2022 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

Define $x \diamond y$ to be $|x - y|$ for all real numbers $x$ and $y.$ What is the value of

$(1 \diamond (2 \diamond 3)) - ((1 \diamond 2) \diamond 3)?$

$-2$

$-1$

$0$

$1$

$2$

Answer: A

Concepts:custom operation absolute value order of operations

Difficulty rating: 890

Small Hint:

Evaluate the inner $\diamond$ first, using $a \diamond b = |a-b|$

Big Hint:

Compute $1 \diamond (2 \diamond 3)$ and $(1 \diamond 2) \diamond 3$ separately, then subtract

Solution:

Since $2 \diamond 3 = |2-3| = 1,$ we get $1 \diamond (2 \diamond 3) = 1 \diamond 1 = 0.$

Since $1 \diamond 2 = |1-2| = 1,$ we get $(1 \diamond 2) \diamond 3 = 1 \diamond 3 = |1-3| = 2.$

The value is $0 - 2 = -2.$

Thus, the correct answer is A.

2.

In rhombus $ABCD,$ point $P$ lies on segment $\overline{AD}$ so that $\overline{BP} \perp \overline{AD},$ $AP = 3,$ and $PD = 2.$ What is the area of $ABCD?$ (Note: the figure is not drawn to scale.)

$3\sqrt5$

$10$

$6\sqrt5$

$20$

$25$

Answer: D

Concepts:rhombus Pythagorean Theorem area

Difficulty rating: 1020

Small Hint:

The side length of the rhombus is $AD = AP + PD$

Big Hint:

In right triangle $APB,$ find $BP,$ then use area $= AD \cdot BP$

Solution:

The side length is $AD = AP + PD = 5,$ so $AB = 5.$ In right triangle $APB,$ $\begin{aligned} BP &= \sqrt{AB^2 - AP^2} \\ &= \sqrt{25 - 9} = 4. \end{aligned}$

Taking $AD$ as the base and $BP$ as the height, the area is $AD \cdot BP = 5 \cdot 4 = 20.$

Thus, the correct answer is D.

3.

How many of the first ten numbers of the sequence $121,$ $11211,$ $1112111,$ $\ldots$ are prime numbers?

$0$

$1$

$2$

$3$

$4$

Answer: A

Concepts:factoring prime

Difficulty rating: 1350

Small Hint:

Look for a factorization of each term into two integers both greater than $1$

Big Hint:

Try writing a term as a repunit times a number of the form $10^n + 1,$ for example $11211 = 111 \cdot 101$

Solution:

The $n$ th term consists of $n$ ones, then a $2,$ then $n$ ones. It factors as a repunit times a number of the form $10^n + 1:$ $\begin{aligned} &121 = 11 \cdot 11, \\ &\quad 11211 = 111 \cdot 101, \\ &\quad 1112111 = 1111 \cdot 1001, \end{aligned}$ and in general the $n$ th term equals $\underbrace{1\cdots1}_{n+1} \cdot (10^n + 1).$

For every $n \ge 1$ both factors exceed $1,$ so every term is composite. None of the ten numbers is prime.

Thus, the correct answer is A.

4.

For how many values of the constant $k$ will the polynomial $x^2 + kx + 36$ have two distinct integer roots?

$6$

$8$

$9$

$14$

$16$

Answer: B

Concepts:Vieta’s Formulas factor systematic listing

Difficulty rating: 1200

Small Hint:

If the roots are $p$ and $q,$ then $pq = 36$ and $k = -(p+q)$

Big Hint:

Count the ordered-independent factor pairs of $36$ with two distinct integers, including negatives

Solution:

If the roots are integers $p$ and $q,$ then $pq = 36$ and $k = -(p+q).$ Distinct roots must have the same sign, so we list factor pairs of $36$ with $p \ne q.$

The positive pairs are $(1,36), (2,18), (3,12), (4,9),$ and the negative pairs are $(-1,-36),$ $(-2,-18),$ $(-3,-12),$ $(-4,-9).$ The pair $(6,6)$ is excluded since the roots must be distinct.

