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2021 AMC12A真题与答案

Category: AMC美国数学竞赛, 国际竞赛, 翰林严选 Date: 2021年2月20日下午4:56

2021 AMC 12A 真题与答案

文末答案

Problem1

What is the value of $2^{1+2+3}-(2^1+2^2+2^3)?$ $\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57$

Problem2

Under what conditions does $\sqrt{a^2+b^2}=a+b$ hold, 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.

Problem3

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? $\textbf{(A) }10,272 \qquad \textbf{(B) }11,700 \qquad \textbf{(C) }13,362 \qquad \textbf{(D) }14,238 \qquad \textbf{(E) }15,462$

Problem4

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.

Problem5

When a student multiplied the number $66$ by the repeating decimal $\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},$ 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 $\underline{a}\underline{b}?$ $\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$

Problem6

A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$ . When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$ . How many cards were in the deck originally? $\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$

Problem7

What is the least possible value of $(xy-1)^2+(x+y)^2$ for all real numbers $x$ and $y?$ $\textbf{(A) }0 \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac12 \qquad \textbf{(D) }1 \qquad \textbf{(E) }2$

Problem8

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

Problem9

Which of the following is equivalent to $(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?$ $\textbf{(A) }3^{127}+2^{127} \qquad \textbf{(B) }3^{127}+2^{127}+2\cdot 3^{63}+3\cdot 2^{63} \qquad \textbf{(C) }3^{128}-2^{128} \qquad \textbf{(D) }3^{128}+2^{128} \qquad \textbf{(E) }5^{127}$

Problem10

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? $[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1*s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]$ $\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$

Problem11

A laser is placed at the point $(3,5)$ . The laser bean 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? $\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5$

Problem12

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

Problem13

Of the following complex numbers $z$ , which one has the property that $z^5$ has the greatest real part? $\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i$

Problem14

What is the value of $\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?$ $\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000$

Problem15

A choir direction 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 $100$ ? $\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad$

Problem16

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

Problem17

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

Problem18

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 $b$ . Furthermore, suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$ . For which of the following numbers $x$ is $f(x)<0$ ? $\textbf{(A) }\frac{17}{32} \qquad \textbf{(B) }\frac{11}{16} \qquad \textbf{(C) }\frac79 \qquad \textbf{(D) }\frac76\qquad \textbf{(E) }\frac{25}{11}$

Problem19

How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$ ? $\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$

Problem20

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

Problem21

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 $\sqrt{\frac mn}$ 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 $\frac ca$ , where $2a$ is the length of the major axis of $E$ and $2c$ is the is the distance between its two foci.) $\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$

Problem22

Suppose that the roots of the polynomial $P(x)=x^3+ax^2+bx+c$ are $\cos \frac{2\pi}7,\cos \frac{4\pi}7,$ and $\cos \frac{6\pi}7$ , where angles are in radians. What is $abc$ ? $\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{\sqrt[3]7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}$

Problem23

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? $\textbf{(A) }\frac{9}{16} \qquad \textbf{(B) }\frac{5}{8} \qquad \textbf{(C) }\frac34 \qquad \textbf{(D) }\frac{25}{32}\qquad \textbf{(E) }\frac{13}{16}$

Problem24

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 $\frac{a\sqrt{b}}{c}$ , 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 $a+b+c$ ? $\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$

Problem25

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

参考答案及详解注册后登录可见：

Solution1

We evaluate the given expression to get that $2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \implies \boxed{\text{(B)}}$

Solution 2

方法1、Square both sides to get $a^{2}+b^{2}=a^{2}+2ab+b^{2}$ . Then, $0=2ab\rightarrow ab=0$ . Also, it is clear that both sides of the equation must be nonnegative. The answer is $\boxed{\textbf{(D)}}$ . 方法2、The left side of the original equation is the arithmetic square root, which is always nonnegative. So, we need $a+b\ge 0,$ which eliminates $\textbf{(B)}$ and $\textbf{(E)}.$ Next, picking $(a,b)=(0,0)$ reveals that $\textbf{(A)}$ is incorrect, and picking $(a,b)=(1,2)$ reveals that $\textbf{(C)}$ is incorrect. By POE (Process of Elimination), the answer is $\boxed{\textbf{(D)}}.$ ~MRENTHUSIASM 方法3、If we graph $\sqrt{x^2+y^2}=x+y,$ then we get the positive $x$ -axis and the positive $y$ -axis, plus the origin. Therefore, the answer is $\boxed{\textbf{(D)}}.$

Solution 3

方法1、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 $11a=17402$ . So $a=1582$ . The difference is $10a-a = 9a$ . So, the answer is $9(1582) = 14238 = \boxed{\textbf{(D)}}$ .
方法2、Since the ones place of a multiple of $10$ is $0$ , this implies the other integer has to end with a $2$ since both integers sum up to a number that ends with a $2$ . Thus, the ones place of the difference has to be $10-2=8$ , and the only answer choice that ends with an $8$ is $\boxed{\textbf{(D)}~14238}$

Solution 4

方法1、We know that purple snakes cannot subtract, thus they cannot add either. Since happy snakes must be able to add, the purple snakes cannot be happy. Therefore, we know that the happy snakes are not purple and the answer is $\boxed{\textbf{(D)}}$ .
方法2、We are given that $\text{(1) Happy}\Rightarrow\text{can add}$ $\text{(2) Purple}\Rightarrow\text{cannot subtract}$ $\text{(3) Cannot subtract}\Rightarrow\text{cannot add}$ Combining $\text{(2)}$ and $\text{(3)}$ into $\text{(*)}$ below, we have $\text{(1) Happy}\Rightarrow\text{can add}$ $\text{(*) Purple}\Rightarrow\text{cannot subtract}\Rightarrow\text{cannot add}$ Clearly, the answer is $\boxed{\textbf{(D)}}.$

