2003 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

Which of the following is the same as

$\frac{2 - 4 + 6 - 8 + 10 - 12 + 14}{3 - 6 + 9 - 12 + 15 - 18 + 21}?$

$-1$

$-\dfrac{2}{3}$

$\dfrac{2}{3}$

$1$

$\dfrac{14}{3}$

Answer: C

Concepts:factoring fraction

Difficulty rating: 840

Small Hint:

Factor a common constant out of the numerator and the denominator

Big Hint:

The two parenthesized sums $1 - 2 + 3 - \cdots + 7$ are identical

Solution:

Factor $2$ from the numerator and $3$ from the denominator: $\frac{2(1 - 2 + 3 - 4 + 5 - 6 + 7)}{3(1 - 2 + 3 - 4 + 5 - 6 + 7)}.$

The equal parenthesized sums cancel, leaving $\dfrac{2}{3}.$

Thus, the correct answer is C.

2.

Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $\$1$ more than a pink pill, and Al’s pills cost a total of $\$546$ for the two weeks. How much does one green pill cost?

$\$7$

$\$14$

$\$19$

$\$20$

$\$39$

Answer: D

Concepts:system of equations linear equation

Difficulty rating: 1020

Small Hint:

Two weeks is $14$ days, so find the cost of one day’s pair of pills first

Big Hint:

If a green pill costs $g,$ then the pink pill costs $g - 1,$ and $g + (g-1)$ is one day’s cost

Solution:

Over $14$ days the daily cost of the two pills is $\frac{546}{14} = 39.$

Let $g$ be the cost of a green pill. The pink pill costs $g - 1,$ so $g + (g - 1) = 39,$ giving $g = 20.$

Thus, the correct answer is D.

3.

Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $\$1$ each, begonias $\$1.50$ each, cannas $\$2$ each, dahlias $\$2.50$ each, and Easter lilies $\$3$ each. What is the least possible cost, in dollars, for her garden?

$108$

$115$

$132$

$144$

$156$

Answer: A

Concepts:optimization area

Difficulty rating: 1170

Small Hint:

Find the area of each of the five regions from the figure

Big Hint:

To minimize the total, put the most expensive flower in the smallest region and so on

Solution:

The five regions have areas $4,$ $6,$ $15,$ $20,$ and $21$ square feet.

To minimize the cost, plant the most expensive flowers in the smallest regions. The least possible cost is $\begin{aligned} &3(4) + 2.5(6) + 2(15) \\ &\quad {}+ 1.5(20) + 1(21) = 108. \end{aligned}$

Thus, the correct answer is A.

4.

Moe uses a mower to cut his rectangular $90$ -foot by $150$ -foot lawn. The swath he cuts is $28$ inches wide, but he overlaps each cut by $4$ inches to make sure that no grass is missed. He walks at the rate of $5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?

$0.75$

$0.8$

$1.35$

$1.5$

$3$

Answer: C

Concepts:rate area unit conversion

Difficulty rating: 1270

Small Hint:

Each pass effectively cuts a strip only $28 - 4 = 24$ inches wide because of the overlap

Big Hint:

A $24$ -inch strip is $2$ feet wide, so each foot walked mows $2$ square feet

Solution:

Because of the overlap, each pass adds a strip $28 - 4 = 24$ inches $= 2$ feet wide. So each foot Moe walks mows $2$ square feet, that is, $2 \cdot 5000 = 10000$ square feet per hour.

The lawn has area $90 \cdot 150 = 13500$ square feet, so the time is $\frac{13500}{10000} = 1.35$ hours.

Thus, the correct answer is C.

5.

Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $4 : 3.$ The horizontal length of a $27$ -inch television screen is closest, in inches, to which of the following?

$20$

$20.5$

$21$

$21.5$

$22$

Answer: D

Concepts:Pythagorean Triple ratio and proportion

Difficulty rating: 1050

Small Hint:

With sides in ratio $4 : 3,$ the height, length, and diagonal are in ratio $3 : 4 : 5$

Big Hint:

The diagonal is $27,$ so the horizontal length is $\dfrac{4}{5}$ of $27$

Solution:

A rectangle with side ratio $4 : 3$ has height, length, and diagonal in ratio $3 : 4 : 5.$ With diagonal $27,$ the horizontal length is $\frac{4}{5}(27) = 21.6,$ which is closest to $21.5.$

Thus, the correct answer is D.

6.

The second and fourth terms of a geometric sequence are $2$ and $6.$ Which of the following is a possible first term?

