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2000 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4:56

2000 AMC 12 真题

答案详细解析请参考文末

Problem 1

In the year $2001$ , the United States will host the International Mathematical Olympiad. Let $I,M,$ and $O$ be distinct positive integers such that the product $I cdot M cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$ ? $textbf{(A)} 23 qquad textbf{(B)} 55 qquad textbf{(C)} 99 qquad textbf{(D)} 111 qquad textbf{(E)} 671$

Problem 2

$2000(2000^{2000}) =$ $textbf{(A)} 2000^{2001} qquad textbf{(B)} 4000^{2000} qquad textbf{(C)} 2000^{4000} qquad textbf{(D)} 4,000,000^{2000} qquad textbf{(E)} 2000^{4,000,000}$

Problem 3

Each day, Jenny ate $20%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally? $textbf{(A)} 40 qquad textbf{(B)} 50 qquad textbf{(C)} 55 qquad textbf{(D)} 60 qquad textbf{(E)} 75$

Problem 4

The Fibonacci sequence $1,1,2,3,5,8,13,21,ldots$ starts with two $1$ 's, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence? $textbf{(A)} 0 qquad textbf{(B)} 4 qquad textbf{(C)} 6 qquad textbf{(D)} 7 qquad textbf{(E)} 9$

Problem 5

If $|x - 2| = p,$ where $x < 2,$ then $x - p =$ $textbf{(A)} -2 qquad textbf{(B)} 2 qquad textbf{(C)} 2-2p qquad textbf{(D)} 2p-2 qquad textbf{(E)} |2p-2|$

Problem 6

Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? $textbf{(A)} 22 qquad textbf{(B)} 60 qquad textbf{(C)} 119 qquad textbf{(D)} 194 qquad textbf{(E)} 231$

Problem 7

How many positive integers $b$ have the property that $log_{b} 729$ is a positive integer? $textbf{(A)} 0 qquad textbf{(B)} 1 qquad textbf{(C)} 2 qquad textbf{(D)} 3 qquad textbf{(E)} 4$

Problem 8

Figures $0$ , $1$ , $2$ , and $3$ consist of $1$ , $5$ , $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]$ $textbf{(A)} 10401 qquadtextbf{(B)} 19801 qquadtextbf{(C)} 20201 qquadtextbf{(D)} 39801 qquadtextbf{(E)} 40801$

Problem 9

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were $71,76,80,82,$ and $91$ . What was the last score Mrs. Walters entered? $textbf{(A)} 71 qquad textbf{(B)} 76 qquad textbf{(C)} 80 qquad textbf{(D)} 82 qquad textbf{(E)} 91$

Problem 10

The point $P = (1,2,3)$ is reflected in the $xy$ -plane, then its image $Q$ is rotated $180^circ$ about the $x$ -axis to produce $R$ , and finally, $R$ is translated $5$ units in the positive- $y$ direction to produce $S$ . What are the coordinates of $S$ ? $textbf {(A) } (1,7, - 3) qquad textbf {(B) } ( - 1,7, - 3) qquad textbf {(C) } ( - 1, - 2,8) qquad textbf {(D) } ( - 1,3,3) qquad textbf {(E) } (1,3,3)$

Problem 11

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$ . Which of the following is a possible value of $frac {a}{b} + frac {b}{a} - ab$ ? $textbf{(A)} - 2 qquad textbf{(B)} frac { - 1}{2} qquad textbf{(C)} frac {1}{3} qquad textbf{(D)} frac {1}{2} qquad textbf{(E)} 2$

Problem 12

Let $A, M,$ and $C$ be nonnegative integers such that $A + M + C=12$ . What is the maximum value of $A cdot M cdot C + A cdot M + M cdot C + A cdot C$ ? $textbf{(A)} 62 qquad textbf{(B)} 72 qquad textbf{(C)} 92 qquad textbf{(D)} 102 qquad textbf{(E)} 112$

Problem 13

One morning each member of Angela’s family drank an $8$ -ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family? $textbf {(A)} 3 qquad textbf {(B)} 4 qquad textbf {(C)} 5 qquad textbf {(D)} 6 qquad textbf {(E)} 7$

