2021 AMC 10B Spring — Problems & Online Timed Exam

Timed

1:15:00

1.

How many integer values of $x$ satisfy $|x| < 3\pi?$

$9$

$10$

$18$

$19$

$20$

Answer: D

Concepts:absolute value counting integers in a range

Difficulty rating: 560

Small Hint:

Use $3\pi$ as a little more than $9$

Big Hint:

Count the integers strictly between $-3\pi$ and $3\pi$

Solution:

Every integer from $-9$ to $9,$ inclusive, works. This yields $9-(-9)+1 = 19$ solutions.

Thus, the correct answer is D .

2.

What is the value of $\begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2}? \end{aligned}$

$0$

$4\sqrt{3}-6$

$6$

$4\sqrt{3}$

$4\sqrt{3}+6$

Answer: D

Concepts:radical absolute value

Difficulty rating: 770

Small Hint:

Remember that $\sqrt{u^2}=|u|$

Big Hint:

Decide the signs of $3-2\sqrt3$ and $3+2\sqrt3$ separately

Solution:

Because $\sqrt{u^2}=|u|,$ the expression equals $|3-2\sqrt3|+|3+2\sqrt3|.$ Since $2\sqrt3>3,$ this becomes $(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3.$

Thus, the correct answer is D .

3.

In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?

$5$

$6$

$8$

$11$

$20$

Answer: C

Concepts:percentage system of equations

Difficulty rating: 870

Small Hint:

Let the numbers of juniors and seniors be variables

Big Hint:

The equal debate-team counts give a second equation

Video solution:

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Written solution:

Let the number of juniors be $j$ and the number of seniors be $s.$ Then, $j+s = 28$ and $0.25j = 0.1s.$ This means $2.5j = s,$ so $3.5j = 28.$ This makes $j=8.$

Thus, the correct answer is C .

4.

At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

$23$

$32$

$37$

$41$

$64$

Answer: B

Concepts:counting pairs

Difficulty rating: 960

Small Hint:

Use the blue-blue pairs to find how many mixed pairs there are

Big Hint:

Every mixed pair uses exactly one yellow-shirted student

Video solution:

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Written solution:

There are $2\cdot 23 = 46$ students with blue shirts that are in a pair with just blue shirts. This means there are $57-46=11$ students in blue shirts who are paired with someone wearing a yellow shirt, meaning exactly $11$ people wearing yellow shirts are paired with someone wearing a blue shirt.

This leaves just $64$ students wearing a yellow shirt who are paired with someone else wearing a yellow shirt. This yields $\dfrac{64}2 = 32$ pairs.

Thus, the answer is B .

5.

The ages of Jonie’s four cousins are distinct single-digit positive integers. Two of the cousins’ ages multiplied together give $24,$ while the other two multiply to $30.$ What is the sum of the ages of Jonie’s four cousins?

$21$

$22$

$23$

$24$

$25$

Answer: B

Concepts:factor ages

Difficulty rating: 900

Small Hint:

List the single-digit factor pairs of $30$ first

Big Hint:

The ages are distinct, so exclude any factor pair that reuses an age

Video solution:

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Written solution:

The only pair of single-digit positive integers with product $30$ is $5,6.$ The single-digit factor pairs of $24$ are $3,8$ and $4,6.$ Because all four ages are distinct, the pair $4,6$ would repeat the age $6,$ so the other pair must be $3,8.$ Thus the ages are $3,5,6,8,$ whose sum is $22.$

Thus, the answer is B .

6.

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84,$ and the afternoon class’s mean score is $70.$ The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}.$ What is the mean of the scores of all the students?

$74$

$75$

$76$

$77$

$78$

Answer: C

Concepts:mean ratio and proportion

Difficulty rating: 900

Small Hint:

Use $3x$ and $4x$ students for the two classes

Big Hint:

Compute the weighted average from the two total score sums

Video solution:

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Written solution:

Let the number of people in the first class be $3x.$ This means the number of people in the second class is $4x.$

Thus, the sum of the scores of the first class is $84\cdot 3x = 252x$ and the sum of the scores for the people in the second class is $70\cdot 4x = 280x.$ This means the total sum is $532x,$ with $7x$ people.

Therefore, the average of all the students is $\dfrac{532x}{7x} = 76.$

Thus, the correct answer is C .