Solution5

It is known that $0.\overline{ab}=\frac{ab}{99}$ and $0.ab=\frac{ab}{100}$ . Let $\overline {ab} = x$ . We have that $66(1+\frac{x}{100})+0.5=66(1+\frac{x}{99})$ . Solving gives that $100x-75=99x$ so $x=\boxed{\text{(E)} 75}$ . ~aop2014

Solution 6

方法1、If the probability of choosing a red card is $\frac{1}{3}$ , the red and black cards are in ratio $1:2$ . This means at the beginning there are $x$ red cards and $2x$ black cards.
After $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\frac{1}{4}$ so the ratio of red to black cards is $1:3$ . This means in the new deck the number of black cards is also $3x$ for the same $x$ red cards.
So, $3x = 2x + 4$ and $x=4$ meaning there are $4$ red cards in the deck at the start and $2(4) = 8$ black cards.
So the answer is $8+4 = 12 = \boxed{\textbf{(C)}}$ .
方法2、For the number of cards, the final deck is $\frac43$ of the original deck. Adding $4$ cards to the original deck is the same as increasing the original deck by $\frac13$ of itself. So, the original deck has $\boxed{\textbf{(C) }12}$ cards.
~MRENTHUSIASM
方法3、Suppose there were $x$ cards in the deck originally. Now, the deck has $x+4$ cards, which must be a multiple of $4.$ Only $12+4=16$ is a multiple of $4.$ So, the answer is $x=\boxed{\textbf{(C) }12}.$

Solution 7

方法1、Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the trivial inequality(all squares are nonnegative) the minimum value for this is $\boxed{\text{(D)} 1}$ , which can be achieved at $x=y=0$ . ~aop2014 方法2、Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ : $\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0$ $\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0$ Thus, there is a local extreme at $(0, 0)$ . Because this is the only extreme, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{\bold{(D)} \ 1}$

Solution8

Making a small chart, we have $\begin{tabular}{c|c|c|c|c|c|c|c|c|c} D0&D1&D2&D3&D4&D5&D6&D7&D8&D9\\\hline 0&0&1&1&1&2&3&4&6&9\\\hline E&E&O&O&O&E&O&E&E&O \end{tabular}$ This starts repeating every 7 terms, so $D_{2021}=D_5=E$ , $D_{2022}=D_6=O$ , and $D_{2023}=D_7=E$ . Thus, the answer is $\boxed{\textbf{(C) }(E, O, E)}$ ~JHawk0224

Solution 9

方法1、All you need to do is multiply the entire equation by $(3-2)$ . Then all the terms will easily simplify by difference of squares and you will get $3^{128}-2^{128}$ or $\boxed{C}$ as your final answer. Notice you don't need to worry about $3-2$ because that's equal to $1$ .
方法2、If you weren't able to come up with the $(3 - 2)$ insight, then you could just notice that the answer is divisible by $(2 + 3) = 5$ , and $(2^2 + 3^2) = 13$ . We can then use Fermat's Little Theorem for $p = 5, 13$ on the answer choices to determine which of the answer choices are divisible by both $5$ and $13$ . This is $\boxed{C}$ .
方法3、After expanding the first few terms, the result after each term appears to be $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$ where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by $2^{2^{n-1}}$ would give $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}$ , and all the previous terms multiplied by $3^{2^{n-1}}$ would give $3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}$ . Their sum is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$ , so the proof is complete. Since $\frac{3^{2^n}-2^{2^n}}{3-2}$ is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$ , the answer is $\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}$ .

Solution 10(Use Tables to Organize Information)

方法1、Initial Scenario $\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & 3 & h_1 & \frac13\pi(3)^2h_1=3\pi h_1 \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & \frac13\pi(6)^2h_2=12\pi h_2 \end{array}$ By similar triangles:
For the narrow cone, the ratio of base radius to height is $\frac{3}{h_1},$ which remains constant.
For the wide cone, the ratio of base radius to height is $\frac{6}{h_2},$ which remains constant.
Equating the initial volumes gives $3\pi h_1=12\pi h_2,$ which simplifies to $\frac{h_1}{h_2}=4.$ Final Scenario (Two solutions follow from here.)

1.1 (Fraction Trick)

Let the base radii of the narrow cone and the wide cone be $3x$ and $6y,$ respectively, where $x,y>1.$ We have the following table: $\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & 3x & h_1x & \frac13\pi(3x)^2h_1=3\pi h_1 x^3 \\ [2ex] \textbf{Wide Cone} & 6y & h_2y & \frac13\pi(6y)^2h_2=12\pi h_2 y^3 \end{array}$ Equating the final volumes gives $3\pi h_1 x^3=12\pi h_2 y^3,$ which simplifies to $x^3=y^3,$ or $x=y.$ Lastly, the requested ratio is $\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4}.$ PS: 1. This problem uses the following fraction trick:
For unequal positive numbers $a,b,c$ and $d,$ if $\frac ab = \frac cd = k,$ then $\frac{a\pm c}{b\pm d}=k.$ Quick Proof From $\frac ab = \frac cd = k,$ we know that $a=bk$ and $c=dk$ . Therefore, $\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.$ 2. The work above shows that, regardless of the shape or the volume of the solid dropped in, as long as the solid sinks to the bottom and is completely submerged without spilling any liquid, the answer will remain unchanged.
~MRENTHUSIASM

