2012 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

A bug crawls along a number line, starting at $-2.$ It crawls to $-6,$ then turns around and crawls to $5.$ How many units does the bug crawl altogether?

$9$

$11$

$13$

$14$

$15$

Answer: E

Concepts:absolute value

Difficulty rating: 770

Small Hint:

Add the two separate crawl distances rather than the net displacement

Big Hint:

The first leg is $|-6-(-2)|$ and the second is $|5-(-6)|$

Solution:

The bug crawls $|-6-(-2)| = 4$ units on the first leg and $|5-(-6)| = 11$ units on the second leg.

The total distance is $4 + 11 = 15.$

Thus, the correct answer is E.

2.

Cagney can frost a cupcake every $20$ seconds and Lacey can frost a cupcake every $30$ seconds. Working together, how many cupcakes can they frost in $5$ minutes?

$10$

$15$

$20$

$25$

$30$

Answer: D

Concepts:rate unit conversion

Difficulty rating: 880

Small Hint:

Convert $5$ minutes to seconds, or count each person’s cupcakes per minute

Big Hint:

In $5$ minutes Cagney frosts $\dfrac{300}{20}$ and Lacey frosts $\dfrac{300}{30}$

Solution:

In $5$ minutes there are $300$ seconds. Cagney frosts $\dfrac{300}{20} = 15$ cupcakes and Lacey frosts $\dfrac{300}{30} = 10$ cupcakes.

Together they frost $15 + 10 = 25$ cupcakes.

Thus, the correct answer is D.

3.

A box $2$ centimeters high, $3$ centimeters wide, and $5$ centimeters long can hold $40$ grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold $n$ grams of clay. What is $n?$

$120$

$160$

$200$

$240$

$280$

Answer: D

Concepts:volume ratio and proportion

Difficulty rating: 880

Small Hint:

The capacity is proportional to the volume

Big Hint:

The volume is multiplied by $2 \times 3 \times 1$

Solution:

The second box has $2 \times 3 \times 1 = 6$ times the volume of the first, so it holds $6$ times as much clay.

Therefore $n = 6 \cdot 40 = 240.$

Thus, the correct answer is D.

4.

In a bag of marbles, $\dfrac{3}{5}$ of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?

$\dfrac{2}{5}$

$\dfrac{3}{7}$

$\dfrac{4}{7}$

$\dfrac{3}{5}$

$\dfrac{4}{5}$

Answer: C

Concepts:fraction ratio and proportion

Difficulty rating: 970

Small Hint:

Start with a convenient total, such as $5$ marbles

Big Hint:

Blue and red begin at $3$ and $2$ ; doubling the red gives $3$ blue and $4$ red

Solution:

Suppose there are $5$ marbles: $3$ blue and $2$ red. Doubling the red gives $4$ red while the blue stays at $3.$

The total is now $3 + 4 = 7,$ so the fraction that is red is $\dfrac{4}{7}.$

Thus, the correct answer is C.

5.

A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of $280$ pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?

$8$

$16$

$25$

$64$

$96$

Answer: D

Concepts:ratio and proportion system of equations

Difficulty rating: 1040

Small Hint:

Express every count in terms of the number of blueberries

Big Hint:

With $b$ blueberries there are $2b$ raspberries, $8b$ cherries, and $24b$ grapes

Solution:

Let $b$ be the number of blueberries. Then there are $2b$ raspberries, $4 \cdot 2b = 8b$ cherries, and $3 \cdot 8b = 24b$ grapes.

The total is $b + 2b + 8b + 24b = 35b = 280,$ so $b = 8$ and there are $8b = 64$ cherries.

Thus, the correct answer is D.

6.

The sums of three whole numbers taken in pairs are $12,$ $17,$ and $19.$ What is the middle number?

