## AMC 10 2021 Spring Test B

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have
**75 minutes**working time to complete the test.

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

$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20$

$\textbf{D}$

Here we have $-3\pi<x<3\pi$, or approximately $-3(3.14)=-9.42<x<3(3.14)=9.42$. So the answer is $9+1+9=19$.

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

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3} + 6$

$\textbf{D}$

$\sqrt3\approx1.732$, so $3-2\sqrt3<0$. Hence, we have $$\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}=2\sqrt3-3+3+2\sqrt3=4\sqrt3$$

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?

$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$

$\textbf{C}$

Let the number of juniors in the program be $x$. Then the number of seniors is $28-x$. Here we have $25\%x=10\%(28-x)\rightarrow x=8$.

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?

$\textbf{(A) }23 \qquad \textbf{(B) }32 \qquad \textbf{(C) }37 \qquad \textbf{(D) }41 \qquad \textbf{(E) }64$

$\textbf{B}$

$23\cdot2=46$ students in blue shirts are in pairs with each other, so $57-46=11$ students in blue shirts are in pairs with 11 students with yellow shirts. Hence, $75-11=64$ students in yellow shirts are in pairs with each other, which means 32 pairs of yellow shirts.

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?

$\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25$

$\textbf{B}$

First look at the two cousins' ages that multiply to $24$. Since the ages must be single-digit, the ages must either be $3 \text{ and } 8$ or $4 \text{ and } 6.$

Next, look at the two cousins' ages that multiply to $30$. Since the ages must be single-digit, the only ages that work are $5 \text{ and } 6.$ Remembering that all the ages must all be distinct, the only solution that works is when the ages are $3, 8$ and $5, 6$.

We are required to find the sum of the ages, which is $3 + 8 + 5 + 6 =22.$

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 $\dfrac34$. What is the mean of the score of all the students?

$\textbf{(A) }74 \qquad \textbf{(B) }75 \qquad \textbf{(C) }76 \qquad \textbf{(D) }77 \qquad \textbf{(E) }78$

$\textbf{C}$

Assuming that there are $3n$ students in the morning class and $4n$ students in the afternoon class. The total number of students is $7n$, and the total score is $3n\cdot84+4n\cdot70=532n$. So the average is $\dfrac{532n}{7n}=76$.

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 $S$?

$\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi$

$\textbf{D}$

Suppose that line $\ell$ is horizontal, and each circle lies either north or south to $\ell.$ We construct the circles one by one:

1. Without the loss of generality, we draw the circle with radius $7$ north to $\ell.$

2. To maximize the area of region $S,$ we draw the circle with radius $5$ south to $\ell.$

3. Now, we need to subtract the circle with radius $3$ at least. The optimal situation is that the circle with radius $3$ encompasses the circle with radius $1,$ in which we do not need to subtract more. That is, the two smallest circles are on the same side of $\ell,$ but can be on either side.

The diagram below shows one possible configuration of the four circles:

Together, the answer is $\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=65\pi.$

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?

$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$

$\textbf{A}$

In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $C$ and $G,$ respectively.

By observations, we proceed as follows:\begin{alignat*}{6} A=15^2=225 \quad &\implies \quad &B &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ \quad &\implies \quad &C &= &&B-1 &= 210& \\ \quad &\implies \quad &D &= &&C-13 &= 197& \\ \quad &\implies \quad &E &= &&D-14 &= 183& \\ \quad &\implies \quad &F &= &&E-13 &= 170& \\ \quad &\implies \quad &G &= &&F-13 &= 157&. \end{alignat*}Therefore, the answer is $C+G=367.$

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 $b-a?$

$\textbf{(A) }1 \qquad \textbf{(B) }3 \qquad \textbf{(C) }5 \qquad \textbf{(D) }7 \qquad \textbf{(E) }9$

$\textbf{D}$

The final coordinate is $(-6,3)$. Assuming the coordinate before reflecting is $(m,n)$. The slope of the line between the two points is $\dfrac{n-3}{m-(-6)}=\dfrac{n-3}{m+6}$. The line should be perpendicular to the line $y=-x$. Hence, we have $$\dfrac{n-3}{m+6}\cdot(-1)=-1$$ The middle point of $(-6.3)$ and $(m,n)$ is $\left(\dfrac{m-6}2,\dfrac{n+3}2\right)$. The middle point must be on the line $y=-x$. So we get $$\dfrac{n+3}{2}=-\dfrac{m-6}{2}$$ Finally we get $m=-3$, $n=6$.