$-\sqrt{3}$

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

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

$\sqrt{3}$

$3$

Answer: B

Concepts:geometric sequence radical

Difficulty rating: 1290

Small Hint:

If the first term is $a$ and the ratio is $r,$ then $ar = 2$ and $ar^3 = 6$

Big Hint:

Dividing the two equations gives $r^2 = 3,$ so $r = \pm\sqrt{3}$

Solution:

Let the first term be $a$ and the common ratio $r.$ Then $ar = 2$ and $ar^3 = 6,$ so $r^2 = 3$ and $r = \pm\sqrt{3}.$

The first term is $a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}.$ The choice $-\dfrac{2\sqrt{3}}{3}$ appears among the options.

Thus, the correct answer is B.

7.

Penniless Pete’s piggy bank has no pennies in it, but it has $100$ coins, all nickels, dimes, and quarters, whose total value is $\$8.35.$ It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?

$0$

$13$

$37$

$64$

$83$

Answer: D

Concepts:system of equations Diophantine Equation bounding to limit cases

Difficulty rating: 1430

Small Hint:

With $n$ nickels, $d$ dimes, $q$ quarters: $n + d + q = 100$ and $5n + 10d + 25q = 835$

Big Hint:

Eliminate $n$ to get $d + 4q = 67,$ then bound $q$ so that $n,d,q \ge 0$ are integers

Solution:

Let $n,$ $d,$ $q$ be the numbers of nickels, dimes, quarters. Then $n + d + q = 100$ and $n + 2d + 5q = 167$ (dividing the value equation by $5$ ).

Subtracting gives $d + 4q = 67,$ so $d = 67 - 4q.$

The largest $d$ is at $q = 0,$ giving $d = 67$ (with $n = 33$ ). The smallest occurs at $q = 16,$ giving $d = 3$ (with $n = 81$ ). The difference is $67 - 3 = 64.$

Thus, the correct answer is D.

8.

Let $\clubsuit(x)$ denote the sum of the digits of the positive integer $x.$ For example, $\clubsuit(8) = 8$ and $\clubsuit(123) = 1 + 2 + 3 = 6.$ For how many two-digit values of $x$ is $\clubsuit(\clubsuit(x)) = 3?$

$3$

$4$

$6$

$9$

$10$

Answer: E

Concepts:digits casework

Difficulty rating: 1390

Small Hint:

Let $y = \clubsuit(x).$ For a two-digit $x,$ $y$ is between $1$ and $18$

Big Hint:

$\clubsuit(y) = 3$ forces $y = 3$ or $y = 12;$ count two-digit $x$ with each digit sum

Solution:

Let $y = \clubsuit(x).$ Since $x \le 99,$ we have $y \le 18.$ Then $\clubsuit(y) = 3$ requires $y = 3$ or $y = 12.$

The two-digit numbers with digit sum $3$ are $12, 21, 30$ ( $3$ values), and those with digit sum $12$ are $39, 48, 57, 66, 75, 84, 93$ ( $7$ values), for $10$ in all.

Thus, the correct answer is E.

9.

Let $f$ be a linear function for which $f(6) - f(2) = 12.$ What is $f(12) - f(2)?$

$12$

$18$

$24$

$30$

$36$

Answer: D

Concepts:slope linear equation

Difficulty rating: 1040

Small Hint:

A linear function has a constant rate of change (slope)

Big Hint:

The slope is $\dfrac{f(6) - f(2)}{6 - 2},$ and the answer is that slope times $12 - 2$

Solution:

The slope of $f$ is $\frac{f(6) - f(2)}{6 - 2} = \frac{12}{4} = 3.$

Therefore $f(12) - f(2) = 3(12 - 2) = 30.$

Thus, the correct answer is D.

10.

Several figures can be made by attaching two equilateral triangles to the regular pentagon $ABCDE$ in two of the five positions shown. How many non-congruent figures can be constructed in this way?

$1$

$2$

$3$

$4$

$5$

Answer: B

Concepts:symmetry casework

Difficulty rating: 1490

Small Hint:

By symmetry, assume one triangle is attached to a fixed side, then place the second

Big Hint:

Only the two sides adjacent versus non-adjacent to the fixed side give different shapes, up to reflection

Solution:

Assume one triangle is attached to side $AB.$ The second triangle can be attached to a side that is one step away or two steps away from $AB.$

Attaching it to $BC$ or $CD$ gives two figures; attaching it to $AE$ or $DE$ gives figures that are mirror images of these across the pentagon’s axis of symmetry.

