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

1.

A bakery owner turns on his doughnut machine at $8\!:\!30$ am. At $11\!:\!10$ am the machine has completed one third of the day’s job. At what time will the doughnut machine complete the job?

$1\!:\!50$ pm

$3\!:\!00$ pm

$3\!:\!30$ pm

$4\!:\!30$ pm

$5\!:\!50$ pm

Concepts:ratio and proportion date and time

Difficulty rating: 800

Small Hint:

Find how long it took to finish the first third of the job

Big Hint:

The whole job takes three times as long as that first third

Solution:

From $8\!:\!30$ am to $11\!:\!10$ am is $2$ hours $40$ minutes, or $160$ minutes, to complete one third of the job.

The whole job then takes $3 \cdot 160 = 480$ minutes, or $8$ hours. Adding $8$ hours to $8\!:\!30$ am gives $4\!:\!30$ pm.

Thus, D is the correct answer.

2.

What is the reciprocal of $\dfrac{1}{2} + \dfrac{2}{3}?$

$\dfrac{6}{7}$

$\dfrac{7}{6}$

$\dfrac{5}{3}$

$3$

$\dfrac{7}{2}$

Concepts:fraction

Difficulty rating: 910

Small Hint:

Add the two fractions using a common denominator

Big Hint:

The reciprocal swaps the numerator and denominator

Solution:

Using a common denominator, $\dfrac{1}{2} + \dfrac{2}{3} = \dfrac{3}{6} + \dfrac{4}{6} = \dfrac{7}{6}.$

The reciprocal of $\dfrac{7}{6}$ is $\dfrac{6}{7}.$

Thus, A is the correct answer.

3.

Suppose that $\tfrac{2}{3}$ of $10$ bananas are worth as much as $8$ oranges. How many oranges are worth as much as $\tfrac{1}{2}$ of $5$ bananas?

$2$

$\dfrac{5}{2}$

$3$

$\dfrac{7}{2}$

$4$

Concepts:ratio and proportion

Difficulty rating: 1100

Small Hint:

First find how many oranges a single banana is worth

Big Hint:

$\tfrac{2}{3}$ of $10$ bananas is $\tfrac{20}{3}$ bananas, and these equal $8$ oranges

Solution:

Since $\tfrac{2}{3}$ of $10$ bananas is $\tfrac{20}{3}$ bananas, worth $8$ oranges, one banana is worth $8 \div \dfrac{20}{3} = \dfrac{24}{20} = \dfrac{6}{5}$ oranges.

Then $\tfrac{1}{2}$ of $5$ bananas is $\tfrac{5}{2}$ bananas, worth $\dfrac{5}{2} \cdot \dfrac{6}{5} = 3$ oranges.

Thus, C is the correct answer.

4.

Which of the following is equal to the product $\begin{aligned} &\dfrac{8}{4} \cdot \dfrac{12}{8} \cdot \dfrac{16}{12} \\ &\quad \cdots \dfrac{4n + 4}{4n} \cdots \dfrac{2008}{2004}? \end{aligned}$

$251$

$502$

$1004$

$2008$

$4016$

Concepts:telescoping

Difficulty rating: 1180

Small Hint:

Each denominator cancels with the numerator of the previous fraction

Big Hint:

Only the last numerator and the first denominator survive

Solution:

Every denominator except the first cancels with the numerator of the preceding fraction, so the product collapses to $\dfrac{2008}{4} = 502.$

Thus, B is the correct answer.

5.

Suppose that $\dfrac{2x}{3} - \dfrac{x}{6}$ is an integer. Which of the following statements must be true about $x?$

It is negative.

It is even, but not necessarily a multiple of $3.$

It is a multiple of $3,$ but not necessarily even.

It is a multiple of $6,$ but not necessarily a multiple of $12.$

It is a multiple of $12.$

Concepts:algebraic manipulation parity

Difficulty rating: 1100

Small Hint:

Combine the two fractions over a common denominator

Big Hint:

Simplify to $\dfrac{x}{2},$ then decide what divisibility condition makes this an integer

Solution:

Combining the fractions, $\dfrac{2x}{3} - \dfrac{x}{6} = \dfrac{4x - x}{6} = \dfrac{x}{2}.$

This is an integer exactly when $x$ is even. The example $x = 4$ is even but not a multiple of $3,$ which rules out every other statement.

Thus, B is the correct answer.