Now consider the rotation. The line connecting $(1,5)$ and $(-3,6)$ must be perpendicular to the line connecting $(1,5)$ and $(a,b)$. So we have $$\dfrac{6-5}{-3-1}\cdot\dfrac{b-5}{a-1}=-1$$ Rotation doesn't change the distance between points, so we have $$\sqrt{(-3-1)^2+(6-5)^2}=\sqrt{(a-1)^2+(b-5)^2}$$ Solving, we find $(a,b)=(0,1)\ \text{or}\ (2,9)$. By the direction of rotation (counterclockwise), we rule out $(0,1)$.

Therefore, the answer is $b-a=9-2=7$.

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

$\textbf{(A) }1.5 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }4.5 \qquad \textbf{(E) }6$

$\textbf{A}$

The volume of water in the cone is $\dfrac12\cdot\pi(12)^2\cdot18=864\pi$. Let the height of water in the cylinder be $h$. Then the volume of water in the cylinder is $\pi(24)^2h=864\pi\rightarrow h=1.5$.

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?

$\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64$

$\textbf{D}$

Grandma cuts the large rectangular pan of brownies into $m\times n$ pieces. The number of pieces along the perimeter is $2m+2n-4$. Hence, we have $2m+2n-4=\dfrac12mn$. Rearranging the equation gives $$(m-4)(n-4)=8$$Since $m$ and $n$ are both positive, we obtain $(m, n) = (5, 12), (6, 8)$ (up to ordering). By inspection, $5 \cdot 12 = 60$ maximizes the number of brownies.

Let $N=34\cdot34\cdot63\cdot270.$ What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N?$

$\textbf{(A) }1:16 \qquad \textbf{(B) }1:15 \qquad \textbf{(C) }1:14 \qquad \textbf{(D) }1:8 \qquad \textbf{(E) }1:3$

$\textbf{C}$

Prime factorize $N$ to get $N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}$. For each odd divisor $n$ of $N$, there exist even divisors $2n, 4n, 8n$ of $N$. Therefore, the ratio is $1:(2+4+8)=1 : 14.$

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 $n + d ?$

$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$

$\textbf{B}$

By converting all the bases to base 10, we have

\begin{align}

3n^2+2n+d&=263\\

3n^2+2n+4&=6^3+6^2+6d+1

\end{align} Subtracting the equations, we get $d-4=10-6d\rightarrow d=2$.

Plugging $d=2$ back equation (1), we find $n=9\ \text{or}\ -\dfrac{29}{3}$. Since the base must be positive, $n=9.$

The answer is $n+d=9+2=11$.

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?

$\textbf{(A) }5\dfrac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\dfrac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\dfrac12$

$\textbf{B}$

Since two parallel chords have the same length $38$, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is $2d + d = 3d$, and the distance between each of the chords is just $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we create two different right triangles:

- One with base $\dfrac{38}{2}= 19$, height $d$, and hypotenuse $r$ ($\triangle RAO$ on the diagram)

- Another with base $\dfrac{34}{2} = 17$, height $3d$, and hypotenuse $r$ ($\triangle LBO$ on the diagram)

By the Pythagorean theorem, we can create the following system of equations: \[19^2 + d^2 = r^2\] \[17^2 + (2d + d)^2 = r^2\] Solving, we find $d = 3$, so $2d = 6$.

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

$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$

$\textbf{B}$

Note that the exponents of $x^{11}-7x^7+x^3$ are an arithmetic sequence, so they are symmetric around the middle term. Thus, we have $$x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})$$ We square $x+\dfrac{1}{x} = \sqrt{5}$ to get $x^2+2+\dfrac{1}{x^2} = 5$, and therefore $x^2+\dfrac{1}{x^2} = 3$. Squaring again and we get $x^4+2+\dfrac{1}{x^4} = 9$, so $x^4-7+\dfrac{1}{x^4} = 0$. Hence, the answer is $x^7(x^4-7+\dfrac{1}{x^4})=0$.

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 $15$?

$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8$

$\textbf{C}$

A number divisible by 15 must be divisible by both 3 and 5. To be divisible by 5, the number should end with 5 or 0. However, the number end with 0 can not be an uphill integer. So the number must end up with 5. To be divisible by 3, we have several cases:

Case 1: sum of digits = 6. There is only one number, $15.$

Case 2: sum of digits = 9. There are two numbers: $45$ and $135.$

Case 3: sum of digits = 12. There are two numbers: $345$ and $1245.$

Case 4: sum of digits = 15. There is only one number, $12345.$

Hence, the answer is $6.$

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?