Each of these $8$ pairs gives a different value of $k.$

Thus, the correct answer is B.

5.

The point $(-1, -2)$ is rotated $270^\circ$ counterclockwise about the point $(3, 1).$ What are the coordinates of its new position?

$(-3, -4)$

$(0, 5)$

$(2, -1)$

$(4, 3)$

$(6, -3)$

Answer: B

Concepts:transformation coordinate geometry

Difficulty rating: 1200

Small Hint:

Translate so the center of rotation is the origin, rotate, then translate back

Big Hint:

A $270^\circ$ counterclockwise rotation sends $(x, y)$ to $(y, -x)$

Solution:

Relative to the center $(3, 1),$ the point is at $(-1 - 3,\, -2 - 1) = (-4, -3).$

A $270^\circ$ counterclockwise rotation sends $(x, y)$ to $(y, -x),$ so $(-4, -3)$ becomes $(-3, 4).$

Translating back gives $(3 - 3,\, 1 + 4) = (0, 5).$

Thus, the correct answer is B.

6.

Consider the following $100$ sets of $10$ elements each:

$\{1,2,3,\ldots,10\},$ $\{11,12,13,\ldots,20\},$ $\{21,22,23,\ldots,30\},$ $\vdots$ $\{991,992,993,\ldots,1000\}.$

How many of these sets contain exactly two multiples of $7?$

$40$

$42$

$43$

$49$

$50$

Answer: B

Concepts:multiple counting integers in a range

Difficulty rating: 1350

Small Hint:

A block of $10$ consecutive integers contains either one or two multiples of $7$

Big Hint:

Count the total multiples of $7$ in $\{1,\ldots,1000\},$ then let $x$ blocks have two and $100-x$ have one

Solution:

Among $1$ to $1000$ there are $\left\lfloor \tfrac{1000}{7} \right\rfloor = 142$ multiples of $7.$ Because $10 \gt 7,$ each block of $10$ consecutive integers contains one or two multiples of $7.$

If $x$ blocks contain two and the remaining $100 - x$ contain one, then $2x + (100 - x) = 142,$ so $x = 42.$

Thus, the correct answer is B.

7.

Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?

$5$

$7$

$9$

$11$

$13$

Answer: D

Concepts:mode median (data)mean optimization

Difficulty rating: 1380

Small Hint:

The mode exceeds the median, so it can only appear among the two largest numbers

Big Hint:

Let the median be $m;$ for a unique mode the two largest must both equal $m+2,$ and the sum is $5(m-2)$

Solution:

List the numbers in increasing order with median $m.$ The mode is $m + 2 \gt m,$ so it can only occur among the two largest entries; for it to be the unique mode, both of them must equal $m + 2.$

The mean is $m - 2,$ so the total is $5(m-2).$ With the two largest equal to $m + 2$ and the median $m,$ the two smallest sum to $5(m-2) - m - 2(m+2)$ $= 2m - 14.$

The two smallest are distinct positive integers, so $2m - 14 \ge 1 + 2 = 3,$ giving $m \ge 9.$ With $m = 9$ the list $1, 3, 9, 11, 11$ works, so the least mode is $m + 2 = 11.$

Thus, the correct answer is D.

8.

What is the graph of $y^4 + 1 = x^4 + 2y^2$ in the coordinate plane?

two intersecting parabolas

two nonintersecting parabolas

two intersecting circles

a circle and a hyperbola

a circle and two parabolas

Answer: D

Concepts:difference of squares circle hyperbola

Difficulty rating: 1440

Small Hint:

Move everything to one side to get $(y^2 - 1)^2 = (x^2)^2$

Big Hint:

Factor as a difference of squares, giving two separate equations

Solution:

Rearranging, $y^4 - 2y^2 + 1 = x^4,$ so $(y^2 - 1)^2 = (x^2)^2.$ This factors as $(y^2 - 1 - x^2)(y^2 - 1 + x^2) = 0.$

Thus either $y^2 - x^2 = 1,$ which is a hyperbola, or $x^2 + y^2 = 1,$ which is a circle.

Thus, the correct answer is D.

9.

