## AMC 12 2021 Fall 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.

What is the value of $1234+2341+3412+4123?$

$(\textbf{A})\: 10{,}000\qquad(\textbf{B}) \: 10{,}010\qquad(\textbf{C}) \: 10{,}110\qquad(\textbf{D}) \: 11{,}000\qquad(\textbf{E}) \: 11{,}110$

$\textbf{E}$

We see that $1, 2, 3,$ and $4$ each appear in the units, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$, we find that the sum is equal to\[10\cdot(1+10+100+1000)= 11{,}110\]

What is the area of the shaded figure shown below?

$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$

$\textbf{B}$

To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$) from the area of the larger triangle (base $4$ and height $5$):\[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=6\]

The expression $\dfrac{2021}{2020} - \dfrac{2020}{2021}$ is equal to the fraction $\dfrac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is $1$. What is $p?$

$\textbf{(A)}\: 1\qquad\textbf{(B)} \: 9\qquad\textbf{(C)} \: 2020\qquad\textbf{(D)} \: 2021\qquad\textbf{(E)} \: 4041$

$\textbf{E}$

\[\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}\] So the answer is $2021+2020=4041$.

At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$

$\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120$

$\textbf{C}$

At 4:00 the temperature difference is $|N-8|=2$. Hence, we get $N=10\ \text{or}\ 6$. So the product is 60.

Let $n=8^{2022}$. Which of the following is equal to $\dfrac{n}{4}?$

$\textbf{(A)}\: 4^{1010}\qquad\textbf{(B)} \: 2^{2022}\qquad\textbf{(C)} \: 8^{2018}\qquad\textbf{(D)} \: 4^{3031}\qquad\textbf{(E)} \: 4^{3032}$

$\textbf{E}$

We have\[n=8^{2022}= \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.\]Therefore,\[\frac{n}4=4^{3032}\]

The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$, where $m$ and $k$ are integers and $6$ is not a divisor of $m$. What is $m+k?$

$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$

$\textbf{B}$

The prime factorization of 2021 is $2021=43\cdot47$. So the positive integer must be in the form of $a^{42}\cdot b^{46}$. To minimize the integer, we set $a=3$ and $b=2$. So the integer is $3^{42} \cdot 2^{46} = 2^4 \cdot 2^{42} \cdot 3^{42} = 16 \cdot 6^{42}$. Hence, we get $m=16$ and $k=42$, so $m+k = 16 + 42 = 58.$

Call a fraction $\dfrac{a}{b}$, not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?

$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$

$\textbf{C}$

The special fractions are\[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\]We rewrite them in the simplest form:\[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\]Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once:\[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\]For the set $\{2,4,14\},$ two elements (not necessarily different) can sum to $4,6,8,16,18,28.$

For the set $\left\{\dfrac{1}{2},1\dfrac{1}{2},6\dfrac{1}{2}\right\},$ two elements (not necessarily different) can sum to $1,2,3,7,8,13.$

For the set $\left\{\dfrac{1}{4},2\dfrac{3}{4}\right\},$ two elements (not necessarily different) can sum to $3.$

Together, there are $11$ distinct integers that can be written as the sum of two, not necessarily different, special fractions:\[1,2,3,4,6,7,8,13,16,18,28.\]

The largest prime factor of $16384$ is $2$ because $16384 = 2^{14}$. What is the sum of the digits of the greatest prime number that is a divisor of $16383$?

$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$

$\textbf{C}$

We have\begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*}Since $129$ is composite, $127$ is the largest prime divisible by $16383$. The sum of $127$'s digits is $1+2+7=10$.

The knights in a certain kingdom come in two colors. $\dfrac{2}{7}$ of them are red, and the rest are blue. Furthermore, $\dfrac{1}{6}$ of the knights are magical, and the fraction of red knights who are magical is $2$ times the fraction of blue knights who are magical. What fraction of red knights are magical?