$\textbf{(A) }\text{Ravon was given card 3.}\\$

$\textbf{(B) }\text{Aditi was given card 3.}\\$

$\textbf{(C) }\text{Ravon was given card 4.}\\$

$\textbf{(D) }\text{Aditi was given card 4.}\\$

$\textbf{(E) }\text{Tyrone was given card 7.}$

$\textbf{C}$

By logical deduction, we consider the scores from lowest to highest:\begin{align*} \text{Oscar's score is 4.} &\implies \text{Oscar is given cards 1 and 3.} \\ &\implies \text{Aditi is given cards 2 and 5.} \\ &\implies \text{Ravon is given cards 4 and 7.} && \\ &\implies \text{Tyrone is given cards 6 and 10.} \\ &\implies \text{Kim is given cards 8 and 9.} \end{align*}Therefore, the answer is C.

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?

$\textbf{(A)} ~\dfrac{1}{120} \qquad\textbf{(B)} ~\dfrac{1}{32} \qquad\textbf{(C)} ~\dfrac{1}{20} \qquad\textbf{(D)} ~\dfrac{3}{20} \qquad\textbf{(E)} ~\dfrac{1}{6}$

$\textbf{C}$

There is a $\dfrac36$ chance for the first number to be even, a $\dfrac{2}{5}$ chance for the second number to be a distinct even number, and a $\dfrac{1}{4}$ chance for the third number to be even and distinct from the first two. After that, we can roll for any times until an odd number appears. So the answer is $\dfrac{3}{6} \cdot \dfrac{2}{5} \cdot \dfrac{1}{4} = \dfrac{1}{20}$.

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 $S$?

$\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37$

$\textbf{D}$

Supposing that there are $n$ numbers in the set $S$. The smallest integer is $x$, and the largest integer is $x+72$. Let $s$ be the sum of all numbers in the set $S$. Now we have

\begin{align*}

s&=32(n-1)+x+72\\

s&=35(n-2)+x+x+72\\

s&=40(n-1)+x

\end{align*} Solving, we find $x=8$, $n=10$, $s=368$. So the average of all integers in the set $S$ is $\dfrac{s}{n}=\dfrac{368}{10}=36.8$.

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

$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$

$\textbf{D}$

Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end.

Note that $\triangle AED$ and $\triangle ABC$ are $120$ - $30$ - $30$ triangles. The area of each of them is $$\dfrac12AB\cdot AC\sin\angle ABC=\dfrac12\cdot2\cdot2\cdot\sin120^\circ=\sqrt3$$ For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$.

Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11}$. So the answer is $m+n=12+11=23$.

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 $\overline{AB}$ at point $E$. Suppose that $C'D = \dfrac{1}{3}$. What is the perimeter of triangle $\bigtriangleup AEC' ?$

$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\dfrac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\dfrac{13}{6} \qquad\textbf{(D)} ~1 + \dfrac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\dfrac{7}{3}$

$\textbf{A}$

We can set the point on $CD$ where the fold occurs as point $F$. Assuming $FD=x$, then $CF=C'F=1-x$. In right triangle $C'DF$, we have \[x^2 + \left(\frac{1}{3}\right)^2 = (1-x)^2 \rightarrow x=\frac{4}{9}\] We know this is a 3-4-5 triangle because the side lengths are $\dfrac{3}{9}, \dfrac{4}{9}\ \text{and}\ \dfrac{5}{9}$. We also know that $EAC'$ is similar to $C'DF$ because angle $EC'F$ is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of $C'DF \times \dfrac{AC'}{DF}$, which is$$\left(\dfrac39+\dfrac49+\dfrac59\right) \times \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} \times \frac{3}{2} = 2$$

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 $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$

$\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542$

$\textbf{D}$

Use complementary counting. Denote $T_n$ as the total number of ways to put $n$ colors to $n$ boxes by 3 people out of which $f_n$ ways are such that no box has uniform color. Notice $T_n = (n!)^3$. From this setup we see the question is asking for $1-\dfrac{f_5}{(5!)^3}$. To find $f_5$ we want to exclude the cases of a) one box of the same color, b) 2 boxes of the same color, c) 3 boxes of same color, d) 4 boxes of the same color, and e) 5 boxes of the same color. Cases d) and e) coincide. From this, we have\[f_5=T_5 -{\binom{5}{1}\binom{5}{1}\cdot f_4 - \binom{5}{2}\binom{5}{2}\cdot 2!\cdot f_3 - \binom{5}{3}\binom{5}{3}\cdot 3!\cdot f_2 - 5!}\]In case b), there are $\dbinom{5}{2}$ ways to choose 2 boxes that have the same color, $\dbinom{5}{2}$ ways to choose the two colors, 2! ways to permute the 2 chosen colors, and $f_3$ ways so that the remaining 3 boxes don’t have the same color. The same goes for cases a) and c). In case e), the total number of ways to permute 5 colors is $5!$. Now, we just need to calculate $f_2$, $f_3$ and $f_4$.