6.

Heather compares the price of a new computer at two different stores. Store A offers $15\%$ off the sticker price followed by a $\$90$ rebate, and store B offers $25\%$ off the same sticker price with no rebate. Heather saves $\$15$ by buying the computer at store A instead of store B. What is the sticker price of the computer, in dollars?

$750$

$900$

$1000$

$1050$

$1500$

Concepts:percentage linear equation

Difficulty rating: 1270

Small Hint:

Write each store’s price in terms of the sticker price $x$

Big Hint:

Store A’s price is $15$ dollars less than store B’s price

Solution:

Let $x$ be the sticker price in dollars. Store A charges $0.85x - 90$ dollars, and store B charges $0.75x$ dollars.

Since store A is $15$ dollars cheaper, $0.85x - 90 = 0.75x - 15,$ so $0.10x = 75$ and $x = 750.$

Thus, A is the correct answer.

7.

While Steve and LeRoy are fishing $1$ mile from shore, their boat springs a leak, and water comes in at a constant rate of $10$ gallons per minute. The boat will sink if it takes in more than $30$ gallons of water. Steve starts rowing toward the shore at a constant rate of $4$ miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?

$2$

$4$

$6$

$8$

$10$

Concepts:rate unit conversion

Difficulty rating: 1310

Small Hint:

Find how many minutes the trip to shore takes

Big Hint:

Water enters for $15$ minutes, and at most $30$ gallons may remain in the boat

Solution:

Rowing $1$ mile at $4$ miles per hour takes $\tfrac{1}{4}$ hour, or $15$ minutes. In that time $15 \cdot 10 = 150$ gallons of water enter the boat.

Since at most $30$ gallons may remain, LeRoy must bail $150 - 30 = 120$ gallons in $15$ minutes, a rate of $\dfrac{120}{15} = 8$ gallons per minute.

Thus, D is the correct answer.

8.

What is the volume of a cube whose surface area is twice that of a cube with volume $1?$

$\sqrt{2}$

$2$

$2\sqrt{2}$

$4$

$8$

Concepts:cube geometry surface area volume

Difficulty rating: 1380

Small Hint:

A cube with volume $1$ has surface area $6$

Big Hint:

If the larger cube has side $s,$ then $6s^2 = 12$

Solution:

The cube with volume $1$ has side $1$ and surface area $6.$ The larger cube has surface area $12,$ so if its side is $s,$ then $6s^2 = 12,$ giving $s = \sqrt{2}.$

Its volume is $(\sqrt{2})^3 = 2\sqrt{2}.$

Thus, C is the correct answer.

9.

Older television screens have an aspect ratio of $4:3.$ That is, the ratio of the width to the height is $4:3.$ The aspect ratio of many movies is not $4:3,$ so they are sometimes shown on a television screen by “letterboxing” — darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $2:1$ and is shown on an older television screen with a $27$ -inch diagonal. What is the height, in inches, of each darkened strip?

$2$

$2.25$

$2.5$

$2.7$

$3$

Concepts:Pythagorean Triple ratio and proportion

Difficulty rating: 1410

Small Hint:

The screen’s width, height, and diagonal are in ratio $4:3:5$

Big Hint:

The lit movie region keeps the full width but has height equal to half that width

Solution:

Since the sides and diagonal are in ratio $3:4:5,$ the height is $\tfrac{3}{5} \cdot 27 = 16.2$ inches and the width is $\tfrac{4}{5} \cdot 27 = 21.6$ inches.

The movie has aspect ratio $2:1,$ so its height is $\tfrac{21.6}{2} = 10.8$ inches.

Each darkened strip therefore has height $\dfrac{16.2 - 10.8}{2} = 2.7$ inches.

Thus, D is the correct answer.

10.

Doug can paint a room in $5$ hours. Dave can paint the same room in $7$ hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let $t$ be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by $t?$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t + 1) = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t + 1 = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1$

$(5 + 7)t = 1$

Concepts:rate linear equation

Difficulty rating: 1380

Small Hint:

Together they paint $\dfrac{1}{5} + \dfrac{1}{7}$ of the room each hour

Big Hint:

Of the $t$ hours, one hour is lunch, so they actually work $t - 1$ hours

Solution:

In one hour Doug paints $\tfrac{1}{5}$ of the room and Dave paints $\tfrac{1}{7},$ so together they paint $\tfrac{1}{5} + \tfrac{1}{7}$ of the room per hour.

