2008 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

A basketball player made $5$ baskets during a game. Each basket was worth either $2$ or $3$ points. How many different numbers could represent the total points scored by the player?

$2$

$3$

$4$

$5$

$6$

Concepts:counting integers in a range

Difficulty rating: 920

Small Hint:

The total is smallest with all $2$ -point baskets and largest with all $3$ -point baskets

Big Hint:

Every integer between $5 \cdot 2$ and $5 \cdot 3$ is attainable

Solution:

The total ranges from $5 \cdot 2 = 10$ (all two-pointers) to $5 \cdot 3 = 15$ (all three-pointers). Swapping one two-pointer for a three-pointer raises the total by exactly $1,$ so every integer in between occurs.

The possible totals are $10, 11, 12, 13, 14, 15,$ which is $6$ values.

Thus, the correct answer is E.

2.

A $4 \times 4$ block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?

$2$

$4$

$6$

$8$

$10$

Concepts:arithmetic sequence pattern recognition

Difficulty rating: 1020

Small Hint:

After reversing rows $2$ and $4,$ the rows become $1\,2\,3\,4,$ $11\,10\,9\,8,$ $15\,16\,17\,18,$ $25\,24\,23\,22$

Big Hint:

One diagonal reads $1, 10, 17, 22$ and the other reads $4, 9, 16, 25$

Solution:

Reversing the second and fourth rows gives the array with rows $1\,2\,3\,4,$ $11\,10\,9\,8,$ $15\,16\,17\,18,$ and $25\,24\,23\,22.$

The main diagonal sums to $1 + 10 + 17 + 22 = 50,$ and the other diagonal sums to $4 + 9 + 16 + 25 = 54.$

The positive difference is $54 - 50 = 4.$

Thus, the correct answer is B.

3.

A semipro baseball league has teams with $21$ players each. League rules state that a player must be paid at least $\$15{,}000,$ and that the total of all players’ salaries for each team cannot exceed $\$700{,}000.$ What is the maximum possible salary, in dollars, for a single player?

$270{,}000$

$385{,}000$

$400{,}000$

$430{,}000$

$700{,}000$

Concepts:optimization extremal argument

Difficulty rating: 1100

Small Hint:

To make one salary as large as possible, pay the other $20$ players the minimum allowed

Big Hint:

The maximum is $\$700{,}000$ minus the total of $20$ minimum salaries

Solution:

One player earns the most when the other $20$ players each receive the minimum salary of $\$15{,}000.$

Thus the maximum salary is $\$700{,}000$ $- 20 \cdot \$15{,}000$ $= \$700{,}000$ $- \$300{,}000$ $= \$400{,}000.$

Thus, the correct answer is C.

4.

On circle $O,$ points $C$ and $D$ are on the same side of diameter $\overline{AB},$ $\angle AOC = 30^\circ,$ and $\angle DOB = 45^\circ.$ What is the ratio of the area of the smaller sector $COD$ to the area of the circle?

$\dfrac{2}{9}$

$\dfrac{1}{4}$

$\dfrac{5}{18}$

$\dfrac{7}{24}$

$\dfrac{3}{10}$

Concepts:sector angle sum

Difficulty rating: 1160

Small Hint:

$\angle AOC,$ $\angle COD,$ and $\angle DOB$ together make the straight angle $\overline{AB}$

Big Hint:

A sector’s fraction of the circle is its central angle divided by $360^\circ$

Solution:

Since $\angle AOC,$ $\angle COD,$ and $\angle DOB$ fill the straight angle over diameter $\overline{AB},$ $\begin{aligned} \angle COD &= 180^\circ - 30^\circ - 45^\circ \\ &= 105^\circ. \end{aligned}$

The sector’s share of the circle is $\dfrac{105^\circ}{360^\circ} = \dfrac{7}{24}.$

Thus, the correct answer is D.

5.