We have $f_2=T_2-2 = (2!)^3 - 2 = 6$, since we subtract the number of cases where the boxes contain uniform colors, which is 2.

In the same way, $f_3=T_3-\left[3!+ \dbinom{3}{1}\dbinom{3}{1}\cdot f_2 \right] = 156$. Again, we must subtract the number of ways at least 1 box has uniform color. There are $3!$ ways if 3 boxes each contains uniform color. Two boxes each contains uniform color is the same as previous. If one box has the same color, there are $\dbinom{3}{1}$ ways to pick a box, and $\dbinom{3}{1}$ ways to pick a color for that box, 1! ways to permute the chosen color, and $f_2$ ways for the remaining 2 boxes to have non-uniform colors.

Similarly, $f_4=(4!)^3-\left[ 4!+ \dbinom{4}{2}\dbinom{4}{2}\cdot 2! \cdot f_2+ \dbinom{4}{1}\dbinom{4}{1}\cdot f_3\right] = 10,872$

Hence, we have $$f_5 = f_5=(5!)^3-\left[\binom{5}{1}\binom{5}{1}\cdot f_4+ \binom{5}{2}\binom{5}{2}\cdot 2!\cdot f_3+\binom{5}{3}\binom{5}{3}\cdot 3!\cdot f_2 + 5!\right] = (5!)^3 - 306,720$$ Thus, the probability is $$\frac{306,720}{(5!)^3} = \dfrac{71}{400}$$ The answer is $71+400=471$.

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 $\dfrac{1}{196}\left(a+b\sqrt{2}+\pi\right)$, where $a$ and $b$ are positive integers. What is $a+b$?

$\textbf{(A)} ~64 \qquad\textbf{(B)} ~66 \qquad\textbf{(C)} ~68 \qquad\textbf{(D)} ~70 \qquad\textbf{(E)} ~72$

$\textbf{C}$

To find the probability, we look at the $\dfrac{\text{success region}}{\text{total possible region}}$. For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least $\dfrac{1}{2}$, as it's the radius of the coin. This implies the $\text{total possible region}$ is a square with side length $8 - \dfrac{1}{2} - \dfrac{1}{2} = 7$, with an area of $49$. Now, we consider cases where needs to land to partially cover a black region. Note that the center of the coin can lie anywhere inside a green region, as shown below.

$\textbf{Near The Center Square}$

We can have the center of the coin land within $\dfrac{1}{2}$ outside of the center square, or inside of the center square. So, we have a region with $\dfrac{1}{2}$ emanating from every point on the exterior of the square, forming four quarter circles and four rectangles. The four quarter circles combine to make a full circle of radius $\dfrac{1}{2}$, so the area is $\dfrac{\pi}{4}$. The area of a rectangle is $2 \sqrt 2 \cdot \dfrac{1}{2} = \sqrt 2$, so $4$ of them combine to an area of $4 \sqrt 2$. The area of the black square is simply $\left(2\sqrt 2\right)^2 = 8$. So, for this case, we have a combined total of $8 + 4\sqrt 2 + \dfrac{\pi}{4}$.

$\textbf{Near A Triangle}$

We can also have the coin land within $\dfrac{1}{2}$ outside of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by $4$. Consider the above diagram. We can draw an altitude from the bottom left corner of the square to hit the hypotenuse of the green triangle. The length of this when passing through the black region is $\sqrt 2$, and when passing through the white region (while being contained in the green triangle) is $\dfrac{1}{2}$. The distance between the right angle vertices of the black triangle and the green triangle is $\dfrac12\sqrt2$. So, the altitude of the green triangle is $\sqrt 2 + \dfrac{1}{2} - \dfrac{\sqrt 2}{2} = \dfrac{\sqrt 2 + 1}{2}$. The area of an isosceles right triangle is $h^2$, where $h$ is the altitude from the right angle. So, squaring this, we get $\dfrac{3 + 2\sqrt 2}{4}$. Now, we have to multiply this by $4$ to account for all of the green triangles, to get $3 + 2\sqrt 2$ as the final area for this case.

Then, to have the coin touching a black region, we add up the area of our successful regions $$8 + 4\sqrt 2 + \dfrac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \dfrac{\pi}{4} = \dfrac{44 + 24\sqrt 2 + \pi}{4}$$ The total region is $49$, so our probability is $$\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}$$ which implies $a+b = 44+24 = 68$.

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?