1.2 (Bash)

Let the base radii of the narrow cone and the wide cone be $r_1$ and $r_2,$ respectively.
Let the rises of the liquid levels of the narrow cone and the wide cone be $\Delta h_1$ and $\Delta h_2,$ respectively. We have the following table: $\begin{array}{cccc} & \textbf{Base} & \textbf{Height} & \textbf{Volume} \\ [2ex] \textbf{Narrow Cone} & r_1 & h_1+\Delta h_1 & \frac13\pi r_1^2(h_1+\Delta h_1) \\ [2ex] \textbf{Wide Cone} & r_2 & h_2+\Delta h_2 & \frac13\pi r_2^2(h_2+\Delta h_2) \end{array}$ By similar triangles discussed above, we have $\begin{array}{cccc} \frac{3}{h_1}=\frac{r_1}{h_1+\Delta h_1} &\Rightarrow &r_1=\frac{3}{h_1}(h_1+\Delta h_1) & \ \ \ \ \ \ \ (1) \\ [2ex] \frac{6}{h_2}=\frac{r_2}{h_2+\Delta h_2} &\Rightarrow &r_2=\frac{6}{h_2}(h_2+\Delta h_2) & \ \ \ \ \ \ \ (2) \end{array}$ The volume of the marble dropped in is $\frac43\pi(1)^3=\frac43\pi.$ Now, we set up an equation for the volume of the narrow cone and solve for $\Delta h_1:$ $\begin{align*} \frac13\pi r_1^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ \frac13\pi{\underbrace{\left(\frac{3}{h_1}(h_1+\Delta h_1)\right)}_{\text{by (1)}}}^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ \frac{3}{h_1^2}(h_1+\Delta h_1)^3 &= 3h_1+\frac43 \\ (h_1+\Delta h_1)^3 &= h_1^3 + \frac{4h_1^2}{9} \\ \Delta h_1 &= \sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1. \end{align*}$ Next, we set up an equation for the volume of the wide cone $\Delta h_2:$ $\frac13\pi r_2^2(h_2+\Delta h_2) = 12\pi h_2+\frac43\pi.$ Using the exact same process from above (but with different numbers), we get $\Delta h_2 = \sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2.$ Recall that $\frac{h_1}{h_2}=4.$ Therefore, the requested ratio is $\begin{align*} \frac{\Delta h_1}{\Delta h_2}&=\frac{\sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{\sqrt[3]{(4h_2)^3 + \frac{4(4h_2)^2}{9}}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{\sqrt[3]{4^3\left(h_2^3 + \frac{h_2^2}{9}\right)}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{4\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\boxed{\textbf{(E) }4}. \end{align*}$

方法2、The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii 3 and 6 and infinitely large height. Then the base area of the wide cylinder is 4 times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{\textbf{(E) } 4}$ times as much.

Solution 11

方法1、Every time the laser bounces off a wall, instead we can imagine it going straight by reflecting it about the wall. Thus, the laser starts at $(3, 5)$ and ends at $(-7, -5)$ , so the path's length is $\sqrt{10^2+10^2}=\boxed{\textbf{(C)} 10\sqrt{2}}$ 方法2、Let $A=(3,5), D=(7,5), B$ be the point where the beam hits the $y$ -axis, and $C$ be the point where the beam hits the $x$ -axis.
Reflecting $\overline{BC}$ about the $y$ -axis gives $\overline{BC'}.$ Then, reflecting $\overline{CD}$ over the $y$ -axis gives $\overline{C'D'}.$ Finally, reflecting $\overline{C'D'}$ about the $x$ -axis gives $\overline{C'D''},$ as shown below.

It follows that $D''=(-7,-5).$ The total distance that the beam will travel is $\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\ &=AB+BC'+C'D'' \\ &=AD'' \\ &=\sqrt{((3-(-7))^2+(5-(-5))^2} \\ &=\sqrt{200} \\ &=\boxed{\textbf{(C) }10\sqrt2}. \end{align*}$ Graph in

方法3、Define points $A,B,C,$ and $D$ as Solution 2 does.
When a line segment hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Geometrically, the rays coincide when reflected about the line perpendicular to that coordinate axis, creating a line symmetry. Let the slope of $\overline{AB}$ be $m.$ It follows that the slope of $\overline{BC}$ is $-m,$ and the slope of $\overline{CD}$ is $m.$ Here, we conclude that $\overline{AB}\parallel\overline{CD}.$ Next, we locate $E$ on $\overline{CD}$ such that $\overline{BE}\parallel\overline{AD},$ thus $ABED$ is a parallelogram, as shown below.

