ is 3, the exponent of $z$ can go up to 2002, so there are 1002 terms.

$vdots$

if the exponent of $y$ is 2005, then $z$ can only be 0, so there is 1 term.

If we add them up, we get $frac{1003cdot1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003cdot1004$ negative terms.

By subtracting this number from 2015028, we obtain $boxed{D}$ or $1008016$ as our answer.

Solution 2

Alternatively, we can use a generating function to solve this problem. The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of $k$ ). In other words, we want to find $f(x)$ where the coefficient of $x^k$ equals the number of unique terms in $(x+y+z)^k + (x-y-z)^k$ .

First, we note that all unique terms in the expression have the form, $Cx^ay^bz^c$ , where $a+b+c=k$ and $C$ is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of $(x+y+z)^k + (x-y-z)^k$ is $[(1+x+x^2+x^3cdots)^3 = frac{1}{(1-x)^3}]$

Secondly, we note that a certain number of terms of the form, $Cx^ay^bz^c$ , do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when $1 equiv b+ctext{ (mod }2text{)}$ because every unique term is of the form: $[binom{k}{a,b,c}x^ay^bz^c+(-1)^{b+c}binom{k}{a,b,c}x^ay^bz^c]$ for all possible $a,b,c$ .

Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of $k$ . With some thought, we see that this desired generating function is the following: $[2(x+x^3+x^5cdots)(1+x^2+x^4cdots)(1+x+x^2+x^3cdots) = frac{2x}{(1-x)^3(1+x)^2}]$

Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in $(x+y+z)^k + (x-y-z)^k$ , our initial goal: $[frac{1}{(1-x)^3}-frac{2x}{(1-x)^3(1+x)^2} = frac{x^2+1}{(1-x)^3(1+x)^2}]$ which equals $[(x^2+1)(1+x+x^2cdots)^3(1-x+x^2-x^3cdots)^2]$

The coefficient of $x^{2006}$ of the above expression equals $[sum_{a=0}^{2006}binom{2+a}{2}binom{1+2006-a}{1}(-1)^a + sum_{a=0}^{2004}binom{2+a}{2}binom{1+2004-a}{1}(-1)^a]$

Evaluating the expression, we get $1008016$ , as expected.

Solution 3

Define $P$ such that $P=y+z$ . Then the expression in the problem becomes: $(x+P)^{2006}+(x-P)^{2006}$ .

Expanding this using binomial theorem gives $(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)$ , where $n=2006$ (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).

Simplifying gives: $2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)$ . Note that two terms that come out of different powers of $P$ cannot combine and simplify, as their exponent of $x$ will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, $P^k=(y+z)^k$ will have $k+1$ terms, so the answer is $1+3+5+...+2007=1004^2=1,008,016 implies boxed{D}$ .

Solution 4

Using stars and bars we know that $(x+y+z)^{2006}$ has ${2006+2choose 2}$ or $2015028$ different terms. Now we need to find out how many of these terms are canceled out by $(x-y-z)^{2006}$ . We know that for any term(let's say $x^{a}(-y)^{b}(-z)^{c}$ ) where $a+b+c=2006$ of the expansion of $(x-y-z)^{2006}$ is only going to cancel out with the corresponding term $x^{a}y^{b}z^{c}$ if only $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. Now let's do some casework to see how many terms fit this criteria:

Case 1: $a$ is even

Now we know that $a$ is even and $a+b+c=2006$ . Thus $b+c$ is also even or both $b$ and $c$ are odd or both $b$ and $c$ are even. This case clearly fails the above criteria and there are 0 possible solutions.

Case 2: $a$ is odd

Now we know that $a$ is odd and $a+b+c=2006$ . Thus $b+c$ is odd and $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. All terms that have $a$ being odd work.

We now need to figure out how many of the terms have $a$ as a odd number. We know that $a$ can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus $frac{1003}{2007}$ of the $2015028$ terms will cancel out and $frac{1004}{2007}$ of the terms will work. Thus there are $(frac{1004}{2007})(2015028)$ terms. This number comes out to be $1,008,016$ $implies boxed{D}$ (Author: David Camacho)
Solution 1

This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem:

How many ways are there to choose $k$ elements from an ordered $n$ element set without choosing two consecutive members?

You want to choose $k$ numbers out of $n$ with no consecutive numbers. For each configuration, we can subtract $i-1$ from the $i$ -th element in your subset. This converts your configuration into a configuration with $k$ elements where the largest possible element is $n-k+1$ , with no restriction on consecutive numbers. Since this process is easily reversible, we have a bijection. Without consideration of the second condition, we have: ${15 choose 1} + {14 choose 2} + {13 choose 3} + ... + {9 choose 7} + {8 choose 8}$

Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than $k$ , which translates to subtracting $k - 1$ from the "top" of each binomial coefficient. Now we have, after we cancel all the terms ${n choose k}$ where $n < k$ , ${15 choose 1} + {13 choose 2} + {11 choose 3} + {9 choose 4} + {7 choose 5}= 15 + 78 + 165 + 126 + 21 = boxed{405} Longrightarrow mathrm{(E)}$

Solution 2

Another way of visualizing the solution above would be to use $|$ 's and $-$ 's. Denote $|$ as the numbers we have chosen and $-$ as other numbers. Taking an example, assuming we are picking two numbers, we imagine the shape $| - |$ . This notation forces a number between the two chosen numbers, which blocks the two numbers we picked from being consecutive. Now we consider the orientations with this shape. We have $15 - 3 = 12$ remaining numbers.

We need to find the number of ways to place the remaining $12 -$ 's. We can find this by utilizing stars and bars, with the following marker being placed to represent groups: *| - *|*. Now, we have to place $12$ numbers within $3$ groups, which is ${14 choose 2}$ . The same concept can be used for the remaining numbers. The rest of the solution continues as above.

Solution by: Everyoneintexas

Solution 3

We have the same setup as in the previous solution.

Note that if $n < 2k - 1$ , the answer will be 0. Otherwise, the $k$ elements we choose define $k + 1$ boxes (which divide the nonconsecutive numbers) into which we can drop the $n - k$ remaining elements, with the caveat that each of the middle $k - 1$ boxes must have at least one element (since the numbers are nonconsecutive). This is equivalent to dropping $n - 2k + 1$ elements into $k + 1$ boxes, where each box is allowed to be empty. And this is equivalent to arranging $n - k + 1$ objects, $k$ of which are dividers, which we can do in $F(n, k) = {n - k + 1 choose k}$ ways.

Now, looking at our original question, we see that the thing we want to calculate is just $F(15, 1) + F(14, 2) + F(13, 3) + F(12, 4) + F(11, 5) = {15choose 1} + {13choose2} + {11choose 3} + {9choose 4} + {7 choose 5} = 15 + 78 + 165 + 126 + 21 = 405 Longrightarrow mathrm{(E)}$

以上解析方式仅供参考