$\textbf{(A) }\dfrac{2}{9}\qquad\textbf{(B) }\dfrac{3}{13}\qquad\textbf{(C) }\dfrac{7}{27}\qquad\textbf{(D) }\dfrac{2}{7}\qquad\textbf{(E) }\dfrac{1}{3}$

$\textbf{C}$

Let the fraction of red knights who are magical be $x$. Hence, the fraction of blue knights who are magical is $\dfrac12x$. So the fraction of knight who are magical is $$\dfrac27x+\dfrac57\cdot\dfrac12x=\dfrac16\rightarrow x=\dfrac{7}{27}$$

Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$

$\textbf{A}$

When Alice says "I can't tell who has the larger number." We know that she got neither 1 nor 40.

When Bob says "I know who has the larger number." We know that he got 1, 2, 39, or 40.

Given that Bob got a prime number, we can make sure that Bot got 2.

Since the only perfect square between 201 and 240 is $225=15^2$, we know the number Alice got is $225-2\cdot100=25$.

So the answer is $25+2=27$.

A regular hexagon of side length $1$ is inscribed in a circle. Each minor arc of the circle determined by a side of the hexagon is reflected over that side. What is the area of the region bounded by these $6$ reflected arcs?

$(\textbf{A})\: \dfrac{5\sqrt{3}}{2} - \pi\qquad(\textbf{B}) \: 3\sqrt{3}-\pi\qquad(\textbf{C}) \: 4\sqrt{3}-\dfrac{3\pi}{2}\qquad(\textbf{D}) \: \pi - \dfrac{\sqrt{3}}{2}\qquad(\textbf{E}) \: \dfrac{\pi + \sqrt{3}}{2}$

$\textbf{B}$

This is the graph of the original Hexagon. After reflecting each minor arc over the sides of the hexagon it will look like this:

This bounded region is the same as the area of the hexagon minus the area of 6 reflect arcs. The area of the hexagon is $6\cdot\dfrac{\sqrt3}{4}(1)^2=\dfrac32\sqrt3$. From the first diagram, the area of 6 arcs is $\pi(1)^2-6\cdot\dfrac{\sqrt3}{4}=\pi-\dfrac32\sqrt3$. So the answer is $\dfrac32\sqrt3-\left(\pi-\dfrac32\sqrt3\right)=3\sqrt3-\pi$.

Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation\[x(x-y)+y(y-z)+z(z-x) = 1?\]

$\textbf{(A)} \: x>y$ and $y=z$

$\textbf{(B)} \: x=y-1$ and $y=z-1$

$\textbf{(C)} \: x=z+1$ and $y=x+1$

$\textbf{(D)} \: x=z$ and $y-1=x$

$\textbf{(E)} \: x+y+z=1$

$\textbf{D}$

According to the given equation, we have $$x^2+y^2+z^2-xy-xz-yz=1$$ $$2x^2+2y^2+2z^2-2xy-2xz-2yz=2$$ $$(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)=2$$ $$(x-y)^2+(y-z)^2+(z-x)^2=2$$ It is obvious $x$, $y$, and $z$ are symmetrical. Because $x, y, z$ are integers, $(x-y)^2$, $(y-z)^2$, and $(z-x)^2$ can only equal $0, 1, 1$. So one variable must equal another, and the third variable is $1$ different from those $2$ equal variables. Therefore, the answer is D.

A square with side length $3$ is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length $2$ has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?

$(\textbf{A})\: 19\dfrac14\qquad(\textbf{B}) \: 20\dfrac14\qquad(\textbf{C}) \: 21 \dfrac34\qquad(\textbf{D}) \: 22\dfrac12\qquad(\textbf{E}) \: 23\dfrac34$

$\textbf{B}$

We see 1 triangle on the top, 2 triangles at the waist, and 2 triangles at the bottom. The triangle at the waist has a base of 0.5 and a height of 2. The triangle at the bottom is similar to the triangle at the waist. Since the height of the bottom triangle is 3, its base is 0.75. Now we know the base of the largest triangle is $0.75+3+0.75=4.5$. By similarity, the height of the largest triangle is $\dfrac{2.25}{0.75}\cdot3=9$. So the area is $\dfrac12\cdot4.5\cdot9=20.25$.