So there are only $2$ non-congruent figures.

Thus, the correct answer is B.

11.

Cassandra sets her watch to the correct time at noon. At the actual time of $1{:}00$ PM, she notices that her watch reads $12{:}57$ and $36$ seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads $10{:}00$ PM?

$10{:}22$ PM and $24$ seconds

$10{:}24$ PM

$10{:}25$ PM

$10{:}27$ PM

$10{:}30$ PM

Answer: C

Concepts:rate ratio and proportion

Difficulty rating: 1390

Small Hint:

In $60$ real minutes the watch records only $57.6$ minutes

Big Hint:

If the watch shows $t$ minutes past noon, the real time is $\dfrac{60}{57.6}\,t = \dfrac{25}{24}\,t$

Solution:

In $60$ real minutes the watch advances only $57$ minutes $36$ seconds $= 57.6$ minutes. So when the watch shows $t$ minutes past noon, the real elapsed time is $\dfrac{60}{57.6}t = \dfrac{25}{24}t$ minutes.

The watch reads $10{:}00$ PM after $600$ recorded minutes, so the real elapsed time is $\frac{25}{24}(600) = 625$ minutes $= 10$ hours $25$ minutes past noon. The actual time is $10{:}25$ PM.

Thus, the correct answer is C.

12.

What is the largest integer that is a divisor of $\begin{aligned} &(n + 1)(n + 3)(n + 5) \\ &\quad {}\cdot (n + 7)(n + 9) \end{aligned}$ for all positive even integers $n?$

$3$

$5$

$11$

$15$

$165$

Answer: D

Concepts:divisibility greatest common divisor

Difficulty rating: 1530

Small Hint:

For even $n,$ the factors are five consecutive odd numbers

Big Hint:

Among any five consecutive odd numbers, one is a multiple of $3$ and one is a multiple of $5$

Solution:

For even $n,$ the five factors are consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by $3$ and exactly one by $5,$ so the product is always divisible by $15.$

No larger divisor always works: the products for $n = 10$ and $n = 20$ are $11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$ and $21 \cdot 23 \cdot 25 \cdot 27 \cdot 29,$ whose greatest common divisor is $15.$

Thus, the correct answer is D.

13.

An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies $75\%$ of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radius?

$2 : 1$

$3 : 1$

$4 : 1$

$16 : 3$

$6 : 1$

Answer: B

Concepts:cone sphere volume

Difficulty rating: 1490

Small Hint:

The sphere and cone share the same radius $r;$ sphere volume is $\dfrac{4}{3}\pi r^3$

Big Hint:

Set $75\%$ of the sphere’s volume equal to the cone’s volume $\dfrac{1}{3}\pi r^2 h$

Solution:

Let $r$ be the common radius and $h$ the cone’s height. The melted ice cream fills the cone, so $\frac{3}{4}\cdot\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h.$

This simplifies to $\pi r^3 = \dfrac{1}{3}\pi r^2 h,$ so $h = 3r,$ a ratio of $3 : 1.$

Thus, the correct answer is B.

14.

In rectangle $ABCD,$ $AB = 5$ and $BC = 3.$ Points $F$ and $G$ are on $\overline{CD}$ so that $DF = 1$ and $GC = 2.$ Lines $AF$ and $BG$ intersect at $E.$ Find the area of $\triangle AEB.$

$10$

$\dfrac{21}{2}$

$12$

$\dfrac{25}{2}$

$15$

Answer: D

Concepts:similarity triangle area

Difficulty rating: 1580

Small Hint:

$FG = 5 - 1 - 2 = 2,$ and $\triangle FEG$ is similar to $\triangle AEB$

Big Hint:

Let the distance from $E$ to line $CD$ be $k;$ the distance to $AB$ is $k + 3,$ with ratio $\dfrac{k}{k+3} = \dfrac{2}{5}$

Solution:

Since $FG = 5 - 1 - 2 = 2$ and $\overline{FG} \parallel \overline{AB},$ triangles $FEG$ and $AEB$ are similar with ratio $\dfrac{FG}{AB} = \dfrac{2}{5}.$

Let the distance from $E$ to line $CD$ be $k.$ Then the distance from $E$ to $AB$ is $k + 3,$ and $\frac{k}{k + 3} = \frac{2}{5},$ giving $k = 2.$

The height of $\triangle AEB$ is $k + 3 = 5,$ so its area is $\frac{1}{2}(5)(5) = \frac{25}{2}.$

Thus, the correct answer is D.