7.

In a plane, four circles with radii $1,$ $3,$ $5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell.$ Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S?$

$24\pi$

$32\pi$

$64\pi$

$65\pi$

$84\pi$

Answer: D

Concepts:tangent circles circle area optimization

Difficulty rating: 1240

Small Hint:

Circles on the same side of the line are nested

Big Hint:

Inside exactly one nested circle means outside the next smaller circle

Video solution:

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Written solution:

On one side of $\ell,$ circles tangent at $A$ are nested. For nested circles with radii $r_1>r_2>\cdots,$ the points inside exactly one of those circles have area $\pi(r_1^2-r_2^2)$ if there are at least two circles; a third smaller nested circle does not count because its points are inside three circles, not exactly one.

To maximize the area, put the circle of radius $7$ alone on one side, and put the circles of radii $5,3,1$ on the other side. This gives

$\begin{aligned} &7^2\pi+(5^2-3^2)\pi \\ &=49\pi+16\pi=65\pi. \end{aligned}$

Thus, the answer is D .

8.

Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?

$367$

$368$

$369$

$379$

$380$

Answer: A

Concepts:pattern recognition perfect square

Difficulty rating: 1420

Small Hint:

The outer $15\times15$ ring ends with $225$ in the upper right

Big Hint:

The second row includes the top row of the inner $13\times13$ spiral

Video solution:

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Written solution:

In the outer $15\times15$ ring, the top row contains $211,212,\ldots,225,$ so the number just below $211$ in the second row is $210.$ This is the greatest number in the second row.

The inner $13\times13$ spiral has $13^2=169$ in its upper-right corner. In the second row of the full grid, the inner-ring entries run from $157$ to $169.$ Thus the least entry in that row is $157.$

The required sum is $210+157=367.$

Thus, the answer is A .

9.

The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x.$ The image of $P$ after these two transformations is at $(-6,3).$ What is $b - a ?$

$1$

$3$

$5$

$7$

$9$

Answer: D

Concepts:transformation coordinate geometry work backwards

Difficulty rating: 1220

Small Hint:

Undo the reflection first

Big Hint:

Undo the rotation by translating the center of rotation to the origin

Video solution:

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Written solution:

Work backward. Reflecting $(-6,3)$ across $y=-x$ gives $(-3,6).$

Now undo the $90^\circ$ counterclockwise rotation by rotating $(-3,6)$ clockwise about $(1,5).$ Relative to $(1,5),$ the point is $(-4,1).$ A clockwise quarter-turn sends this to $(1,4),$ and translating back gives $(2,9).$

Thus $a=2,$ $b=9,$ and $b-a=7.$

Thus, the answer is D .

10.

An inverted cone with base radius $12 \mathrm{ cm}$ and height $18 \mathrm{ cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of $24 \mathrm{ cm}.$ What is the height in centimeters of the water in the cylinder?

$1.5$

$3$

$4$

$4.5$

$6$

Answer: A

Concepts:volume cone cylinder

Difficulty rating: 1020

Small Hint:

Set the cone volume equal to the cylinder volume

Big Hint:

The cylinder radius is twice the cone radius

Video solution:

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Written solution:

The volumes must be the same since the water is poured from one to another. The volume of the cone is $\frac{r^2h \pi}3 = \frac{12^2\cdot 18 \pi}3 = 864 \pi.$

The volume of the cylinder is $r^2h \pi =24^2h \pi = 576h \pi.$ This makes $576 h \pi = 864 \pi,$ so $h = 1.5.$

Thus, the answer is A .

11.

Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?

$24$

$30$

$48$

$60$

$64$

Answer: D

Concepts:Simon’s Favorite Factoring Trick optimization

Difficulty rating: 1420

Small Hint:

Let the pan be cut into an $l\times w$ grid of pieces

Big Hint:

Set the number of interior pieces equal to the number of perimeter pieces

Video solution:

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Written solution:

Suppose the cuts make an $l\times w$ grid of pieces. The number of interior pieces is $(l-2)(w-2),$ and the total number of pieces is $lw.$ Since the number of interior pieces equals the number of perimeter pieces, the interior pieces make up half the total:

$(l-2)(w-2)=\frac{lw}{2}.$

Multiplying out gives $lw-4l-4w+8=0,$ or

$(l-4)(w-4)=8.$

The positive factor pairs of $8$ give $(l,w)=(5,12)$ or $(6,8),$ up to order. These produce $60$ or $48$ pieces, respectively, so the greatest possible number is $60.$

Thus, the answer is D .