Una rolls $6$ standard $6$-sided dice simultaneously and calculates the product of the $6{ }$ numbers obtained. What is the probability that the product is divisible by $4?$

$\textbf{(A)}\: \dfrac34\qquad\textbf{(B)} \: \dfrac{57}{64}\qquad\textbf{(C)} \: \dfrac{59}{64}\qquad\textbf{(D)} \: \dfrac{187}{192}\qquad\textbf{(E)} \: \dfrac{63}{64}$

$\textbf{C}$

We will use complementary counting to find the probability that the product is not divisible by $4$. Then, we can find the probability that we want by subtracting this from 1. We split this into $2$ cases.

Case 1: The product is not divisible by $2$. We need every number to be odd, and since the chance we roll an odd number is $\dfrac12,$ our probability is $\left(\dfrac12\right)^6=\dfrac1{64}.$

Case 2: The product is divisible by $2$, but not by $4$. We need $5$ numbers to be odd, and one to be divisible by $2$, but not by $4$. There is a $\dfrac12$ chance that an odd number is rolled, a $\dfrac13$ chance that we roll a number satisfying the second condition (only $2$ and $6$ work), and $6$ ways to choose the order in which the even number appears. Our probability is $\left(\dfrac12\right)^5\left(\dfrac13\right)\cdot6=\dfrac1{16}.$

Therefore, the probability the product is not divisible by $4$ is $\dfrac1{64}+\dfrac1{16}=\dfrac{5}{64}$.

Our answer is $1-\dfrac{5}{64}= \dfrac{59}{64}$.

In square $ABCD$, points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$, respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$, with $BR=6$ and $PR=7$. What is the area of the square?

$(\textbf{A})\: 85\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 100\qquad(\textbf{D}) \: 117\qquad(\textbf{E}) \: 125$

$\textbf{D}$

We see that $\angle PAB=\angle QBC=90^\circ$, $AB=BC$, and $\angle ABP=\angle BCQ$. So $\triangle APB \cong \triangle BQC.$ Thus, $QC = PB = PR + RB = 7 + 6 = 13.$ Let $x$ be the length of side $CR,$ then $RQ = 13-x.$ Because $\overline{BR}$ is the altitude of $\triangle BCQ$, we can use the property that $QR \cdot RC = BR^2.$ Substituting the given lengths, we have\[(13-x) \cdot x = 36.\] which gives $x = 4$ and $x = 9.$ Since $CR=\sqrt{BC^2-BR^2}$ and $QR=\sqrt{BQ^2-BR^2}$, We eliminate the possibility of $x=4$ because $RC > QR.$ Thus, the side length of the square, by Pythagorean Theorem, is\[\sqrt{9^2 +6^2} == \sqrt{117}.\]Hence, the area of the square is $(\sqrt{117})^2 = 117.$

Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?

$(\textbf{A})\: 1.6\qquad(\textbf{B}) \: 1.8\qquad(\textbf{C}) \: 2.0\qquad(\textbf{D}) \: 2.2\qquad(\textbf{E}) \: 2.4$

$\textbf{D}$

After the first swap, we do casework on the next swap.

Case 1: Silva swaps the two balls that were just swapped. There is only one way for Silva to do this, and it leaves 5 balls occupying their original position.

Case 2: Silva swaps one ball that has just been swapped with one that hasn't swapped. There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions.

Case 3: Silva swaps two balls that have not been swapped. There are two ways for Silva to do this, and it leaves 1 ball occupying their original positions.

So the expected number of balls that occupy their original positions is $\dfrac15\cdot5+\dfrac25\cdot2+\dfrac25\cdot1=2.2$.

