2004 AMC 12A Solutions | LIVE by Po-Shen Loh

1.

Alicia earns $\$20$ per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia’s wages are used to pay local taxes?

$0.0029$

$0.029$

$0.29$

$2.9$

$29$

Concepts:percentage unit conversion

Difficulty rating: 840

Small Hint:

Convert $\$20$ into cents before taking the percentage

Big Hint:

$\$20 = 2000$ cents, and $1.45\%$ of that is the tax

Solution:

Since $\$20$ is $2000$ cents, the tax is $0.0145 \times 2000 = 29$ cents per hour.

Thus, the correct answer is E.

2.

On the AMC $12,$ each correct answer is worth $6$ points, each incorrect answer is worth $0$ points, and each problem left unanswered is worth $2.5$ points. If Charlyn leaves $8$ of the $25$ problems unanswered, how many of the remaining problems must she answer correctly in order to score at least $100?$

$11$

$13$

$14$

$16$

$17$

Concepts:inequality floor and ceiling functions

Difficulty rating: 1020

Small Hint:

The $8$ blanks already contribute $2.5 \times 8$ points

Big Hint:

Find the smallest number of correct answers whose $6$ points each cover the remaining points

Solution:

The $8$ unanswered problems are worth $2.5 \times 8 = 20$ points, so Charlyn needs at least $100 - 20 = 80$ more points from correct answers.

Each correct answer is worth $6$ points, and the smallest multiple of $6$ that is at least $80$ is $84 = 6 \times 14.$ So she needs at least $14$ correct answers.

Thus, the correct answer is C.

3.

For how many ordered pairs of positive integers $(x, y)$ is $x + 2y = 100?$

$33$

$49$

$50$

$99$

$100$

Concepts:Diophantine Equation counting integers in a range

Difficulty rating: 1080

Small Hint:

Solve for $x$ in terms of $y$

Big Hint:

$x = 100 - 2y$ must be positive, which bounds how large the positive integer $y$ can be

Solution:

Writing $x = 100 - 2y,$ the value of $x$ is a positive integer precisely when $y$ is a positive integer with $1 \le y \le 49.$

This gives $49$ valid ordered pairs.

Thus, the correct answer is B.

4.

Bertha has $6$ daughters and no sons. Some of her daughters have $6$ daughters, and the rest have none. Bertha has a total of $30$ daughters and granddaughters, and no great-granddaughters. How many of Bertha’s daughters and granddaughters have no daughters?

$22$

$23$

$24$

$25$

$26$

Concepts:basic counting

Difficulty rating: 1150

Small Hint:

First count the granddaughters, then determine how many daughters have children

Big Hint:

Every woman with daughters has exactly $6$ of them

Solution:

Bertha has $30 - 6 = 24$ granddaughters, none of whom have daughters.

These granddaughters are the children of $\frac{24}{6} = 4$ of Bertha’s daughters, so exactly $4$ women have daughters.

Therefore the number of women with no daughters is $30 - 4 = 26.$

Thus, the correct answer is E.

5.

The graph of a line $y = mx + b$ is shown. Which of the following is true?

$mb \lt -1$

$-1 \lt mb \lt 0$

$mb = 0$

$0 \lt mb \lt 1$

$mb \gt 1$

Concepts:slope coordinate geometry

Difficulty rating: 1120

Small Hint:

Read the sign and size of the slope $m$ and the $y$ -intercept $b$ separately

Big Hint:

Both $m$ and $b$ have absolute value less than $1,$ but opposite signs

Solution:

The $y$ -intercept of the line is between $0$ and $1,$ so $0 \lt b \lt 1.$

The slope is negative and shallow, between $-1$ and $0,$ so $-1 \lt m \lt 0.$

The product $mb$ is therefore negative with absolute value less than $1,$ giving $-1 \lt mb \lt 0.$

Thus, the correct answer is B.

6.

Let $U = 2 \cdot 2004^{2005},$ $V = 2004^{2005},$ $W = 2003 \cdot 2004^{2004},$ $X = 2 \cdot 2004^{2004},$ $Y = 2004^{2004}$ and $Z = 2004^{2003}.$ Which of the following is largest?