12.

Let $N = 34 \cdot 34 \cdot 63 \cdot 270.$ What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N?$

$1 : 16$

$1 : 15$

$1 : 14$

$1 : 8$

$1 : 3$

Answer: C

Concepts:sum of factors prime factorization parity

Difficulty rating: 1140

Small Hint:

Separate divisors by their power of $2$

Big Hint:

Each odd divisor corresponds to even divisors $2d,4d,8d$

Solution:

Using prime factorization, we get $N =2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.$

If we have an odd divisor $x$ of $N,$ then $2x,4x,8x$ are divisors of $N,$ which has a combined sum of $14x.$ If we take the sum of every odd divisor, then the even divisors must have a sum which is $14$ times the sum of the odd divisors. Therefore, the requested ratio is $1:14.$

Thus, the answer is C .

13.

Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263,$ and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$

$10$

$11$

$13$

$15$

$16$

Answer: B

Concepts:number base system of equations

Difficulty rating: 1280

Small Hint:

Translate each base numeral into base ten

Big Hint:

Subtract the two equations to solve for the digit first

Video solution:

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Written solution:

The first statement means $3n^2 + 2n+d = 263.$

Similarly, the second statement means $3n^2 +2n +4$ $= 6^3 +6^2 +6d+1$ $= 253 + 6d.$

Subtracting these shows us that $4-d = 6d-10$ $7d = 14$ $d=2.$ Therefore, $3n^2 + 2n + 2 = 263,$ so $n(3n+2) = 261.$

This implies $n=9.$ Therefore, $n+d = 11.$

Thus, the answer is B .

14.

Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,$ $38,$ and $34.$ What is the distance between two adjacent parallel lines?

$5\frac12$

$6$

$6\frac12$

$7$

$7\frac12$

Answer: B

Concepts:chord circle Pythagorean Theorem

Difficulty rating: 1540

Small Hint:

Equal chord lengths are equally far from the circle center

Big Hint:

If the equal chords are adjacent lines, the third line is three half-spacings from the center

Video solution:

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Written solution:

The two chords of length $38$ are equally far from the center of the circle. They cannot lie on the two outer lines: then the center would lie on the middle line, whose chord would be a diameter longer than $38,$ not the given $34.$ Therefore, the equal chords lie on adjacent lines, with the center halfway between them. Let that half-distance be $d.$ Then each $38$ -chord is distance $d$ from the center, and the $34$ -chord is distance $3d$ from the center.

If the circle has radius $r,$ then

$r^2=19^2+d^2=17^2+(3d)^2.$

Thus $19^2-17^2=8d^2,$ so $72=8d^2,$ and $d=3.$ The distance between adjacent parallel lines is $2d=6.$

Thus, the answer is B .

15.

The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}.$ What is the value of $x^{11}-7x^{7}+x^3?$

$-1$

$0$

$1$

$2$

$\sqrt{5}$

Answer: B

Concepts:algebraic manipulation symmetry (algebra)

Difficulty rating: 1340

Small Hint:

Square the given equation to find $x^2+x^{-2}$

Big Hint:

Square again to find a relation involving $x^4+x^{-4}$

Video solution:

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Written solution:

Since $x+\frac{1}{x} = \sqrt{5},$ squaring yields $x^2 + 2 + \frac 1{x^2} = 5$ $x^2 + \frac 1{x^2} = 3.$ Squaring again yields $x^4 + 2 + \frac 1{x^4} = 9$ $x^4 -7 + \frac 1{x^4} = 0.$ Multiplying by $x^7$ yields $x^{11}-7x^{7}+x^3=0.$

Thus, the answer is B .

16.

Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357,$ $89,$ and $5$ are all uphill integers, but $32,$ $1240,$ and $466$ are not. How many uphill integers are divisible by $15?$

$4$

$5$

$6$

$7$

$8$

Answer: C

Concepts:subsets divisibility

Difficulty rating: 1480

Small Hint:

The last digit must be $5$

Big Hint:

Choose a subset of ${1,2,3,4}$ whose digit sum gives divisibility by $3$

Video solution:

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Written solution:

If a number is divisible by $15,$ it has a units digit of $0$ or $5.$ If the units digit is $0$ and the digits are strictly increasing, then the number is $0,$ which isn’t positive. Therefore, we can just look at numbers with a units digit of $5.$

Next, we need to find uphill integers that are a multiple of $3.$ This means the other digits are a subset of $\{1,2,3,4\}.$ Taking the sum of the set must have a remainder of $1$ when divided by $3.$ Also, having or taking out $3$ wouldn’t affect the remainder, so we can take the number of subsets without a $3$ and multiply it by $2.$ There are only $3$ such subsets, namely $\{1\}, \{4\},$ and $\{1,2,4\}.$ Thus, there are $6$ total subsets.

Thus, the correct answer is C .

17.

Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given $2$ cards out of a set of $10$ cards numbered $1,$ $2,$ $3,$ $\dots,$ $10.$ The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon-- $11,$ Oscar-- $4,$ Aditi-- $7,$ Tyrone-- $16,$ Kim-- $17.$ Which of the following statements is true?

Ravon was given card $3.$

Aditi was given card $3.$

Ravon was given card $4.$

Aditi was given card $4.$

Tyrone was given card $7.$

Answer: C

Concepts:logical deduction

Difficulty rating: 1420

Small Hint:

Oscar’s cards are forced by his score

Big Hint:

After Oscar and Aditi are determined, use the remaining possible sums

Video solution:

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Written solution:

Oscar’s score of $4$ forces his cards to be $1$ and $3.$ From the remaining cards, the only pair with sum $7$ is $2$ and $5,$ so those are Aditi’s cards.

The unused cards are now $4,6,7,8,9,10.$ Tyrone’s score of $16$ can come only from $6+10$ or $7+9.$ If Tyrone had $7$ and $9,$ then the remaining cards $4,6,8,10$ could not be paired to give Ravon’s score $11.$ Thus Tyrone has $6$ and $10.$

The remaining cards are $4,7,8,9.$ Ravon must have $4$ and $7$ to score $11,$ leaving $8$ and $9$ for Kim’s score of $17.$ Therefore, Ravon was given card $4.$

Thus, the answer is C .

18.

A fair $6$ -sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?

$\dfrac{1}{120}$

$\dfrac{1}{32}$

$\dfrac{1}{20}$

$\dfrac{3}{20}$

$\dfrac{1}{6}$

Answer: C

Concepts:dice (probability)basic probability

Difficulty rating: 1220

Small Hint:

Ignore repeated rolls until a new face type appears

Big Hint:

The first three distinct outcomes must be the three even numbers before any odd number appears

Video solution:

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Written solution:

The probability that the first number is even is $\frac 36.$

The probability that the second distinct number is even is $\frac 25.$

The probability that the third distinct number is even is $\frac 14.$

The combined probability is $\dfrac{3\cdot 2\cdot 1}{6\cdot 5\cdot 4} = \dfrac 1{20}.$

Thus, the answer is C .

19.

Suppose that $S$ is a finite set of positive integers.

If the greatest integer in $S$ is removed from $S,$ then the average value (arithmetic mean) of the integers remaining is $32.$ If the least integer in $S$ is also removed, then the average value of the integers remaining is $35.$ If the greatest integer is then returned to the set, the average value of the integers rises to $40.$ The greatest integer in the original set $S$ is $72$ greater than the least integer in $S.$

What is the average value of all the integers in the set $S?$

$36.2$

$36.4$

$36.6$

$36.8$

$37$

Answer: D

Concepts:mean system of equations

Difficulty rating: 1540

Small Hint:

Write equations for the three averages using the total sum, least value, greatest value, and set size

Big Hint:

Subtract the two averages with denominator $n-1$ to use the difference $72$

Video solution:

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Written solution:

Let the sum of all the integers be $s,$ the greatest number be $g,$ the least number be $l,$ and the size of $S$ be $n.$

The two averages of sets with $n-1$ elements give $\begin{aligned} s-l&=40(n-1),\\ s-g&=32(n-1). \end{aligned}$ Subtracting and using $g-l=72$ gives $72=8(n-1),$ so $n=10.$

Removing both extremes leaves $8$ numbers with sum $8\cdot35=280.$ On the other hand, removing only the least number leaves a sum of $9\cdot40=360,$ so $g=360-280=80.$ Hence $l=80-72=8,$ and the original sum is $s=280+80+8=368.$

The required average is $\frac{s}{n}=\frac{368}{10}=36.8.$

Thus, the answer is D .

