## AMC 12 2021 Spring Test A

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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\[2^{1+2+3}-(2^1+2^2+2^3)?\]

$\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57$

$\textbf{B}$

\[2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \]

Under what conditions is $\sqrt{a^2+b^2}=a+b$ true, where $a$ and $b$ are real numbers?

$\textbf{(A) }$ It is never true.

$\textbf{(B) }$ It is true if and only if $ab=0$.

$\textbf{(C) }$ It is true if and only if $a+b\ge 0$.

$\textbf{(D) }$ It is true if and only if $ab=0$ and $a+b\ge 0$.

$\textbf{(E) }$ It is always true.

$\textbf{D}$

According to the given information, we get $a^2+b^2=(a+b)^2\rightarrow ab=0$. So the answer is D.

The sum of two natural numbers is $17{,}402$. One of the two numbers is divisible by $10$. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?

$\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426$

$\textbf{D}$

The units digit of a multiple of $10$ will always be $0$. We add a $0$ whenever we multiply by $10$. So, removing the units digit is equal to dividing by $10$.

Let the smaller number (the one we get after removing the units digit) be $a$. This means the bigger number would be $10a$.

We know the sum is $10a+a = 11a$ so $11a=17402$. Hence, we get $a=1582$. The difference is $10a-a = 9a=9(1582) =14{,}238$.

Tom has a collection of $13$ snakes, $4$ of which are purple and $5$ of which are happy. He observes that

$\quad\bullet$ all of his happy snakes can add,

$\quad\bullet$ none of his purple snakes can subtract, and

$\quad\bullet$ all of his snakes that can't subtract also can't add.

Which of these conclusions can be drawn about Tom's snakes?

$\textbf{(A) }$ Purple snakes can add.

$\textbf{(B) }$ Purple snakes are happy.

$\textbf{(C) }$ Snakes that can add are purple.

$\textbf{(D) }$ Happy snakes are not purple.

$\textbf{(E) }$ Happy snakes can't subtract.

$\textbf{D}$

According to statements 2 and 3, all purple snakes can neither subtract nor add. Given that all happy snakes can add, we know that happy snakes are not purple.

When a student multiplied the number $66$ by the repeating decimal,\[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\]where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$-digit number $\underline{a} \ \underline{b}?$

$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$

$\textbf{E}$

It is known that $\underline{0}.\overline{\underline{a} \ \underline{b}}=\dfrac{\underline{a} \ \underline{b}}{99}$ and $\underline{0}.\underline{a} \ \underline{b}=\dfrac{\underline{a} \ \underline{b}}{100}.$

Let $x=\underline{a} \ \underline{b}.$ We have\[66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5.\]Expanding and simplifying give $\dfrac{x}{150}=0.5,$ so $x=75.$

A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\dfrac13$. When $4$ black cards are added to the deck, the probability of choosing red becomes $\dfrac14$. How many cards were in the deck originally?

$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$

$\textbf{C}$

Let the original number of cards be $x$. The number of red cards is $\dfrac13x=\dfrac14(x+4)\rightarrow x=12$.

What is the least possible value of $(xy-1)^2+(x+y)^2$ for all real numbers $x$ and $y?$

$\textbf{(A) }0 \qquad \textbf{(B) }\dfrac14 \qquad \textbf{(C) }\dfrac12 \qquad \textbf{(D) }1 \qquad \textbf{(E) }2$

$\textbf{D}$

We expand the original expression, then factor the result by grouping:\begin{align*} (xy-1)^2+(x+y)^2&=\left(x^2y^2-2xy+1\right)+\left(x^2+2xy+y^2\right) \\ &=x^2y^2+x^2+y^2+1 \\ &=x^2\left(y^2+1\right)+\left(y^2+1\right) \\ &=\left(x^2+1\right)\left(y^2+1\right) \end{align*}Clearly, both factors are positive. Hence, we have\[\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=1\]Note that the least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$

A sequence of numbers is defined by $D_0=0,D_1=0,D_2=1$ and $D_n=D_{n-1}+D_{n-3}$ for $n\ge 3$. What are the parities (evenness or oddness) of the triple of numbers $(D_{2021},D_{2022},D_{2023})$, where $E$ denotes even and $O$ denotes odd?