$4$

$5$

$6$

$7$

$8$

Answer: D

Concepts:system of equations

Difficulty rating: 1080

Small Hint:

Adding all three pairwise sums counts each number twice

Big Hint:

The total $a+b+c$ is half of $12+17+19$ ; subtract each pair sum to get the third number

Solution:

Let the numbers be $a \lt b \lt c.$ Adding the three pairwise sums gives $2(a+b+c) = 12+17+19$ $= 48,$ so $a+b+c = 24.$

Then $a = 24-19 = 5,$ $b = 24-17 = 7,$ and $c = 24-12 = 12.$ The middle number is $7.$

Thus, the correct answer is D.

7.

Mary divides a circle into $12$ sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

$5$

$6$

$8$

$10$

$12$

Answer: C

Concepts:arithmetic sequence optimization

Difficulty rating: 1240

Small Hint:

The twelve angles sum to $360^\circ$

Big Hint:

With first term $a$ and common difference $d,$ $12a + 66d = 360,$ so $2a + 11d = 60$

Solution:

Let $a$ be the smallest angle and $d \ge 0$ the common difference. The sum of the angles is $12a + 66d = 360,$ so $2a + 11d = 60.$

To make $a$ small, take $d$ large. Since $11d$ must be even, $d$ is even, and $d = 4$ gives $2a = 60 - 44 = 16,$ so $a = 8.$ A larger even $d$ makes $a$ non-positive.

Thus, the correct answer is C.

8.

An iterative average of the numbers $1,$ $2,$ $3,$ $4,$ and $5$ is computed in the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?

$\dfrac{31}{16}$

$2$

$\dfrac{17}{8}$

$3$

$\dfrac{65}{16}$

Answer: C

Concepts:mean optimization

Difficulty rating: 1480

Small Hint:

Work out the weight each position contributes to the final average

Big Hint:

For the order $a,b,c,d,e$ the result is $\dfrac{a+b+2c+4d+8e}{16}$ ; the last entries matter most

Solution:

For the order $a, b, c, d, e,$ the iterative average is $\frac{a + b + 2c + 4d + 8e}{16}.$ The later positions carry the most weight.

The largest value uses $(a,b,c,d,e) = (1,2,3,4,5),$ giving $\dfrac{65}{16},$ and the smallest uses $(5,4,3,2,1),$ giving $\dfrac{31}{16}.$

The difference is $\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.$

Thus, the correct answer is C.

9.

A year is a leap year if and only if the year number is divisible by $400$ (such as $2000$ ) or is divisible by $4$ but not by $100$ (such as $2012$ ). The $200$ th anniversary of the birth of novelist Charles Dickens was celebrated on February $7,$ $2012,$ a Tuesday. On what day of the week was Dickens born?

Friday

Saturday

Sunday

Monday

Tuesday

Answer: A

Concepts:modular arithmetic date and time

Difficulty rating: 1540

Small Hint:

Count the total number of days between the two February $7$ th dates, working modulo $7$

Big Hint:

In $200$ years there are $200 \cdot 365$ ordinary days plus one extra for each leap year; $1900$ is not a leap year

Solution:

From February $7,$ $1812$ to February $7,$ $2012$ there are $200 \cdot 365 = 73000$ ordinary days plus one for each leap day.

One quarter of the $200$ years contain a leap day, except $1900,$ giving $\tfrac14 \cdot 200 - 1 = 49$ leap days. So the span is $73049$ days.

Since $73049 = 7 \cdot 10435 + 4,$ the birth day was $4$ days before a Tuesday, which is a Friday.

Thus, the correct answer is A.

10.