The sequence $a_0,$ $a_1,$ $a_2,$ $\cdots$ is a strictly increasing arithmetic sequence of positive integers such that $2^{a_7} = 2^{27} \cdot a_7.$ What is the minimum possible value of $a_2?$

$8$

$12$

$16$

$17$

$22$

Answer: B

Concepts:arithmetic sequence exponent optimization

Difficulty rating: 1530

Small Hint:

Divide both sides by $2^{27}$ to get $2^{a_7 - 27} = a_7$

Big Hint:

After finding $a_7,$ write $a_2 = a_7 - 5d$ and make the common difference $d$ as large as possible

Solution:

Dividing by $2^{27},$ we need $2^{a_7 - 27} = a_7.$ Thus $a_7=2^j$ and $2^j-j=27,$ where $j=a_7-27\ge0.$ The left side is strictly increasing for $j\ge1,$ and $j=5$ works, so uniquely $a_7=32.$

With common difference $d \ge 1,$ we have $a_7 = a_0 + 7d = 32$ and $a_2 = a_0 + 2d = 32 - 5d.$ To minimize $a_2$ we maximize $d;$ since $a_0 = 32 - 7d \ge 1,$ the largest choice is $d = 4$ (giving $a_0 = 4$ ).

Then $a_2 = 32 - 20 = 12.$

Thus, the correct answer is B.

10.

Regular hexagon $ABCDEF$ has side length $2.$ Let $G$ be the midpoint of $\overline{AB},$ and let $H$ be the midpoint of $\overline{DE}.$ What is the perimeter of $GCHF?$

$4\sqrt3$

$8$

$4\sqrt5$

$4\sqrt7$

$12$

Answer: D

Concepts:regular polygon distance formula coordinate geometry

Difficulty rating: 1500

Small Hint:

Place the hexagon on coordinates with its center at the origin

Big Hint:

By symmetry all four sides of $GCHF$ are equal; compute one distance with the distance formula

Solution:

Place the hexagon with center at the origin: $A = (-1, \sqrt3),$ $B = (1, \sqrt3),$ $C = (2, 0),$ $D = (1, -\sqrt3),$ $E = (-1, -\sqrt3),$ $F = (-2, 0).$

Then $G = (0, \sqrt3)$ and $H = (0, -\sqrt3).$ By symmetry all four sides of $GCHF$ are equal, and $GC = \sqrt{2^2 + (\sqrt3)^2} = \sqrt7.$

The perimeter is $4\sqrt7.$

Thus, the correct answer is D.

11.

Let $\begin{aligned} f(n) &= \left(\dfrac{-1 + i\sqrt3}{2}\right)^n \\ &\quad {}+ \left(\dfrac{-1 - i\sqrt3}{2}\right)^n, \end{aligned}$ where $i = \sqrt{-1}.$ What is $f(2022)?$

$-2$

$-1$

$0$

$\sqrt3$

$2$

Answer: E

Concepts:roots of unity complex number

Difficulty rating: 1570

Small Hint:

The two bases are the primitive cube roots of unity, $\omega$ and $\omega^2$

Big Hint:

Then $f(n) = \omega^n + \omega^{-n};$ reduce the exponent $2022$ modulo $3$

Solution:

The two bases are the primitive cube roots of unity, $\omega = e^{\frac{2\pi i}{3}}$ and its conjugate $\omega^2 = e^{-\frac{2\pi i}{3}}.$ So $f(n) = \omega^n + \omega^{-n} = 2\cos\dfrac{2\pi n}{3}.$

Since $2022$ is a multiple of $3,$ $\omega^{2022} = 1,$ so $f(2022) = 1 + 1 = 2.$

Thus, the correct answer is E.

12.