$U - V$

$V - W$

$W - X$

$X - Y$

$Y - Z$

Concepts:factoring exponent

Difficulty rating: 1220

Small Hint:

Factor each difference to write it as a multiple of a power of $2004$

Big Hint:

$U - V = 2004^{2005},$ while every other choice is a multiple of the smaller $2004^{2004}$

Solution:

Compute each difference by factoring:

$U - V = 2004^{2005},$ $V - W = 2004^{2004},$ $W - X = 2001 \cdot 2004^{2004},$ $X - Y = 2004^{2004},$ and $Y - Z = 2003 \cdot 2004^{2003}.$

Since $2004^{2005} = 2004 \cdot 2004^{2004}$ exceeds each of the others, none of which reaches $2004^{2005},$ the difference $U - V$ is the largest.

Thus, the correct answer is A.

7.

A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players $A,$ $B,$ and $C$ start with $15,$ $14,$ and $13$ tokens, respectively. How many rounds will there be in the game?

$36$

$37$

$38$

$39$

$40$

Concepts:process simulation invariant

Difficulty rating: 1390

Small Hint:

Track the token counts over one full cycle of three rounds

Big Hint:

Every three rounds each player loses exactly one token, so find when the leader hits $0$

Solution:

After three rounds the players $A,$ $B,$ and $C$ have $14,$ $13,$ and $12$ tokens, respectively. Every subsequent three rounds reduces each player’s supply by one token.

After $36$ rounds they have $3,$ $2,$ and $1$ tokens. In the $37$ th round player $A,$ who has the most, gives away all three of their tokens and runs out, ending the game.

Thus, the correct answer is B.

8.

In the figure, $\angle EAB$ and $\angle ABC$ are right angles, $AB = 4,$ $BC = 6,$ $AE = 8,$ and $\overline{AC}$ and $\overline{BE}$ intersect at $D.$ What is the difference between the areas of $\triangle ADE$ and $\triangle BDC?$

$2$

$4$

$5$

$8$

$9$

Concepts:triangle area area decomposition

Difficulty rating: 1370

Small Hint:

Add the shared triangle $\triangle ABD$ to each of the two triangles

Big Hint:

The difference equals $[\triangle ABE] - [\triangle ABC]$

Solution:

Let $x,$ $y,$ and $z$ be the areas of $\triangle ADE,$ $\triangle BDC,$ and $\triangle ABD,$ respectively.

Then $\triangle ABE$ has area $\tfrac12 \cdot 4 \cdot 8 = 16 = x + z,$ and $\triangle ABC$ has area $\tfrac12 \cdot 4 \cdot 6 = 12 = y + z.$

The requested difference is $\begin{aligned} x - y &= (x + z) - (y + z) \\ &= 16 - 12 = 4. \end{aligned}$

Thus, the correct answer is B.

9.

A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume, by what percent must the height be decreased?

$10$

$25$

$36$

$50$

$60$

Concepts:cylinder power scaling of length, area, and volume percentage

Difficulty rating: 1310

Small Hint:

The base area scales with the square of the diameter

Big Hint:

Multiplying the diameter by $\tfrac54$ multiplies the base area by $\left(\tfrac54\right)^2$

Solution:

Multiplying the diameter by $\tfrac54$ multiplies the base area by $\left(\tfrac54\right)^2 = \tfrac{25}{16}.$

To keep the volume fixed, the height must be multiplied by $\tfrac{16}{25} = 0.64.$ That is a decrease of $1 - 0.64 = 0.36,$ or $36\%.$

Thus, the correct answer is C.

10.

The sum of $49$ consecutive integers is $7^5.$ What is their median?

$7$

$7^2$

$7^3$

$7^4$

$7^5$

Concepts:arithmetic sequence mean

Difficulty rating: 1370

Small Hint:

For consecutive integers the median equals the mean

Big Hint:

The median is the sum divided by the number of terms, $\dfrac{7^5}{49}$

Solution:

The sum of a set of consecutive integers equals the number of terms times their mean, and for consecutive integers the mean equals the median.

