2012 AMC 10A — Problems & Online Timed Exam

Timed

1:15:00

1.

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: 560

Small Hint:

Convert each frosting rate to cupcakes per minute

Big Hint:

Add the two rates, then multiply by $5$ minutes

Solution:

Cagney can make $60 \div 20 = 3$ cupcakes per minute, and Lacey can make $60 \div 30 = 2.$

In $5$ minutes, they can make $5(2 + 3) = 5 \cdot 5 = 25.$

Thus, D is the correct answer.

2.

A square with side length $8$ is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?

$2 \text{ by } 4$

$2 \text{ by } 6$

$2 \text{ by } 8$

$4 \text{ by } 4$

$4 \text{ by } 8$

Answer: E

Concepts:square (geometry)rectangle

Difficulty rating: 450

Small Hint:

One side of each rectangle stays $8$

Big Hint:

The other side is half of the square side length

Solution:

Note that the one of the sides remains the same as the original square.

This means that one dimension is $8.$ The other dimension is the original side cut in half, which is $4.$

Thus, E is the correct answer.

3.

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: 560

Small Hint:

Add the distance from $-2$ to $-6$ and from $-6$ to $5$

Big Hint:

Distance on a number line uses absolute differences

Solution:

It crawls a distance of $|-2 - (-6)| = |4| = 4$ when it moves from $-2$ to $-6.$ It then travels a distance of $|-6 - 5| = |-11| = 11$ as it moves from $-6$ to $5.$

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

Thus, E is the correct answer.

4.

Let $\angle ABC = 24^\circ$ and $\angle ABD = 20^\circ .$ What is the smallest possible degree measure for $\angle CBD?$

$0$

$2$

$4$

$6$

$12$

Answer: C

Concepts:angle chasing optimization

Difficulty rating: 770

Small Hint:

Both angles share ray $BA$

Big Hint:

The smallest angle occurs when the two given angles overlap as much as possible

Solution:

Note that both of the angles share the ray $AB.$ To minimize the desired degree, we want $D$ to be between $A$ and $C.$

This would make $\begin{aligned} \angle CBD &= \angle ABC - \angle ABD \\ &= 4^{\circ}. \end{aligned}$

Thus, C is the correct answer.

5.

Last year $100$ adult cats, half of whom were female, were brought into the Smallville Animal Shelter. Half of the adult female cats were accompanied by a litter of kittens. The average number of kittens per litter was $4.$ What was the total number of cats and kittens received by the shelter last year?

$150$

$200$

$250$

$300$

$400$

Answer: B

Concepts:fraction mean

Difficulty rating: 870

Small Hint:

Half the adult cats are female, and half of those have litters

Big Hint:

Multiply the number of litters by the average litter size

Solution:

We have that there are $100 \div 2 = 50$ female cats. We then have that $50 \div 2 = 25$ cats that have kittens.

Since the average number of kittens per litter is $4,$ the total number of kittens is $4 \cdot 25 = 100.$

The total number of cats and kittens is then $100 + 100 = 200.$

Thus, B is the correct answer.

6.

The product of two positive numbers is $9.$ The reciprocal of one of these numbers is $4$ times the reciprocal of the other number. What is the sum of the two numbers?

$\dfrac{10}{3}$

$\dfrac{20}{3}$

$7$

$\dfrac{15}{2}$

$8$

Answer: D

Concepts:system of equations rational equation

Difficulty rating: 1070

Small Hint:

Let the two numbers be $x$ and $y$

Big Hint:

Use $xy=9$ and the reciprocal condition to relate $x$ and $y$

Solution:

Let the two numbers be $x$ and $y$ such that $xy = 9 \text{ and } \dfrac{1}{x} = \dfrac{4}{y}.$

We get that $y = \dfrac{9}{x}$ $\dfrac{1}{x} = \dfrac{4x}{9}.$ $x = \dfrac{3}{2},$ since $x$ is positive.