Distinct lines $\ell$ and $m$ lie in the $xy$-plane. They intersect at the origin. Point $P(-1, 4)$ is reflected about line $\ell$ to point $P'$, and then $P'$ is reflected about line $m$ to point $P''$. The equation of line $\ell$ is $5x - y = 0$, and the coordinates of $P''$ are $(4,1)$. What is the equation of line $m?$

$(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad$ $(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0$

$\textbf{D}$

We know that the equation of line $\ell$ is $y = 5x$. Line $PP'$ is perpendicular to line $\ell$, so it has slope $-\dfrac{1}{5}$. Then the equation of this perpendicular line is $y = -\dfrac{1}{5}x + c$, and plugging in $P(-1,4)$ for $x$ and $y$ yields $c = \dfrac{19}{5}$.

The midpoint of $P'$ and $P$ lies at the intersection of $y = 5x$ and $y = -\dfrac{1}{5}x + \dfrac{19}{5}$. Solving, we get the intersection coordinate $\left(\dfrac{19}{26},\dfrac{95}{26}\right)$. Let the coordinate of $P'$ be $\left(x',y'\right)$. Then by the midpoint formula, $\dfrac{x' - 1}{2} = \dfrac{19}{26} \implies x' = \dfrac{32}{13}$. We can find the y-value of $P'$ the same way, so $P' = (\dfrac{32}{13},\dfrac{43}{13})$.

Now we have to reflect $P'$ over $m$ to get to $(4,1)$. The midpoint of $P'$ and $P''$ will lie on $m$, and this midpoint is, by the midpoint formula, $(\dfrac{42}{13},\dfrac{28}{13})$. $y = mx$ must satisfy this point, so $m = \dfrac{28}{13}\div\dfrac{42}{13} =\dfrac{2}{3}$.

Hence, the equation of line $m$ is $y = \dfrac{2}{3}x \implies 2x-3y = 0 $.

Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$-sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c$?

$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

$\textbf{E}$

Let $A$ be the center of the figure, and $B$, $C$, $D$ be the vertices on the sides. The $24$-sided polygon is made out of $24$ shapes like $\triangle ABC$. $\angle BAC=360^\circ/24=15^\circ$, $\angle EAC = 45^\circ$, so $\angle{EAB} = 30^{\circ}$. Then $EB=AE\tan 30^\circ = \sqrt{3}$; therefore $BC=EC-EB=3-\sqrt{3}$. So the area of $\triangle ABC$ is\[\dfrac12AE\cdot BC=\dfrac12(3)(3-\sqrt3)=\dfrac32(3-\sqrt3)\]and the required area is $24\cdot\dfrac32(3-\sqrt3) =108-36\sqrt{3}$. Hence, $a+b+c=108+36+3=147$.

Let $N$ be the positive integer $7777\ldots777$, a $313$-digit number where each digit is a $7$. Let $f(r)$ be the leading digit of the $r{ }$th root of $N$. What is

$$f(2) + f(3) + f(4) + f(5)+ f(6)?$$

$(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$

$\textbf{A}$

We can rewrite $N$ as $\dfrac{7}{9}\cdot 9999\ldots999 = \dfrac{7}{9}\cdot(10^{313}-1)$. When approximating values, as we will shortly do, the minus one will become negligible so we can ignore it. When we take the power of ten out of the square root, we’ll be multiplying by another power of ten, so the leading digit will not change. Thus the leading digit of $f(r)$ will be equal to the leading digit of $\sqrt[r]{\dfrac{7}{9} \cdot 10^{313\pmod r}}$.

Then $f(2)$ is the first digit of $\sqrt{\dfrac{7}{9}\cdot(10)} = \sqrt{\dfrac{70}{9}} = \sqrt{7.\ldots} \approx 2$

$f(3): \sqrt[3]{\dfrac{7}{9} \cdot 10} = \sqrt[3]{\dfrac{70}{9}} = \sqrt[3]{7.\ldots} \approx 1$.

$f(4): \sqrt[4]{\dfrac{7}{9} \cdot 10} = \sqrt[4]{\dfrac{70}{9}} = \sqrt[4]{7.\ldots} \approx 1$.

$f(5): \sqrt[5]{\dfrac{7}{9} \cdot 1000} = \sqrt[5]{\dfrac{7000}{9}} = \sqrt[5]{777.\ldots} \approx 3$.