So the median is $\dfrac{7^5}{49} = \dfrac{7^5}{7^2} = 7^3 = 343.$

Thus, the correct answer is C.

11.

The average value of all the pennies, nickels, dimes, and quarters in Paula’s purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?

$0$

$1$

$2$

$3$

$4$

Concepts:mean linear equation

Difficulty rating: 1420

Small Hint:

Let $n$ be the current number of coins and write the total value two ways after adding a quarter

Big Hint:

Solve $20n + 25 = 21(n + 1)$ for $n,$ then figure out which coins total $20n$ cents

Solution:

If Paula has $n$ coins, their total value is $20n$ cents. Adding a quarter gives $n + 1$ coins worth $20n + 25$ cents, which must also equal $21(n + 1)$ cents.

So $20n + 25 = 21(n + 1),$ giving $n = 4.$

Four coins totalling $80$ cents must be three quarters and one nickel, so the number of dimes is $0.$

Thus, the correct answer is A.

12.

Let $A = (0, 9)$ and $B = (0, 12).$ Points $A'$ and $B'$ are on the line $y = x,$ and $\overline{AA'}$ and $\overline{BB'}$ intersect at $C = (2, 8).$ What is the length of $\overline{A'B'}?$

$2$

$2\sqrt{2}$

$3$

$2 + \sqrt{2}$

$3\sqrt{2}$

Concepts:coordinate geometry distance formula

Difficulty rating: 1480

Small Hint:

Find the equations of lines $AC$ and $BC,$ then intersect each with $y = x$

Big Hint:

$A'$ and $B'$ both lie on $y = x,$ so each has equal coordinates

Solution:

Line $AC$ passes through $(0, 9)$ with slope $\tfrac{8 - 9}{2 - 0} = -\tfrac12,$ so its equation is $y = -\tfrac12 x + 9.$ Setting $y = x$ gives $A' = (6, 6).$

Line $BC$ passes through $(0, 12)$ with slope $-2,$ so $y = -2x + 12.$ Setting $y = x$ gives $B' = (4, 4).$

Then $\begin{aligned} A'B' &= \sqrt{(6 - 4)^2 + (6 - 4)^2} \\ &= 2\sqrt{2}. \end{aligned}$

Thus, the correct answer is B.

13.

Let $S$ be the set of points $(a, b)$ in the coordinate plane, where each of $a$ and $b$ may be $-1,$ $0,$ or $1.$ How many distinct lines pass through at least two members of $S?$

$8$

$20$

$24$

$27$

$36$

Concepts:lattice point combinations

Difficulty rating: 1540

Small Hint:

Start from the $\binom{9}{2}$ pairs of points, each of which determines a line

Big Hint:

Some lines pass through three collinear points and get counted $\binom{3}{2} = 3$ times

Solution:

There are $\binom{9}{2} = 36$ pairs of points, and each pair determines a line.

However, there are three horizontal, three vertical, and two diagonal lines that each pass through three collinear points of $S.$ Each such line is counted $3$ times, an overcount of $2$ per line.

With $8$ such lines, the number of distinct lines is $36 - 2 \cdot 8 = 20.$

Thus, the correct answer is B.

14.

A sequence of three real numbers forms an arithmetic progression with a first term of $9.$ If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

$1$

$4$

$36$

$49$

$81$

Concepts:arithmetic sequence geometric sequence quadratic

Difficulty rating: 1630

Small Hint:

Write the arithmetic terms as $9,$ $9 + d,$ $9 + 2d$

Big Hint:

In a geometric progression the middle term squared equals the product of the outer terms

Solution:

The arithmetic progression is $9,$ $9 + d,$ $9 + 2d.$ After the additions, the geometric progression is $9,$ $11 + d,$ $29 + 2d.$

The geometric condition gives $(11 + d)^2 = 9(29 + 2d),$ which simplifies to $d^2 + 4d - 140 = 0,$ so $d = 10$ or $d = -14.$

The corresponding third terms $29 + 2d$ are $49$ and $1,$ so the smallest possible value is $1.$

Thus, the correct answer is A.