15.

A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF?$

$1 - \dfrac{\sqrt{2}}{2}$

$\dfrac{\sqrt{2}}{4}$

$\sqrt{2} - 1$

$\dfrac{1}{2}$

$\dfrac{1 + \sqrt{2}}{4}$

Answer: D

Concepts:regular polygon area decomposition symmetry

Difficulty rating: 1740

Small Hint:

Let $O$ be the center of the octagon, which is the midpoint of diagonal $AE$

Big Hint:

$\triangle AOB$ is one of the eight triangles from the center, so it has area $\dfrac{1}{8}$

Solution:

Let $O$ be the center of the octagon. Joining $O$ to the vertices splits the octagon into $8$ congruent triangles, so $\triangle AOB$ has area $\dfrac{1}{8}.$

Since $O$ is the midpoint of $\overline{AE},$ triangles $AOB$ and $BOE$ have equal areas, so $\triangle ABE$ has area $\dfrac{1}{4}.$

The rectangle $ABEF$ is split by diagonal $\overline{BE}$ into two congruent triangles, so $\triangle ABE$ is half of it. Hence $ABEF$ has area $\dfrac{1}{2}.$

Thus, the correct answer is D.

16.

Three semicircles of radius $1$ are constructed on diameter $\overline{AB}$ of a semicircle of radius $2.$ The centers of the small semicircles divide $\overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

$\pi - \sqrt{3}$

$\pi - \sqrt{2}$

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

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

$\dfrac{7}{6}\pi - \dfrac{\sqrt{3}}{2}$

Answer: E

Concepts:sector equilateral triangle area decomposition

Difficulty rating: 1680

Small Hint:

The large semicircle has area $\dfrac{1}{2}\pi(2)^2 = 2\pi$

Big Hint:

Use inclusion-exclusion: subtract the two overlaps from the total area of the three small semicircles

Solution:

The large semicircle has area $\dfrac{1}{2}\pi(2)^2 = 2\pi.$

The three small semicircles have total area $\dfrac{3\pi}{2}$ before their overlaps are accounted for. The centers of each adjacent pair are $1$ unit apart, so their intersection is bounded by two $60^\circ$ sectors and an equilateral triangle. Each of the two overlaps therefore has area $\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}.$

The shaded area is $\begin{aligned} &2\pi - \left[\frac{3\pi}{2} - 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)\right] \\ &= \frac{7}{6}\pi - \frac{\sqrt{3}}{2}. \end{aligned}$

Thus, the correct answer is E.

17.

If $\log(xy^3) = 1$ and $\log(x^2y) = 1,$ what is $\log(xy)?$

$-\dfrac{1}{2}$

$0$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$1$

Answer: D

Concepts:logarithm system of equations

Difficulty rating: 1540

Small Hint:

Let $X = \log x$ and $Y = \log y;$ the equations become linear in $X$ and $Y$

Big Hint:

Solve $X + 3Y = 1$ and $2X + Y = 1,$ then compute $X + Y$

Solution:

Let $X = \log x$ and $Y = \log y.$ Then $X + 3Y = 1$ and $2X + Y = 1.$

Solving gives $X = \dfrac{2}{5}$ and $Y = \dfrac{1}{5},$ so $\log(xy) = X + Y = \frac{3}{5}.$

Thus, the correct answer is D.

18.

Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^c b^d.$ What is $a + b + c + d?$

$30$

$31$

$32$

$33$

$34$

Answer: B

Concepts:prime factorization modular arithmetic

Difficulty rating: 1710

Small Hint:

The minimum $x$ uses only the primes $7$ and $11,$ so write $x = 7^c 11^d$

Big Hint:

Match exponents of $7$ and $11:$ need $5c + 1 \equiv 0 \pmod{13}$ and $5d \equiv 1 \pmod{13}$

Solution:

For the minimum $x,$ neither $x$ nor $y$ has prime factors other than $7$ and $11.$ Write $x = 7^c 11^d,$ so $7x^5 = 7^{5c+1} 11^{5d}.$ Writing $y = 7^m 11^n,$ we need $7^{5c+1}11^{5d} = 7^{13m}11^{13n+1}.$

Matching exponents: $5c + 1 \equiv 0 \pmod{13}$ gives the least $c = 5,$ and $5d \equiv 1 \pmod{13}$ gives the least $d = 8.$ So $a = 7,$ $b = 11,$ and $\begin{aligned} &a + b + c + d \\ &= 7 + 11 + 5 + 8 = 31. \end{aligned}$

Thus, the correct answer is B.