Problem 14

When the mean, median, and mode of the list $[10,2,5,2,4,2,x]$ are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of $x$ ? $textbf {(A)} 3 qquad textbf {(B)} 6 qquad textbf{(C)} 9 qquad textbf {(D)} 17 qquad textbf {(E)} 20$

Problem 15

Let $f$ be a function for which $f(x/3) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$ . $textbf {(A)} -1/3 qquad textbf {(B)} -1/9 qquad textbf {(C)} 0 qquad textbf {(D)} 5/9 qquad textbf {(E)} 5/3$

Problem 16

A checkerboard of $13$ rows and $17$ columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered $1,2,ldots,17$ , the second row $18,19,ldots,34$ , and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1,2,ldots,13$ , the second column $14,15,ldots,26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system). $textbf {(A)} 222 qquad textbf {(B)} 333qquad textbf {(C)} 444 qquad textbf {(D)} 555 qquad textbf {(E)} 666$

Problem 17

A circle centered at $O$ has radius $1$ and contains the point $A$ . The segment $AB$ is tangent to the circle at $A$ and $angle AOB = theta$ . If point $C$ lies on $overline{OA}$ and $overline{BC}$ bisects $angle ABO$ , then $OC =$ $textbf {(A)} sec^2 theta - tan theta qquad textbf {(B)} frac 12 qquad textbf {(C)} frac{cos^2 theta}{1 + sin theta}qquad textbf {(D)} frac{1}{1+sintheta} qquad textbf {(E)} frac{sin theta}{cos^2 theta}$

Problem 18

In year $N$ , the $300$ th day of the year is a Tuesday. In year $N+1$ , the $200$ th day is also a Tuesday. On what day of the week did the $100$ th day of year $N-1$ occur? $textbf {(A)} text{Thursday} qquad textbf {(B)} text{Friday}qquad textbf {(C)} text{Saturday}qquad textbf {(D)} text{Sunday}qquad textbf {(E)} text{Monday}$

Problem 19

In triangle $ABC$ , $AB = 13$ , $BC = 14$ , $AC = 15$ . Let $D$ denote the midpoint of $overline{BC}$ and let $E$ denote the intersection of $overline{BC}$ with the bisector of angle $BAC$ . Which of the following is closest to the area of the triangle $ADE$ ? $textbf {(A)} 2 qquad textbf {(B)} 2.5 qquad textbf {(C)} 3 qquad textbf {(D)} 3.5 qquad textbf {(E)} 4$

Problem 20

If $x,y,$ and $z$ are positive numbers satisfying $x + frac{1}{y} = 4, y + frac{1}{z} = 1,$ and $z + frac{1}{x} = frac73,$ then what is the value of $xyz$ ? $textbf {(A)} 2/3 qquad textbf {(B)} 1 qquad textbf {(C)} 4/3 qquad textbf {(D)} 2 qquad textbf {(E)} 7/3$

Problem 21

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is $textbf {(A)} frac{1}{2m+1} qquad textbf {(B)} m qquad textbf {(C)} 1-m qquad textbf {(D)} frac{1}{4m} qquad textbf {(E)} frac{1}{8m^2}$

Problem 22

The graph below shows a portion of the curve defined by the quartic polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d$ . Which of the following is the smallest? $textbf{(A)} P(-1)\ textbf{(B)} text{The product of the zeros of } P\ textbf{(C)} text{The product of the non-real zeros of } P \ textbf{(D)} text{The sum of the coefficients of } P \ textbf{(E)} text{The sum of the real zeros of } P$

Problem 23

Professor Gamble buys a lottery ticket, which requires that he pick six different integers from $1$ through $46$ , inclusive. He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property— the sum of the base-ten logarithms is an integer. What is the probability that Professor Gamble holds the winning ticket? $textbf {(A)} 1/5 qquad textbf {(B)} 1/4 qquad textbf {(C)} 1/3 qquad textbf {(D)} 1/2 qquad textbf {(E)} 1$