Of the total time $t,$ one hour is spent at lunch, so they work for $t - 1$ hours. The fraction painted must equal $1,$ giving $\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1.$

Thus, D is the correct answer.

11.

Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum?

$154$

$159$

$164$

$167$

$189$

Concepts:cube geometry optimization

Difficulty rating: 1560

Small Hint:

On each cube the opposite faces are $1$ & $32,$ $2$ & $16,$ and $4$ & $8$

Big Hint:

The two lower cubes each hide a pair of opposite faces; the top cube hides only its bottom face

Solution:

The six faces of each cube sum to $1 + 2 + 4 + 8 + 16 + 32 = 63.$ From the pattern, the pairs of opposite faces are $1$ & $32,$ $2$ & $16,$ and $4$ & $8.$

Each of the two lower cubes hides a pair of opposite faces (top and bottom); hiding the pair $4 + 8 = 12$ is best. The top cube hides only its bottom face, so hide the $1.$

The greatest sum is $\begin{aligned} &3 \cdot 63 - 2 \cdot 12 - 1 \\ &= 189 - 24 - 1 \\ &= 164. \end{aligned}$

Thus, C is the correct answer.

12.

A function $f$ has domain $[0, 2]$ and range $[0, 1].$ (The notation $[a, b]$ denotes $\{x : a \le x \le b\}.$ ) What are the domain and range, respectively, of the function $g$ defined by $g(x) = 1 - f(x + 1)?$

$[-1, 1],$ $[-1, 0]$

$[-1, 1],$ $[0, 1]$

$[0, 2],$ $[-1, 0]$

$[1, 3],$ $[-1, 0]$

$[1, 3],$ $[0, 1]$

Concepts:function

Difficulty rating: 1620

Small Hint:

$f(x + 1)$ is defined when $0 \le x + 1 \le 2$

Big Hint:

Subtracting $f(x + 1) \in [0, 1]$ from $1$ reverses the interval but keeps it $[0, 1]$

Solution:

The value $f(x + 1)$ is defined when $0 \le x + 1 \le 2,$ that is, $-1 \le x \le 1,$ so the domain of $g$ is $[-1, 1].$

As $f(x + 1)$ ranges over $[0, 1],$ the value $1 - f(x + 1)$ ranges over $[0, 1]$ as well, so the range of $g$ is $[0, 1].$

Thus, B is the correct answer.

13.

Points $A$ and $B$ lie on a circle centered at $O,$ and $\angle AOB = 60^\circ.$ A second circle is internally tangent to the first and tangent to both $OA$ and $OB.$ What is the ratio of the area of the smaller circle to that of the larger circle?

$\dfrac{1}{16}$

$\dfrac{1}{9}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

Concepts:tangent circles special right triangle area ratio

Difficulty rating: 1620

Small Hint:

The center of the small circle lies on the bisector of $\angle AOB$

Big Hint:

Dropping a radius to $OA$ makes a $30$ - $60$ - $90$ triangle in which $OE = 2r$

Solution:

Let $r$ and $R$ be the radii of the smaller and larger circles, and let $E$ be the center of the smaller circle. By symmetry $E$ lies on the bisector of $\angle AOB,$ so $OE$ makes a $30^\circ$ angle with $OA.$

Dropping the radius $ED$ perpendicular to $OA$ gives a $30$ - $60$ - $90$ triangle with $OE = 2 \cdot ED = 2r.$ Since the circles are internally tangent, $OE = R - r.$

Then $R - r = 2r,$ so $R = 3r$ and $\tfrac{r}{R} = \tfrac{1}{3}.$ The ratio of areas is $\left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}.$

Thus, B is the correct answer.

14.

What is the area of the region defined by the inequality $|3x - 18| + |2y + 7| \le 3?$

$3$

$\dfrac{7}{2}$

$4$

$\dfrac{9}{2}$

$5$

Concepts:absolute value rhombus area

Difficulty rating: 1660

Small Hint:

The region is a rhombus centered where $3x - 18 = 0$ and $2y + 7 = 0$

Big Hint:

Its horizontal diagonal spans $x \in [5, 7]$ and its vertical diagonal spans $y \in [-5, -2]$

Solution:

The region is a rhombus centered at $\left(6, -\tfrac{7}{2}\right).$ Setting $2y + 7 = 0$ gives $|3x - 18| \le 3,$ so $x \in [5, 7],$ a horizontal diagonal of length $2.$

Setting $3x - 18 = 0$ gives $|2y + 7| \le 3,$ so $y \in [-5, -2],$ a vertical diagonal of length $3.$

The area of the rhombus is half the product of its diagonals, $\dfrac{1}{2} \cdot 2 \cdot 3 = 3.$

Thus, A is the correct answer.