Let $B=(0,b).$ By the property of slopes, we get $E=(4,b).$ By symmetry, we obtain $C=(2,0).$ Applying the slope formula on $\overline{AB}$ and $\overline{DC}$ gives $m=\frac{5-b}{3-0}=\frac{5-0}{7-2}.$ Equating the last two expressions gives $b=2.$ By the Distance Formula, $AB=3\sqrt2,BC=2\sqrt2,$ and $CD=5\sqrt2.$ The total distance that the beam will travel is $AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.$
方法4、Define points $A,B,C,$ and $D$ as Solution 2 does.
Since choices $\textbf{(B)}, \textbf{(C)},$ and $\textbf{(D)}$ all involve $\sqrt2,$ we suspect that one of them is the correct answer. We take a guess in faith that $\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$ form $45^\circ$ angles with the coordinate axes, then we get that $B=(0,2)$ and $C=(2,0).$ This result verifies our guess. Following the penultimate paragraph of Solution 3 gives the answer $\boxed{\textbf{(C) }10\sqrt2}.$

Solution 12

方法1、By Vieta's formulae, the sum of the 6 roots is 10 and the product of the 6 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, $B = -32 - 48 - 8 = \boxed{\textbf{(A)} -88}$ .
方法2、Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain $B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{\textbf{(A)} -88}.$

Solution 13

方法1、First, $\textbf{(B)} = 2\text{cis}(150), \textbf{(C)} =2\text{cis}(135)$ $, \textbf{(D)} =2\text{cis}(120)$ .
Taking the real part of the 5th power of each we have: $\textbf{(A): }(-2)^5=-32$ , $\textbf{(B): }32\cos(650)=32\cos(30)=16\sqrt{3}$ $\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}$ $\textbf{(D): }32\cos(600)=32\cos(240)$ which is negative $\textbf{(E): }(2i)^5$ which is zero
Thus, the answer is $\boxed{\textbf{(B)}}$ . 方法2、For every complex number $z=a+bi,$ where $a$ and $b$ are real numbers and $i=\sqrt{-1},$ its magnitude is $|z|=\sqrt{a^2+b^2}.$ For each choice, we get that the magnitude is $2.$ Rewriting each choice to the polar form $z=re^{i\theta},$ we know that by the De Moivre's Theorem, the real part of $z^5$ is $\text{Re}(z^5)=r^5\cos{(5\theta)}.$ We make a table as follows: $\begin{array}{c|c|c} \textbf{Choice} & \boldsymbol{\theta} & \textbf{Re}\boldsymbol{(z^5)} \\ \hline & & \\ [-1ex] \textbf{(A)} & \pi & 32\cos{(5\pi)}=32\cos\pi=32(-1) \\ [2ex] \textbf{(B)} & \frac{5\pi}{6} & 32\cos{\frac{25\pi}{6}}=32\cos{\frac{\pi}{6}}=32\left(\frac{\sqrt3}{2}\right) \\ [2ex] \textbf{(C)} & \frac{3\pi}{4} & 32\cos{\frac{15\pi}{4}}=32\cos{\frac{7\pi}{4}}=32\left(\frac{\sqrt2}{2}\right) \\ [2ex] \textbf{(D)} & \frac{2\pi}{3} & 32\cos{\frac{10\pi}{3}}=32\cos{\frac{4\pi}{3}}=32\left(-\frac{1}{2}\right) \\ [2ex] \textbf{(E)} & \frac{\pi}{2} & 32\cos{\frac{5\pi}{2}}=32\cos{\frac{\pi}{2}}=32\left(0\right) \end{array}$ Clearly, the answer is $\boxed{\textbf{(B) }-\sqrt3+i}.$

Solution 14

方法1、This equals $\left(\sum_{k=1}^{20}k\log_5(3)\right)\left(\sum_{k=1}^{100}\log_9(25)\right)=\frac{20\cdot21}{2}\cdot\log_5(3)\cdot100\log_3(5)=\boxed{\textbf{(E)} 21000}$ 方法2、We use the following property of logarithms: $\log_{p^n}{(q^n)}=\log_{p}{q}.$ We can prove it quickly using the Change of Base Formula: $\log_{p^n}{(q^n)}=\frac{\log_{p}{(q^n)}}{\log_{p}{(p^n)}}=\frac{n\log_{p}{q}}{n\log_{p}{p}}=\frac{\log_{p}{q}}{1}=\log_{p}{q}.$ Now, we simplify the expressions inside the summations: $\begin{align*} \log_{5^k}{{3^k}^2}&=\log_{5^k}{(3^k)^k} \\ &=k\log_{5^k}{3^k} \\ &=k\log_{5}{3}, \end{align*}$ and $\begin{align*} \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ &=\log_{3}{5}. \end{align*}$ Using these results, we evaluate the original expression: $\begin{align*} \left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\left(\sum_{k=1}^{100} \log_{3}{5}\right) \\ &= \left(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\left(\log_{3}{5}\cdot\sum_{k=1}^{100} 1\right) \\ &= \left(\sum_{k=1}^{20} k\right)\left(\sum_{k=1}^{100} 1\right) \\ &= \left(\frac{21\cdot20}{2}\right)\left(100\right) \\ &= \boxed{\textbf{(E) }21,000}. \end{align*}$ 方法3、First, we can get rid of the $k$ exponents using properties of logarithms: $\left(\log_{5^k} 3^{k^2}\right) = k^2 * \frac{1}{k} * \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k$ (Leaving the single $k$ in the exponent will come in handy later). Similarly, $\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \log_{9} 25 = \log_{9} 5^2$ Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms: $\left(\sum_{k=1}^{20} \log_{5} 3^k\right) = \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} = \log_{5} 3^{(1 + 2 + \dots + 20)}$ $\left(\sum_{k=1}^{100} \log_{9} 5^2\right) = \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2= \log_{9} 5^{2(100)} = \log_{9} 5^{200}$ To evaluate the exponent of the $3$ in the first logarithm, we use the triangular numbers equation: $1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210$ Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: $\log_{a} b\log_{x} y = \log_{a} y\log_{x} b$ Thus, $\left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) = \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right)$ $= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}$ 方法4、In $\sum_{k=1}^{20} \log_{5^k} 3^{k^2},$ note that the addends are greater than $1$ for all $k\geq2.$ In $\sum_{k=1}^{100} \log_{9^k} 25^k,$ note that the addends are greater than $1$ for all $k\geq1.$ By a rough approximation, $\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=1}^{20} 1\right)\left(\sum_{k=1}^{100} 1\right)=(20)(100)=2,000,$ from which we eliminate choices $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}.$ We get the answer $\boxed{\textbf{(E) }21,000}$ by either an educated guess or continued estimation: Since $3^3=27\approx25,$ it follows that $9^{3/2}\approx25.$ By a (very) rough approximation, $\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)\approx\left(\sum_{k=1}^{20} 1\right)\left(\sum_{k=1}^{100} \frac{3}{2}\right)=(20)(150)=3,000.$ From here, it should be safe to guess that the answer is $\textbf{(E)}.$ As an extra guaranty, note that $\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=1}^{20} 1 = 20.$ Therefore, we must have $\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>>3,000.$