15.

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run $100$ meters. They next meet after Sally has run $150$ meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

$250$

$300$

$350$

$400$

$500$

Concepts:relative speed distance rate and time ratio and proportion

Difficulty rating: 1540

Small Hint:

By the first meeting the two together have covered half the track

Big Hint:

Between the first and second meetings they together cover a full track, so Brenda runs twice her first leg

Solution:

Starting at opposite ends, when they first meet they have together run half the track. Between the first and second meetings, they together run a full track length.

Since Brenda runs at a constant speed and covered $100$ meters before the first meeting, she covers $2 \cdot 100 = 200$ meters between the two meetings.

Adding Sally’s $150$ meters over that same interval gives a track length of $200 + 150 = 350$ meters.

Thus, the correct answer is C.

16.

The set of all real numbers $x$ for which $\small \log_{2004}(\log_{2003}(\log_{2002}(\log_{2001} x)))$ is defined is $\{x \mid x \gt c\}.$ What is the value of $c?$

$0$

$2001^{2002}$

$2002^{2003}$

$2003^{2004}$

$2001^{2002^{2003}}$

Concepts:logarithm inequality

Difficulty rating: 1660

Small Hint:

Work from the outside in: each logarithm requires its argument to be positive

Big Hint:

$\log_{2004}(\cdots)$ is defined only when $\log_{2003}(\cdots) \gt 0,$ i.e. the next argument exceeds $1$

Solution:

The expression is defined if and only if $\log_{2003}(\log_{2002}(\log_{2001} x)) \gt 0,$ that is, $\log_{2002}(\log_{2001} x) \gt 1.$

This holds if and only if $\log_{2001} x \gt 2002,$ which is equivalent to $x \gt 2001^{2002}.$

Therefore $c = 2001^{2002}.$

Thus, the correct answer is B.

17.

Let $f$ be a function with the following properties:

(i) $f(1) = 1,$ and

(ii) $f(2n) = n \cdot f(n)$ for any positive integer $n.$

What is the value of $f(2^{100})?$

$1$

$2^{99}$

$2^{100}$

$2^{4950}$

$2^{9999}$

Concepts:functional equation recursion summation

Difficulty rating: 1720

Small Hint:

Apply the rule repeatedly with $n = 2^{k}$ to unwind $f(2^{100})$

Big Hint:

The exponent accumulates as $1 + 2 + \cdots + 99$

Solution:

Applying $f(2n) = n \cdot f(n)$ with $n = 2^{k},$ we get $f(2^{k+1}) = 2^{k} \cdot f(2^{k}).$

Unwinding from $f(2^1) = f(2) = 1 \cdot f(1) = 2^0,$ the exponents accumulate: $\begin{aligned} f(2^n) &= 2^{0 + 1 + 2 + \cdots + (n-1)} \\ &= 2^{\frac{n(n-1)}{2}}. \end{aligned}$

Therefore $f(2^{100}) = 2^{\frac{100 \cdot 99}{2}} = 2^{4950}.$

Thus, the correct answer is D.

18.

Square $ABCD$ has side length $2.$ A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E.$ What is the length of $\overline{CE}?$

$\dfrac{2 + \sqrt{5}}{2}$

$\sqrt{5}$

$\sqrt{6}$

$\dfrac{5}{2}$

$5 - \sqrt{5}$

Concepts:tangent line power of a point Pythagorean Theorem

Difficulty rating: 1860

Small Hint:

Two tangent segments from a common external point to a circle are equal

Big Hint:

With $x = AE,$ the tangent lengths give $CE = 2 + x,$ then apply the Pythagorean theorem to $\triangle CDE$

Solution:

Let $F$ be the point where $CE$ touches the semicircle. Since $CB$ and $CF$ are both tangents from $C,$ we have $CF = CB = 2.$ Similarly, with $x = AE,$ the tangents from $E$ give $EF = EA = x.$

Thus $CE = CF + FE = 2 + x.$ In right triangle $CDE,$ where $CD = 2$ and $DE = 2 - x,$ $(2 - x)^2 + 2^2 = (2 + x)^2.$

Expanding gives $8x = 4,$ so $x = \tfrac12$ and $CE = 2 + \tfrac12 = \tfrac52.$

Thus, the correct answer is D.