15.

Let $k = 2008^2 + 2^{2008}.$ What is the units digit of $k^2 + 2^k?$

$0$

$2$

$4$

$6$

$8$

Concepts:units digit modular exponentiation

Difficulty rating: 1740

Small Hint:

The units digit of $2^n$ cycles $2, 4, 8, 6$ with period $4$

Big Hint:

Find the units digit of $k$ first, then note that $k$ is a multiple of $4$

Solution:

The units digit of $2008^2$ is $4.$ Since $2008$ is a multiple of $4,$ the units digit of $2^{2008}$ is $6.$ Thus $k$ has units digit $0,$ and so does $k^2.$

Both $2008^2$ and $2^{2008}$ are multiples of $4,$ so $k$ is a multiple of $4.$ Therefore the units digit of $2^k$ is $6.$

The units digit of $k^2 + 2^k$ is then $0 + 6 = 6.$

Thus, D is the correct answer.

16.

The numbers $\log(a^3 b^7),$ $\log(a^5 b^{12}),$ and $\log(a^8 b^{15})$ are the first three terms of an arithmetic sequence, and the $12$ th term of the sequence is $\log(b^n).$ What is $n?$

$40$

$56$

$76$

$112$

$143$

Concepts:arithmetic sequence logarithm

Difficulty rating: 1800

Small Hint:

Write each term as $p \log a + q \log b$ and set the consecutive differences equal

Big Hint:

Equal differences force $\log a = 2 \log b,$ reducing every term to a multiple of $\log b$

Solution:

The three terms are $3\log a + 7\log b,$ $5\log a + 12\log b,$ and $8\log a + 15\log b.$ Setting the two consecutive differences equal, $\begin{aligned} &2\log a + 5\log b \\ &= 3\log a + 3\log b, \end{aligned}$ so $\log a = 2\log b.$

The first term is then $(3 \cdot 2 + 7)\log b = 13\log b,$ and the common difference is $(2 \cdot 2 + 5)\log b = 9\log b.$

The $12$ th term is $\begin{aligned} (13 + 11 \cdot 9)\log b &= 112\log b \\ &= \log(b^{112}), \end{aligned}$ so $n = 112.$

Thus, D is the correct answer.

17.

Let $a_1,$ $a_2,$ $\ldots$ be a sequence of integers determined by the rule $a_n = \frac{a_{n-1}}{2}$ if $a_{n-1}$ is even and $a_n = 3a_{n-1} + 1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \le 2008$ is it true that $a_1$ is less than each of $a_2,$ $a_3,$ and $a_4?$

$250$

$251$

$501$

$502$

$1004$

Concepts:recursion modular arithmetic casework

Difficulty rating: 1870

Small Hint:

If $a_1$ is even then $a_2 \lt a_1,$ so $a_1$ must be odd

Big Hint:

Split the odd case according to whether $a_1 \equiv 1$ or $a_1 \equiv 3 \pmod 4$

Solution:

If $a_1$ is even, then $a_2 = \frac{a_1}{2} \lt a_1,$ so the condition fails.

If $a_1 \equiv 1 \pmod 4,$ then $a_2 = 3a_1 + 1$ is a multiple of $4,$ so $a_3 = \frac{3a_1 + 1}{2}$ and $a_4 = \frac{3a_1 + 1}{4} \le a_1,$ and again the condition fails.

If $a_1 \equiv 3 \pmod 4,$ then $a_2$ is even but not a multiple of $4,$ so $a_3 = \frac{3a_1 + 1}{2} \gt a_1,$ and $a_3$ is odd, giving $a_4 = 3a_3 + 1 \gt a_3 \gt a_1.$ The condition holds.

Exactly $\tfrac{2008}{4} = 502$ values of $a_1 \le 2008$ satisfy $a_1 \equiv 3 \pmod 4.$

Thus, D is the correct answer.

18.