A triangle has area $30,$ one side of length $10,$ and the median to that side of length $9.$ Let $\theta$ be the acute angle formed by that side and the median. What is $\sin\theta?$

$\dfrac{3}{10}$

$\dfrac{1}{3}$

$\dfrac{9}{20}$

$\dfrac{2}{3}$

$\dfrac{9}{10}$

Answer: D

Concepts:triangle area median (geometry)trigonometry

Difficulty rating: 1610

Small Hint:

A median splits a triangle into two pieces of equal area

Big Hint:

One half-triangle has sides $5$ and $9$ with included angle $\theta$ and area $15$

Solution:

The median divides the triangle into two triangles of equal area $15.$ One of them has the two sides of length $5$ (half the base) and $9$ (the median) meeting at angle $\theta.$

Its area is $\tfrac12 \cdot 5 \cdot 9 \sin\theta = 15,$ so $\sin\theta = \dfrac{2 \cdot 15}{5 \cdot 9} = \dfrac{2}{3}.$

Thus, the correct answer is D.

11.

Alex, Mel, and Chelsea play a game that has $6$ rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is $\dfrac12,$ and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?

$\dfrac{5}{72}$

$\dfrac{5}{36}$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$1$

Answer: B

Concepts:basic probability independent events multiset permutations

Difficulty rating: 1540

Small Hint:

First find each player’s per-round win probability

Big Hint:

Since $P(\text{Mel}) + P(\text{Chelsea}) = \dfrac12$ and $P(\text{Mel}) = 2P(\text{Chelsea}),$ multiply by the number $\dfrac{6!}{3!\,2!\,1!}$ of arrangements

Solution:

Since Alex wins with probability $\tfrac12,$ the others share the remaining $\tfrac12.$ With Mel twice as likely as Chelsea, $P(\text{Mel}) = \tfrac13$ and $P(\text{Chelsea}) = \tfrac16.$

The number of orderings of the wins $AAAMMC$ is $\dfrac{6!}{3!\,2!\,1!} = 60.$ The probability is $\begin{aligned} 60 \cdot \left(\tfrac12\right)^3 \left(\tfrac13\right)^2 \left(\tfrac16\right) &= \frac{60}{432} \\ &= \frac{5}{36}. \end{aligned}$

Thus, the correct answer is B.

12.

A square region $ABCD$ is externally tangent to the circle with equation $x^2 + y^2 = 1$ at the point $(0, 1)$ on the side $CD.$ Vertices $A$ and $B$ are on the circle with equation $x^2 + y^2 = 4.$ What is the side length of this square?

$\dfrac{\sqrt{10} + 5}{10}$

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

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

$\dfrac{2\sqrt{19} - 4}{5}$

$\dfrac{9 - \sqrt{17}}{5}$

Answer: D

Concepts:coordinate geometry quadratic symmetry

Difficulty rating: 1770

Small Hint:

Use the symmetry across the $y$ -axis to write $A = (a, b)$ and $B = (-a, b)$

Big Hint:

The side length is $2a$ horizontally and $b - 1$ vertically, so $b = 2a + 1$ ; substitute into $a^2 + b^2 = 4$

Solution:

By symmetry let $A = (a, b)$ with $a \gt 0$ and $B = (-a, b).$ The square sits on the tangent point $(0,1),$ so its horizontal width is $2a$ and its height is $b - 1.$

Since these are equal, $2a = b - 1,$ giving $b = 2a + 1.$

Substituting into $a^2 + b^2 = 4$ yields $5a^2 + 4a - 3 = 0.$ The positive root is $a = \dfrac{\sqrt{19} - 2}{5},$ so the side length is $2a = \dfrac{2\sqrt{19} - 4}{5}.$

Thus, the correct answer is D.

13.

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at $8{:}00$ AM and all three always take the same amount of time to eat lunch. On Monday the three of them painted $50\%$ of a house, quitting at $4{:}00$ PM. On Tuesday, when Paula wasn’t there, the two helpers painted only $24\%$ of the house and quit at $2{:}12$ PM. On Wednesday Paula worked by herself and finished the house by working until $7{:}12$ PM. How long, in minutes, was each day’s lunch break?