Then $y = 9 \div \dfrac{3}{2} = 6.$

The desired sum is then $6 + \dfrac{3}{2} = \dfrac{15}{2}.$

Thus, D is the correct answer.

7.

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:ratio and proportion fraction

Difficulty rating: 870

Small Hint:

Start with a $3:2$ ratio of blue to red marbles

Big Hint:

Doubling the red marbles changes the ratio to $3:4$

Solution:

WLOG, let the number of marbles in the bag be $5.$ Since we only care about ratios, we can do this. Then there are $3$ blue marbles and $2$ red marbles.

Doubling the red marbles gives us $4$ of them. Then the fraction of red marbles is $\dfrac{4}{4 + 3} = \dfrac{4}{7}.$

Thus, C is the correct answer.

8.

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: 1020

Small Hint:

Add the three pair sums

Big Hint:

The total of the three pair sums is twice the sum of the three numbers

Solution:

Let the three numbers be $a, b, c$ where $a \lt b \lt c.$ None of them are equal, since all three sums are different.

Then $a + b = 12, a + c = 17,$ and $b + c = 19.$

Adding all three equations together gives us $2(a + b + c) = 48.$

Then $a + b + c = 24,$ from which we can subtract $a + c = 17,$ to get $b = 7.$

Thus, D is the correct answer.

9.

A pair of six-sided fair dice are labeled so that one die has only even numbers (two each of $2,$ $4,$ and $6$ ), and the other die has only odd numbers (two each of $1,$ $3,$ and $5$ ). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is $7?$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: D

Concepts:dice (probability)casework

Difficulty rating: 960

Small Hint:

List the even-odd pairs that sum to $7$

Big Hint:

Each listed value appears on two faces of its die

Solution:

The pairs of numbers that sum to $7$ are $(2, 5), (4, 3), \text{ and } (6, 1).$

There is a $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}$ chance that we get any of these pairs.

There are $3$ pairs, which means that the total probability that the rolls sum to $7$ is $3 \cdot \dfrac{1}{9} = \dfrac{1}{3}.$

Thus, D is the correct answer.

10.

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 Diophantine Equation optimization

Difficulty rating: 1420

Small Hint:

Let the arithmetic sequence be $a,a+d,\ldots,a+11d$

Big Hint:

Use $12a+66d=360$ and maximize $d$ while keeping $a$ positive

Solution:

Let $a$ be the smallest possible sector angle and $d$ be the difference in the arithmetic sequence.

Then we have that $\dfrac{12}{2}(a + a + 11d) = 12a + 66d$ is the sum of the arithmetic sequence.

We have that this sums to $360,$ so $12a + 66d = 360$ $2a + 11d = 60.$

We want to minimize $a,$ so we maximize $d.$ If $d = 5,$ then $a$ is not an integer, so $d = 4$ and $a = 8.$

Thus, C is the correct answer.

11.

Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3,$ respectively. A line externally tangent to both circles intersects ray $AB$ at point $C.$ What is $BC?$

$4$

$4.8$

$10.2$

$12$

$14.4$

Answer: D

Concepts:tangent circles tangent line similarity

Difficulty rating: 1420

Small Hint:

Draw radii to the two tangent points

Big Hint:

The two right triangles with vertex $C$ are similar

Solution:

Let $x$ be $BC.$ Note that $\triangle CEB$ and $\triangle CDA$ are similar due to angle-angle (tangent lines are perpendicular to radii).

Then $\dfrac{x}{3} = \dfrac{8 + x}{5}.$ Cross-multiplying gives us $5x = 24 + 3x$ $x = 12.$

Thus, D is the correct answer.

12.

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 $100$ (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:date and time modular arithmetic

Difficulty rating: 1540

Small Hint:

Count day shifts modulo $7$

Big Hint:

In $200$ years, include leap days but exclude $1900$

Solution:

On a typical year with $365$ days, moving one year into the future moves the day of the week forward one since $365 = 52 \cdot 7 + 1.$

On a leap year however, the day of the week gets moved forward twice since there is an extra day.