19.

Circles $A,$ $B,$ and $C$ are externally tangent to each other and internally tangent to circle $D.$ Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D.$ What is the radius of circle $B?$

$\dfrac{2}{3}$

$\dfrac{\sqrt{3}}{2}$

$\dfrac{7}{8}$

$\dfrac{8}{9}$

$\dfrac{1 + \sqrt{3}}{3}$

Concepts:tangent circles Pythagorean Theorem system of equations

Difficulty rating: 1920

Small Hint:

Since $A$ passes through the center of $D$ and is internally tangent, circle $D$ has radius $2$

Big Hint:

Place the center of $D$ at the origin; use right triangles formed by the centers and the axis of symmetry

Solution:

Circle $A$ has radius $1$ and passes through the center of $D$ while being internally tangent to $D,$ so $D$ has radius $2.$

Place the center of $D$ at the origin, with $A$ centered at $(-1, 0).$ By symmetry, $B$ has center $(x, y)$ and radius $r,$ with $C$ its mirror image across the horizontal axis, so the two congruent circles touch on that axis and $y = r.$

Internal tangency to $D$ gives $x^2 + y^2 = (2 - r)^2,$ and external tangency to $A$ gives $(x + 1)^2 + y^2 = (1 + r)^2.$

Subtracting and using $y = r$ yields $x = \tfrac23$ and $r = \tfrac89.$ The radius of circle $B$ is $\tfrac89.$

Thus, the correct answer is D.

20.

Select numbers $a$ and $b$ between $0$ and $1$ independently and at random, and let $c$ be their sum. Let $A,$ $B,$ and $C$ be the results when $a,$ $b,$ and $c,$ respectively, are rounded to the nearest integer. What is the probability that $A + B = C?$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

Concepts:geometric probability casework

Difficulty rating: 1990

Small Hint:

Model $(a, b)$ as a random point in the unit square and split by whether each of $a, b$ is below $\tfrac12$

Big Hint:

The condition fails only when $a, b \ge \tfrac12$ but $a + b \lt \tfrac32,$ or $a, b \lt \tfrac12$ but $a + b \ge \tfrac12$

Solution:

Represent the choices as a point $(a, b)$ in the unit square. Each of $a$ and $b$ rounds to $0$ if below $\tfrac12$ and to $1$ otherwise, while $c = a + b$ rounds based on $\tfrac12$ and $\tfrac32.$

The equation fails in exactly two regions. If $a, b \lt \tfrac12,$ it fails when $a + b \ge \tfrac12;$ this is a right triangle of area $\tfrac18.$ If $a, b \ge \tfrac12,$ it fails when $a + b \lt \tfrac32;$ this is another right triangle of area $\tfrac18.$ When exactly one of $a, b$ is at least $\tfrac12,$ the equation always holds.

Thus the failure probability is $\tfrac18 + \tfrac18 = \tfrac14,$ so the requested probability is $1 - \tfrac14 = \tfrac34.$

Thus, the correct answer is E.

21.

If $\sum_{n=0}^{\infty} \cos^{2n} \theta = 5,$ what is the value of $\cos 2\theta?$

$\dfrac{1}{5}$

$\dfrac{2}{5}$

$\dfrac{\sqrt{5}}{5}$

$\dfrac{3}{5}$

$\dfrac{4}{5}$

Concepts:geometric sequence trigonometric identity

Difficulty rating: 1820

Small Hint:

The series is geometric with ratio $\cos^2 \theta$

Big Hint:

Sum to get $\dfrac{1}{\sin^2 \theta} = 5,$ then use $\cos 2\theta = 1 - 2\sin^2 \theta$

Solution:

The series is geometric with first term $1$ and ratio $\cos^2 \theta,$ so its sum is $\dfrac{1}{1 - \cos^2 \theta} = \dfrac{1}{\sin^2 \theta} = 5.$

Thus $\sin^2 \theta = \tfrac15,$ and $\begin{aligned} \cos 2\theta &= 1 - 2\sin^2 \theta \\ &= 1 - \tfrac25 = \tfrac35. \end{aligned}$

Thus, the correct answer is D.