$30$

$36$

$42$

$48$

$60$

Answer: D

Concepts:rate system of equations

Difficulty rating: 1810

Small Hint:

Let the lunch break be $m$ minutes and let $p$ and $h$ be the painting rates (percent per minute)

Big Hint:

Working minutes are $480 - m,$ $372 - m,$ and $672 - m$ ; set up $(p+h)(480-m) = 50,$ $h(372-m) = 24,$ $p(672-m) = 26$

Solution:

Let $m$ be the lunch length in minutes. The three worked $480 - m$ minutes Monday, the helpers $372 - m$ minutes Tuesday, and Paula $672 - m$ minutes Wednesday.

If Paula paints $p\%$ per minute and the helpers together paint $h\%$ per minute, then $\begin{aligned} (p+h)(480-m) &= 50, \\ h(372-m) &= 24, \\ p(672-m) &= 26. \end{aligned}$

Adding the last two equations and subtracting from the first gives $108h - 192p = 0,$ so $h = \tfrac{16}{9}p.$ Solving the system gives $p = \tfrac{1}{24}$ and $m = 48.$

Thus, the correct answer is D.

14.

The closed curve in the figure is made up of $9$ congruent circular arcs each of length $\dfrac{2\pi}{3},$ where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side $2.$ What is the area enclosed by the curve?

$2\pi + 6$

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

$3\pi + 4$

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

$\pi + 6\sqrt{3}$

Answer: E

Concepts:area decomposition sector regular polygon

Difficulty rating: 1880

Small Hint:

Each arc is a $120^\circ$ sector of a unit circle, since arc length $\dfrac{2\pi}{3}$ means radius $1$

Big Hint:

The outward and inward sectors rearrange into one full unit circle added to the hexagon of side $2$

Solution:

Each arc has length $\dfrac{2\pi}{3}$ on a unit circle, so it is a $120^\circ$ sector. The nine equal sectors can be reassembled so that the enclosed region equals the regular hexagon of side $2$ plus one full circle of radius $1.$

A regular hexagon of side $2$ splits into $6$ equilateral triangles of side $2,$ so its area is $6 \cdot \dfrac{\sqrt3}{4} \cdot 2^2 = 6\sqrt3.$

Adding the unit circle’s area $\pi$ gives $\pi + 6\sqrt3.$

Thus, the correct answer is E.

15.

A $3 \times 3$ square is partitioned into $9$ unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated $90^\circ$ clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability that the grid is now entirely black?

$\dfrac{49}{512}$

$\dfrac{7}{64}$

$\dfrac{121}{1024}$

$\dfrac{81}{512}$

$\dfrac{9}{32}$

Answer: A

Concepts:basic probability symmetry casework

Difficulty rating: 1930

Small Hint:

Treat the four corners, the four edges, and the center as three independent groups

Big Hint:

A square ends black unless it is white and the square rotated into its position was also white; the center must start black

Solution:

The four corners form one cycle under the rotation, the four edge squares form another, and the center is fixed. These three groups are independent.

A position remains white exactly when both it and the square rotated into it were originally white. Thus the corners end black exactly when their cyclic string has no adjacent pair of whites. The allowed strings are the all-black string, the $4$ strings with one white, and the $2$ strings with two opposite whites: $7$ of the $2^4=16$ possibilities. Hence the corner probability is $\frac{7}{16}.$ The same argument applies to the four edge squares.

The center is black at the end only if it started black, with probability $\dfrac12.$ Multiplying, the whole grid is black with probability $\frac12 \cdot \left(\frac{7}{16}\right)^2 = \frac{49}{512}.$

Thus, the correct answer is A.

16.