Fifty of the years between $2012$ and $1812$ are multiples of $4,$ but we have to discard $1900,$ which leaves us with $49$ leap years.

Therefore, moving back $200$ years means we have to go back $200 + 49 = 249 = 35 \cdot 7 + 4$ days, which means we move back $4$ days in the week.

This takes us back to Friday, which is the day that corresponds to February $7,$ $1812.$

Thus, A is the correct answer.

13.

An iterative average of the numbers $1,$ $2,$ $3,$ $4,$ and $5$ is computed 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 algebraic manipulation optimization

Difficulty rating: 1600

Small Hint:

Write the final average as a weighted sum

Big Hint:

Later entries receive larger weights

Solution:

Let the order of the numbers be $a, b, c, d, e.$

Then the iterative average is $\dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}$ $= \dfrac{a + b + 2c + 4d + 8e}{16}.$

To minimize this, we make the order $5, 4, 3, 2, 1,$ which gives us a sum of $\dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.$

To maximize it, we have to reverse this order to get an average of $\dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}.$

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

Thus, C is the correct answer.

14.

Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?

$480$

$481$

$482$

$483$

$484$

Answer: B

Concepts:basic counting parity pattern recognition

Difficulty rating: 1020

Small Hint:

Rows alternate between $16$ and $15$ black squares

Big Hint:

There are $16$ rows of one type and $15$ of the other

Solution:

Note that there are $15$ rows with $15$ black tiles and $16$ rows with $16$ black tiles.

This can be seen by observing that the first row has $16$ black tiles, and all the other rows alternate with $15$ and $16$ tiles.

Then, due to the alternating pattern, there will be a total of $16$ rows with $16$ tiles, and the other rows have $15$ tiles.

The total number of black squares is then $15^2 + 16^2 = 481.$

Thus, B is the correct answer.

15.

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC?$

$\dfrac16$

$\dfrac15$

$\dfrac29$

$\dfrac13$

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

Answer: B

Concepts:coordinate geometry triangle area

Difficulty rating: 1480

Small Hint:

Place $A$ and $B$ on coordinate axes

Big Hint:

Find the intersection of the two drawn lines

Solution:

We can use coordinate geometry to figure out where the intersection of the two lines occurs.

Let $A$ be the origin and $B = (1, 0).$ Then the slope of the line through $A$ is $-\dfrac{1}{2},$ which makes the equation of the line $y = -\dfrac{1}{2}x.$ The slope of the line through $B$ is $2.$ The $y$ -intercept is $-2.$ This makes the equation of this line $y = 2x - 2.$

Equating the equations, we get $2x - 2 = -\dfrac{1}{2}x$ $x = \dfrac{4}{5}.$

This makes the $y$ -coordinate of $C$ $-\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}.$

The area of triangle $ABC$ is then $\dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}.$

Thus, B is the correct answer.

16.

Three runners start running simultaneously from the same point on a $500$ -meter circular track. They each run clockwise around the course maintaining constant speeds of $4.4,$ $4.8,$ and $5.0$ meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?

$1{,}000$

$1{,}250$

$2{,}500$

$5{,}000$

$10{,}000$

Answer: C

Concepts:relative speed least common multiple distance rate and time

Difficulty rating: 1600

Small Hint:

Use relative speeds on the circular track

Big Hint:

Find when the middle and fast runners lap the slow runner by whole laps

Solution:

Let us find the amount of time that it takes for the runner running at $4.8$ meters per second to lap the slowest person.

We must have that $4.8x - 4.4x = 500$ $x = 1250,$ where $x$ is the amount of time it takes for the faster runner to lap the other.

Note that $4.4 \cdot 1250 = 5500,$ which means that these two runners always intersect at the starting line.

We now have to find the least time, $t,$ such that $t$ is a multiple of $1250$ and the fastest runner ends up at the starting line.