A class collects $\$50$ to buy flowers for a classmate who is in the hospital. Roses cost $\$3$ each, and carnations cost $\$2$ each. No other flowers are to be used. How many different bouquets could be purchased for exactly $\$50?$

$1$

$7$

$9$

$16$

$17$

Concepts:Diophantine Equation parity

Difficulty rating: 1270

Small Hint:

If there are $r$ roses and $c$ carnations, then $3r + 2c = 50$

Big Hint:

Since $2c$ and $50$ are even, $3r$ is even, so $r$ must be even; find how many even $r$ work

Solution:

Let $r$ be the number of roses and $c$ the number of carnations, so $3r + 2c = 50$ with $r, c \ge 0.$

Because $2c$ and $50$ are even, $3r$ must be even, forcing $r$ to be even. The largest possible $r$ is $16$ (since $3 \cdot 17 \gt 50$ ), so $r \in \{0, 2, 4, \ldots, 16\}.$

That gives $9$ values of $r,$ each determining a bouquet.

Thus, the correct answer is C.

6.

Postman Pete has a pedometer to count his steps. The pedometer records up to $99999$ steps, then flips over to $00000$ on the next step. Pete plans to determine his mileage for a year. On January $1$ Pete sets the pedometer to $00000.$ During the year, the pedometer flips from $99999$ to $00000$ forty-four times. On December $31$ the pedometer reads $50000.$ Pete takes $1800$ steps per mile. Which of the following is closest to the number of miles Pete walked during the year?

$2500$

$3000$

$3500$

$4000$

$4500$

Concepts:place value estimation unit conversion

Difficulty rating: 1350

Small Hint:

Each flip of the pedometer represents $100000$ steps; add the final reading of $50000$

Big Hint:

Divide the total number of steps by $1800$ to convert to miles

Solution:

Each flip counts $100000$ steps, so the year’s steps total $\begin{aligned} &44 \cdot 100000 \\ &\quad {}+ 50000 = 4{,}450{,}000. \end{aligned}$

At $1800$ steps per mile, the mileage is $\dfrac{4{,}450{,}000}{1800} \approx 2472,$ which is closest to $2500.$

Thus, the correct answer is A.

7.

For real numbers $a$ and $b,$ define $a \$ b = (a - b)^2.$ What is $(x - y)^2 \$ (y - x)^2?$

$0$

$x^2 + y^2$

$2x^2$

$2y^2$

$4xy$

Concepts:custom operation symmetry

Difficulty rating: 1250

Small Hint:

Note that $(y - x)^2 = (x - y)^2$

Big Hint:

So the expression is $t \$ t$ with $t = (x - y)^2,$ and $t \$ t = (t - t)^2$

Solution:

Since $(y - x)^2 = (x - y)^2,$ both inputs to the operation are the same value $t = (x - y)^2.$

Therefore $(x - y)^2 \$ (y - x)^2$ $= t \$ t$ $= (t - t)^2 = 0.$

Thus, the correct answer is A.

8.

Points $B$ and $C$ lie on $\overline{AD}.$ The length of $\overline{AB}$ is $4$ times the length of $\overline{BD},$ and the length of $\overline{AC}$ is $9$ times the length of $\overline{CD}.$ The length of $\overline{BC}$ is what fraction of the length of $\overline{AD}?$

$\dfrac{1}{36}$

$\dfrac{1}{13}$

$\dfrac{1}{10}$

$\dfrac{5}{36}$

$\dfrac{1}{5}$

Concepts:ratio and proportion

Difficulty rating: 1350

Small Hint:

From $AB = 4\,BD$ and $AB + BD = AD,$ get $BD = \tfrac{1}{5}AD$

Big Hint:

Similarly $CD = \tfrac{1}{10}AD,$ and $BC = BD - CD$

Solution:

Since $AB = 4\,BD$ and $AB + BD = AD,$ we have $5\,BD = AD,$ so $BD = \tfrac{1}{5}AD.$

Likewise $AC = 9\,CD$ with $AC + CD = AD$ gives $CD = \tfrac{1}{10}AD.$

Because $B$ and $C$ both measure from $A,$ $\;BC = BD - CD$ $= \tfrac{1}{5}AD - \tfrac{1}{10}AD$ $= \tfrac{1}{10}AD.$

Thus, the correct answer is C.

9.

Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6.$ Point $C$ is the midpoint of the minor arc $AB.$ What is the length of the line segment $AC?$

$\sqrt{10}$

$\dfrac{7}{2}$

$\sqrt{14}$

$\sqrt{15}$

$4$

Concepts:chord perpendicular bisector Pythagorean Theorem

Difficulty rating: 1500

Small Hint:

The radius through the arc midpoint $C$ perpendicularly bisects chord $AB,$ meeting it at $D$ with $AD = 3$

Big Hint:

Find the center-to-chord distance $OD,$ then $DC = 5 - OD,$ and apply the Pythagorean theorem to $\triangle ADC$

Solution:

Let $O$ be the center and $D$ the point where $\overline{OC}$ meets $\overline{AB}.$ Since $C$ is the midpoint of arc $AB,$ $\overline{OC}$ is the perpendicular bisector of the chord, so $AD = 3.$

In right triangle $ADO,$ $OD = \sqrt{5^2 - 3^2} = 4,$ so $DC = OC - OD = 5 - 4 = 1.$

Then in right triangle $ADC,$ $AC = \sqrt{AD^2 + DC^2}$ $= \sqrt{3^2 + 1^2}$ $= \sqrt{10}.$

Thus, the correct answer is A.

10.

Bricklayer Brenda would take $9$ hours to build a chimney alone, and bricklayer Brandon would take $10$ hours to build it alone. When they work together, they talk a lot, and their combined output is decreased by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?

$500$

$900$

$950$

$1000$

$1900$

Concepts:rate linear equation

Difficulty rating: 1530

Small Hint:

If the chimney has $n$ bricks, Brenda lays $\tfrac{n}{9}$ per hour and Brandon lays $\tfrac{n}{10}$ per hour

Big Hint:

Together they lay $\tfrac{n}{9} + \tfrac{n}{10} - 10$ bricks per hour, and $5$ hours of this equals $n$

Solution:

Let $n$ be the number of bricks. Alone, Brenda lays $\tfrac{n}{9}$ bricks per hour and Brandon lays $\tfrac{n}{10}.$ Together, their rate is $\tfrac{n}{9} + \tfrac{n}{10} - 10.$

Working for $5$ hours completes the chimney: $5\left(\tfrac{n}{9} + \tfrac{n}{10} - 10\right) = n.$ Expanding, $\tfrac{5n}{9} + \tfrac{5n}{10} - 50 = n,$ so $\tfrac{95n}{90} - n = 50,$ giving $\tfrac{5n}{90} = 50.$

Hence $n = 900.$

Thus, the correct answer is B.

11.

A cone-shaped mountain has its base on the ocean floor and has a height of $8000$ feet. The top $\tfrac{1}{8}$ of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain, in feet?

$4000$

$2000(4 - \sqrt{2})$

$6000$

$6400$

$7000$

Concepts:cone similarity power scaling of length, area, and volume

Difficulty rating: 1570

Small Hint:

The above-water part is a smaller cone similar to the whole mountain

Big Hint:

The volume ratio $\tfrac{1}{8}$ is the cube of the height ratio; the depth is the total height minus the above-water cone’s height

Solution:

The part above the water is a cone similar to the whole mountain, with volume $\tfrac{1}{8}$ of the total. Since volume scales as the cube of length, the above-water cone’s height is $\sqrt[3]{\tfrac{1}{8}} = \tfrac{1}{2}$ of the full height.

So the above-water height is $8000 \cdot \tfrac{1}{2} = 4000$ feet.

The ocean depth at the base is the submerged height, $8000 - 4000 = 4000$ feet.

Thus, the correct answer is A.

12.

For each positive integer $n,$ the mean of the first $n$ terms of a sequence is $n.$ What is the $2008$ th term of the sequence?

$2008$

$4015$

$4016$

$4{,}030{,}056$

$4{,}032{,}064$

Concepts:mean summation

Difficulty rating: 1500

Small Hint:

If the mean of the first $n$ terms is $n,$ then their sum is $n^2$

Big Hint:

The $n$ th term equals the sum of the first $n$ minus the sum of the first $n - 1$

Solution:

Since the mean of the first $n$ terms is $n,$ their sum is $n \cdot n = n^2.$

The $n$ th term is the difference of consecutive sums, $n^2 - (n-1)^2 = 2n - 1.$

For $n = 2008,$ the term is $2 \cdot 2008 - 1 = 4015.$

Thus, the correct answer is B.

13.