$f(6): \sqrt[6]{\dfrac{7}{9} \cdot 10} = \sqrt[6]{\dfrac{70}{9}} = \sqrt[6]{7.\ldots} \approx 1$.

The final answer is therefore $2+1+1+3+1 =8$.

In a particular game, each of $4$ players rolls a standard $6{ }$-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo's first roll was a $5,$ given that he won the game?

$(\textbf{A})\: \dfrac{61}{216}\qquad(\textbf{B}) \: \dfrac{367}{1296}\qquad(\textbf{C}) \: \dfrac{41}{144}\qquad(\textbf{D}) \: \dfrac{185}{648}\qquad(\textbf{E}) \: \dfrac{11}{36}$

$\textbf{C}$

The conditional probability formula states that $P(A|B) = \dfrac{P(A\cap B)}{P(B)}$, where $A|B$ means A given B and $A\cap B$ means A and B. Therefore the probability that Hugo rolls a five given he won is $\dfrac{P(A \cap B)}{P(B)}$, where A is the probability that he rolls a five and B is the probability that he wins. In written form,\[\text{P(Hugo rolled a 5 given he won)}=\frac{\text{P(Hugo rolls a 5 and wins)}}{\text{P(Hugo wins)}}.\]

The probability that Hugo wins is $\dfrac{1}{4}$ by symmetry since there are four people playing and there is no bias for any one player. The probability that he gets a 5 and wins is more difficult; we will have to consider cases on how many players tie with Hugo...

$\textbf{Case 1:}$ No Players Tie

In this case, all other players must have numbers from 1 through four.

There is a $\left(\dfrac{4}{6}\right)^{3} = \dfrac{8}{27}$ chance of this happening.

$\textbf{Case 2:}$ One Player Ties

In this case, there are $\dbinom31$ $= 3$ ways to choose which other player ties with Hugo, and the probability that this happens is $\dfrac{1}{6} \cdot \left(\dfrac{4}{6}\right)^2$. The probability that Hugo wins on his next round is then $\dfrac{1}{2}$ because there are now two players rolling die.

Therefore the total probability in this case is $3 \cdot \dfrac{1}{2} \cdot \dfrac{1}{6} \cdot \left(\dfrac{4}{6}\right)^2= \dfrac{1}{9}$.

$\textbf{Case 3:}$ Two Players Tie

In this case, there are $\dbinom32$ $= 3$ ways to choose which other players tie with Hugo, and the probability that this happens is $\left(\dfrac{1}{6}\right)^2 \cdot \dfrac{4}{6}$. The probability that Hugo wins on his next round is then $\dfrac{1}{3}$ because there are now three players rolling the die.

Therefore the total probability in this case is $3 \cdot \dfrac{1}{3} \cdot \left(\dfrac{1}{6}\right)^2 \cdot \dfrac{4}{6} = \dfrac{1}{54}$.

$\textbf{Case 4:}$ All Three Players Tie

In this case, the probability that all three players tie with Hugo is $\left(\dfrac{1}{6}\right)^3$. The probability that Hugo wins on the next round is $\dfrac{1}{4}$, so the total probability is $\dfrac{1}{4} \cdot \left(\dfrac{1}{6}\right)^3 = \dfrac{1}{864}$.

Finally, Hugo has a $\dfrac{1}{6}$ probability of rolling a five himself, so the total probability is\[\frac{1}{6}\left(\frac{8}{27} + \frac{1}{9} + \frac{1}{54} + \frac{1}{864}\right) = \frac{1}{6}\left(\frac{369}{864}\right) = \frac{1}{6}\left(\frac{41}{96}\right)\]Finally, the total probability is this probability divided by $\dfrac{1}{4}$. the final answer is \[4 \cdot \frac{1}{6}\left(\frac{41}{96}\right) = \frac{2}{3} \cdot \frac{41}{96} = \frac{41}{48 \cdot 3} = \frac{41}{144}\]

Regular polygons with $5$, $6$, $7$, and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?