Problem 24

If circular arcs $AC$ and $BC$ have centers at $B$ and $A$ , respectively, then there exists a circle tangent to both $stackrel{frown}{AC}$ and $stackrel{frown}{BC}$ , and to $overline{AB}$ . If the length of $stackrel{frown}{BC}$ is $12$ , then the circumference of the circle is $textbf {(A)} 24 qquad textbf {(B)} 25 qquad textbf {(C)} 26 qquad textbf {(D)} 27 qquad textbf {(E)} 28$

Problem 25

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.) $textbf {(A)} 210 qquad textbf {(B)} 560 qquad textbf {(C)} 840 qquad textbf {(D)} 1260 qquad textbf {(E)} 1680$ $[asy] import three; import math; unitsize(1.5cm); currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); [/asy]$

2000 AMC12 真题答案详细解析

1.The sum is the highest if two factors are the lowest.
So, $1 cdot 3 cdot 667 = 2001$ and $1+3+667=671 Longrightarrow boxed{text{(E)}}$ .
$2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2001} Rightarrow boxed{A}$
Since Jenny eats $20%$ of her jelly beans per day, $80%=frac{4}{5}$ of her jelly beans remain after one day.Let $x$ be the number of jelly beans in the jar originally. $frac{4}{5}cdotfrac{4}{5}cdot x=32$ $frac{16}{25}cdot x=32$ $x=frac{25}{16}cdot32= 50 Rightarrow boxed{B}$
Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $bmod{10}$ : $1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$ The last digit to appear in the units position of a number in the Fibonacci sequence is $6 Longrightarrow boxed{mathrm{C}}$ .
When $x < 2,$ $x-2$ is negative so $|x - 2| = 2-x = p$ and $x = 2-p$ .Thus $x-p = (2-p)-p = 2-2p$ . $boxed{text{C}}$
Solution 1

All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B, D, and A. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can eliminate E. So, the answer must be $boxed{textbf{(C) }119}$ .

Solution 2

Let the two primes be $p$ and $q$ . We wish to obtain the value of $pq-(p+q)$ , or $pq-p-q$ . Using Simon's Favorite Factoring Trick, we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$ . Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$ , we see that the answer is $boxed{textbf{(C) }119}$ .

Solution 3

The answer must be in the form $pq - p - q$ = $(p - 1)(q - 1) - 1$ . Since $p - 1$ and $q - 1$ are both even, $(p - 1)(q - 1) - 1$ is $3 pmod 4$ , and the only answer that is $3 pmod 4$ is $boxed{textbf{(C) }119}$ .
If $displaystyle log_{b} 729 = n$ , then $b^n = 729$ . Since $729 = 3^6$ , $displaystyle b$ must be $3$ to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of $displaystyle b$ $Longrightarrow mathrm{E}$ .
Figures $0$ , $1$ , $2$ , and $3$ consist of $1$ , $5$ , $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]$ $mathrm{(A)} 10401 qquadmathrm{(B)} 19801 qquadmathrm{(C)} 20201 qquadmathrm{(D)} 39801 qquadmathrm{(E)} 40801$

Solution 1

We can divide up figure $n$ to get the sum of the sum of the first $n+1$ odd numbers and the sum of the first $n$ odd numbers. If you do not see this, here is the example for $n=3$ : $[asy] draw((3,0)--(4,0)--(4,7)--(3,7)--cycle); draw((0,3)--(7,3)--(7,4)--(0,4)--cycle); draw((2,1)--(5,1)--(5,6)--(2,6)--cycle); draw((1,2)--(6,2)--(6,5)--(1,5)--cycle); draw((3,0)--(3,7)); [/asy]$ The sum of the first $n$ odd numbers is $n^2$ , so for figure $n$ , there are $(n+1)^2+n^2$ unit squares. We plug in $n=100$ to get $boxed{textbf{(C) }20201}$ .

Solution 2

Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$ . Thus, the original sequence can be generated from a quadratic function.
If $f(n) = an^2 + bn + c$ , and $f(0) = 1$ , $f(1) = 5$ , and $f(2) = 13$ , we get a system of three equations in three variables: $f(0) = 1$ gives $c = 1$ $f(1) = 5$ gives $a + b + c = 5$ $f(2) = 13$ gives $4a + 2b + c = 13$ Plugging in $c=1$ into the last two equations gives $a + b = 4$ $4a + 2b = 12$ Dividing the second equation by 2 gives the system: $a + b = 4$ $2a + b = 6$ Subtracting the first equation from the second gives $a = 2$ , and hence $b = 2$ . Thus, our quadratic function is: $f(n) = 2n^2 + 2n + 1$ Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$ , which is choice $boxed{textbf{(C) }20201}$ .