$\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)$

$\textbf{C}$

We construct the following table:

Note that $(D_7,D_8,D_9)$ and $(D_0,D_1,D_2)$ have the same parities, so the parity is periodic with period $7.$ Since the remainders of $(2021\div7,2022\div7,2023\div7)$ are $(5,6,7),$ we conclude that $(D_{2021},D_{2022},D_{2023})$ and $(D_5,D_6,D_7)$ have the same parities, namely $(E,O,E).$

Which of the following is equivalent to\[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\]

$\textbf{(A) }3^{127}+2^{127} \qquad \textbf{(B) }3^{127}+2^{127}+2\cdot 3^{63}+3\cdot 2^{63} \qquad \textbf{(C) }3^{128}-2^{128} \qquad \textbf{(D) }3^{128}+2^{128} \qquad \textbf{(E) }5^{127}$

$\textbf{C}$

By multiplying the entire equation by $3-2=1$, all the terms will simplify by difference of squares, and the final answer is $3^{128}-2^{128}$.

Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$

$\textbf{E}$

Let the heights of the narrow cone and the wide cone be $h_1$ and $h_2,$ respectively. The volume of the liquid in the narrow cone is $\dfrac13\pi(3)^2h_1=3\pi h_1$. The volume of the liquid in the wide cone is $\dfrac13\pi(6)^2h_2=12\pi h_2$. Since the volume of liquid is the same, we have $$3\pi h_1=12\pi h_2$$ After dropping the marble, let the rises of the liquid levels in the narrow cone and the wide cone be $\Delta h_1$ and $\Delta h_2$. Now the volume is $$3\pi(h_1+\Delta h_1)=12\pi(h_2+\Delta h_2)$$ Now we have \begin{align*}3\pi\Delta h_1&=12\pi\Delta h_2\\ h_1:h_2&=4:1\end{align*}

A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path?

$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5$

$\textbf{C}$

When the object located at point $(3,5)$ is reflected by the y-axis, the image locates at $(-3,5)$. Then the image is reflected ny the x-axis, the new image locates at $(-3,-5)$. The total optical distance is the distance between $(-3,-5)$ and $(7,5)$. So the answer is $\sqrt{10^2+10^2}=10\sqrt2$.

All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$?

$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$

$\textbf{A}$

By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$. By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$, so the function is $$(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$$ Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = -88$.

Of the following complex numbers $z$, which one has the property that $z^5$ has the greatest real part?

$\textbf{(A) }{-}2 \qquad \textbf{(B) }{-}\sqrt3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i$

$\textbf{B}$

We rewrite each answer choice to the polar form\[z=r\operatorname{cis}\theta=r(\cos\theta+i\sin\theta),\]where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$

By De Moivre's Theorem, the real part of $z^5$ is\[\operatorname{Re}\left(z^5\right)=r^5\cos{(5\theta)}.\]We construct a table as follows:

Clearly, the answer is B.

What is the value of\[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?\]

$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2{,}200\qquad \textbf{(E) }21{,}000$

$\textbf{E}$

We can simplify the expressions inside the summations: $$\log_{5^k}{3^{k^2}}=\dfrac{k^2}{k}\log_53=k\log_53$$ $$\log_{9^k}25^k=\log_{3^{2k}}5^{2k}=\dfrac{2k}{2k}\log_35=\log_{3}{5}$$ So the original expression is \begin{align*} \left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\cdot\left(\sum_{k=1}^{100} \log_{3}{5}\right) \\ &= \left(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\cdot\left(\log_{3}{5}\cdot\sum_{k=1}^{100} 1\right) \\ &= \left(\sum_{k=1}^{20} k\right)\cdot\left(\sum_{k=1}^{100} 1\right) \\ &= \frac{21\cdot20}{2}\cdot100 \\ &= 21{,}000. \end{align*}

A choir director must select a group of singers from among his $6$ tenors and $8$ basses. The only requirements are that the difference between the numbers of tenors and basses must be a multiple of $4$, and the group must have at least one singer. Let $N$ be the number of different groups that could be selected. What is the remainder when $N$ is divided by $100$?