Solution 15

方法1、We know the choose function and we know the pair multiplication $MN$ so we do the multiplications and additions. $\binom{6}{0}\left(\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{1}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{2}\left(\binom{8}{2}+\binom{8}{6}\right)+\binom{6}{3}\left(\binom{8}{3}+\binom{8}{7}\right)+\\\binom{6}{4}\left(\binom{8}{0}+\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{5}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{6}\left(\binom{8}{2}+\binom{8}{6}\right) = 4095\equiv\boxed{(D) 95}\pmod{100}$
方法2、The problem can be done using a roots of unity filter. Let $f(x,y)=(1+x)^8(1+y)^6$ . By expanding the binomials and distributing, $f(x,y)$ is the generating function for different groups of basses and tenors. That is, $f(x,y)=\sum_{m=0}^8\sum_{n=0}^6 a_{mn}x^my^n$ where $a_{mn}$ is the number of groups of $m$ basses and $n$ tenors. What we want to do is sum up all values of $a_{mn}$ for which $4\mid m-n$ except for $a_{00}=1$ . To do this, define a new function $g(x)=f(x,x^{-1})=\sum_{m=0}^8\sum_{n=0}^6 a_{mn}x^{m-n}=(1+x)^8(1+x^{-1})^6.$ Now we just need to sum all coefficients of $g(x)$ for which $4\mid m-n$ . Consider a monomial $h(x)=x^k$ . If $4\mid k$ , $h(i)+h(-1)+h(-i)+h(1)=1+1+1+1=4$ otherwise, $h(i)+h(-1)+h(-i)+h(1)=0.$ $g(x)$ is a sum of these monomials so this gives us a method to determine the sum we're looking for: $\frac{g(i)+g(-1)+g(-i)+g(1)}{4}=2^{12}=4096$ (since $g(-1)=0$ and it can be checked that $g(i)=-g(-i)$ ). Hence, the answer is $4096-1$ with the $-1$ for $a_{00}$ which gives $\boxed{95}$ . ~lawliet163 方法3、 $\begin{array}{c|c|c|c} \textbf{\# of Tenors} & \textbf{\# of Basses} & \textbf{\# of Ways} & \textbf{Rewrite \# of Ways} \\ [1ex] \hline\hline & & & \\ [-1ex] 0 & 8 & \binom{6}{0}\binom{8}{8} & \\ [1ex] 1 & 1 & \binom{6}{1}\binom{8}{1} & \binom{6}{1}\binom{8}{7}\\ [1ex] 2 & 2 & \binom{6}{2}\binom{8}{2} & \binom{6}{2}\binom{8}{6}\\ [1ex] 3 & 3 & \binom{6}{3}\binom{8}{3} & \binom{6}{3}\binom{8}{5}\\ [1ex] 4 & 4 & \binom{6}{4}\binom{8}{4} & \\ [1ex] 5 & 5 & \binom{6}{5}\binom{8}{5} & \binom{6}{5}\binom{8}{3}\\ [1ex] 6 & 6 & \binom{6}{6}\binom{8}{6} & \binom{6}{6}\binom{8}{2}\\ [1ex] \hline & & & \\ [-1ex] 0 & 4 & \binom{6}{0}\binom{8}{4} & \\ [1ex] 1 & 5 & \binom{6}{1}\binom{8}{5} & \binom{6}{1}\binom{8}{3}\\ [1ex] 2 & 6 & \binom{6}{2}\binom{8}{6} & \binom{6}{2}\binom{8}{2}\\ [1ex] 3 & 7 & \binom{6}{3}\binom{8}{7} & \binom{6}{3}\binom{8}{1}\\ [1ex] 4 & 8 & \binom{6}{4}\binom{8}{8} & \binom{6}{4}\binom{8}{0}\\ [1ex] \hline & & & \\ [-1ex] 4 & 0 & \binom{6}{4}\binom{8}{0} & \binom{6}{2}\binom{8}{0}\\ [1ex] 5 & 1 & \binom{6}{5}\binom{8}{1} & \binom{6}{1}\binom{8}{1}\\ [1ex] 6 & 2 & \binom{6}{6}\binom{8}{2} & \binom{6}{0}\binom{8}{2} \end{array}$ We will use the Vandermonde's Identity to find the requested sum: $\begin{align*} \left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{8-k}\right]+\left[\sum_{k=0}^{4}\binom{6}{k}\binom{8}{4-k}\right]+\left[\sum_{k=0}^{2}\binom{6}{k}\binom{8}{2-k}\right]&=\binom{14}{8}+\binom{14}{4}+\binom{14}{2} \\ &=\binom{14}{6}+\binom{14}{4}+\binom{14}{2} \\ &=3003+1001+91 \\ &=4095 \\ &\equiv\boxed{\textbf{(D) } 95}\pmod{100}. \end{align*}$

方法4、We claim that if the empty group is allowed, then there are $\mathbf{2^{12}}$ ways to choose the singers satisfying the requirements.