Solution 3

We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $sum_{n=1}^{100} n = 5050$ squares. $4 cdot 5050 = 20200$ . Adding in the original center box we have $20200 + 1 = boxed{textbf{(C) }20201}$ .

Solution 4

Let $a_n$ be the number of squares in figure $n$ . We can easily see that $[a_0=4cdot 0+1]$ $[a_1=4cdot 1+1]$ $[a_2=4cdot 3+1]$ $[a_3=4cdot 6+1.]$ Note that in $a_n$ , the number multiplied by the 4 is the $n$ th triangular number. Hence, $a_{100}=4cdot frac{100cdot 101}{2}+1=boxed{textbf{(C) }20201}$ .

Solution 5

Let $f_n$ denote the number of unit cubes in a figure. We have $[f_0=1]$ $[f_1=5]$ $[f_2=13]$ $[f_3=25]$ $[f_4=41]$ $[...]$ Computing the difference between the number of cubes in each figure yields $[4,8,12,16,...]$ It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the difference being $4$ . Let this sequence be $a_n$ From $f_0$ to $f_{100}$ , the sequence will have $100$ terms. Using the arithmetic sum formula yields $[S_{100}=frac{100[2cdot 4+(100-1)4]}{2}]$ $[=50(2cdot 4+99cdot 4)]$ $[=50(101cdot 4)]$ $[=200cdot 101]$ $[=20200]$ So $f_{100}=1+20200=boxed{textbf{(C) }20201}$ unit cubes.
Solution 1

The first number is divisible by 1.
The sum of the first two numbers is even.
The sum of the first three numbers is divisible by 3.
The sum of the first four numbers is divisible by 4.
The sum of the first five numbers is 400.
Since 400 is divisible by 4, the last score must also be divisible by 4. Therefore, the last score is either 76 or 80.
Case 1: 76 is the last number entered.
Since $400equiv 76equiv 1pmod{3}$ , the fourth number must be divisible by 3, but none of the scores are divisible by 3.
Case 2: 80 is the last number entered.
Since $80equiv 2pmod{3}$ , the fourth number must be $2pmod{3}$ . That number is 71 and 71 only. The next number must be 91, since the sum of the first two numbers is even. So the only arrangement of the scores $76, 82, 91, 71, 80$ $Rightarrow text{(C)}$

Solution 2

We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers $pmod{3}$ , which gives 2,1,2,1,1, respectively. Clearly, the only way to get a number divisible by 3 by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by 4, 71 must be next. That leaves 80 for last, so the answer is $mathrm{C}$ .
Step 1: Reflect in the xy plane. Replace z with its additive inverse: $(1,2,-3)$ Step 2: Rotate around x-axis 180 degrees. Replace y and z with their respective additive inverses. $(1, -2, 3)$ Step 3: Translate 5 units in positive-y direction. Replace y with y+5. $(1,3,3) Rightarrow text {(E) }$
$frac {a}{b} + frac {b}{a} - ab = frac{a^2 + b^2}{ab} - (a - b) = frac{a^2 + b^2}{a-b} - frac{(a-b)^2}{(a-b)} = frac{2ab}{a-b} = frac{2(a-b)}{a-b} =2 Rightarrow text{(E)}$ .Another way is to solve the equation for $b,$ giving $b = frac{a}{a+1};$ then substituting this into the expression and simplifying gives the answer of $2.$

Solution 2

This simplifies to $ab+b-a=0 Rightarrow (a+1)(b-1) = -1$ . The two integer solutions to this are $(-2,2)$ and $(0,0)$ . The problem states than $a$ and $b$ are non-zero, so we consider the case of $(-2,2)$ . So, we end up with $frac{-2}{2} + frac{2}{-2} - 2 cdot -2 = 4 - 1 - 1 = boxed{2}~ textbf{E}$
Solution 1

It is not hard to see that $[(A+1)(M+1)(C+1)=]$ $[AMC+AM+AC+MC+A+M+C+1]$ Since $A+M+C=12$ , we can rewrite this as $[(A+1)(M+1)(C+1)=]$ $[AMC+AM+AC+MC+13]$ So we wish to maximize $[(A+1)(M+1)(C+1)-13]$ Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$ , we set $A=M=C=4$ Which gives us $[(4+1)(4+1)(4+1)-13=112]$ so the answer is $boxed{text{E}}$ .