Vertex $E$ of equilateral $\triangle ABE$ is in the interior of unit square $ABCD.$ Let $R$ be the region consisting of all points inside $ABCD$ and outside $\triangle ABE$ whose distance from $\overline{AD}$ is between $\tfrac{1}{3}$ and $\tfrac{2}{3}.$ What is the area of $R?$

$\dfrac{12 - 5\sqrt{3}}{72}$

$\dfrac{12 - 5\sqrt{3}}{36}$

$\dfrac{\sqrt{3}}{18}$

$\dfrac{3 - \sqrt{3}}{9}$

$\dfrac{\sqrt{3}}{12}$

Concepts:coordinate geometry equilateral triangle calculus

Difficulty rating: 1730

Small Hint:

Take $\overline{AD}$ as a side of the square; “distance from $\overline{AD}$ between $\tfrac13$ and $\tfrac23$ ” is a vertical strip of width $\tfrac13$ and area $\tfrac13$

Big Hint:

From that strip’s area $\tfrac13,$ subtract the part of the strip that lies inside $\triangle ABE$

Solution:

Place $A = (0,0),$ $B = (1,0),$ $C = (1,1),$ $D = (0,1),$ so $\overline{AD}$ lies along the $y$ -axis and distance from $\overline{AD}$ is the $x$ -coordinate. The region lies in the strip $\tfrac13 \le x \le \tfrac23,$ which within the square has area $\tfrac13.$

Equilateral $\triangle ABE$ has $E = \left(\tfrac12, \tfrac{\sqrt3}{2}\right),$ with side $AE$ on $y = \sqrt3\,x$ and side $BE$ on $y = \sqrt3(1 - x).$ The area of the triangle inside the strip is $\begin{aligned} &\int_{\frac{1}{3}}^{\frac{1}{2}} \sqrt3\,x\,dx \\ &\quad {}+ \int_{\frac{1}{2}}^{\frac{2}{3}} \sqrt3(1 - x)\,dx \\ &= 2\int_{\frac{1}{3}}^{\frac{1}{2}}\sqrt3\,x\,dx \\ &= \frac{5\sqrt3}{36}. \end{aligned}$

Therefore $[R] = \frac13 - \frac{5\sqrt3}{36} = \frac{12 - 5\sqrt3}{36}.$

Thus, the correct answer is B.

14.

A circle has a radius of $\log_{10}(a^2)$ and a circumference of $\log_{10}(b^4).$ What is $\log_a b?$

$\dfrac{1}{4\pi}$

$\dfrac{1}{\pi}$

$\pi$

$2\pi$

$10^{2\pi}$

Concepts:logarithm circumference

Difficulty rating: 1630

Small Hint:

Circumference equals $2\pi$ times the radius

Big Hint:

Use $\log_{10}(a^2) = 2\log_{10} a$ and $\log_{10}(b^4) = 4\log_{10} b,$ then form $\dfrac{\log_{10} b}{\log_{10} a}$

Solution:

The circumference is $2\pi$ times the radius, so $\log_{10}(b^4) = 2\pi \log_{10}(a^2).$

Rewriting, $4\log_{10} b = 4\pi \log_{10} a,$ hence $\log_{10} b = \pi \log_{10} a.$

Therefore $\log_a b = \dfrac{\log_{10} b}{\log_{10} a} = \pi.$

Thus, the correct answer is C.

15.

On each side of a unit square, an equilateral triangle of side length $1$ is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length $1$ is constructed. The interiors of the square and the $12$ triangles have no points in common. Let $R$ be the region formed by the union of the square and all the triangles, and let $S$ be the smallest convex polygon that contains $R.$ What is the area of the region that is inside $S$ but outside $R?$

$\dfrac{1}{4}$

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

$1$

$\sqrt{3}$

$2\sqrt{3}$

Concepts:triangle area angle chasing trigonometry

Difficulty rating: 1660

Small Hint:

The gap between $S$ and $R$ is four small triangles, one at each corner of the square

Big Hint:

Each gap triangle has two sides of length $1$ meeting at angle $360^\circ - 90^\circ - 4 \cdot 60^\circ;$ use $\tfrac12 ab \sin C$

Solution:

The convex hull $S$ differs from $R$ only near the four corners of the square, where a small triangular gap forms. Each gap triangle has two sides of length $1$ (outer edges of adjacent triangles).