20.

The figure below is constructed from $11$ line segments, each of which has length $2.$ The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n},$ where $m$ and $n$ are positive integers. What is $m + n ?$

$20$

$21$

$22$

$23$

$24$

Answer: D

Concepts:area decomposition equilateral triangle special right triangle

Difficulty rating: 1950

Small Hint:

Use the equal length segments to identify equilateral-triangle altitudes

Big Hint:

Split the pentagon into two side pieces and the central isosceles triangle

Solution:

Let $F$ be the unlabeled point joined to $A,B,$ and $C.$ Because all the drawn segments have length $2,$ triangles $ABF$ and $CBF$ are equilateral and lie on opposite sides of $BF.$ Hence $\angle ABC=120^\circ,$ so

$[ABC]=\frac12\cdot2\cdot2\sin120^\circ=\sqrt3.$

The same reasoning on the other side gives $[ADE]=\sqrt3.$ Also, the Law of Cosines in $\triangle ABC$ gives $AC^2=12,$ and similarly $AD^2=12.$ Thus the altitude of isosceles triangle $ACD$ to its base $CD=2$ is

$\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.$

Therefore $[ACD]=\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}.$ The pentagon’s total area is

$\sqrt{12}+\sqrt{11},$

so $m+n=12+11=23.$

Thus, the answer is D .

21.

A square piece of paper has side length $1$ and vertices $A,$ $B,$ $C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C',$ and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E.$ Suppose that $C'D = \frac{1}{3}.$ What is the perimeter of $\triangle AEC'?$

$2$

$1+\dfrac{2}{3}\sqrt{3}$

$\dfrac{13}{6}$

$1 + \dfrac{3}{4}\sqrt{3}$

$\dfrac{7}{3}$

Answer: A

Concepts:paper folding coordinate geometry

Difficulty rating: 2230

Small Hint:

A fold is a reflection, so the crease is the perpendicular bisector of $CC'$

Big Hint:

Find $AE,$ $AC',$ and $EC'$ from the folded right triangle

Video solution:

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Written solution:

Use coordinates with $A=(0,1),$ $B=(0,0),$ $C=(1,0),$ and $D=(1,1).$ Since $C'D=\frac13,$ we have $C'=(\frac23,1),$ so $AC'=\frac23.$

The fold reflects $C$ to $C',$ so the image of side $BC$ is the line through $C'$ and $E.$ Reflecting $B=(0,0)$ across the perpendicular bisector of $CC'$ gives $(-\frac{2}{15},\frac25).$ The line through this point and $C'$ meets $AB$ at $E=(0,\frac12).$

Thus $AE=\frac12,$ and

$EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56.$

The perimeter of $\triangle AEC'$ is

$\frac12+\frac23+\frac56=2.$

Thus, the answer is A .

22.

Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$

$47$

$94$

$227$

$471$

$542$

Answer: D

Concepts:inclusion-exclusion permutations basic probability

Difficulty rating: 2150

Small Hint:

Fix Ang’s placement and count Ben’s and Jasmin’s permutations

Big Hint:

Use inclusion-exclusion on the boxes where both other placements match Ang’s color

Video solution:

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Written solution:

Fix Ang’s placement and label each box by the color Ang put in it. Ben and Jasmin each choose a permutation of the five colors, so there are $(5!)^2$ equally likely pairs of placements.

For a specified set of $k$ boxes to receive three blocks of the same color, both Ben and Jasmin must match Ang in those $k$ boxes. This can happen in $((5-k)!)^2$ ways. By inclusion-exclusion, the number of successful placement pairs is

$\begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2. \end{aligned}$

This equals

$2880-360+40-5+1=2556.$

Therefore the probability is

$\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.$

Thus $m+n=71+400=471.$

Thus, the answer is D .

23.

A square with side length $8$ is colored white except for $4$ black isosceles right triangular regions with legs of length $2$ in each corner of the square and a black diamond with side length $2\sqrt{2}$ in the center of the square, as shown in the diagram.