Solution 2

If you know that to maximize your result you have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make $A,M$ and $C$ as close as possible. In this case, they would all be equal to $4$ , so $AMC+AM+AC+MC=64+16+16+16=112$ , giving you the answer of $boxed{text{E}}$ .
Solution 1

Let $c$ be the total amount of coffee, $m$ of milk, and $p$ the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so $[left(frac{c}{6} + frac{m}{4}right)p = c + m]$ Regrouping, we get $2c(6-p)=3m(p-4)$ . Since both $c,m$ are positive, it follows that $6-p$ and $p-4$ are also positive, which is only possible when $p = 5 mathrm{(C)}$ .

Solution 2 (less rigorous)

One could notice that (since there are only two components to the mixture) Angela must have more than her "fair share" of milk and less then her "fair share" of coffee in order to ensure that everyone has $8$ ounces. The "fair share" is $1/p.$ So, $[frac{1}{6} < frac{1}{p}<frac{1}{4}]$ Which requires that $p$ be $p = 5 mathrm{(C)},$ since $p$ is a whole number.
- The mean is $frac{10+2+5+2+4+2+x}{7} = frac{25+x}{7}$ .
- Arranged in increasing order, the list is $2,2,2,4,5,10$ , so the median is either $2,4$ or $x$ depending upon the value of $x$ .
- The mode is $2$ , since it appears three times.
Solution 1

Let $y = frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -frac{1}{3}, frac{2}{9}$ . These sum up to $boxed{textbf{(B) }-frac19}$ . (We can also use Vieta's formulas to find the sum more quickly.)

Solution 2

Set $f(frac{x}{3}) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $frac{x}{3} = 3z$ gives $x = 9z$ ), thus the sum of the roots in the equation $f(3z)=7$ is $-frac{1}{9}$ or $boxed{textbf{(B) }-frac19}$ .

Solution 3

Since we have $f(x/3)$ , $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$ . We set this equal to 7: $81z^2 + 9z +1 = 7 Longrightarrow 81z^2 + 9z - 6 = 0$ . For any quadratic $ax^2 + bx +c = 0$ , the sum of the roots is $-frac{b}{a}$ . Thus, the sum of the roots of this equation is $-frac{9}{81} = boxed{textbf{(B) }-frac19}$ .
Index the rows with $i = 1, 2, 3, ..., 13$ Index the columns with $j = 1, 2, 3, ..., 17$ For the first row number the cells $1, 2, 3, ..., 17$ For the second, $18, 19, ..., 34$ and so onSo the number in row = $i$ and column = $j$ is $f(i, j) = 17(i-1) + j = 17i + j - 17$ Similarly, numbering the same cells column wise we find the number in row = $i$ and column = $j$ is $g(i, j) = i + 13j - 13$ So we need to solve $f(i, j) = g(i, j)$ $17i + j - 17 = i + 13j - 13$ $16i = 4 + 12j$ $4i = 1 + 3j$ $i = (1 + 3j)/4$ We get $(i, j) = (1, 1), f(i, j) = g(i, j) = 1$ $(i, j) = (4, 5), f(i, j) = g(i, j) = 56$ $(i, j) = (7, 9), f(i, j) = g(i, j) = 111$ $(i, j) = (10, 13), f(i, j) = g(i, j) = 166$ $(i, j) = (13, 17), f(i, j) = g(i, j) = 221$ $boxed{D}$ $555$
Solution 1

Since $overline{AB}$ is tangent to the circle, $triangle OAB$ is a right triangle. This means that $OA = 1$ , $AB = tan theta$ and $OB = sec theta$ . By the Angle Bisector Theorem, $[frac{OB}{OC} = frac{AB}{AC} Longrightarrow AC sec theta = OC tan theta]$ We multiply both sides by $cos theta$ to simplify the trigonometric functions, $[AC=OC sin theta]$ Since $AC + OC = 1$ , $1 - OC = OC sin theta Longrightarrow$ $OC = dfrac{1}{1+sin theta}$ . Therefore, the answer is $boxed{textbf{(D)} dfrac{1}{1+sin theta}}$ .