Triangle $ABC,$ with sides of length $5,$ $6,$ and $7,$ has one vertex on the positive $x$ -axis, one on the positive $y$ -axis, and one on the positive $z$ -axis. Let $O$ be the origin. What is the volume of tetrahedron $OABC?$

$\sqrt{85}$

$\sqrt{90}$

$\sqrt{95}$

$10$

$\sqrt{105}$

Concepts:3D geometry system of equations volume

Difficulty rating: 1910

Small Hint:

Let the vertices be $(a, 0, 0),$ $(0, b, 0),$ $(0, 0, c)$ and write the three side-length equations

Big Hint:

Adding the three equations gives $a^2 + b^2 + c^2,$ and the volume is $\tfrac{1}{6}abc$

Solution:

Let $A = (a, 0, 0),$ $B = (0, b, 0),$ $C = (0, 0, c).$ Assigning the sides, $\begin{aligned} a^2 + b^2 &= 25, \\ b^2 + c^2 &= 36, \\ a^2 + c^2 &= 49. \end{aligned}$

Adding gives $a^2 + b^2 + c^2 = 55,$ so $a^2 = 19,$ $b^2 = 6,$ and $c^2 = 30.$

The volume is $\begin{aligned} \dfrac{1}{6}abc &= \dfrac{1}{6}\sqrt{19 \cdot 6 \cdot 30} \\ &= \dfrac{1}{6}\sqrt{3420} \\ &= \sqrt{95}. \end{aligned}$

Thus, C is the correct answer.

19.

In the expansion of $\begin{aligned} &\left(1 + x + x^2 + \cdots + x^{27}\right) \\ &\quad {}\cdot \left(1 + x + x^2 + \cdots + x^{14}\right)^2, \end{aligned}$ what is the coefficient of $x^{28}?$

$195$

$196$

$224$

$378$

$405$

Concepts:generating functions partitions and compositions

Difficulty rating: 1930

Small Hint:

A term has the form $x^{a+b+c}$ with $0 \le a \le 27$ and $0 \le b, c \le 14$

Big Hint:

For each of the $15^2$ pairs $(b, c),$ there is a unique valid $a,$ except when $b = c = 0$

Solution:

Each term is $x^{a + b + c}$ with $0 \le a \le 27$ and $0 \le b, c \le 14.$ To get $x^{28}$ we need $a = 28 - b - c.$

There are $(14 + 1)^2 = 225$ choices for $(b, c).$ For every choice except $(b, c) = (0, 0),$ the required $a = 28 - b - c$ lies in $[0, 27],$ giving a valid term.

The coefficient of $x^{28}$ is therefore $225 - 1 = 224.$

Thus, C is the correct answer.

20.

Triangle $ABC$ has $AC = 3,$ $BC = 4,$ and $AB = 5.$ Point $D$ is on $AB,$ and $CD$ bisects the right angle. The inscribed circles of $\triangle ADC$ and $\triangle BCD$ have radii $r_a$ and $r_b,$ respectively. What is $\frac{r_a}{r_b}?$

$\dfrac{1}{28}(10 - \sqrt{2})$

$\dfrac{3}{56}(10 - \sqrt{2})$

$\dfrac{1}{14}(10 - \sqrt{2})$

$\dfrac{5}{56}(10 - \sqrt{2})$

$\dfrac{3}{28}(10 - \sqrt{2})$

Concepts:angle bisector theorem incircle, incenter, and inradius area ratio

Difficulty rating: 2100

Small Hint:

By the Angle Bisector Theorem, $AD:DB = CA:CB = 3:4$

Big Hint:

For each small triangle $r = \frac{\text{area}}{s};$ the two triangles share the base $CD$

Solution:

By the Angle Bisector Theorem, $AD:DB = CA:CB = 3:4,$ so $AD = \tfrac{15}{7}$ and $BD = \tfrac{20}{7}.$ The areas of $\triangle ADC$ and $\triangle BCD$ share base $CD,$ so they are in ratio $3:4,$ namely $\tfrac{18}{7}$ and $\tfrac{24}{7}.$

Splitting $\triangle ABC$ along $CD,$ which meets each leg at $45^\circ,$ gives $\dfrac{3 \cdot CD}{2\sqrt{2}} + \dfrac{4 \cdot CD}{2\sqrt{2}} = 6,$ so $CD = \tfrac{12\sqrt{2}}{7}.$

The two semiperimeters are $\begin{aligned} s_a &= \dfrac{6}{7}(3 + \sqrt{2}), \\ s_b &= \dfrac{6}{7}(4 + \sqrt{2}). \end{aligned}$ Using $r = \frac{\text{area}}{s},$ $\begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} \\ &= \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}}, \end{aligned}$

Rationalizing, $\dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} = \dfrac{10 - \sqrt{2}}{7},$ so $\begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{3}{4} \cdot \dfrac{10 - \sqrt{2}}{7} \\ &= \dfrac{3}{28}(10 - \sqrt{2}). \end{aligned}$

Thus, E is the correct answer.