The angle between those two sides is $360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ,$ so each gap has area $\tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14.$

The total area is $4 \cdot \tfrac14 = 1.$

Thus, the correct answer is C.

16.

A rectangular floor measures $a$ feet by $b$ feet, where $a$ and $b$ are positive integers with $b \gt a.$ An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half the area of the entire floor. How many possibilities are there for the ordered pair $(a, b)?$

$1$

$2$

$3$

$4$

$5$

Concepts:area Simon’s Favorite Factoring Trick factor

Difficulty rating: 1660

Small Hint:

The painted rectangle is $(a - 2) \times (b - 2)$ and equals half the floor, so $ab = 2(a - 2)(b - 2)$

Big Hint:

Rearrange to $(a - 4)(b - 4) = 8$ and count factor pairs with $b \gt a \gt 0$

Solution:

The painted rectangle measures $(a - 2)$ by $(b - 2)$ and has half the area of the floor, so $ab = 2(a - 2)(b - 2).$

Expanding gives $0 = ab - 4a - 4b + 8,$ and adding $8$ yields $(a - 4)(b - 4) = 8.$

With $b \gt a \gt 0,$ the only valid factor pairs of $8$ are $(a - 4, b - 4) = (1, 8)$ and $(2, 4),$ giving $(a, b) = (5, 12)$ and $(6, 8).$

There are $2$ possibilities.

Thus, the correct answer is B.

17.

Let $A,$ $B$ and $C$ be three distinct points on the graph of $y = x^2$ such that line $AB$ is parallel to the $x$ -axis and $\triangle ABC$ is a right triangle with area $2008.$ What is the sum of the digits of the $y$ -coordinate of $C?$

$16$

$17$

$18$

$19$

$20$

Concepts:parabola slope triangle area digits

Difficulty rating: 1800

Small Hint:

Write $A = (a, a^2)$ and $B = (-a, a^2);$ since $A, B, C$ have distinct $x$ -coordinates, the right angle is at $C$

Big Hint:

Perpendicular legs $CA$ and $CB$ force $a^2 - c^2 = 1,$ the height above $\overline{AB};$ get $|a|$ from the area, then $c^2$

Solution:

Since $AB$ is horizontal, take $A = (a, a^2)$ and $B = (-a, a^2),$ and let $C = (c, c^2).$ The right angle cannot be at $A$ or $B$ (that would need $c = \pm a$ ), so it is at $C.$

Put $d = a^2-c^2.$ Taking the dot product of $\overrightarrow{CA}$ and $\overrightarrow{CB}$ gives $d(d-1)=0.$ The points are distinct, so $d \ne 0;$ hence $a^2-c^2=1.$ This value is the height of the triangle above $\overline{AB}.$

The area is $\tfrac12 \cdot AB \cdot \text{height}$ $= \tfrac12 (2|a|)(1)$ $= |a| = 2008,$ so $a^2 = 2008^2 = 4{,}032{,}064$ and the $y$ -coordinate of $C$ is $c^2 = a^2 - 1 = 4{,}032{,}063.$

Its digit sum is $4 + 0 + 3 + 2 + 0 + 6 + 3 = 18.$

Thus, the correct answer is C.

18.

A pyramid has a square base $ABCD$ and vertex $E.$ The area of square $ABCD$ is $196,$ and the areas of $\triangle ABE$ and $\triangle CDE$ are $105$ and $91,$ respectively. What is the volume of the pyramid?

$392$

$196\sqrt{6}$

$392\sqrt{2}$

$392\sqrt{3}$

$784$

Concepts:pyramid volume Heron’s Formula

Difficulty rating: 1910

Small Hint:

The base has side $14;$ the given triangle areas give slant distances $EF = \tfrac{2 \cdot 105}{14}$ and $EG = \tfrac{2 \cdot 91}{14}$ to sides $AB$ and $CD$

Big Hint:

Triangle $EFG$ has sides $15, 13, 14$ and stands perpendicular to the base; its altitude to $FG = 14$ is the pyramid’s height

Solution:

The square has side $\sqrt{196} = 14.$ Let $F$ and $G$ be the feet of the perpendiculars from $E$ to $AB$ and $CD.$ Then $FG = 14,$ $EF = \tfrac{2 \cdot 105}{14} = 15,$ and $EG = \tfrac{2 \cdot 91}{14} = 13.$

Triangle $EFG$ lies in a plane perpendicular to the base, so its altitude to $FG$ is the pyramid’s height. By Heron’s formula with $s = 21,$ its area is $\sqrt{21 \cdot 6 \cdot 8 \cdot 7} = 84,$ so the altitude to $FG$ is $\tfrac{2 \cdot 84}{14} = 12.$

The volume is $\tfrac13 \cdot 196 \cdot 12 = 784.$

Thus, the correct answer is E.