Solution 2

Alternatively, one could notice that OC approaches the value 1/2 as theta gets close to 90 degrees. The only choice that is consistent with this is (D).

Solution 3 (with minimal trig)

Let's assign a value to $theta$ so we don't have to use trig functions to solve. $60$ is a good value for $theta$ , because then we have a $30-60-90 triangle$ -- $angle BAC=90$ because $AB$ is tangent to Circle $O$ .
Using our special right triangle, since $AO=1$ , $OB=2$ , and $AB=sqrt{3}$ .
Let $OC=x$ . Then $CA=1-x$ . since $BC$ bisects $angle ABO$ , we can use the angle bisector theorem: $frac{2}{x}=frac{sqrt{3}}{1-x}$ $2-2x=sqrt{3}x$ $2=(sqrt{3}+2)x$ $x=frac{2}{sqrt{3}+2}$ .
Now, we only have to use a bit of trig to guess and check: the only trig facts we need to know to finish the problem is: $sintheta =frac{Opposite}{Hypotenuse}$ $costheta =frac{Adjacent}{Hypotenuse}$ $tantheta =frac{Opposite}{Adjacent}$ .
With a bit of guess and check, we get that the answer is $boxed{D}$ .
There are either $[65 + 200 = 265]$ or $[66 + 200 = 266]$ days between the first two dates depending upon whether or not year $N$ is a leap year. Since $7$ divides into $266$ but not $265$ , for both days to be a Tuesday, year $N$ must be a leap year.Hence, year $N-1$ is not a leap year, and so since there are $[265 + 300 = 565]$ days between the date in years $N,text{ }N-1$ , this leaves a remainder of $5$ upon division by $7$ . Since we are subtracting days, we count 5 days before Tuesday, which gives us $boxed{mathbf{(A)} text{Thursday}.}$
Solution 1

The answer is exactly $3$ , choice $mathrm{(C)}$ . We can find the area of triangle $ADE$ by using the simple formula $frac{bh}{2}$ . Dropping an altitude from $A$ , we see that it has length $12$ ( we can split the large triangle into a $9-12-15$ and a $5-12-13$ triangle). Then we can apply the Angle Bisector Theorem on triangle $ABC$ to solve for $BE$ . Solving $frac{13}{BE}=frac{15}{14-BE}$ , we get that $BE=frac{13}{2}$ . $D$ is the midpoint of $BC$ so $BD=7$ . Thus we get the base of triangle $ADE, DE$ , to be $frac{1}{2}$ units long. Applying the formula $frac{bh}{2}$ , we get $frac{12*frac{1}{2}}{2}=3$ .

Solution 2

The area of $ADE$ is $frac{DEcdot h}{2}=frac{DE}{BC} cdot frac{BCcdot h}{2}=frac{DE}{BC}[ABC]$ where $h$ is the height of triangle $ABC$ . Using Angle Bisector Theorem, we find $frac{13}{BE}=frac{15}{14-BE}$ , which we solve to get $BE=frac{13}{2}$ . $D$ is the midpoint of $BC$ so $BD=7$ . Thus we get the base of triangle $ADE, DE$ , to be $frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$ . $[s=frac{AB+BC+AC}{2}=21]$ $[[ABC]=sqrt{(s)(s-AB)(s-BC)(s-AC)}=sqrt{(21)(8)(7)(6)}=84]$ $[frac{DE}{BC}[ABC]=frac{frac{1}{2}}{14}cdot 84=3]$ Therefore, the answer is $mathrm{C}$ .
Solution 1