$(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68$

$\textbf{E}$

Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice. (We can see this because to have a vertex of the $m$-gon on an arc subtended by a side of the $n$-gon, there will be one intersection to “enter” the arc and one to “exit” the arc.)

This means that we will end up with $2$ times the number of sides in the polygon with fewer sides.

If we have polygons with $5,$ $6,$ $7,$ and $8$ sides, we need to consider each possible pair of polygons and count their intersections.

Throughout $6$ of these pairs, the $5$-sided polygon has the least number of sides $3$ times, the $6$-sided polygon has the least number of sides $2$ times, and the $7$-sided polygon has the least number of sides $1$ time.

Therefore the number of intersections is $2\cdot(3\cdot5+2\cdot6+1\cdot7)=68$.

For each integer $n\geq 2$, let $S_n$ be the sum of all products $jk$, where $j$ and $k$ are integers and $1\leq j<k\leq n$. What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$?

$\textbf{(A)}\ 196 \qquad\textbf{(B)}\ 197 \qquad\textbf{(C)}\ 198 \qquad\textbf{(D)}\ 199 \qquad\textbf{(E)}\ 200$

$\textbf{B}$

To get from $S_n$ to $S_{n+1}$, we add $1(n+1)+2(n+1)+\cdots +n(n+1)=(1+2+\cdots +n)(n+1)=\dfrac{n(n+1)^2}{2}$.

Now, we can look at the different values of $n$ mod $3$. For $n\equiv 0\pmod{3}$ and $n\equiv 2\pmod{3}$, we have $\dfrac{n(n+1)^2}{2}\equiv 0\pmod{3}$. However, for $n\equiv 1\pmod{3}$, we have $\dfrac{1\cdot {2}^2}{2}\equiv 2\pmod{3}.$

Clearly, $S_2\equiv 2\pmod{3}.$ Using the above result, we have

\begin{align*}

S_2\equiv 2\pmod{3}\\

S_3\equiv 2\pmod{3}\\

S_4\equiv 2\pmod{3}\\

S_5\equiv 1\pmod{3}\\

S_6\equiv 1\pmod{3}\\

S_7\equiv 1\pmod{3}\\

S_8\equiv 0\pmod{3}\\

S_9\equiv 0\pmod{3}\\

S_{10}\equiv 0\pmod{3}

\end{align*}

We find a period of 3. After $3\cdot 3=9$, we have $S_{17}$, $S_{18}$, and $S_{19}$ all divisible by $3$, as well as $S_{26}, S_{27}, S_{28}$, and $S_{35}$. Thus, our answer is $8+9+10+17+18+19+26+27+28+35=197$.

Each of the $5$ sides and the $5$ diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?

$(\textbf{A})\: \dfrac23\qquad(\textbf{B}) \: \dfrac{105}{128}\qquad(\textbf{C}) \: \dfrac{125}{128}\qquad(\textbf{D}) \: \dfrac{253}{256}\qquad(\textbf{E}) \: 1$

$\textbf{D}$

Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear. After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.

$\textbf{Case 1}$: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection.

$\textbf{Case 2}$: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.

$\textbf{Case 3}$: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are $\dbinom42$ ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of $6\cdot 2 = 12$ ways to color the pentagon so that no such triangle has the same color for all of its sides.

These are all the cases, and there are a total of $2^{10}$ ways to color the pentagon. Therefore the answer is $1-\dfrac{12}{1024} = 1-\dfrac{3}{256} = \dfrac{253}{256}$.

A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)

$(\textbf{A})\: 7\qquad(\textbf{B}) \: 8\qquad(\textbf{C}) \: 9\qquad(\textbf{D}) \: 10\qquad(\textbf{E}) \: 11$

$\textbf{A}$

This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory. We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube.

Now the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence. For example, in Figure $(1)$, as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube.

Here is how the $4$ blue unit cubes are arranged:

4 blue unit cubes in a plane: Figure (1).