Every $1250$ seconds, the fastest runner runs $1250 \cdot 5 = 6250$ meters. Then in $2500$ seconds, the fastest runner runs $12500$ meters, which is a whole number of laps.

Thus, C is the correct answer.

17.

Let $a$ and $b$ be relatively prime integers with $a > b > 0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.$ What is $a-b?$

$1$

$2$

$3$

$4$

$5$

Answer: C

Concepts:factoring rational equation quadratic

Difficulty rating: 1670

Small Hint:

Factor $a^3-b^3$

Big Hint:

The equation becomes a quadratic in $\frac{a}{b}$

Solution:

Recall that we can factor $a^3 - b^3 = (a - b)(a^2 + ab + b^2).$ Canceling out this factor gives us that $\dfrac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \dfrac{73}{3}.$

Cross-multiplying and rearranging gives us $70a^2 - 149ab + 70b^2 = 0.$ Since $b \neq 0,$ we can divide through by $b^2$ to get $70\left(\dfrac{a}{b}\right)^2 - 149\dfrac{a}{b} + 70 = 0.$

Applying the quadratic formula and noting that $a \gt b$ gives us that $\dfrac{a}{b} = \dfrac{10}{7}.$

Since $a$ and $b$ are relatively prime, we have that $a = 10$ and $b = 7.$ Their difference is $10 - 7 = 3.$

Thus, C is the correct answer.

18.

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:sector regular polygon area decomposition

Difficulty rating: 1930

Small Hint:

Each arc is a $120^\circ$ arc of a unit circle

Big Hint:

Rearrange the sector pieces around the regular hexagon

Solution:

Each arc has radius $1$ and length $\frac{2\pi}{3},$ so the corresponding circular sectors all have the same area. As the diagram shows, the sector pieces may be rearranged: everything added to the regular hexagon combines to exactly one unit circle.

The regular hexagon is made of six equilateral triangles of side $2,$ so its area is $6\left(\dfrac{\sqrt3}{4}\cdot2^2\right)=6\sqrt3.$ Adding the unit circle gives a total enclosed area of $\pi+6\sqrt3.$

Thus, E is the correct answer.

19.

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 $7:12$ P.M.

How long, in minutes, was each day’s lunch break?

$30$

$36$

$42$

$48$

$60$

Answer: D

Concepts:rate system of equations

Difficulty rating: 2060

Small Hint:

Let the lunch break be $m$ minutes

Big Hint:

Use Monday for $p+h$ , Tuesday for $h$ , and Wednesday for $p$

Solution:

Let the lunch break be $m$ minutes, Paula’s rate be $p$ percent per minute, and the helpers’ combined rate be $h$ percent per minute.

Monday gives $(p+h)(480-m)=50$ . Tuesday gives $h(372-m)=24$ . Since the remaining work on Wednesday was $26$ percent, Wednesday gives $p(672-m)=26$ .

Adding the Tuesday and Wednesday equations and subtracting the Monday equation gives $108h-192p=0$ , so $h=\dfrac{16}{9}p$ .

Substituting into Monday gives $\dfrac{25}{9}p(480-m)=50$ , while Wednesday gives $p(672-m)=26$ . Solving these two equations gives $m=48$ .

Thus, D is the correct answer.

20.

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 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 independent events symmetry casework

Difficulty rating: 1980

Small Hint:

Corner squares form a $4$ -cycle under rotation

Big Hint:

For each $4$ -cycle, count colorings with no white square receiving a white predecessor

Solution:

The center square must initially be black, contributing probability $\dfrac12$ . The four corner squares form one cycle under the rotation, and the four edge-middle squares form another identical cycle.

For one $4$ -cycle, the final four positions are all black unless a white square is rotated into a position that was also white. The successful initial colorings are $BBBB$ , the four rotations of $BBBW$ , and the two rotations of $BWBW$ , for $7$ colorings out of $16$ .

The same count applies to the edge-middle cycle, independently. Therefore the probability is $\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}$ .

Thus, A is the correct answer.