19.

A function $f$ is defined by $f(z) = (4 + i)z^2 + \alpha z + \gamma$ for all complex numbers $z,$ where $\alpha$ and $\gamma$ are complex numbers and $i^2 = -1.$ Suppose that $f(1)$ and $f(i)$ are both real. What is the smallest possible value of $|\alpha| + |\gamma|?$

$1$

$\sqrt{2}$

$2$

$2\sqrt{2}$

$4$

Concepts:complex number optimization

Difficulty rating: 1990

Small Hint:

Write $\alpha = a + bi$ and $\gamma = c + di;$ requiring $f(1)$ and $f(i)$ to be real gives two linear conditions

Big Hint:

They force $a = 1 - d$ and $b = -1 - d,$ so $|\alpha| + |\gamma|$ $= \sqrt{2 + 2d^2} + \sqrt{c^2 + d^2};$ minimize over $c, d$

Solution:

Let $\alpha = a + bi$ and $\gamma = c + di.$ Then $f(1) = (4 + a + c)$ $+ (1 + b + d)i$ and $f(i) = (-4 - b + c)$ $+ (-1 + a + d)i.$

Both being real forces $1 + b + d = 0$ and $-1 + a + d = 0,$ i.e. $a = 1 - d$ and $b = -1 - d.$

Hence $\begin{aligned} &|\alpha| + |\gamma| \\ &= \sqrt{(1 - d)^2 + (1 + d)^2} \\ &\quad {}+ \sqrt{c^2 + d^2} \\ &= \sqrt{2 + 2d^2} \\ &\quad {}+ \sqrt{c^2 + d^2}, \end{aligned}$ which is smallest when $c = d = 0,$ giving $\sqrt{2}.$

Thus, the correct answer is B.

20.

Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck travels at $10$ feet per second in the same direction as Michael and stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?

$4$

$5$

$6$

$7$

$8$

Concepts:relative speed distance rate and time process simulation

Difficulty rating: 1860

Small Hint:

Number the pails so Michael starts at pail $0$ and the truck at pail $1;$ Michael reaches pail $n$ at time $40n$ seconds

Big Hint:

The truck leaves pail $n$ at $50(n - 1)$ and arrives at $50(n - 1) - 30;$ find the pails where they coincide, then check for a crossing while the truck moves between them

Solution:

Number the pails so Michael is at pail $0$ and the truck at pail $1$ at time $0.$ Michael reaches pail $n$ at $40n$ seconds. The truck spends $20$ seconds between pails and $30$ stopped, so it leaves pail $n$ at $50(n - 1)$ seconds and (for $n \ge 2$ ) arrives at $50(n - 1) - 30.$

Michael is at pail $n$ while the truck is there exactly when $50(n-1) - 30 \le 40n$ $\le 50(n-1),$ which simplifies to $5 \le n \le 8.$ So they meet at pail $5$ (at $t = 200,$ as the truck departs), pail $6$ ( $t = 240$ ), pail $7$ ( $t = 280$ ), and pail $8$ ( $t = 320,$ as the truck arrives).

Between pails $6$ and $7$ the truck (moving at $10$ ft/s) pulls ahead of and is then overtaken by Michael once more, adding one crossing. In all, they meet $5$ times.

Thus, the correct answer is B.

21.

Two circles of radius $1$ are to be constructed as follows. The center of circle $A$ is chosen uniformly and at random from the line segment joining $(0, 0)$ to $(2, 0).$ The center of circle $B$ is chosen uniformly and at random, and independently of the first choice, from the line segment joining $(0, 1)$ to $(2, 1).$ What is the probability that circles $A$ and $B$ intersect?