3 blue unit cubes in a plane: Figure (2) to (5). Note that Figure (3) and (4) are not equivalent because one can not be achieved by rotating another one.

2 blue unit cubes in a plane: Figure (6) and (7).

So the answer is $7$.

A rectangle with side lengths $1{ }$ and $3,$ a square with side length $1,$ and a rectangle $R$ are inscribed inside a larger square as shown. The sum of all possible values for the area of $R$ can be written in the form $\dfrac mn$, where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$(\textbf{A})\: 14\qquad(\textbf{B}) \: 23\qquad(\textbf{C}) \: 46\qquad(\textbf{D}) \: 59\qquad(\textbf{E}) \: 67$

$\textbf{E}$

We see that the polygon bounded by the small square, large square, and rectangle of known lengths is an isosceles triangle. Let’s draw a perpendicular from the vertex of this triangle to its opposing side;

We see that this creates two congruent triangles. Let the smaller side of the triangle have length $a$ and let the larger side of the triangle have length $b$. Now we see by AAS congruency that if we draw perpendiculars that surround the smaller square, each outer triangle will be congruent to these two triangles.

Now notice that these small triangles are also similar to the large triangle bounded by the bigger square and the rectangle by AA, and the ratio of the sides are 1:3, so we can fill in the lengths of that triangle. Similarly, the small triangle on the right bounded by the rectangle and the square is also congruent to the other small triangles by AAS, so we can fill in those sides;

Since the larger square by definition has all equal sides, we can set the sum of the lengths of the sides equal to each other. $3a+b+b+a = 3b+a \implies 3a = b$. Now let's draw some more perpendiculars and rename the side lengths.

By AA similarity, when we draw a perpendicular from the intersection of the two rectangles to the large square, we create a triangle below that is similar to the small congruent triangles with length $a,3a$. Since we don't know its scale, we'll label its sides $c,3c$.

The triangle that is created above the perpendicular is congruent to the triangle on the opposite of the rectangle with unknown dimensions because they share the same hypotenuse and have two angles in common. Thus we can label these two triangles accordingly.

The side length of the big square is $10a$, so we can find the remaining dimensions of the triangle bounded by the rectangle with unknown dimensions and the large square in terms of $a$ and $c$:

This triangle with side lengths $4a-c$ and $6a-3c$ is similar to the triangle directly below it with side lengths $3a$ and $3c$ by AA similarity, so we can set up a ratio equation: $$\frac{3a}{3c} = \frac{6a-3c}{4a-c} \implies 4a^2-ac = -3c^2 + 6ac \implies 4a^2 - 7ac + 3c^2 = 0 \implies (4a-3c)(a-c) = 0$$ There are two solutions to this equation; $c = \dfrac{4}{3}a$ and $c = a$. For the first solution, the triangle in the corner has sides $2a$ and $\dfrac{8}{3}a$. Using Pythagorean theorem on that triangle, the hypotenuse has length $\dfrac{10}{3}a$. The triangle directly below has side lengths $3a$ and $4a$ in this case, so special right triangle yields the hypotenuse to be $5a$. The area of the rectangle is thus $5a\cdot\dfrac{10}{3}a = \dfrac{50}{3}a^2$. For the second solution, the side lengths of the corner triangle are $3a$ and $3a$, so the hypotenuse of the triangle is $3\sqrt{2}a$. The triangle below that also has side lengths $3a$ and $3a$, so its hypotenuse is the same. Then the area of the rectangle is $(3\sqrt{2}a)^2 = 18a^2$.

The sum of the possible areas of the rectangle is therefore $18a^2+\dfrac{50}{3}a^2 = \dfrac{104}{3}a^2$.

Using Pythagorean theorem on the original small congruent triangles, $a^2+9a^2 = 1$ or $a^2 = \dfrac{1}{10}$. Therefore the sum of the possible areas of the rectangle is $\dfrac{104}{3}\cdot\dfrac{1}{10} = \dfrac{52}{15}$. Therefore $m = 52$, $n = 15$, and $m + n = 67$.