Kayla rolls four fair $6$ -sided dice. What is the probability that at least one of the numbers Kayla rolls is greater than $4$ and at least two of the numbers she rolls are greater than $2?$

$\dfrac{2}{3}$

$\dfrac{19}{27}$

$\dfrac{59}{81}$

$\dfrac{61}{81}$

$\dfrac{7}{9}$

Answer: D

Concepts:dice (probability)complementary counting inclusion-exclusion

Difficulty rating: 1630

Small Hint:

Sort each die into low $\{1,2\},$ mid $\{3,4\},$ or high $\{5,6\},$ each with probability $\tfrac13$

Big Hint:

You need at least one high die and at least two dice that are not low; count the good category patterns out of $3^4$

Solution:

Sort each die into low $\{1,2\},$ mid $\{3,4\},$ or high $\{5,6\};$ each has probability $\tfrac13,$ so the $3^4 = 81$ category patterns are equally likely.

We need at least one high die (a number greater than $4$ ) and at least two dice that are greater than $2$ (mid or high). The two bad events are having no high die and having at most one non-low die.

There are $2^4 = 16$ patterns of the first kind, $9$ of the second kind, and $5$ in their intersection. By inclusion-exclusion the count of good patterns is $81 - 16 - 9 + 5 = 61.$

The probability is $\dfrac{61}{81}.$

Thus, the correct answer is D.

13.

The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5.$ Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

$15\dfrac18$

$15\dfrac38$

$15\dfrac12$

$15\dfrac58$

$15\dfrac78$

Answer: D

Concepts:coordinate geometry area decomposition Pythagorean Triple

Difficulty rating: 1660

Small Hint:

Set up coordinates; the tilt of the square makes $3$ - $4$ - $5$ right triangles at the corners

Big Hint:

The whole square lies inside the rectangle except one triangle sticking out past the top edge; subtract that triangle from $25$

Solution:

Place the rectangle as $[0,8] \times [0,4].$ The tilted square, using the $3$ - $4$ - $5$ right triangles, has vertices $(4, 0),$ $(0, 3),$ $(3, 7),$ and $(7, 4).$

The entire square lies inside the rectangle except for the triangle poking above the top edge $y = 4.$ That triangle has vertices $(0.75, 4),$ $(3, 7),$ and $(7, 4),$ with area $\dfrac12 \cdot (7 - 0.75) \cdot (7 - 4) = \dfrac{75}{8}.$

The region inside both is $25 - \dfrac{75}{8} = \dfrac{125}{8} = 15\dfrac58.$

Thus, the correct answer is D.

14.

The graph of $y = x^2 + 2x - 15$ intersects the $x$ -axis at points $A$ and $C$ and the $y$ -axis at point $B.$ What is $\tan(\angle ABC)?$

$\dfrac17$

$\dfrac14$

$\dfrac37$

$\dfrac12$

$\dfrac47$

Answer: E

Concepts:coordinate geometry vector trigonometry

Difficulty rating: 1570

Small Hint:

Factor to find the intercepts: $A$ and $C$ on the $x$ -axis, $B$ on the $y$ -axis

Big Hint:

With vectors $\vec{BA}$ and $\vec{BC},$ use $\tan\theta = \dfrac{|\vec{BA} \times \vec{BC}|}{\vec{BA} \cdot \vec{BC}}$

Solution:

Factoring, $x^2 + 2x - 15 = (x+5)(x-3),$ so $A = (-5, 0)$ and $C = (3, 0),$ and the $y$ -intercept is $B = (0, -15).$

Then $\vec{BA} = (-5, 15)$ and $\vec{BC} = (3, 15).$ Using the cross and dot products, $\begin{gathered} \tan(\angle ABC) = \scriptsize \dfrac{|(-5)(15) - (15)(3)|}{(-5)(3) + (15)(15)} \\ = \dfrac{120}{210} \\ = \dfrac47. \end{gathered}$

Thus, the correct answer is E.

15.

One of the following numbers is not divisible by any prime number less than $10.$ Which is it?

$2^{606} - 1$

$2^{606} + 1$

$2^{607} - 1$

$2^{607} + 1$

$2^{607} + 3^{607}$

Answer: C

Concepts:modular arithmetic multiplicative order divisibility

Difficulty rating: 1730

Small Hint:

All five numbers are odd, so test divisibility only by $3, 5,$ and $7$

Big Hint:

Use the cycles of $2^n$ modulo $3, 5, 7:$ for example $2^n \equiv 1 \pmod 3$ when $n$ is even

Solution:

Every option is odd, so only the primes $3, 5, 7$ need checking.