$\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad$

$\textbf{D}$

Suppose that we choose $a$ tensors and $b$ basses. $a-b$ is a multiple of 4. So $a-b=4, 0, -4, \text{or} -8$.

Case 1: $a-b=4$. Now $(a,b)$ could be $(4,0)$, $(5,1)$ or $(6,2)$. The number of arrangements is $$\dbinom64\dbinom80+\dbinom65\dbinom81+\dbinom66\dbinom82=91$$ Case 2: $a-b=0$. Now $(a,b)$ could be $(1,1)$, $(2,2)$, $(3,3)$, $(4,4)$, $(5,5)$ or $(6,6)$. The number of arrangements is $$\dbinom61\dbinom81+\dbinom62\dbinom82+\dbinom63\dbinom83+\dbinom64\dbinom84+\dbinom65\dbinom85+\dbinom66\dbinom86=3002$$ Case 3: $a-b=-4$. Now $(a.b)$ could be $(0,4)$, $(1,5)$, $(2,6)$, $(3,7)$ or $(4,8)$. The number of arrangements is $$\dbinom60\dbinom84+\dbinom61\dbinom85+\dbinom62\dbinom86+\dbinom63\dbinom87+\dbinom64\dbinom88=1001$$ Case 4: $a-b=-8$. Now $(a,b)$ is $(0,8)$. The number of arrangements is $$\dbinom60\dbinom88=1$$ The total number of different groups is $N=91+3002+1001+1=4095$. The reminder when $N$ is divided by 100 is 95.

In the following list of numbers, the integer $n$ appears $n$ times in the list for $1\le n \le 200$.\[1,2,2,3,3,3,4,4,4,...,200,200,...,200\]What is the median of the numbers in this list?

$\textbf{(A) }100.5 \qquad \textbf{(B) }134 \qquad \textbf{(C) }142 \qquad \textbf{(D) }150.5\qquad \textbf{(E) }167$

$\textbf{C}$

There are $1+2+\dots+199+200=\dfrac{(200)(201)}{2}=20100$ numbers in total. Let the median be $k$. We want to find the median $k$ such that\[\dfrac{k(k+1)}{2}=\dfrac{20100}2,\]or\[k(k+1)=20100.\]Note that $\sqrt{20100}\approx100\sqrt2 \approx 142$. Plugging this value in as $k$ gives\[\frac{1}{2}(142)(143)=10153.\]$10153-142<10050$, so $142$ is the $10050$nd and $10051$rd numbers. Hence, the median is 142.

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?

$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$

$\textbf{D}$

Since $BC=CD$ and $P$ is the midpoint of $BD$, we know that $CP\perp BD$. Given that $AP\parallel CD$, we find that $\triangle ABD\sim\triangle CDP$. So $\dfrac{AD}{CP}=\dfrac{BD}{DP}=2$. We also found that $\triangle AOD\sim\triangle COP$, so $\dfrac{OD}{OP}=\dfrac{AD}{CP}=2\rightarrow OD=2OP=22$. By the Pythagorean theorem, $CP=\sqrt{CD^2-DP^2}=\sqrt{43^2-(11+22)^2}=2\sqrt{190}$. So $AD=2CP=4\sqrt{190}$. The answer is $m+n=4+190=194$.