First, we set one tenor and one bass aside. We argue that each group from the $12$ remaining singers (of any size, including $0$ ) corresponds to exactly one desired group from the original $14$ singers.
The $12$ remaining singers can form $2^{12}=\sum_{\substack{t=0 \\ b=0}}^{\substack{t=5 \\ b=7}}\binom5t\binom7b$ groups. The left side counts directly, while the right side uses casework (selecting $t$ tenors and $b$ basses for each group). Now, we map each group from the $12$ to a group from the $14.$ By casework: $(1) \ |b-t|\equiv\pm1\pmod{4}$ Clearly, the mapping is satisfied. For each group from the $12,$ we can obtain a desired group from the $14$ by adding one tenor or one bass accordingly. $(2) \ |b-t|\equiv0\pmod{4}$ Since $\binom7b=\binom{7}{7-b},$ we can select $7-b$ basses instead of $b$ basses, without changing the number of groups. Therefore, we have the absolute difference $|(7-b)-t|=|7-(b+t)|.$ Since $b\equiv t\pmod{4},$ we conclude that $|7-(b+t)|\equiv 1\text{ or }3\pmod{4},$ and the mapping is satisfied by case $(1).$ $(2) \ |b-t|\equiv2\pmod{4}$ By the same reasoning as Case $(2),$ we select $7-b$ basses instead of $b$ basses. The absolute difference also is $|7-(b+t)|.$ Since $b-t$ is even, it follows that $b+t$ is also even, and $|7-(b+t)|\equiv 1\text{ or }3\pmod{4}.$ The mapping is satisfied by case $(1).$

Solution 16

方法1、There are $1+2+..+199+200=\frac{(200)(201)}{2}=20100$ numbers in total. Let the median be $k$ . We want to find the median $k$ such that $\frac{k(k+1)}{2}=20100/2,$ or $k(k+1)=20100.$ Note that $\sqrt{20100} \approx 142$ . Plugging this value in as $k$ gives $\frac{1}{2}(142)(143)=10153.$ $10153-142<10050$ , so $142$ is the $152$ nd and $153$ rd numbers, and hence, our desired answer. $\fbox{(C) 142}$ .
Note that we can derive $\sqrt{20100} \approx 142$ through the formula $\sqrt{n} = \sqrt{a+b} \approx \sqrt{a} + \frac{b}{2\sqrt{a} + 1},$ where $a$ is a perfect square less than or equal to $n$ . We set $a$ to $19600$ , so $\sqrt{a} = 140$ , and $b = 500$ . We then have $n \approx 140 + \frac{500}{2(140)+1} \approx 142$ .
方法2、The $x$ th number of this sequence is $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$ . This is approximately the number divided by $\sqrt{2}$ . $\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{(C) 142}$ 方法3、We can look at answer choice $C$ , which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$ . The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{(C) \text{ } 142}.$

Solution 17

方法1、Angle chasing reveals that $\triangle BPC\sim\triangle BDA$ , therefore $2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43}$ $AB=86$ Additional angle chasing shows that $\triangle ABO \sim\triangle CDO$ , therefore $2=\frac{AB}{CD}=\frac{BP}{PD}=\frac{\frac{BD}{2}+11}{\frac{BD}{2}-11}$ $BD=66$ Since $\triangle ADB$ is right, the Pythagorean theorem implies that $AD=\sqrt{86^2-66^2}$ $AD=4\sqrt{190}$ $4\sqrt{190}\implies 4 + 190 = \boxed{\textbf{D) } 194}$
方法2、Since $\triangle BCD$ is isosceles with legs $\overline{CB}$ and $\overline{CD},$ it follows that the median $\overline{CP}$ is also an altitude of $\triangle BCD.$ Let $DO=x$ and $CP=h.$ We have $PB=x+11.$ Since $\triangle ADO\sim\triangle CPO$ by AA, we have $AD=CP\cdot\frac{DO}{PO}=h\cdot\frac{x}{11}.$ Let the brackets denote areas. Notice that $[ADO]=[BCO]$ (By the same base/height, $[ADC]=[BCD].$ Subtracting $[OCD]$ from both sides gives $[ADO]=[BCO].$ ). Doubling both sides, we have $\begin{align*} 2[ADO]&=2[BCO] \\ \frac{x^2 h}{11}&=(x+22)h \\ x^2&=11x+11\cdot22 \\ (x-22)(x+11)&=0 \\ x&=22. \end{align*}$ In $\triangle CPB,$ we have $h=\sqrt{43^2-33^2}=\sqrt{76\cdot10}=2\sqrt{190}$ and $AD=h\cdot\frac{x}{11}=4\sqrt{190}.$ Finally, $4+190=\boxed{\textbf{(D) }194}.$ 方法3、Let $CP = y$ and $CP$ is perpendicular bisector of $DB.$ Let $DO = x,$ so $DP = PB = 11+x.$ (1) $\triangle CPO \sim \triangle ADO,$ so we get $\frac{AD}{x} = \frac{y}{11},$ or $AD = \frac{xy}{11}.$ (2) pythag on $\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$ (3) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y.$ Thus, $xy/11 = 2y,$ or $x = 22.$ And $y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},$ so $AD = 4 \sqrt{190}$ and the answer is $\boxed{194}.$ 方法4、Observe that $\triangle BPC$ is congruent to $\triangle DPC$ ; both are similar to $\triangle BDA$ . Let's extend $\overline{AD}$ and $\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$ . Observe that $\angle BDE$ is $90$ degrees, and that $\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA$ . Thus, by ASA, we know that $\triangle ABD \cong \triangle EBD$ , thus, $AD = ED$ , meaning $D$ is the midpoint of $AE$ . Let $M$ be the midpoint of $\overline{DE}$ . Note that $\triangle CME$ is congruent to $\triangle BPC$ , thus $BC = CE$ , meaning $C$ is the midpoint of $\overline{BE}.$ Therefore, $\overline{AC}$ and $\overline{BD}$ are both medians of $\triangle ABE$ . This means that $O$ is the centroid of $\triangle ABE$ ; therefore, because the centroid divides the median in a 2:1 ratio, $\frac{BO}{2} = DO = \frac{BD}{3}$ . Recall that $P$ is the midpoint of $BD$ ; $DP = \frac{BD}{2}$ . The question tells us that $OP = 11$ ; $DP-DO=11$ ; we can write this in terms of $DB$ ; $\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66$ .
We are almost finished. Each side length of $\triangle ABD$ is twice as long as the corresponding side length $\triangle CBP$ or $\triangle CPD$ , since those triangles are similar; this means that $AB = 2 \cdot 43 = 86$ . Now, by Pythagorean theorem on $\triangle ABD$ , $AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$ . $4+190 = \boxed{194, \textbf{D}}$