Circle $C_1$ has its center $O$ lying on circle $C_2.$ The two circles meet at $X$ and $Y.$ Point $Z$ in the exterior of $C_1$ lies on circle $C_2$ and $XZ = 13,$ $OZ = 11,$ and $YZ = 7.$ What is the radius of circle $C_1?$

$5$

$\sqrt{26}$

$3\sqrt{3}$

$2\sqrt{7}$

$\sqrt{30}$

Answer: E

Concepts:law of cosines chord circle

Difficulty rating: 1870

Small Hint:

Let $r$ be the radius, so $OX = OY = r$ ; these equal chords of $C_2$ subtend equal angles at $Z$

Big Hint:

Apply the Law of Cosines to triangles $XZO$ and $YZO$ and set the two expressions for $\cos\angle Z$ equal

Solution:

Let $r$ be the radius of $C_1,$ so $OX = OY = r.$ These are equal chords of $C_2,$ so they subtend equal angles at $Z:$ $\angle XZO = \angle OZY.$

Applying the Law of Cosines to triangles $XZO$ and $YZO,$ $\begin{aligned} &\frac{13^2 + 11^2 - r^2}{2 \cdot 13 \cdot 11} \\ &= \frac{7^2 + 11^2 - r^2}{2 \cdot 7 \cdot 11}. \end{aligned}$

Clearing denominators and solving gives $r^2 = 30,$ so $r = \sqrt{30}.$

Thus, the correct answer is E.

17.

Let $S$ be a subset of $\{1, 2, 3, \ldots, 30\}$ with the property that no pair of distinct elements in $S$ has a sum divisible by $5.$ What is the largest possible size of $S?$

$10$

$13$

$15$

$16$

$18$

Answer: B

Concepts:modular arithmetic subsets extremal argument

Difficulty rating: 1800

Small Hint:

Sort the numbers by their remainder modulo $5$ ; there are $6$ numbers in each class

Big Hint:

Residues $1$ and $4$ cannot both appear, nor can $2$ and $3,$ and at most one multiple of $5$ is allowed

Solution:

Group $\{1, \ldots, 30\}$ by residue modulo $5;$ each class has $6$ numbers. A sum is divisible by $5$ when the residues are $0{+}0,$ $1{+}4,$ or $2{+}3.$

So $S$ can use at most one number $\equiv 0,$ and only one of the classes $\{1\}, \{4\}$ and only one of $\{2\}, \{3\}.$ That allows at most $1 + 6 + 6 = 13$ numbers.

The set $\{1, 2, 6, 7, 11, 12,$ $16, 17, 21, 22,$ $26, 27, 30\}$ achieves $13,$ so the maximum is $13.$

Thus, the correct answer is B.

18.

Triangle $ABC$ has $AB = 27,$ $AC = 26,$ and $BC = 25.$ Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC.$ What is $BI?$

$15$

$5 + \sqrt{26} + 3\sqrt{3}$

$3\sqrt{26}$

$\dfrac{2}{3}\sqrt{546}$

$9\sqrt{3}$

Answer: A

Concepts:incircle, incenter, and inradius Heron’s Formula Pythagorean Theorem

Difficulty rating: 1980

Small Hint:

Drop perpendiculars from $I$ to the sides; the tangent segment from $B$ has length $s - b$

Big Hint:

Find the inradius from area $= rs$ and Heron’s formula, then use the right triangle with legs $r$ and $BD$

Solution:

Let $D$ be the foot of the perpendicular from the incenter $I$ to $BC.$ The tangent length $BD = s - AC,$ where $s = \tfrac12(25 + 26 + 27) = 39,$ so $BD = 39 - 26 = 13.$

By Heron’s formula the area is $\sqrt{39 \cdot 14 \cdot 13 \cdot 12},$ and the inradius satisfies $r^2 = \dfrac{(s-a)(s-b)(s-c)}{s}$ $= \dfrac{14 \cdot 13 \cdot 12}{39}$ $= 56.$

In right triangle $BDI,$ $BI^2 = r^2 + BD^2$ $= 56 + 169$ $= 225,$ so $BI = 15.$

Thus, the correct answer is A.

19.

Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?