Let $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b)=f(a)+f(b)$ for all positive rational numbers $a$ and $b$. Suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$. For which of the following numbers $x$ is $f(x)<0$?

$\textbf{(A) }\dfrac{17}{32} \qquad \textbf{(B) }\dfrac{11}{16} \qquad \textbf{(C) }\dfrac79 \qquad \textbf{(D) }\dfrac76\qquad \textbf{(E) }\dfrac{25}{11}$

$\textbf{E}$

We know that $f(p) = f(p \cdot 1) = f(p) + f(1)$. By transitive, we have\[f(p) = f(p) + f(1).\]Subtracting $f(p)$ from both sides gives $0 = f(1).$ Also\[f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2\]\[f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3\]\[f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11\]In $\textbf{(A)}$ we have $f\left(\dfrac{17}{32}\right)=17+5f\left(\dfrac{1}{2}\right)=17-5(2)=7$.

In $\textbf{(B)}$ we have $f\left(\dfrac{11}{16}\right)=11+4f\left(\dfrac{1}{2}\right)=11-4(2)=3$.

In $\textbf{(C)}$ we have $f\left(\dfrac{7}{9}\right)=7+2f\left(\dfrac{1}{3}\right)=7-2(3)=1$.

In $\textbf{(D)}$ we have $f\left(\dfrac{7}{6}\right)=7+f\left(\dfrac{1}{2}\right)+f\left(\dfrac{1}{3}\right)=7-2-3=2$.

In $\textbf{(E)}$ we have $f\left(\dfrac{25}{11}\right)=10+f\left(\dfrac{1}{11}\right)=10-11=-1$.

Thus, our answer is $\textbf{(E) }$.

How many solutions does the equation $\sin \left( \dfrac{\pi}2 \cos x\right)=\cos \left( \dfrac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$?

$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$

$\textbf{C}$

By the Cofunction Identity $\cos\theta=\sin\left(\dfrac{\pi}{2}-\theta\right),$ we rewrite the given equation:\[\sin \left(\frac{\pi}2 \cos x\right) = \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right)\] When the value of $x$ varies $0\rightarrow\dfrac{\pi}{2}\rightarrow\pi$, the value of $\dfrac{\pi}{2}\cos x$ varies $\dfrac{\pi}{2}\rightarrow0\rightarrow-\dfrac{\pi}{2}$, and the value of $\dfrac{\pi}{2}\left(1-\sin x\right)$ varies $\dfrac{\pi}{2}\rightarrow0\rightarrow\dfrac{\pi}{2}$. Recall that if $\sin\theta=\sin\phi,$ then $\theta=\phi+2n\pi$ or $\theta=\pi-\phi+2n\pi$ for some integer $n.$ Therefore, we have two cases:

$\textbf{Case 1}$: when $x\in\left[0,\dfrac{\pi}{2}\right]$, $\dfrac{\pi}{2}\cos x=\dfrac{\pi}{2}\left(1-\sin x\right)$

Here we have $\cos x+\sin x=1$. This only happens at $x=0$ and $x=\dfrac{\pi}{2}$.

$\textbf{Case 2}$: $x\in\left(\dfrac{\pi}{2},1\right]$

Now $\dfrac{\pi}{2}\cos x\in\left(0,-\dfrac{\pi}{2}\right]$, $\dfrac{\pi}{2}\left(1-\sin x\right)\in\left(0,\dfrac{\pi}{2}\right]$. So $\sin\left(\dfrac{\pi}{2}\cos x\right)<0$, $\sin\left(\dfrac{\pi}{2}-\dfrac{\pi}{2}\sin x\right)>0$. They can't be equal.

In conclusion, the answer is 2.