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

$\dfrac{3\sqrt{3} + 2}{8}$

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

$\dfrac{2 + \sqrt{3}}{4}$

$\dfrac{4\sqrt{3} - 3}{4}$

Concepts:geometric probability distance formula

Difficulty rating: 2040

Small Hint:

With centers $(a, 0)$ and $(b, 1),$ the circles intersect iff $\sqrt{(a - b)^2 + 1} \le 2,$ i.e. $|a - b| \le \sqrt{3}$

Big Hint:

In the square $[0, 2]^2$ of area $4,$ find the area with $|a - b| \le \sqrt{3}$ by removing the two corner triangles where $|a - b| \gt \sqrt{3}$

Solution:

Let the centers be $(a, 0)$ and $(b, 1)$ with $a, b \in [0, 2].$ The circles (radius $1$ each) intersect iff the distance between centers is at most $2:$ $\begin{aligned} &\sqrt{(a - b)^2 + 1} \le 2 \\ &\iff |a - b| \le \sqrt{3}. \end{aligned}$

The pairs $(a, b)$ fill the square $[0, 2]^2$ of area $4.$ The failing region $|a - b| \gt \sqrt3$ is two right triangles, each with legs $2 - \sqrt3,$ of total area $(2 - \sqrt3)^2 = 7 - 4\sqrt3.$

So the favorable area is $4 - (7 - 4\sqrt3) = 4\sqrt3 - 3,$ and the probability is $\frac{4\sqrt3 - 3}{4}.$

Thus, the correct answer is E.

22.

A parking lot has $16$ spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires $2$ adjacent spaces. What is the probability that she is able to park?

$\dfrac{11}{20}$

$\dfrac{4}{7}$

$\dfrac{81}{140}$

$\dfrac{3}{5}$

$\dfrac{17}{28}$

Concepts:complementary counting combinations arrangements with restrictions

Difficulty rating: 2110

Small Hint:

The $4$ empty spaces are equally likely to be any $4$ of the $16;$ she fails exactly when no two empties are adjacent

Big Hint:

The number of ways to choose $4$ empty spaces among $16$ with no two adjacent is $\binom{13}{4};$ subtract that probability from $1$

Solution:

After the $12$ cars park, $4$ spaces are empty, equally likely to be any $4$ of the $16,$ for $\binom{16}{4} = 1820$ equally likely sets.

Auntie Em fails exactly when no two empty spaces are adjacent. The number of ways to place $4$ non-adjacent empties among $16$ is $\binom{13}{4} = 715.$

So the probability she can park is $1 - \frac{715}{1820} = \frac{1105}{1820} = \frac{17}{28}.$

Thus, the correct answer is E.

23.

The sum of the base- $10$ logarithms of the divisors of $10^n$ is $792.$ What is $n?$

$11$

$12$

$13$

$14$

$15$

Concepts:logarithm factor counting

Difficulty rating: 1860

Small Hint:

The sum of the logs of the divisors equals the log of the product of all divisors

Big Hint:

$10^n$ has $(n + 1)^2$ divisors, and the product of the divisors of $N$ is $N^{\frac{d(N)}{2}};$ this gives $\tfrac12 n(n + 1)^2 = 792$

Solution:

The sum of the base- $10$ logs of the divisors is the log of their product. A number $N$ with $d(N)$ divisors has divisor product $N^{\frac{d(N)}{2}}.$

Here $N = 10^n$ has $d(N) = (n + 1)^2$ divisors, so the product is $(10^n)^{\frac{(n+1)^2}{2}}$ and its log is $\frac{n(n + 1)^2}{2} = 792.$

Thus $n(n + 1)^2 = 1584$ $= 11 \cdot 144$ $= 11 \cdot 12^2,$ giving $n = 11.$

Thus, the correct answer is A.

24.