Option A: $2^{606} \equiv 1 \pmod 3,$ so $2^{606} - 1$ is divisible by $3.$ Option B: $2^{606} \equiv 4 \pmod 5,$ so $2^{606} + 1$ is divisible by $5.$ Option D: $2^{607} \equiv 2 \pmod 3,$ so $2^{607} + 1$ is divisible by $3.$ Option E: modulo $5,$ $2^{607} + 3^{607} \equiv 3 + 2 = 5 \equiv 0.$

For $2^{607} - 1:$ it is $\equiv 1 \pmod 3,$ $\equiv 2 \pmod 5,$ and (since $2^3 \equiv 1 \pmod 7$ and $607 \equiv 1 \pmod 3$ ) $\equiv 1 \pmod 7.$ So it is not divisible by any prime below $10.$

Thus, the correct answer is C.

16.

Suppose $x$ and $y$ are positive real numbers such that

$x^y = 2^{64}$ and $(\log_2 x)^{\log_2 y} = 2^7.$

What is the greatest possible value of $\log_2 y?$

$3$

$4$

$3 + \sqrt2$

$4 + \sqrt3$

$7$

Answer: C

Concepts:logarithm substitution quadratic

Difficulty rating: 1800

Small Hint:

Let $a = \log_2 x$ and $b = \log_2 y;$ take $\log_2$ of both equations

Big Hint:

You get $a \cdot 2^b = 2^6$ and $a = 2^{\frac{7}{b}};$ combine to form a quadratic in $b$

Solution:

Let $a = \log_2 x$ and $b = \log_2 y.$ Taking $\log_2$ of $x^y = 2^{64}$ gives $y \log_2 x = 64,$ i.e. $a \cdot 2^b = 2^6.$

Taking $\log_2$ of the second equation gives $b \log_2 a = 7,$ so $a = 2^{\frac{7}{b}}.$ Substituting, $2^{\frac{7}{b}} \cdot 2^b = 2^6,$ so $b + \dfrac7b = 6,$ i.e. $b^2 - 6b + 7 = 0.$

Thus $b = 3 \pm \sqrt2,$ and the greatest value of $\log_2 y$ is $3 + \sqrt2.$

Thus, the correct answer is C.

17.

How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1,$ $2,$ $3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1,$ $2,$ $3,$ and $4,$ in some order? For example, the array $\begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \end{bmatrix}$ satisfies the condition.

$144$

$240$

$336$

$576$

$624$

Answer: D

Concepts:multiplication principle casework

Difficulty rating: 1840

Small Hint:

The row of sum $4$ is all $1$ s and the column of sum $4$ is all $1$ s

Big Hint:

Delete the sum- $4$ column; the remaining $4 \times 3$ array has row sums $0,1,2,3$ and column sums $1,2,3$

Solution:

The row with sum $4$ is all $1$ s and the column with sum $4$ is all $1$ s. There are $4!$ ways to assign the row sums $1, 2, 3, 4$ to the four rows, and $4$ choices for which column has sum $4.$

Delete that column. The remaining $4 \times 3$ array has row sums $0, 1, 2, 3$ and must have column sums $1, 2, 3.$ The all-zero and all-one rows are forced; the rows of reduced sum $1$ and $2$ can be placed in $6$ ways to produce column sums $1, 2, 3$ in some order.

The total is $24 \cdot 4 \cdot 6 = 576.$

Thus, the correct answer is D.

18.

Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:

Any filled square with two or three filled neighbors remains filled. Any empty square with exactly three filled neighbors becomes a filled square. All other squares remain empty or become empty.

A sample transformation is shown in the figure below.

Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)

$14$

$18$

$22$

$26$

$30$

Answer: C

Concepts:process simulation casework

Difficulty rating: 2000

Small Hint:

Only the inner $3 \times 3$ can be filled; the center becoming filled needs exactly $3$ filled neighbors (if it was empty) or $2$ - $3$ (if it was filled)

Big Hint:

The binding constraint is that no border square gains exactly three filled neighbors; check that every valid configuration has exactly three filled cells

Solution:

Only the inner $3 \times 3$ squares can start filled. For the center to be filled afterward, if it began empty it needs exactly $3$ filled neighbors, and if it began filled it needs $2$ or $3.$

Every other square must end empty. The key restriction is that no border square may acquire exactly three filled neighbors, which rules out filling all three squares along an outer edge of the $3 \times 3.$

If the center starts empty, exactly $3$ of its $8$ neighbors must be filled. Checking these triples up to square symmetry leaves four types. With ring coordinates centered at $(0,0),$ representatives are $\begin{gathered} \{(-1,-1),(-1,0),(1,1)\},\\ \{(-1,-1),(-1,1),(1,-1)\},\\ \{(-1,-1),(-1,1),(1,0)\},\\ \{(-1,-1),(0,1),(1,0)\}. \end{gathered}$ Their symmetry-orbit sizes are $8,4,4,4,$ giving $20$ configurations.

If the center starts filled, the same neighbor check leaves only the two configurations in which the other filled cells are opposite corner neighbors. Hence the total is $20+2=22.$

Thus, the correct answer is C.

19.

In $\triangle ABC$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle AGE$ is equilateral. Then $\cos(C)$ can be written as $\dfrac{m\sqrt p}{n},$ where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m + n + p?$

$44$

$48$

$52$

$56$

$60$

Answer: A

Concepts:median (geometry)centroid law of cosines

Difficulty rating: 2020

Small Hint:

The centroid splits $AD$ so $AG = \tfrac23 AD,$ and $GE = \tfrac13 BE;$ also $AE = \tfrac12 AC$

Big Hint:

Set all three equal and use the median-length formula to relate the sides, then apply the law of cosines

Solution:

Let $a = BC,$ $b = CA,$ $c = AB.$ Since $E$ is the midpoint of $AC,$ $AE = \tfrac{b}{2}.$ The centroid gives $AG = \tfrac23 m_a$ and $GE = \tfrac13 m_b,$ where $m_a, m_b$ are the medians from $A$ and $B.$

Equilateral $\triangle AGE$ means $AG = GE = AE.$ From $\tfrac23 m_a = \tfrac{b}{2}$ we get $m_a = \tfrac34 b,$ which with $m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}$ gives $2c^2 - a^2 = \tfrac{b^2}{4}.$ From $\tfrac13 m_b = \tfrac{b}{2}$ we get $m_b = \tfrac32 b,$ giving $a^2 + c^2 = 5b^2.$

Solving, $c^2 = \tfrac{7b^2}{4}$ and $a^2 = \tfrac{13b^2}{4}.$ Taking $b = 2$ gives $a^2 = 13,$ $c^2 = 7,$ so $\begin{aligned} \cos C &= \dfrac{a^2 + b^2 - c^2}{2ab} \\ &= \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} \\ &= \dfrac{5\sqrt{13}}{26}. \end{aligned}$

Then $m + n + p = 5 + 26 + 13 = 44.$

Thus, the correct answer is A.

20.

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1,$ the remainder is $x + 2,$ and when $P(x)$ is divided by the polynomial $x^2 + 1,$ the remainder is $2x + 1.$ There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

$10$

$13$

$19$

$20$

$23$

Answer: E

Concepts:polynomial Chinese Remainder Theorem

Difficulty rating: 2020

Small Hint:

The least-degree polynomial is a cubic; write $P(x) = x + 2$ $+ (x^2 + x + 1)(px + q)$

Big Hint:

Reduce that expression modulo $x^2 + 1$ using $x^2 \equiv -1,$ and match it to $2x + 1$

Solution:

No linear polynomial can have the two different remainders. If $P(x)=ax^2+bx+c$ were quadratic, comparing its remainders would give both $b-a=1,\ c-a=2$ and $b=2,\ c-a=1,$ a contradiction. Thus the least possible degree is $3.$ Write $P(x) = (x + 2)$ $+ (x^2 + x + 1)(px + q),$ which has remainder $x + 2$ upon division by $x^2 + x + 1.$

Reducing modulo $x^2 + 1$ (so $x^2 \equiv -1$ ) gives remainder $(q + 1)x + (2 - p).$ Setting this equal to $2x + 1$ gives $q = 1$ and $p = 1.$

Then $P(x) = x^3 + 2x^2 + 3x + 3,$ and the sum of the squares of the coefficients is $1 + 4 + 9 + 9 = 23.$

Thus, the correct answer is E.