22.

Three mutually tangent spheres of radius $1$ rest on a horizontal plane. A sphere of radius $2$ rests on them. What is the distance from the plane to the top of the larger sphere?

$3 + \dfrac{\sqrt{30}}{2}$

$3 + \dfrac{\sqrt{69}}{3}$

$3 + \dfrac{\sqrt{123}}{4}$

$\dfrac{52}{9}$

$3 + 2\sqrt{2}$

Concepts:sphere centroid Pythagorean Theorem

Difficulty rating: 2150

Small Hint:

The three small centers form an equilateral triangle of side $2$ at height $1$

Big Hint:

The big center lies above the triangle’s centroid; use the distance between centers $1 + 2 = 3$ to find its height

Solution:

Let the centers of the three unit spheres be $A,$ $B,$ $C,$ forming an equilateral triangle of side $2$ at height $1$ above the plane, and let $E$ be the center of the large sphere directly above the centroid $D$ of $\triangle ABC.$

The distance from a vertex to the centroid is $AD = \tfrac{2\sqrt3}{3},$ and $AE = 1 + 2 = 3,$ so $\begin{aligned} DE &= \sqrt{3^2 - \left(\tfrac{2\sqrt3}{3}\right)^2} \\ &= \sqrt{9 - \tfrac{4}{3}} \\ &= \dfrac{\sqrt{69}}{3}. \end{aligned}$

Since $D$ is $1$ unit above the plane and the top of the large sphere is $2$ units above $E,$ the total height is $1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$

Thus, the correct answer is B.

23.

A polynomial $\begin{aligned} &P(x) = c_{2004} x^{2004} + c_{2003} x^{2003} \\ &\quad {}+ \cdots + c_1 x + c_0 \end{aligned}$ has real coefficients with $c_{2004} \ne 0$ and $2004$ distinct complex zeros $z_k = a_k + b_k i,$ $1 \le k \le 2004$ with $a_k$ and $b_k$ real, $a_1 = b_1 = 0,$ and $\sum_{k=1}^{2004} a_k = \sum_{k=1}^{2004} b_k.$ Which of the following quantities can be a nonzero number?

$c_0$

$c_{2003}$

$b_2 b_3 \ldots b_{2004}$

$\displaystyle\sum_{k=1}^{2004} a_k$

$\displaystyle\sum_{k=1}^{2004} c_k$

Concepts:complex number Vieta’s Formulas polynomial

Difficulty rating: 2350

Small Hint:

Since $z_1 = 0$ is a root, $c_0 = P(0) = 0;$ nonreal roots come in conjugate pairs

Big Hint:

$\sum c_k = P(1),$ which is not forced to be zero; check each other option must vanish

Solution:

Since $z_1 = a_1 + b_1 i = 0$ is a root, $c_0 = P(0) = 0.$

The nonreal zeros occur in conjugate pairs, so $\sum b_k = 0,$ and the hypothesis then forces $\sum a_k = 0.$ The coefficient $c_{2003}$ equals $-c_{2004}$ times the sum of the roots $\sum a_k + i \sum b_k = 0,$ so $c_{2003} = 0.$

Because the degree is even, at least one of $z_2, \ldots, z_{2004}$ is real, making one $b_k = 0,$ so $b_2 b_3 \cdots b_{2004} = 0.$ Thus (A) through (D) all must be $0.$

On the other hand, $\sum_{k=1}^{2004} c_k = P(1),$ and a valid polynomial such as $P(x) = x(x - 2)(x - 3) \cdots$ $\cdot (x - 2003)$ $\cdot \left(x + \sum_{k=2}^{2003} k\right)$ has $P(1) \ne 0.$ So only $\sum c_k$ can be nonzero.