21.

Let points $A=(0,0,0),$ $B=(1,0,0),$ $C=(0,2,0),$ and $D=(0,0,3).$ Points $E,$ $F,$ $G,$ and $H$ are midpoints of line segments $\overline{BD},$ $\text{ } \overline{AB},$ $\text{ } \overline {AC},$ and $\overline{DC}$ respectively. What is the area of $EFGH?$

$\sqrt{2}$

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

$\dfrac{3\sqrt{5}}{4}$

$\sqrt{3}$

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

Answer: C

Concepts:3D geometry midpoint distance formula

Difficulty rating: 1730

Small Hint:

Use midsegments in triangles $ABD$ and $ABC$

Big Hint:

The quadrilateral is a rectangle with side lengths from $AD$ and $BC$

Solution:

Note that $EF = \dfrac{1}{2}AD$ since it is a midsegment of $\triangle ABD.$ Similarly, $HG = \dfrac{1}{2}AD$ and $FG = \dfrac{1}{2}BC.$

We also have that $\overline{EF}$ and $\overline{HG}$ are perpendicular to the $xy$ -plane, which means that they are perpendicular to $\overline{FG}$ and $\overline{EH}.$

This tells us that $EFGH$ is rectangle since $EF = HG.$ We have $EF = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}.$

We also have that $FG = \dfrac{1}{2} \sqrt{1^2 + 2^2} = \dfrac{\sqrt{5}}{2}.$

The area of $EFGH$ is then $EF \cdot FG = \dfrac{3}{2} \cdot \dfrac{\sqrt{5}}{2} = \dfrac{3\sqrt{5}}{4}.$

Thus, C is the correct answer.

22.

The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n?$

$255$

$256$

$257$

$258$

$259$

Answer: A

Concepts:summation difference of squares Diophantine Equation

Difficulty rating: 2260

Small Hint:

Use that the first $m$ odd numbers sum to $m^2$

Big Hint:

Turn the condition into $(2m+p)(2m-p)=847$

Solution:

The first $m$ positive odd integers sum to $m^2$ , and the first $n$ positive even integers sum to $n(n+1)$ . Thus $m^2=n(n+1)+212$ .

As a quadratic in $n$ , this has discriminant $1-4(212-m^2)=4m^2-847$ , which must be an odd square. Let $p^2=4m^2-847$ . Then $(2m+p)(2m-p)=847$ .

The positive factor pairs of $847$ are $847\cdot1$ , $121\cdot7$ , and $77\cdot11$ . They give $p=423,57,33$ , respectively.

Because $n=\dfrac{-1+p}{2}$ , the possible values of $n$ are $211,28,16$ . Their sum is $255$ .

Thus, A is the correct answer.

23.

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 casework bijection

Difficulty rating: 2200

Small Hint:

Case on the common number of friends

Big Hint:

Use complements to pair the $1$ -friend and $4$ -friend cases, and the $2$ -friend and $3$ -friend cases

Solution:

We case on the value of friends that each person has. This value ranges from $1$ to $4$ , since the graph is neither empty nor complete.

Note that the cases for $1$ and $2$ friends correspond with the case for $4$ and $3$ friends, since choosing who are friends determines who are not friends.

Case $1:$ everyone has $1$ friend

This means that the $6$ people must split up into $3$ pairs where the people in each pair are friends.

There are $5$ choices for the friend for the first person. This leaves $4$ people remaining.

There are then $3$ choices for the friend of the next unpaired person. The remaining $2$ people are then forced to be friends.

Therefore, there are $3 \cdot 5 = 15$ possibilities for this case.

Case $2:$ everyone has $2$ friends

There are two possibilities for this case. There could be two triples where everyone in a triple is friends with each other.

For this possibility, there are $\binom{6}{3} = 20$ ways to choose the people in the first triple. We have to divide by $2$ since we can swap the pairs. This gives us $20 \div 2 = 10$ configurations.

The second possibility is that the friends form one $6$ -cycle.