$60$

$170$

$290$

$320$

$660$

Answer: B

Concepts:graph theory complementary counting casework

Difficulty rating: 2090

Small Hint:

Model the friendships as a graph on $6$ vertices where everyone has the same degree $n,$ with $1 \le n \le 4$

Big Hint:

The cases $n = 1$ and $n = 4$ are complements of each other, as are $n = 2$ and $n = 3;$ count the two $2$ -regular graph types separately

Solution:

Model people as vertices of a graph, with edges for friendships. Everyone has the same degree $n$ with $1 \le n \le 4.$ The cases $n$ and $6 - 1 - n$ are complementary graphs, so $n = 1$ pairs with $n = 4$ and $n = 2$ with $n = 3.$

For $n = 1$ the graph is a perfect matching: $5 \cdot 3 = 15$ ways. Thus $n = 4$ also gives $15.$

For $n = 2$ the graph is a union of cycles: either two triangles $\left(\binom{5}{2} = 10\right)$ or one hexagon $\left(\dfrac{6!}{12} = 60\right),$ totaling $70.$ Thus $n = 3$ also gives $70.$

The total is $15 + 15 + 70 + 70 = 170.$

Thus, the correct answer is B.

20.

Consider the polynomial $\begin{aligned} P(x) &= \prod_{k=0}^{10}\left(x^{2^k} + 2^k\right) \\ &= (x+1)(x^2+2)(x^4+4) \\ &\quad \cdots (x^{1024}+1024). \end{aligned}$

The coefficient of $x^{2012}$ is equal to $2^a.$ What is $a?$

$5$

$6$

$7$

$10$

$24$

Answer: B

Concepts:polynomial power of 2 number base

Difficulty rating: 2220

Small Hint:

Each factor contributes either its $x^{2^k}$ term or its constant $2^k$ ; the chosen exponents must sum to $2012$

Big Hint:

There is only one way to write $2012$ as a sum of distinct powers of two, so exactly one term produces $x^{2012}$

Solution:

Expanding the product, a term of degree $2012$ comes from choosing $x^{2^k}$ from some factors so that the exponents sum to $2012.$ Since powers of two are distinct, this corresponds to the binary representation $2012 = 11111011100_2.$

That representation is unique, so exactly one term gives $x^{2012},$ and its coefficient is the product of the constants $2^k$ from the remaining factors: those with $k \in \{0, 1, 5\}.$

The coefficient is $2^0 \cdot 2^1 \cdot 2^5 = 2^6,$ so $a = 6.$

Thus, the correct answer is B.

21.

Let $a,$ $b,$ and $c$ be positive integers with $a \ge b \ge c$ such that $a^2 - b^2 - c^2 + ab = 2011$ and $\begin{aligned} &a^2 + 3b^2 + 3c^2 \\ &\quad {}- 3ab - 2ac - 2bc = -1997. \end{aligned}$

What is $a?$

$249$

$250$

$251$

$252$

$253$

Answer: E

Concepts:algebraic manipulation system of equations Diophantine Equation

Difficulty rating: 2090

Small Hint:

Add the two equations to get a clean symmetric expression

Big Hint:

The sum simplifies to $(a-b)^2$ $+ (b-c)^2$ $+ (c-a)^2 = 14,$ and $14 = 3^2 + 2^2 + 1^2$ uniquely

Solution:

Adding the two equations gives $2a^2 + 2b^2 + 2c^2$ $- 2ab - 2bc - 2ca = 14,$ that is, $\begin{aligned} &(a-b)^2 + (b-c)^2 \\ &\quad {}+ (c-a)^2 = 14. \end{aligned}$

The only way to write $14$ as a sum of three squares is $9 + 4 + 1.$ Since $a \ge b \ge c,$ we get $a - c = 3,$ with either $(a-b, b-c) = (2,1)$ or $(1,2).$

Substituting $(a, b, c) = (c+3, c+1, c)$ into the first equation gives $3(2c+3) + 2(c+1) = 2011,$ so $c = 250$ and $(a, b, c) = (253, 251, 250).$ The other case has no integer solution.