Suppose that on a parabola with vertex $V$ and a focus $F$ there exists a point $A$ such that $AF=20$ and $AV=21$. What is the sum of all possible values of the length $FV?$

$\textbf{(A) }13 \qquad \textbf{(B) }\dfrac{40}3 \qquad \textbf{(C) }\dfrac{41}3 \qquad \textbf{(D) }14\qquad \textbf{(E) }\dfrac{43}3$

$\textbf{B}$

Let $\mathcal{P}$ be the parabola, and $V$ be the origin. $F$ lies on the positive $y$ axis, and $d=FV$. The equation of the parabola is then $x^{2}=4dy$. If the coordinates of $A$ are $(p, q),$ we have $$p^2=4dq$$ then $$p^{2}+q^{2}=441$$ since the distance from the origin to $A$ is $21$. Note also that the parabola is the set of all points equidistant from $F$ and a line known as its directrix, which in this case is a horizontal line $d$ units below the origin. Since the distance from $A$ to its directrix is equal to $AF,$ then $A$ is $20$ units above this line and therefore $q=20-d$. Substituting for $p$ and $q$ yields $$4d(20-d)+(20-d)^{2}=441$$which simplifies to $$3d^{2}-40d+41=0$$ Therefore the sum of all possible values of $d$ is $-\dfrac{b}{a}=\dfrac{40}{3}$ by Viète's Formulas.

The five solutions to the equation\[(z-1)(z^2+2z+4)(z^2+4z+6)=0\]may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$. The eccentricity of $\mathcal E$ can be written in the form $\sqrt{\dfrac mn}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\dfrac ca$, where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)

$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$

$\textbf{A}$

Completing the square in the original equation, we have\[(z-1)\left((z+1)^2+3\right)\left((z+2)^2+2\right)=0,\]from which $z=1,-1\pm\sqrt{3}i,-2\pm\sqrt{2}i.$

Now, we will find the equation of an ellipse $\mathcal E$ that passes through $(1,0),\left(-1,\pm\sqrt3\right),$ and $\left(-2,\pm\sqrt2\right)$ in the $xy$-plane. Note that since these five points are symmetric about the $x$-axis, so must $\mathcal E$.

The formula of $\mathcal E$ is\[\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1 \]with the center $(h,0)$ and the axes' lengths $2a$ and $2b.$

Plugging the points $(1,0),\left(-1,\sqrt3\right),$ and $\left(-2,\sqrt2\right)$ into, respectively, we have the following system of equations:\begin{align} \frac{(1-h)^2}{a^2}&=1 \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1 \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1 \end{align} Eliminating the variable $b$ from (2) and (3), we have \begin{align} \dfrac{3(h+2)^2}{a^2}-\dfrac{2(h+1)^2}{a^2}=1 \end{align} Eliminating the variable $a$ from (1) and (4), we have \begin{align} 3(h+2)^2-2(h+1)^2=(h-1)^2 \end{align} which can be simplified as $$h^2+8h+10=h^2-2h+1$$ so $h=-\dfrac{9}{10}$. Therefore, $a^2=\dfrac{361}{100}$, $b^2=\dfrac{361}{120}$, $c^2=a^2-b^2=\dfrac{361}{600}$. The eccentricity is $$\dfrac{c}{a}=\sqrt{\dfrac{c^2}{a^2}}=\sqrt{\dfrac16}$$ Hence, the answer is $m+n=1+6=7$.

Suppose that the roots of the polynomial $P(x)=x^3+ax^2+bx+c$ are $\cos \dfrac{2\pi}7,\cos \dfrac{4\pi}7,$ and $\cos \dfrac{6\pi}7$, where angles are in radians. What is $abc$?