19.

Let $S$ be the set of permutations of the sequence $1,$ $2,$ $3,$ $4,$ $5$ for which the first term is not $1.$ A permutation is chosen randomly from $S.$ The probability that the second term is $2,$ in lowest terms, is $\frac{a}{b}.$ What is $a + b?$

$5$

$6$

$11$

$16$

$19$

Answer: E

Concepts:permutations conditional probability

Difficulty rating: 1620

Small Hint:

Count $S$ directly: $4$ choices for the first term, then $4!$ for the rest

Big Hint:

Count permutations in $S$ whose second term is $2:$ the first term is $3, 4,$ or $5$

Solution:

The set $S$ contains $4 \cdot 4! = 96$ permutations, since the first term has $4$ choices and the remaining four terms can be arranged in $4!$ ways.

For the second term to be $2,$ the first term must be $3, 4,$ or $5$ (not $1,$ not $2$ ), giving $3$ choices, and the remaining three terms can be arranged in $3!$ ways: $3 \cdot 3! = 18.$

The probability is $\dfrac{18}{96} = \dfrac{3}{16},$ so $a + b = 3 + 16 = 19.$

Thus, the correct answer is E.

20.

Part of the graph of $f(x) = ax^3 + bx^2 + cx + d$ is shown. What is $b?$

$-4$

$-2$

$0$

$2$

$4$

Answer: B

Concepts:polynomial substitution

Difficulty rating: 1580

Small Hint:

Read off $f(-1) = 0,$ $f(1) = 0,$ and $f(0) = 2$ from the graph

Big Hint:

Adding $f(1) + f(-1)$ eliminates $a$ and $c,$ leaving $2b + 2d = 0$

Solution:

The graph passes through $(-1, 0),$ $(1, 0),$ and $(0, 2).$ So $f(0) = d = 2.$

Adding $\begin{aligned} &f(1) + f(-1) \\ &= (a + b + c + d) \\ &\quad {}+ (-a + b - c + d) \\ &= 2b + 2d = 0, \end{aligned}$ so $b = -d = -2.$

Thus, the correct answer is B.

21.

An object moves $8$ cm in a straight line from $A$ to $B,$ turns at an angle $\alpha,$ measured in radians and chosen at random from the interval $(0, \pi),$ and moves $5$ cm in a straight line to $C.$ What is the probability that $AC \lt 7?$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: D

Concepts:law of cosines geometric probability trigonometry

Difficulty rating: 1910

Small Hint:

Let $\beta = \pi - \alpha$ be the interior angle at $B;$ apply the Law of Cosines to $\triangle ABC$

Big Hint:

$AC^2 = 89 - 80\cos\beta,$ so $AC \lt 7$ is equivalent to $\cos\beta \gt \dfrac{1}{2}$

Solution:

Let $\beta = \pi - \alpha$ be the interior angle of $\triangle ABC$ at $B.$ By the Law of Cosines, $\begin{aligned} &AC^2 = 8^2 + 5^2 \\ &\quad {}- 2(8)(5)\cos\beta \\ &= 89 - 80\cos\beta. \end{aligned}$

Then $AC \lt 7$ means $89 - 80\cos\beta \lt 49,$ i.e. $\cos\beta \gt \dfrac{1}{2},$ i.e. $\beta \lt \dfrac{\pi}{3}.$

As $\alpha$ is uniform on $(0, \pi),$ so is $\beta.$ The probability is $\frac{\frac{\pi}{3}}{\pi} = \frac{1}{3}.$

Thus, the correct answer is D.

22.

Let $ABCD$ be a rhombus with $AC = 16$ and $BD = 30.$ Let $N$ be a point on $\overline{AB},$ and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD},$ respectively. Which of the following is closest to the minimum possible value of $PQ?$

$6.5$

$6.75$

$7$

$7.25$

$7.5$

Answer: C

Concepts:rhombus altitude optimization

Difficulty rating: 2020

Small Hint:

Let $O$ be the center; the diagonals meet at right angles, so $OPNQ$ is a rectangle and $PQ = ON$

Big Hint:

$ON$ is smallest when $N$ is the foot of the altitude from $O$ in right triangle $AOB$ with legs $8$ and $15$

Solution:

Let $O$ be the intersection of the diagonals. Then $\triangle AOB$ is right-angled at $O$ with legs $OA = 8$ and $OB = 15.$ Quadrilateral $OPNQ$ has right angles at $O,$ $P,$ and $Q,$ so it is a rectangle and $PQ = ON.$

The minimum of $ON$ is the altitude from $O$ to $\overline{AB}$ in $\triangle AOB.$ Since $AB = \sqrt{8^2 + 15^2} = 17,$ equating the two area expressions gives $\begin{aligned} ON &= \frac{OA \cdot OB}{AB} \\ &= \frac{8 \cdot 15}{17} \\ &= \frac{120}{17} \approx 7.06. \end{aligned}$

This is closest to $7.$

Thus, the correct answer is C.