Thus, the correct answer is E.

22.

Distinct planes $p_1,$ $p_2,$ $\ldots,$ $p_k$ intersect the interior of a cube $Q.$ Let $S$ be the union of the faces of $Q$ and let $P = \bigcup_{j=1}^{k} p_j.$ The intersection of $P$ and $S$ consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of $Q.$ What is the difference between the maximum and the minimum possible values of $k?$

$8$

$12$

$20$

$23$

$24$

Answer: C

Concepts:cube geometry 3D geometry casework

Difficulty rating: 2460

Small Hint:

On each face the drawn segments join midpoints of edges; classify a plane’s cross-section as a square, rectangle, triangle, or hexagon

Big Hint:

For the maximum, count all planes of each type; for the minimum, cover all $24$ short and $12$ long segments with as few planes as possible

Solution:

On every face, the required segments join midpoints of edges. A plane cutting the cube meets the faces in one of four symmetric shapes: a square through midpoints ( $3$ such planes), a rectangle per edge ( $12$ planes), a triangle per vertex ( $8$ planes), or a regular hexagon per pair of opposite vertices ( $4$ planes).

Using all of them gives the maximum $k = 3 + 12 + 8 + 4 = 27.$

The full figure consists of $24$ short segments and $12$ long segments. A square plane contains $(0,4),$ a rectangle $(2,2),$ a triangle $(3,0),$ and a hexagon $(6,0)$ short and long segments, respectively. Give each short segment weight $1$ and each long segment weight $\frac{3}{2}.$ Every plane then covers weight at most $6,$ whereas the required segments have total weight $24+(\frac{3}{2})12=42.$ Thus at least $7$ planes are needed. The $4$ hexagon planes cover all $24$ short segments, and the $3$ square planes cover all $12$ long segments, so $k=7$ is attainable.

The difference is $27 - 7 = 20.$

Thus, the correct answer is C.

23.

Let $S$ be the square one of whose diagonals has endpoints $(0.1, 0.7)$ and $(-0.1, -0.7).$ A point $v = (x, y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0 \le x \le 2012$ and $0 \le y \le 2012.$ Let $T(v)$ be a translated copy of $S$ centered at $v.$ What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coordinates in its interior?

$0.125$

$0.14$

$0.16$

$0.25$

$0.32$

Answer: C

Concepts:geometric probability lattice point area

Difficulty rating: 2340

Small Hint:

The square $S$ has diagonals of length $\sqrt2$ and area $1;$ by periodicity, reduce to one unit cell of the lattice

Big Hint:

$T(v)$ contains a lattice point exactly when $v$ lies in the copy of $S$ centered there; containing two requires $v$ in the overlap of copies centered at adjacent lattice points

Solution:

The diagonal from $(0.1, 0.7)$ to $(-0.1, -0.7)$ has length $\sqrt{0.2^2 + 1.4^2} = \sqrt2,$ so $S$ is a square of area $1.$ The translate $T(v)$ contains a lattice point exactly when $v$ lies inside the copy of $S$ centered at that point.

Containing exactly two interior lattice points requires $v$ to lie in the overlap of two copies centered at adjacent lattice points. By periodicity the answer is the total such overlap area within one unit cell.

Consider copies centered at $(0,0)$ and $(1,0).$ Their overlap is a rectangle. A quarter-turn of the listed half-diagonal gives the relevant vertices $(0.7,-0.1)$ and $(0.3,0.1).$ Following the two pairs of parallel sides through their intersections gives side lengths $0.4$ and $0.2,$ so the overlap area is $0.08.$ Diagonally centered copies do not overlap, and horizontal and vertical adjacencies contribute equally. Accounting for the cell boundaries gives probability $2\cdot0.08=0.16.$

Thus, the correct answer is C.

24.

Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1 = 0.201,$ $a_2 = (0.2011)^{a_1},$ $a_3 = (0.20101)^{a_2},$ and $a_4 = (0.201011)^{a_3},$ and more generally $a_k = \begin{cases} \tiny \left(0.\underbrace{20101\ldots0101}_{k+2 \text{ digits}}\right)^{a_{k-1}}, & \tiny \text{if } k \text{ is odd,} \\ \tiny \left(0.\underbrace{20101\ldots01011}_{k+2 \text{ digits}}\right)^{a_{k-1}}, & \tiny \text{if } k \text{ is even.} \end{cases}$

Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}.$ What is the sum of all the integers $k,$ $1 \le k \le 2011,$ such that $a_k = b_k?$

$671$

$1006$

$1341$

$2011$

$2012$

Answer: C

Concepts:exponent inequality

Difficulty rating: 2460

Small Hint:

Each base is between $0$ and $1,$ so raising to a power reverses order; compare consecutive terms

Big Hint:

The values interleave: $1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010}$ $\gt a_{2011} \gt a_{2009}$ $\gt \cdots \gt a_1;$ a term is fixed when its rank equals its index

Solution:

Because each base lies strictly between $0$ and $1,$ the function $t \mapsto (\text{base})^t$ is decreasing, while $t \mapsto t^b$ is increasing for $b \gt 0.$ Comparing terms shows the sequence orders as $\begin{aligned} &1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010} \\ &\gt a_{2011} \gt a_{2009} \\ &\gt \cdots \gt a_1 \gt 0. \end{aligned}$

So in the decreasing arrangement, the even-indexed terms come first, then the odd-indexed terms in reverse. A term satisfies $a_k = b_k$ exactly when its position equals its index, which for the descending odd tail requires $2(k - 1006) = 2011 - k.$

Solving gives $3k = 4023,$ so $k = 1341,$ the unique fixed index, and the sum is $1341.$

Thus, the correct answer is C.

25.

Let $f(x) = |2\{x\} - 1|$ where $\{x\}$ denotes the fractional part of $x.$ The number $n$ is the smallest positive integer such that the equation $nf(xf(x)) = x$ has at least $2012$ real solutions $x.$ What is $n?$

Note: the fractional part of $x$ is a real number $y = \{x\},$ such that $0 \le y \lt 1$ and $x - y$ is an integer.

$30$

$31$

$32$

$62$

$64$

Answer: C

Concepts:floor and ceiling functions counting intersections

Difficulty rating: 2720

Small Hint:

$f$ is a periodic triangular wave with $0 \le f(x) \le 1,$ so all solutions lie in $[0, n]$

Big Hint:

On each unit interval the graph of $y = f(xf(x))$ oscillates a growing number of times; the total number of intersections with $y = \dfrac{x}{n}$ is $2n^2$

Solution:

Since $0 \le f(x) \le 1,$ every solution lies in $[0,n].$ The function $f$ is a triangular wave of period $1.$ Put $g(x)=xf(x).$ For each integer $a\ge1,$ the function $g$ decreases from $a$ to $0$ on $[a,a+\tfrac12),$ while it increases from $0$ to $a+1$ on $[a+\tfrac12,a+1).$ The first interval $[0,\tfrac12)$ is exceptional, but it contributes no intersection with $y=\frac{x}{n}.$

Counting the oscillations, on the intervals $[a, a + \tfrac12)$ and $[a + \tfrac12, a+1)$ the curve $y = f(g(x))$ meets the line $y = \tfrac{x}{n}$ a total of $2a$ and $2(a+1)$ times. Summing over $a = 0, \ldots, n-1$ gives $\sum_{a=0}^{n-1}\bigl(2a + 2(a+1)\bigr) = 2n^2$ real solutions.

The smallest $n$ with $2n^2 \ge 2012$ is $n = 32,$ since $2 \cdot 31^2 = 1922$ and $2 \cdot 32^2 = 2048.$

Thus, the correct answer is C.