$\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)$

$\textbf{B}$

Supposing that there are several walls, where each wall is composed of 1 or 2 bricks, and the numbers of both 1-brick walls and 2-brick walls are even. This is called a symmetrical configuration. For example, $(2,2,2,2,1,1)$ is a symmetrical configuration. The second player will win for any symmetrical configuration if he repeats the movements of the first player and keeps a symmetrical configuration. Come back to the example of $(2,2,2,2,1,1)$. If the first player removes a 1-brick wall, the second player should remove another 1-brick wall to keep symmetry. If the first play removes a 2-brick wall, the second player should remove another 2-brick wall. If the first player removes 1 brick from a 2-brick wall, the second player also removes 1 brick from another 2-brick wall. Hence, the second player can guarantee his success.

Now we turn to the choices:

$(6,1,1)$ can be turned into $(2,2,1,1)$ by Arjun, which is symmetric, so Beth will lose.

$(6,3,1)$ can be turned into $(3,1,3,1)$ by Arjun, which is symmetric, so Beth will lose.

$(6,2,2)$ can be turned into $(2,2,2,2)$ by Arjun, which is symmetric, so Beth will lose.

$(6,3,2)$ can be turned into $(3,2,3,2)$ by Arjun, which is symmetric, so Beth will lose.

So the only possible answer is $(6,2,1)$.

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 $\dfrac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$

$\textbf{(A) }31 \qquad \textbf{(B) }47 \qquad \textbf{(C) }62\qquad \textbf{(D) }72 \qquad \textbf{(E) }85$

$\textbf{E}$

First, we find a numerical representation for the number of lattice points in $S$ that are under the line $y=mx.$ For any value of $x,$ the highest lattice point under $y=mx$ is $\lfloor mx \rfloor.$ Because every lattice point from $(x, 1)$ to $(x, \lfloor mx \rfloor)$ is under the line, the total number of lattice points under the line is $\sum_{x=1}^{30}(\lfloor mx \rfloor).$

Now, we proceed by finding lower and upper bounds for $m.$ To find the lower bound, we start with an approximation. If $300$ lattice points are below the line, then around $\dfrac{1}{3}$ of the area formed by $S$ is under the line. By using the formula for a triangle's area, we find that when $x=30, y \approx 20.$ Solving for $m$ assuming that $(30, 20)$ is a point on the line, we get $m = \dfrac{2}{3}.$ Plugging in $m$ to $\sum_{x=1}^{30}(\lfloor mx \rfloor),$ we get:

\[\sum_{x=1}^{30}(\lfloor \frac{2}{3}x \rfloor) = 0 + 1 + 2 + 2 + 3 + \cdots + 18 + 18 + 19 + 20\]

We have a repeat every $3$ values (every time $y=\dfrac{2}{3}x$ goes through a lattice point). Thus, we can use arithmetic sequences to calculate the value above:

\begin{align*}

\sum_{x=1}^{30}(\lfloor \frac{2}{3}x \rfloor)& = 0 + 1 + 2 + 2 + 3 + \cdots + 18 + 18 + 19 + 20\\

&=\frac{20(21)}{2} + 2 + 4 + 6 + \cdots + 18\\

&=210 + \frac{2+18}{2}\cdot 9\\

&=300

\end{align*} This means that $\dfrac{2}{3}$ is a possible value of $m.$ Furthermore, it is the lower bound for $m.$ This is because $y=\dfrac{2}{3}x$ goes through many points (such as $(21, 14)$). If $m$ was lower, $y=mx$ would no longer go through some of these points, and there would be less than $300$ lattice points under it.

Now, we find an upper bound for $m.$ Imagine increasing $m$ slowly and rotating the line $y=mx,$ starting from the lower bound of $m=\dfrac{2}{3}.$The upper bound for $m$ occurs when $y=mx$ intersects a lattice point again

In other words, we are looking for the first $m > \dfrac{2}{3}$ that is expressible as a ratio of positive integers $\dfrac{p}{q}$ with $q \le 30.$ For each $q=1,\dots,30$, the smallest multiple of $\dfrac{1}{q}$ which exceeds $\dfrac{2}{3}$ is $$1, \frac{2}{2}, \frac{3}{3}, \frac{3}{4}, \frac{4}{5}, \cdots , \frac{19}{27}, \frac{19}{28}, \frac{20}{29}, \frac{21}{30}$$ respectively, and the smallest of these is $\dfrac{19}{28}.$

The lower bound is $\dfrac{2}{3}$ and the upper bound is $\dfrac{19}{28}.$ Their difference is $\dfrac{1}{84},$ so the answer is $1 + 84 =85.$