21.

Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^2 + y^2 = 4,$ $x^2 + y^2 = 64,$ and $(x - 5)^2 + y^2 = 3.$ What is the sum of the areas of all circles in $S?$

$48\pi$

$68\pi$

$96\pi$

$102\pi$

$136\pi$

Answer: E

Concepts:tangent circles circle area casework

Difficulty rating: 2170

Small Hint:

A circle tangent to both concentric circles (radii $2$ and $8$ ) has radius $3$ or radius $5$

Big Hint:

The radius- $3$ circles have centers on the circle of radius $5;$ the radius- $5$ circles on the circle of radius $3.$ Count how many of each are also tangent to the third circle

Solution:

The first two circles are concentric with radii $2$ and $8.$ A circle tangent to both either has radius $3$ with center at distance $5$ from the origin, or radius $5$ with center at distance $3$ from the origin.

The third circle has center $(5, 0)$ and radius $\sqrt3.$ For each candidate radius $s\in\{3,5\},$ tangency requires the center’s distance from $(5,0)$ to be $s+\sqrt3$ or $s-\sqrt3.$ Each of these two distance circles intersects the appropriate center-locus in two symmetric points. Hence exactly four radius- $3$ circles and four radius- $5$ circles work.

The sum of the areas is $4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2$ $= 36\pi + 100\pi = 136\pi.$

Thus, the correct answer is E.

22.

Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n = 1,$ $2,$ $3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0, 1).$ During the $n$ th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$ th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1?$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$\dfrac34$

$\dfrac56$

Answer: C

Concepts:geometric probability independent events

Difficulty rating: 2110

Small Hint:

Since each $t_n \lt 1,$ she always takes at least two steps; she takes exactly two iff $t_1 + t_2 \gt 1$

Big Hint:

The times are independent of the increments; use $P(x_1 + x_2 \gt 1) = \tfrac12$ and $P(x_1 + x_2 + x_3 \gt 1) = \tfrac56$

Solution:

Because each $t_n \lt 1,$ Amelia always completes at least two steps. She stops after exactly two steps when $t_1 + t_2 \gt 1,$ which happens with probability $\tfrac12;$ otherwise she takes all three steps.

The increments are independent of the times. If she takes two steps, her position is $x_1 + x_2,$ and $P(x_1 + x_2 \gt 1) = \tfrac12.$ If she takes three, her position is $x_1 + x_2 + x_3,$ and $P(x_1 + x_2 + x_3 \gt 1)$ $= 1 - \tfrac16 = \tfrac56.$

The answer is $\tfrac12 \cdot \tfrac12 + \tfrac12 \cdot \tfrac56 = \tfrac14 + \tfrac{5}{12} = \tfrac23.$

Thus, the correct answer is C.

23.

Let $x_0,$ $x_1,$ $x_2,$ $\ldots$ be a sequence of numbers, where each $x_k$ is either $0$ or $1.$ For each positive integer $n,$ define $S_n = \sum_{k=0}^{n-1} x_k 2^k.$ Suppose $7 S_n \equiv 1 \pmod{2^n}$ for all $n \ge 1.$ What is the value of the sum $x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?$

$6$

$7$

$12$

$14$

$15$

Answer: A

Concepts:modular arithmetic number base pattern recognition

Difficulty rating: 2270

Small Hint:

$S_n$ is the number formed by the low $n$ bits; the condition says $S_n \equiv 7^{-1} \pmod{2^n}$

Big Hint:

So the bits $x_k$ are the binary (2-adic) digits of $\tfrac17;$ find their repeating pattern

Solution:

Since $S_n$ is the integer formed by the low $n$ bits, the condition $7S_n \equiv 1$ means $S_n \equiv 7^{-1} \pmod{2^n}$ for every $n.$ Thus the digits $x_k$ are the base- $2$ digits of $\tfrac17$ as a $2$ -adic number.