We multiply all given expressions to get: $[(1)xyz + x + y + z + frac{1}{x} + frac{1}{y} + frac{1}{z} + frac{1}{xyz} = frac{28}{3}]$ Adding all the given expressions gives that $[(2) x + y + z + frac{1}{x} + frac{1}{y} + frac{1}{z} = 4 + frac{7}{3} + 1 = frac{22}{3}]$ We subtract $(2)$ from $(1)$ to get that $xyz + frac{1}{xyz} = 2$ . Hence, by inspection, $boxed{xyz = 1 rightarrow B}$ . $[]$ ~AopsUser101

Solution 2

We have a system of three equations and three variables, so we can apply repeated substitution. $[4 = x + frac{1}{y} = x + frac{1}{1 - frac{1}{z}} = x + frac{1}{1-frac{1}{7/3-1/x}} = x + frac{7x-3}{4x-3}]$ Multiplying out the denominator and simplification yields $4(4x-3) = x(4x-3) + 7x - 3 Longrightarrow (2x-3)^2 = 0$ , so $x = frac{3}{2}$ . Substituting leads to $y = frac{2}{5}, z = frac{5}{3}$ , and the product of these three variables is $1$ .
Solution 1

$[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W); [/asy]$ WLOG, let a side of the square be $1$ . Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = frac{1}{2}bh = frac{h}{2}$ , the height of the triangle with area $m$ is $2m$ . Therefore $frac{2m}{1} = frac{1}{x}$ where $x$ is the base of the other triangle. $x = frac{1}{2m}$ , and the area of that triangle is $frac{1}{2} cdot 1 cdot frac{1}{2m} = frac{1}{4m} text{(D)}$ .

Solution 2

$[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$b$",(2.5,0),S); label("$a$",(0,1.5),W); label("$c$",(2.5,1),W); label("$A$",(0.5,2.5),W); label("$B$",(3.5,0.75),W); label("$C$",(1,1),W); [/asy]$ From the diagram from the previous solution, we have $a$ , $b$ as the legs and $c$ as the side length of the square. WLOG, let the area of triangle $A$ be $m$ times the area of square $C$ .
Since triangle $A$ is similar to the large triangle, it has $h_A = a(frac{c}{b}) = frac{ac}{b}$ , $b_A = c$ and $[[A] = frac{bh}{2} = frac{ac^2}{2b} = m[C] = mc^2]$ Thus $frac{a}{2b} = m$ Now since triangle $B$ is similar to the large triangle, it has $h_B = c$ , $b_B = bfrac{c}{a} = frac{bc}{a}$ and $[[B] = frac{bh}{2} = frac{bc^2}{2a} = nc^2 = n[C]]$ Thus $n = frac{b}{2a} = frac{1}{4(frac{a}{2b})} = frac{1}{4m}$ . $text{(D)}$ .
Note that there are 3 maxima/minima. Hence we know that the rest of the graph is greater than 10. We approximate each of the above expressions:
1. According to the graph, $P(-1) > 4$
2. The product of the roots is $d$ by Vieta’s formulas. Also, $d = P(0) > 5$ according to the graph.
3. The product of the real roots is $>5$ , and the total product is $<6$ (from above), so the product of the non-real roots is $< frac{6}{5}$ .
4. The sum of the coefficients is $P(1) > 1.5$
5. The sum of the real roots is $> 5$ .
Clearly $mathrm{(C)}$ is the smallest.
The product of the numbers have to be a power of $10$ in order to have an integer base ten logarithm. Thus all of the numbers must be in the form $2^m5^n$ . Listing out such numbers from $1$ to $46$ , we find $1,2,4,5,8,10,16,20,25,32,40$ are the only such numbers. Immediately it should be noticed that there are a larger number of powers of $2$ than of $5$ . Since a number in the form of $10^k$ must have the same number of $2$ s and $5$ s in its factorization, we require larger powers of $5$ than we do of $2$ . To see this, for each number subtract the power of $5$ from the power of $2$ . This yields $0,1,2,-1,3,0,4,1,-2,5,2$ , and indeed the only non-positive terms are $0,0,-1,-2$ . Since there are only two zeros, the largest number that Professor Gamble could have picked would be $2$ .Thus Gamble picks numbers which fit $-2 + -1 + 0 + 0 + 1 + 2$ , with the first four having already been determined to be ${25,5,1,10}$ . The choices for the $1$ include ${2,20}$ and the choices for the $2$ include ${4,40}$ . Together these give four possible tickets, which makes Professor Gamble’s probability $1/4 mathrm{(B)}$ .
Solution 1