21.

A permutation $(a_1, a_2, a_3, a_4, a_5)$ of $(1, 2, 3, 4, 5)$ is heavy-tailed if $a_1 + a_2 \lt a_4 + a_5.$ What is the number of heavy-tailed permutations?

$36$

$40$

$44$

$48$

$52$

Concepts:permutations symmetry complementary counting

Difficulty rating: 2050

Small Hint:

By symmetry, $a_1 + a_2 \lt a_4 + a_5$ and $a_1 + a_2 \gt a_4 + a_5$ occur equally often

Big Hint:

Count the balanced permutations $a_1 + a_2 = a_4 + a_5;$ here $a_3 \in \{1, 3, 5\}$

Solution:

Call a permutation balanced if $a_1 + a_2 = a_4 + a_5.$ Reversing the entries swaps the two strict cases, so heavy-tailed and heavy-headed permutations are equally numerous.

The total $1 + 2 + 3 + 4 + 5 = 15$ is odd, so in a balanced permutation $a_3$ must be odd, one of $1, 3, 5.$ For each choice, the remaining four numbers split uniquely into two equal-sum pairs.

Any of the four can be $a_1$ (fixing $a_2$ ), and either remaining number can be $a_4$ (fixing $a_5$ ), giving $3 \cdot 4 \cdot 2 = 24$ balanced permutations.

The other $120 - 24 = 96$ permutations split evenly, so there are $\tfrac{96}{2} = 48$ heavy-tailed permutations.

Thus, D is the correct answer.

22.

A round table has radius $4.$ Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being endpoints of the same side of length $x.$ Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x?$

$2\sqrt{5} - \sqrt{3}$

$3$

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

$2\sqrt{3}$

$\dfrac{5 + 2\sqrt{3}}{2}$

Concepts:circle isosceles triangle Pythagorean Theorem

Difficulty rating: 2120

Small Hint:

For one mat, let $P, Q$ be its outer corners and $R$ the point diametrically opposite $P;$ then $\triangle PQR$ is right-angled with $PR = 8$

Big Hint:

The inner corners form isosceles triangles with vertex angle $120^\circ,$ contributing a segment of length $\sqrt{3}\,x$

Solution:

Take one mat with outer corners $P$ and $Q,$ and let $R$ be the point of the table’s edge diametrically opposite $P.$ Then $PR = 8$ is a diameter, so $\triangle PQR$ has a right angle at $Q,$ with $PQ = x.$

Along $QR,$ the inner corners of neighboring mats meet in an isosceles triangle with two sides of length $x$ and vertex angle $120^\circ,$ whose base is $\sqrt{3}\,x.$ Hence $QR = \sqrt{3}\,x + 2.$

The Pythagorean Theorem gives $x^2 + \left(\sqrt{3}\,x + 2\right)^2 = 64,$ which simplifies to $x^2 + \sqrt{3}\,x - 15 = 0.$

Taking the positive root, $x = \dfrac{-\sqrt{3} + \sqrt{63}}{2} = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$

Thus, C is the correct answer.

23.

The solutions of the equation $z^4 + 4z^3 i - 6z^2 - 4zi - i = 0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon?

$2^{\frac{5}{8}}$

$2^{\frac{3}{4}}$

$2$

$2^{\frac{5}{4}}$

$2^{\frac{3}{2}}$

Concepts:binomial theorem complex number roots of unity

Difficulty rating: 2240

Small Hint:

Add $1 + i$ to both sides to recognize a perfect fourth power

Big Hint:

The four values of $z + i$ lie equally spaced on a circle of radius $2^{\frac{1}{8}},$ forming a square

Solution:

Adding $1 + i$ to both sides, the left side becomes $\begin{aligned} &z^4 + 4z^3 i - 6z^2 - 4zi + 1 \\ &= (z + i)^4, \end{aligned}$ so $(z + i)^4 = 1 + i.$

The four solutions for $w = z + i$ are equally spaced on a circle of radius $|1 + i|^{\frac{1}{4}} = (2^{\frac{1}{2}})^{\frac{1}{4}} = 2^{\frac{1}{8}},$ and they form a square. Subtracting $i$ merely translates it.