$\textbf{(A) }{-}\dfrac{3}{49} \qquad \textbf{(B) }{-}\dfrac{1}{28} \qquad \textbf{(C) }\dfrac{\sqrt[3]7}{64} \qquad \textbf{(D) }\dfrac{1}{32}\qquad \textbf{(E) }\dfrac{1}{28}$

$\textbf{D}$

The equation can be rewrote as $$\left(x-\cos\dfrac{2\pi}{7}\right)\left(x-\cos\dfrac{4\pi}{7}\right)\left(x-\cos\dfrac{6\pi}{7}\right)=0$$ We solve for $a,b,$ and $c$ separately:

$\textbf{Solve for a:}$ By Vieta's Formulas, we have $a = - \left( \cos \dfrac{2\pi}7 + \cos \dfrac{4\pi}7 + \cos \dfrac{6\pi}7 \right).$

Let $z=e^{\frac{2\pi i}{7}}.$ Since $z$ is a $7$th root of unity, we have $z^7=1.$ The real parts of the $7$th roots of unity are $1, \cos \dfrac{2\pi}7, \cos \dfrac{4\pi}7, \cos \dfrac{6\pi}7, \cos \dfrac{8\pi}7, \cos \dfrac{10\pi}7, \cos \dfrac{12\pi}7$ and they sum to $0.$

Note that $\cos\theta=\cos(2\pi-\theta)$ for all $\theta.$ Excluding $1,$ the other six roots add to\[2\left(\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7\right) = -1,\]from which\[\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 = -\frac12.\]Therefore, we get $a = -\left(-\dfrac12\right) = \dfrac12.$

$\textbf{Solve for b:}$ By Vieta's Formulas, we have $b = \cos \dfrac{2\pi}7 \cos \dfrac{4\pi}7 + \cos \dfrac{2\pi}7 \cos \dfrac{6\pi}7 + \cos \dfrac{4\pi}7 \cos \dfrac{6\pi}7.$

Note that $\cos \alpha \cos \beta = \dfrac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}$ for all $\alpha$ and $\beta.$ Therefore, we get\[b=\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2=\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7=-\frac12.\]

$\textbf{Solve for c:}$ By Vieta's Formulas, we have $c = -\cos \dfrac{2\pi}7 \cos \dfrac{4\pi}7 \cos \dfrac{8\pi}7.$

We multiply both sides by $8 \sin{\dfrac{2\pi}{7}},$ then repeatedly apply the angle addition formula for sine:\begin{align*} c \cdot 8 \sin{\frac{2\pi}{7}} &= -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7 \\ &= -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7 \\ &= -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7 \\ &= -\sin \frac{16\pi}7 \\ &= -\sin \frac{2\pi}7. \end{align*}Therefore, we get $c = -\dfrac18.$

Finally, the answer is $abc=\dfrac12\cdot\left(-\dfrac12\right)\cdot\left(-\dfrac18\right)=\dfrac{1}{32}.$

Frieda the frog begins a sequence of hops on a $3\times3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

$\textbf{(A) }\dfrac{9}{16} \qquad \textbf{(B) }\dfrac{5}{8} \qquad \textbf{(C) }\dfrac34 \qquad \textbf{(D) }\dfrac{25}{32}\qquad \textbf{(E) }\dfrac{13}{16}$

$\textbf{D}$

We can find the probability that the frog doesn't land on the corner within 4 steps, then subtracting it by 1 to get the answer.

After the first step, the frog lands on the middle edge.

After the second step, the frog has a probability of $\dfrac12$ to land on the corner, $\dfrac14$ to land on the center, and $\dfrac14$ to land on the middle edge. Here we have 2 cases to discuss.

$\textbf{Case 1:}$ the frog lands on the center with a probability of $\dfrac14$. After the third step, the frog lands on the middle edge. For the fourth step, the frog has a probability of $\dfrac12$ to not land on the corner. So in this case, the probability that the frog doesn't land on the corner within 4 steps is $\dfrac14\cdot\dfrac12=\dfrac18$.

$\textbf{Case 2:}$ the frog lands on the middle edge with a probability of $\dfrac14$. For the third step, she has a probability of $\dfrac14$ to land on the center, which means she can't land on the corner in the fourth step, or a probability of $\dfrac14$ to land on the middle edge, which means she has a probability of $\dfrac12$ to not land on the corner in the fourth step. So in this case, the probability that the frog doesn't land on the corner within 4 steps is $\dfrac14\left(\dfrac14+\dfrac14\cdot\dfrac12\right)=\dfrac{3}{32}$.