Solution 18

方法1、Looking through the solutions we can see that $f(\frac{25}{11})$ can be expressed as $f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})$ so using the prime numbers to piece together what we have we can get $10=11+f(\frac{25}{11})$ , so $f(\frac{25}{11})=-1$ or $\boxed{E}$ .
-Lemonie $f(\frac{25}{11} \cdot 11) = f(25) = f(5) + f(5) = 10$ 方法2、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 $f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7$ .
In $\textbf{(B)}$ we have $f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3$ .
In $\textbf{(C)}$ we have $f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1$ .
In $\textbf{(D)}$ we have $f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2$ .
In $\textbf{(E)}$ we have $f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1$ .
Thus, our answer is $\boxed{\textbf{(E)} \frac{25}{11}}$ 方法3、Consider the rational $\frac{a}{b}$ , for $a,b$ integers. We have $f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)$ . So $f\left(\frac{a}{b}\right)=f(a)-f(b)$ . Let $p$ be a prime. Notice that $f(p^k)=kf(p)$ . And $f(p)=p$ . So if $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ , $f(a)=a_1p_1+a_2p_2+....+a_kp_k$ . We simply need this to be greater than what we have for $f(b)$ . Notice that for answer choices $A,B,C,$ and $D$ , the numerator $(a)$ has less prime factors than the denominator, and so they are less likely to work. We check $E$ first, and it works, therefore the answer is $\boxed{\textbf{(E)}}$ .

方法4、We have the following important results: $(1) \ f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)$ for all positive integers $k$ $(2) \ f\left(a^n\right)=nf(a)$ for all positive rational numbers $a$ $(3) \ f(1)=0$ $(4) \ f\left({\frac 1a}\right)=-f(a)$ for all positive rational numbers $a$ Proofs Result $(1)$ can be shown by induction.
Result $(2):$ Since powers are just repeated multiplication, we will use result $(1)$ to prove result $(2):$ $f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a).$ Result $(3):$ For all positive rational numbers $a,$ we have $f(a)=f(a\cdot1)=f(a)+f(1).$ Therefore, we get $f(1)=0.$ So, result $(3)$ is true.
Result $(4):$ For all positive rational numbers $a,$ we have $f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.$ It follows that $f\left({\frac 1a}\right)=-f(a),$ and result $(4)$ is true.
For all positive integers $x$ and $y,$ suppose $\prod_{k=1}^{m}p_k^{e_k}$ and $\prod_{k=1}^{n}q_k^{d_k}$ are their prime factorizations, respectively, we have $\begin{align*} f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) \\ &=f(x)-f(y) \\ &=f\left(\prod_{k=1}^{m}p_k^{e_k}\right)-f\left(\prod_{k=1}^{n}q_k^{d_k}\right) \\ &=\left[\sum_{k=1}^{m}f\left(p_k^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left(q_k^{d_k}\right)\right] \\ &=\left[\sum_{k=1}^{m}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] \\ &=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. \end{align*}$ We apply function $f$ on each fraction in the choices: $\begin{array}{cccccccc} \textbf{(A) } & f\left(\frac{17}{32}\right) & = & f\left(\frac{17^1}{2^5}\right) & = & [1(17)]-[5(2)] & = & 7 \\ [2ex] \textbf{(B) } & f\left(\frac{11}{16}\right) & = & f\left(\frac{11^1}{2^4}\right) & = & [1(11)]-[4(2)] & = & 3 \\ [2ex] \textbf{(C) } & f\left(\frac{7}{9}\right) & = & f\left(\frac{7^1}{3^2}\right) & = & [1(7)]-[2(3)] & = & 1 \\ [2ex] \textbf{(D) } & f\left(\frac{7}{6}\right) & = & f\left(\frac{7^1}{2^1\cdot3^1}\right) & = & [1(7)]-[1(2)+1(3)] & = & 2 \\ [2ex] \textbf{(E) } & f\left(\frac{25}{11}\right) & = & f\left(\frac{5^2}{11^1}\right) & = & [2(5)]-[1(11)] & = & -1. \end{array}$ Therefore, the answer is $\boxed{\textbf{(E) }\frac{25}{11}}.$