A circular coin with diameter $1$ is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the black region of the square can be written as $\frac{1}{196}\left(a+b\sqrt{2}+\pi\right),$ where $a$ and $b$ are positive integers. What is $a+b?$

$64$

$66$

$68$

$70$

$72$

Answer: C

Concepts:geometric probability area decomposition

Difficulty rating: 2390

Small Hint:

The coin center ranges over a $7\times7$ square

Big Hint:

Count the area within distance $\frac{1}{2}$ of the shaded corner triangles and the shaded center diamond

Solution:

The coin has radius $\frac12,$ so its center is uniformly distributed over a $7\times7$ square of area $49.$

A shaded corner triangle contributes the set of center positions within distance $\frac12$ of that triangle, inside the allowed center square. For each corner this is a right isosceles triangle whose altitude is $\frac{1+\sqrt2}{2},$ so its area is

$\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}.$

All four corners contribute $3+2\sqrt2.$

The center shaded diamond is a square of side $2\sqrt2.$ Expanding it by distance $\frac12$ adds four rectangles of total area $4\sqrt2$ and four quarter-circles of total area $\frac\pi4,$ in addition to the diamond’s area $8.$ Thus the center contribution is

$8+4\sqrt2+\frac\pi4.$

The favorable area is

$\begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4. \end{aligned}$

The probability is

$\begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196}. \end{aligned}$

So $a+b=44+24=68.$

Thus, the answer is C .

24.

Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one “wall” among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),$ $(2,1,2),$ $(4),$ $(4,1),$ $(2,2),$ or $(1,1,2).$

Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?

$(6,1,1)$

$(6,2,1)$

$(6,2,2)$

$(6,3,1)$

$(6,3,2)$

Answer: B

Concepts:combinatorial game

Difficulty rating: 2390

Small Hint:

Compute the game value for a single wall of each relevant size

Big Hint:

A position is losing for the player to move exactly when the xor of the wall values is $0$

Solution:

For a single wall of length $n,$ compute its Sprague-Grundy value from the possible moves. For the wall lengths needed here, the values are

$\begin{aligned} &g(1)=1,\quad g(2)=2, \\ &g(3)=3,\quad g(4)=1, \\ &g(5)=4,\quad g(6)=3. \end{aligned}$

For several walls, the position is losing for the player to move exactly when the xor of the wall values is $0.$ Evaluating the choices gives

$(6,1,1): 3\oplus1\oplus1=3,$

$(6,2,1): 3\oplus2\oplus1=0,$

$(6,2,2): 3\oplus2\oplus2=3,$

$(6,3,1): 3\oplus3\oplus1=1,$

$(6,3,2): 3\oplus3\oplus2=2.$

Only $(6,2,1)$ is losing for the player to move, so Beth has a guaranteed win exactly for that starting configuration.

Thus, the answer is B .

25.

Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$

$31$

$47$

$62$

$72$

$85$

Answer: E

Concepts:lattice point floor and ceiling functions

Difficulty rating: 2390

Small Hint:

Near the relevant slopes, the number of points is $\sum_{x=1}^{30}\lfloor mx\rfloor$

Big Hint:

Find the slope where the $300$ th lattice-point ratio occurs and the next larger possible ratio

Solution:

For a fixed slope $m,$ the number of points in $S$ on or below $y=mx$ is

$\sum_{x=1}^{30}\lfloor mx\rfloor,$

for the slopes near the answer.

At $m=\frac23,$ grouping $x=3k+1,3k+2,3k+3$ for $k=0,1,\ldots,9$ gives

$\lfloor \frac{2x}{3}\rfloor=2k,\ 2k+1, \ 2k+2,$

whose sum over each block is $6k+3.$ Thus the total is

$\sum_{k=0}^9(6k+3)=270+30=300.$

If $m<\frac23,$ the ten points with ratios $\frac{y}{x}=\frac{2}{3}$ are no longer counted, so the count is less than $300.$ Therefore the lower end is $\frac23.$

The next possible ratio $\frac{y}{x}$ greater than $\frac23,$ with $1\le x,y\le30,$ is minimized by checking $x$ modulo $3.$ The best candidates are

$\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10},$

and the smallest is $\frac{19}{28}.$ Hence the interval length is

$\frac{19}{28}-\frac23=\frac1{84}.$

Thus $a+b=1+84=85.$

Thus, the answer is E .