Long division in base $2$ gives digits $x_0 = x_1 = x_2 = 1,$ and thereafter the block repeats with period $3:$ for $k \ge 1,$ $x_k = 0$ exactly when $k$ is divisible by $3,$ and $x_k = 1$ otherwise.

Since $2019$ and $2022$ are both divisible by $3,$ while $2020 \equiv 1$ and $2021 \equiv 2 \pmod 3,$ we get $x_{2019} = 0,$ $x_{2020} = 1,$ $x_{2021} = 1,$ $x_{2022} = 0.$ The sum is $0 + 2 + 4 + 0 = 6.$

Thus, the correct answer is A.

24.

The figure below depicts a regular $7$ -gon inscribed in a unit circle.

What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?

$49$

$98$

$147$

$168$

$196$

Answer: C

Concepts:chord roots of unity trigonometric identity

Difficulty rating: 2370

Small Hint:

A chord between vertices $d$ apart has squared length $2 - 2\cos\dfrac{2\pi d}{7};$ there are $7$ of each type $d = 1, 2, 3$

Big Hint:

Use $\sum_{d=1}^{3} \cos\dfrac{2\pi d}{7} = -\dfrac12$ to evaluate $\sum \left(2 - 2\cos\dfrac{2\pi d}{7}\right)^2$

Solution:

A chord joining two vertices $d$ steps apart has squared length $2 - 2\cos\dfrac{2\pi d}{7},$ and there are $7$ chords for each of $d = 1, 2, 3.$ The required sum is $7 \sum_{d=1}^{3} \left(2 - 2\cos\tfrac{2\pi d}{7}\right)^2.$

Using $\displaystyle\sum_{d=1}^{3} \cos\tfrac{2\pi d}{7} = -\tfrac12$ and $\displaystyle\sum_{d=1}^{3} \cos^2\tfrac{2\pi d}{7} = \tfrac54,$ the inner sum expands to $4\left(3 + 1 + \tfrac54\right) = 21.$

Therefore the total is $7 \cdot 21 = 147.$

Thus, the correct answer is C.

25.

Four regular hexagons surround a square with a side length $1,$ each one sharing an edge with the square, as shown in the figure below. The area of the resulting $12$ -sided outer nonconvex polygon can be written as $m\sqrt n + p,$ where $m,$ $n,$ and $p$ are integers and $n$ is not divisible by the square of any prime. What is $m + n + p?$

$-12$

$-4$

$4$

$24$

$32$

Answer: B

Concepts:regular polygon coordinate geometry shoelace formula

Difficulty rating: 2520

Small Hint:

Each hexagon shares an edge with the square and reaches across to the far side; set up coordinates with the square centered at the origin

Big Hint:

The outer boundary has flat edges at distance $\sqrt3 - \tfrac12$ and reflex notches where adjacent hexagons’ edges cross, at $\left(\tfrac52 - \sqrt3,\ \tfrac52 - \sqrt3\right)$

Solution:

Center the square at the origin with vertices $\left(\pm\tfrac12, \pm\tfrac12\right).$ Each hexagon shares one edge with the square and extends across to the opposite side; the hexagon on the bottom edge, for instance, has its far (top) edge from $\left(-\tfrac12, \sqrt3 - \tfrac12\right)$ to $\left(\tfrac12, \sqrt3 - \tfrac12\right).$

The outer boundary is a $12$ -gon with flat edges at distance $\sqrt3 - \tfrac12$ from the center, convex vertices such as $\left(\sqrt3 - \tfrac12, \tfrac12\right),$ and four reflex notches where adjacent hexagons’ slanted edges meet, at $\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right)$ and its symmetric images.

Applying the shoelace formula to these $12$ vertices gives area $16\sqrt3 - 23,$ so $m = 16,$ $n = 3,$ $p = -23,$ and $m + n + p = -4.$

Thus, the correct answer is B.