In conclusion, the probability that the frog doesn't land on the corner within 4 steps is $\dfrac18+\dfrac{3}{32}=\dfrac{7}{32}$. Our answer is $1-\dfrac{7}{32}=\dfrac{25}{32}$.

Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\dfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$?

$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$

$\textbf{D}$

Let $O=\Gamma$ be the center of the semicircle and $X=\Omega$ be the center of the circle.

Since $\angle QPR=60^\circ$, we have $\angle QXR=120^\circ$. $\triangle QXR$ is an isosceles triangle with base $QR=3\sqrt3$ and $\angle QXR=120^\circ$. So $XQ=XR=3$.

Let $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Bisector Converse, we have $\overline{OM}\perp\overline{QR}$ and $\overline{XM}\perp\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear. $RM=\dfrac{3\sqrt3}{2}, RX=3,$ and $XM=\dfrac{3}{2}.$ By the Pythagorean Theorem on right $\triangle ORM,$ we get $OM=\dfrac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on right $\triangle OXP,$ we get $OP=4.$

Let $C$ be the foot of the perpendicular from $P$ to $\overline{QR},$ and $D$ be the foot of the perpendicular from $X$ to $\overline{PC},$ as shown below:

Clearly, quadrilateral $XDCM$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $\triangle XPD\sim\triangle OXP$ by the AA Similarity, with the ratio of similitude $\dfrac{XP}{OX}=\dfrac 35.$ Therefore, we get $PD=\dfrac 95$ and $PC=PD+DC=PD+XM=\dfrac 95 + \dfrac 32 = \dfrac{33}{10}.$

The area of $\triangle PQR$ is\[\frac12\cdot QR\cdot PC=\frac12\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},\]from which the answer is $99+3+20=122.$

Let $d(n)$ denote the number of positive integers that divide $n$, including $1$ and $n$. For example, $d(1)=1,d(2)=2,$ and $d(12)=6$. (This function is known as the divisor function.) Let\[f(n)=\frac{d(n)}{\sqrt [3]n}.\]There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$. What is the sum of the digits of $N?$

$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$

$\textbf{E}$

We consider the prime factorization of $n:$\[n=\prod_{i=1}^{k}p_i^{e_i}.\]By the Multiplication Principle, we have\[d(n)=\prod_{i=1}^{k}(e_i+1).\]Now, we rewrite $f(n)$ as\[f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_i+1}{p_i^{{e_i}/3}}.\]As $f(n)>0$ for all positive integers $n,$ note that $f(a)>f(b)$ if and only if $f(a)^3>f(b)^3$ for all positive integers $a$ and $b.$ So, $f(n)$ is maximized if and only if\[f(n)^3=\prod_{i=1}^{k}\frac{(e_i+1)^3}{p_i^{{e_i}}}\]is maximized.

For each independent factor $\dfrac{(e_i+1)^3}{p_i^{e_i}}$ with a fixed prime $p_i,$ where $1\leq i\leq k,$ the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. Therefore, for each prime $p_i$ with $\left(p_1,p_2,p_3,p_4,\ldots\right)=\left(2,3,5,7,\ldots\right),$ we look for the nonnegative integer $e_i$ such that $\dfrac{(e_i+1)^3}{p_i^{e_i}}$ is a relative maximum:

Finally, the positive integer we seek is $N=2^3\cdot3^2\cdot5^1\cdot7^1=2520.$ The sum of its digits is $2+5+2+0=9.$

Alternatively, once we notice that $3^2$ is a factor of $N,$ we can conclude that the sum of the digits of $N$ must be a multiple of $9.$ Only choice $\textbf{(E)}$ is possible.