A square inscribed in a circle of radius $2^{\frac{1}{8}}$ has diagonal $2 \cdot 2^{\frac{1}{8}} = 2^{\frac{9}{8}},$ so its side is $\tfrac{2^{\frac{9}{8}}}{\sqrt{2}} = 2^{\frac{5}{8}}.$

The area is $\left(2^{\frac{5}{8}}\right)^2 = 2^{\frac{5}{4}}.$

Thus, D is the correct answer.

24.

Triangle $ABC$ has $\angle C = 60^\circ$ and $BC = 4.$ Point $D$ is the midpoint of $BC.$ What is the largest possible value of $\tan(\angle BAD)?$

$\dfrac{\sqrt{3}}{6}$

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

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

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

$1$

Concepts:coordinate geometry trigonometric identity optimization

Difficulty rating: 2380

Small Hint:

Place $C = (0, 0),$ $B = (2, 2\sqrt{3}),$ and $A = (x, 0)$ with $x \gt 0$

Big Hint:

Using slopes gives $\tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8};$ maximize over $x$

Solution:

Place $C = (0, 0),$ $B = (2, 2\sqrt{3})$ so that $\angle C = 60^\circ$ and $BC = 4,$ and let $A = (x, 0)$ with $x \gt 0.$ Then $D = (1, \sqrt{3})$ is the midpoint of $BC.$

The vectors $\overrightarrow{AB} = (2-x, 2\sqrt{3})$ and $\overrightarrow{AD} = (1-x, \sqrt{3})$ have cross-product magnitude $\sqrt{3}\,x$ and dot product $x^2 - 3x + 8,$ which is always positive. Hence $\tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8}.$

The derivative has the sign of $8-x^2,$ so the unique maximum occurs at $x = 2\sqrt{2}.$ Substituting, $\begin{aligned} \tan(\angle BAD) &= \dfrac{2\sqrt{6}}{16 - 6\sqrt{2}} \\ &= \dfrac{\sqrt{6}}{8 - 3\sqrt{2}} \\ &= \dfrac{\sqrt{3}}{4\sqrt{2} - 3}. \end{aligned}$

Thus, D is the correct answer.

25.

A sequence $(a_1, b_1),$ $(a_2, b_2),$ $(a_3, b_3),$ $\ldots$ of points in the coordinate plane satisfies $\begin{aligned} &(a_{n+1}, b_{n+1}) \\ &= \left(\sqrt{3}\,a_n - b_n,\; \sqrt{3}\,b_n + a_n\right) \\ &\quad (n = 1, 2, 3, \ldots) \end{aligned}$ Suppose that $(a_{100}, b_{100}) = (2, 4).$ What is $a_1 + b_1?$

$-\dfrac{1}{2^{97}}$

$-\dfrac{1}{2^{99}}$

$0$

$\dfrac{1}{2^{98}}$

$\dfrac{1}{2^{96}}$

Concepts:complex number De Moivre’s Theorem

Difficulty rating: 2440

Small Hint:

Write $z_n = a_n + b_n i;$ the recurrence becomes $z_{n+1} = z_n(\sqrt{3} + i)$

Big Hint:

Since $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),$ apply De Moivre’s theorem to $(\sqrt{3} + i)^{99}$

Solution:

Let $z_n = a_n + b_n i.$ Then $\begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i), \end{aligned}$ so $z_{n+1} = z_n(\sqrt{3} + i)$ and $z_{100} = z_1(\sqrt{3} + i)^{99}.$

Since $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),$ De Moivre’s theorem gives $(\sqrt{3} + i)^{99}$ $= 2^{99}(\cos 2970^\circ + i\sin 2970^\circ).$ As $2970^\circ$ is coterminal with $90^\circ,$ this equals $2^{99} i.$

Thus $2 + 4i = z_1 \cdot 2^{99} i,$ so $z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}.$

Then $a_1 = \tfrac{4}{2^{99}}$ and $b_1 = -\tfrac{2}{2^{99}},$ so $a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}.$

Thus, D is the correct answer.