23.

The number of $x$ -intercepts on the graph of $y = \sin(\frac{1}{x})$ in the interval $(0.0001, 0.001)$ is closest to

$2900$

$3000$

$3100$

$3200$

$3300$

Answer: A

Concepts:counting integers in a range floor and ceiling functions trigonometry

Difficulty rating: 1950

Small Hint:

$\sin(\frac{1}{x}) = 0$ when $\frac{1}{x} = k\pi$ for a nonzero integer $k,$ i.e. $x = \dfrac{1}{k\pi}$

Big Hint:

Count integers $k$ with $\dfrac{1000}{\pi} \lt k \lt \dfrac{10000}{\pi}$

Solution:

The intercepts occur where $\frac{1}{x} = k\pi,$ that is $x = \dfrac{1}{k\pi}$ for a nonzero integer $k.$

The condition $0.0001 \lt \dfrac{1}{k\pi} \lt 0.001$ becomes $\frac{1000}{\pi} \lt k \lt \frac{10000}{\pi}.$

The number of such integers is $\begin{aligned} &\left\lfloor \frac{10000}{\pi} \right\rfloor - \left\lfloor \frac{1000}{\pi} \right\rfloor \\ &= 3183 - 318 = 2865, \end{aligned}$ closest to $2900.$

Thus, the correct answer is A.

24.

Positive integers $a,$ $b,$ and $c$ are chosen so that $a \lt b \lt c,$ and the system of equations $2x + y = 2003$ and $\begin{aligned} &y = |x - a| + |x - b| \\ &\quad {}+ |x - c| \end{aligned}$ has exactly one solution. What is the minimum value of $c?$

$668$

$669$

$1002$

$2003$

$2004$

Answer: C

Concepts:absolute value system of equations

Difficulty rating: 2160

Small Hint:

The graph of $y = |x-a| + |x-b| + |x-c|$ is piecewise linear with slopes $-3, -1, 1, 3$

Big Hint:

The line has slope $-2;$ a single intersection forces it to touch the corner at $x = a,$ where $y = b + c - 2a$

Solution:

The function $y = |x-a| + |x-b| + |x-c|$ is piecewise linear with slopes $-3, -1, 1, 3$ and corners at $x = a, b, c.$ The line $2x + y = 2003$ has slope $-2.$

A line of slope $-2$ meets this graph exactly once only if it passes through the leftmost corner $(a,\, b + c - 2a),$ where the graph’s slope jumps from $-3$ to $-1.$ Substituting, $2a + (b + c - 2a) = 2003,$ so $b + c = 2003.$

Since $b \lt c,$ we need $c \gt \dfrac{2003}{2},$ so $c \ge 1002.$ This bound is attained: take $a = 1,$ $b = 1001,$ and $c = 1002.$ Therefore the minimum is $1002.$

Thus, the correct answer is C.

25.

Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?

$\dfrac{1}{36}$

$\dfrac{1}{24}$

$\dfrac{1}{18}$

$\dfrac{1}{12}$

$\dfrac{1}{9}$

Answer: D

Concepts:geometric probability arc chord

Difficulty rating: 2270

Small Hint:

A chord is shorter than the radius exactly when its arc measures less than $60^\circ$

Big Hint:

All three pairwise chords are short precisely when the three points lie within some $60^\circ$ arc

Solution:

A chord has length less than the radius exactly when the arc it subtends is less than $60^\circ,$ since a chord of a $60^\circ$ arc equals the radius.

All three pairwise chords are shorter than the radius precisely when the three points all lie within some arc of $60^\circ.$

For any successful configuration, exactly one of the three points is the counterclockwise endpoint of such a containing arc (apart from probability-zero boundary cases). Choose that endpoint in $3$ ways; each of the other two points independently has probability $\dfrac{1}{6}$ of lying in the next $60^\circ.$ Hence the probability is $3\left(\frac{1}{6}\right)^2 = \frac{1}{12}.$

Thus, the correct answer is D.