Let $A_0 = (0, 0).$ Distinct points $A_1,$ $A_2,$ $\ldots$ lie on the $x$ -axis, and distinct points $B_1,$ $B_2,$ $\ldots$ lie on the graph of $y = \sqrt{x}.$ For every positive integer $n,$ $A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n \ge 100?$

$13$

$15$

$17$

$19$

$21$

Concepts:recursion equilateral triangle summation

Difficulty rating: 2270

Small Hint:

If $c_n = A_{n-1}A_n,$ the apex $B_n$ sits at height $\tfrac{\sqrt3}{2}c_n$ above the midpoint; lying on $y = \sqrt{x}$ gives $\tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}$

Big Hint:

Subtracting consecutive relations yields $c_n = c_{n-1} + \tfrac23,$ so $c_n = \tfrac{2n}{3}$ and $A_0A_n = \tfrac{n(n+1)}{3}$

Solution:

Let $a_n = A_0A_n$ and $c_n = a_n - a_{n-1}$ be the base of the $n$ th equilateral triangle. Its apex $B_n$ lies above the midpoint at height $\tfrac{\sqrt3}{2}c_n,$ and being on $y = \sqrt{x}$ gives $\left(\tfrac{\sqrt3}{2}c_n\right)^2 = a_{n-1} + \tfrac{c_n}{2},$ i.e. $\tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}.$

Writing the same relation for the previous triangle and subtracting gives $c_n = c_{n-1} + \tfrac23,$ and with $c_1 = \tfrac23$ we get $c_n = \tfrac{2n}{3}.$ Summing, $a_n = \sum_{k=1}^n \frac{2k}{3} = \frac{n(n + 1)}{3}.$

We need $\tfrac{n(n + 1)}{3} \ge 100,$ i.e. $n(n + 1) \ge 300.$ Since $16 \cdot 17 = 272$ and $17 \cdot 18 = 306,$ the least such $n$ is $17.$

Thus, the correct answer is C.

25.

Let $ABCD$ be a trapezoid with $AB \parallel CD,$ $AB = 11,$ $BC = 5,$ $CD = 19,$ and $DA = 7.$ Bisectors of $\angle A$ and $\angle D$ meet at $P,$ and bisectors of $\angle B$ and $\angle C$ meet at $Q.$ What is the area of hexagon $ABQCDP?$

$28\sqrt{3}$

$30\sqrt{3}$

$32\sqrt{3}$

$35\sqrt{3}$

$36\sqrt{3}$

Concepts:trapezoid angle bisector law of cosines

Difficulty rating: 2230

Small Hint:

$\angle A + \angle D = 180^\circ,$ so their bisectors meet at right angles, and the midpoint of $\overline{AD}$ is equidistant from $A, D, P$

Big Hint:

This puts $M, P, Q, N$ on the midline with $PQ = \tfrac{AB + CD - AD - BC}{2};$ get the height from $\triangle ADE$ ( $AE \parallel BC,$ sides $7, 5, 8$ )

Solution:

Because $AB \parallel CD,$ $\angle A + \angle D = 180^\circ,$ so the bisectors of $\angle A$ and $\angle D$ meet at right angles, $\angle APD = 90^\circ.$ Then the midpoint $M$ of $\overline{AD}$ is the circumcenter of right triangle $APD,$ giving $MP = MA = MD.$ Thus $\angle MPA = \angle PAM = \angle PAB,$ where the last equality uses the angle bisector at $A.$ Therefore $MP \parallel AB.$ The same argument gives $QN \parallel AB$ for the midpoint $N$ of $\overline{BC}.$ Hence $M, P, Q, N$ are collinear on the midline.

The midline has length $\tfrac{AB + CD}{2} = 15,$ while $MP = \tfrac{AD}{2} = \tfrac72$ and $QN = \tfrac{BC}{2} = \tfrac52.$ Hence $PQ = 15 - \tfrac72 - \tfrac52 = 9.$

Drawing $AE \parallel BC$ with $E$ on $\overline{CD}$ gives $AE = 5$ and $DE = CD - AB = 8.$ In $\triangle ADE,$ $\cos(\angle AED) = \tfrac{8^2 + 5^2 - 7^2}{2 \cdot 8 \cdot 5} = \tfrac12,$ so $\angle AED = 60^\circ$ and the trapezoid’s height is $AF = 5\sin 60^\circ = \tfrac{5\sqrt3}{2}.$

The segment $PQ$ sits at half the height, so the hexagon splits into two trapezoids and $\begin{aligned} &[ABQCDP] \\ &= \frac{AF}{4}\bigl(AB + CD + 2\,PQ\bigr) \\ &= \frac{\frac{5\sqrt3}{2}}{4}(11 + 19 + 18) \\ &= 30\sqrt3. \end{aligned}$

Thus, the correct answer is B.