Thus, the correct answer is E.

24.

A plane contains points $A$ and $B$ with $AB = 1.$ Let $S$ be the union of all disks of radius $1$ in the plane that cover $\overline{AB}.$ What is the area of $S?$

$2\pi + \sqrt{3}$

$\dfrac{8\pi}{3}$

$3\pi - \dfrac{\sqrt{3}}{2}$

$\dfrac{10\pi}{3} - \sqrt{3}$

$4\pi - 2\sqrt{3}$

Concepts:area decomposition sector circle

Difficulty rating: 2350

Small Hint:

A radius- $1$ disk covers $\overline{AB}$ exactly when its center lies within $1$ of both $A$ and $B$

Big Hint:

Find that lens-shaped region $R$ of centers, then $S$ is all points within $1$ of $R$

Solution:

A radius- $1$ disk covers segment $\overline{AB}$ exactly when its center is within $1$ of both $A$ and $B.$ That region $R$ is the lens where the two unit circles centered at $A$ and $B$ overlap.

Each unit circle passes through the other’s center, so the lens is bounded by two $120^\circ$ arcs. Two $120^\circ$ sectors of area $\tfrac{\pi}{3}$ overlap in two equilateral triangles of total area $\tfrac{\sqrt3}{2},$ giving $R$ area $\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}.$

The set $S$ consists of all points within $1$ of $R.$ Beyond $R$ itself, this adds two $60^\circ$ sectors of radius $1$ (each area $\tfrac{\pi}{6}$ ) and two $120^\circ$ annuli of outer radius $2$ and inner radius $1$ (each area $\pi$ ).

Therefore the area of $S$ is $\begin{aligned} &\left(\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}\right) + 2 \cdot \tfrac{\pi}{6} \\ &\quad {}+ 2\pi = 3\pi - \tfrac{\sqrt3}{2}. \end{aligned}$

Thus, the correct answer is C.

25.

For each integer $n \ge 4,$ let $a_n$ denote the base- $n$ number $0.\overline{133}_n.$ The product $a_4 a_5 \ldots a_{99}$ can be expressed as $\dfrac{m}{n!},$ where $m$ and $n$ are positive integers and $n$ is as small as possible. What is the value of $m?$

$98$

$101$

$132$

$798$

$962$

Concepts:number base telescoping sum and difference of cubes

Difficulty rating: 2440

Small Hint:

A repeating base- $n$ fraction $0.\overline{133}_n$ equals $\dfrac{n^2 + 3n + 3}{n^3 - 1}$

Big Hint:

Note $n^2 + 3n + 3 = \dfrac{(n+1)^3 - 1}{n},$ which makes the product telescope

Solution:

Since $n^3 \cdot a_n = 133.\overline{133}_n$ $= a_n + n^2 + 3n + 3,$ we get $\begin{aligned} a_n &= \dfrac{n^2 + 3n + 3}{n^3 - 1} \\ &= \dfrac{(n+1)^3 - 1}{n(n^3 - 1)}. \end{aligned}$

Writing $n^3 - 1 = (n - 1)(n^2 + n + 1)$ and $(n+1)^3 - 1$ $= n\big((n+1)^2 + (n+1) + 1\big),$ the product $a_4 a_5 \cdots a_{99}$ telescopes to $\begin{gathered} \dfrac{3!}{99!} \cdot \dfrac{100^3 - 1}{4^3 - 1} \\ {}= \dfrac{3!}{99!} \cdot \dfrac{99(100^2 + 100 + 1)}{63}. \end{gathered}$

This simplifies to $\dfrac{(2)(10101)}{(21)(98!)} = \dfrac{962}{98!}.$ If $n \le 97,$ then rewriting this fraction with denominator $n!$ would require $98$ to divide $962,$ which it does not. Hence the smallest possible $n$ is $98,$ and $m = 962.$

Thus, the correct answer is E.