Since $AB,BC,AC$ are all radii, it follows that $triangle ABC$ is an equilateral triangle.
Draw the circle with center $A$ and radius $overline{AB}$ . Then let $D$ be the point of tangency of the two circles, and $E$ be the intersection of the smaller circle and $overline{AD}$ . Let $F$ be the intersection of the smaller circle and $overline{AB}$ . Also define the radii $r_1 = AB, r_2 = frac{DE}{2}$ (note that $DE$ is a diameter of the smaller circle, as $D$ is the point of tangency of both circles, the radii of a circle is perpendicular to the tangent, hence the two centers of the circle are collinear with each other and $D$ ).
By the Power of a Point Theorem, $[AF^2 = AE cdot AD Longrightarrow left(frac {r_1}2right)^2 = (AD - 2r_2) cdot AD.]$ Since $AD = r_1$ , then $frac{r_1^2}{4} = r_1 (r_1 - 2r_2) Longrightarrow r_2 = frac{3r_1}{8}$ . Since $ABC$ is equilateral, $angle BAC = 60^{circ}$ , and so $stackrel{frown}{BC} = 12 = frac{60}{360} 2pi r_1 Longrightarrow r_1 = frac{36}{pi}$ . Thus $r_2 = frac{27}{2pi}$ and the circumference of the circle is $27 mathrm{(D)}$ .
(Alternatively, the Pythagorean Theorem can also be used to find $r_2$ in terms of $r_1$ . Notice that since AB is tangent to circle $O$ , $overline{OF}$ is perpendicular to $overline{AF}$ . Therefore, $[AF^2 + OF^2 = AO^2]$ $[left(frac {r_1}{2}right)^2 + r_2^2 = (r_1 - r_2)^2]$ After simplification, $r_2 = frac{3r_1}{8}$ .

Solution 2 (Pythagorean Theorem)

First, note the triangle $ABC$ is equilateral. Next, notice that since the arc $BC$ has length 12, it follows that we can find the radius of the sector centered at $A$ . $frac {1}{6}({2}{pi})AB=12 implies AB=36/{pi}$ . Next, connect the center of the circle to side $AB$ , and call this length $r$ , and call the foot $M$ . Since $ABC$ is equilateral, it follows that $MB=18/{pi}$ , and $OA$ (where O is the center of the circle) is $36/{pi}-r$ . By the pythagorean theorem, you get $r^2+(18/{pi})^2=(36/{pi}-r)^2 implies r=27/2{pi}$ . Finally, we see that the circumference is $2{pi}*27/2{pi}=boxed{(D)27}$ .
Solution 1

Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red. $[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);[/asy]$ There are $7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is $7!/3 = 1680$ . $[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); triple right=(0,1,0); picture p = new picture, r = new picture, s = new picture; draw(p,A--B--E--cycle); draw(p,A--C--D--cycle); draw(p,F--C--B--cycle); draw(p,F--D--E--cycle,dotted+linewidth(0.7)); draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*p); draw(r,A--B--E--cycle); draw(r,A--C--D--cycle); draw(r,F--C--B--cycle); draw(r,F--D--E--cycle,dotted+linewidth(0.7)); draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(2*right)*r); draw(s,A--B--E--cycle); draw(s,A--C--D--cycle); draw(s,F--C--B--cycle); draw(s,F--D--E--cycle,dotted+linewidth(0.7)); draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(4*right)*s); [/asy]$

Solution 2

We consider the dual of the octahedron, the cube; a cube can be inscribed in an octahedron with each of its vertices at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube.
Select any vertex and call it $A$ ; there are $8$ color choices for this vertex, but this vertex can be rotated to any of $8$ locations. After fixing $A$ , we pick another vertex $B$ adjacent to $A$ . There are seven color choices for $B$ , but there are only three locations to which $B$ can be rotated to (since there are three edges from $A$ ). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is $frac{8}{8} cdot frac{7}{3} cdot 6! = 1680 Rightarrow mathrm{(E)}$ .

Solution 3

There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and $8!/24 = 1680 Rightarrow mathrm{(E)}$