Every ordering of the six people around a cycle gives such a graph. Each graph is counted $6$ times by the choice of starting person and $2$ times by the direction of traversal, so there are $\frac{6!}{6\cdot2}=60$ distinct $6$ -cycles. Together with the $10$ pairs of triangles, this case has $10+60=70$ configurations.

The total number of arrangements is then $2(15 + 70) = 170.$

Thus, B is the correct answer.

24.

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 $a^2+3b^2+3c^2-3ab-2ac-2bc$ $=-1997.$ What is $a?$

$249$

$250$

$251$

$252$

$253$

Answer: E

Concepts:algebraic manipulation Diophantine Equation casework

Difficulty rating: 2350

Small Hint:

Add the two equations first

Big Hint:

The three squared differences must be $9,4,1$ in some order

Solution:

Adding together the equations gives us $2(a^2 + b^2 + c^2) - 2(ab + ac + bc)$ $= 14.$

We can group terms and factor this to get $(a - b)^2 + (a - c)^2 + (b - c)^2$ $= 14.$

Note that every term on the left hand side is a nonnegative square integer. The only triple of squares that add to $14$ is $9, 4,$ and $1.$

We have that $a - c$ is the biggest difference among the three pairs. Therefore, $a - c = 3.$

We cannot discern which of the other terms we can match with the other squares. Let us try $a - b = 1$ and $b - c = 2.$

Plugging in these values into the first equation gives us $a^2 - (a - 1)^2 - (a - 3)^2$ $+ a(a - 1) = 2011.$ Simplifying yields $7a = 2021.$ Since $2021$ is not divisible by $7,$ we have that $a - b = 2$ and $b - c = 1.$

In the other case, $b=a-2$ and $c=a-3.$ The first equation becomes $\begin{aligned} a^2-(a-2)^2-(a-3)^2\\ {}+a(a-2)&=2011, \end{aligned}$ or $8a-13=2011.$ Hence $a=253.$

Thus, E is the correct answer.

25.

Real numbers $x,$ $y,$ and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n.$ The probability that no two of $x,$ $y,$ and $z$ are within $1$ unit of each other is greater than $\dfrac{1}{2}.$ What is the smallest possible value of $n?$

$7$

$8$

$9$

$10$

$11$

Answer: D

Concepts:geometric probability volume symmetry

Difficulty rating: 2460

Small Hint:

Order the three random numbers first

Big Hint:

Subtract $1$ from the middle number and $2$ from the largest number

Solution:

This problem lends itself to geometric probability since we can view the interval as a range on an axis.

WLOG, let $n \geq x \geq y \geq z \geq 0.$

Then we have that the points $(x, y, z)$ which satisfy this restriction form a tetrahedron.

The height of this tetrahedron is $n,$ and the base has an area of $\dfrac{n^2}{2}.$ This makes the volume $\dfrac{1}{3} \cdot \dfrac{n^2}{2} \cdot n = \dfrac{n^3}{6}.$

Now we have to apply the restrictions from the problem statement. We need to find the region where $|x - y|, |x - z|, |y - z| \geq 1.$

From our ordering condition that we imposed, these inequalities reduce to $x - y \geq 1 \text{ and } y - z \geq 1.$

These two restrictions form another tetrahedron as shown below.

Note that in the new tetrahedron, all the dimensions have been reduced by $2.$ This makes the height $n - 2$ and the base $\dfrac{(n - 2)^2}{2}.$

The volume is then $\dfrac{1}{3} \cdot \dfrac{(n - 2)^2}{2} \cdot (n - 2)$ $= \dfrac{(n - 2)^3}{6}.$

The desired probability is then $\dfrac{(n - 2)^3}{6} \div \dfrac{n^3}{6} = \dfrac{(n - 2)^3}{n^3}.$

Plugging in all the answer choices, we get that the smallest value such that this fraction is greater than $\dfrac{1}{2}$ is $10.$

Thus, D is the correct answer.