Solution 19

方法1、 $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ The ranges of $\frac{\pi}2 \sin x$ and $\frac{\pi}2 \cos x$ are both $\left[-\frac{\pi}2, \frac{\pi}2 \right]$ , which is included in the range of $\arcsin$ , so we can use it with no issues. $\frac{\pi}2 \cos x=\arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right)$ $\frac{\pi}2 \cos x=\frac{\pi}2 - \frac{\pi}2 \sin x$ $\cos x = 1 - \sin x$ $\cos x + \sin x = 1$ This only happens at $x = 0, \frac{\pi}2$ on the interval $[0,\pi]$ , because one of $\sin$ and $\cos$ must be $1$ and the other $0$ . Therefore, the answer is $\boxed{C) 2}$ 方法2、By the cofunction identity $\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)$ for all $\theta,$ we simplify the given equation: $\begin{align*} \sin \left( \frac{\pi}2 \cos x\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\ \cos \left(\frac{\pi}2-\frac{\pi}2 \cos x\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\ \cos \left(\frac{\pi}2 \left(1 - \cos x \right)\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\ \frac{\pi}2 \left(1 - \cos x \right) &= \frac{\pi}2 \sin x + 2n\pi, \end{align*}$ for some integer $n.$ We keep simplifying: $\begin{align*} 1 - \cos x &= \sin x + 4n \\ 1 - 4n &= \sin x + \cos x. \end{align*}$ By rough constraints, we know that $-2 < \sin x + \cos x < 2,$ so that $-2 < 1 - 4n < 2.$ The only possibility is $n=0.$ From here, we get $\begin{align*} \sin x + \cos x &= 1 \ \ \ \ \ (*) \\ \sin^2 x + \cos^2 x + 2\sin x \cos x &= 1 \\ 2\sin x \cos x &= 0 \\ \sin(2x) &= 0 \\ 2x &= k\pi \\ x &= \frac{k\pi}{2}, \end{align*}$ for some integer $k.$ The possible solutions in $[0,\pi]$ are $x=0,\frac{\pi}{2},\pi,$ but only $x=0,\frac{\pi}{2}$ check the original equation (Note that $x=\pi$ is an extraneous solution formed by squaring $(*)$ above.). Therefore, the answer is $\boxed{\textbf{(C) }2}.$

方法3、Let $f(x)=\sin\left(\frac{\pi}{2}\cos x\right)$ and $g(x)=\cos \left( \frac{\pi}2 \sin x\right).$ This problem is equivalent to counting the intersections of the graphs of $f(x)$ and $g(x)$ in the closed interval $[0,\pi].$ We make a table of values, as shown below: $\begin{array}{c|ccc} & \boldsymbol{x=0} & \boldsymbol{x=\frac{\pi}{2}} & \boldsymbol{x=\pi} \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] \boldsymbol{f(x)} & 1 & 0 & -1 \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] \boldsymbol{g(x)} & 1 & 0 & 1 \end{array}$ The graph of $f(x)$ in $[0,\pi]$ (from left to right) is the same as the graph of $\sin x$ in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (from right to left). The output is from $1$ to $-1$ (from left to right), inclusive, and strictly decreasing.
The graph of $g(x)$ in $[0,\pi]$ (from left to right) has two parts: $(1) \ \cos x$ in $\left[0,\frac{\pi}{2}\right]$ (from left to right). The output is from $1$ to $0$ (from left to right), inclusive, and strictly decreasing. $(2) \ \cos x$ in $\left[0,\frac{\pi}{2}\right]$ (from right to left). The output is from $0$ to $1$ (from left to right), inclusive, and strictly increasing.
If $x\in\left(\frac{\pi}{2},\pi\right],$ then $f(x)<0$ and $g(x)>0.$ So, their graphs do not intersect.
If $x\in\left[0,\frac{\pi}{2}\right],$ then $0\leq f(x),g(x)\leq1.$ Clearly, the graphs intersect at $x=0$ and $x=\frac{\pi}{2}$ (at points $(0,1)$ and $\left(\frac{\pi}{2},0\right),$ respectively), but we will prove/disprove that they are the only points of intersection:
Let $A=\frac{\pi}{2}\cos x$ and $B=\frac{\pi}{2}\sin x.$ It follows that $A,B\in\left[0,\frac{\pi}{2}\right].$ Since $\sin A = \cos B,$ we know that $A+B=\frac{\pi}{2}$ by the cofunction identity: $\begin{align*} \frac{\pi}{2}\cos x + \frac{\pi}{2}\sin x &= \frac{\pi}{2} \\ \cos x + \sin x &=1. \end{align*}$ Applying Solution 2's argument (starts from its last block of equations) to deduce that $(0,1)$ and $\left(\frac{\pi}{2},0\right)$ are the only points of intersection. So, the answer is $\boxed{\textbf{(C) }2}.$

Solution20

Let $\ell$ be the directrix of $\mathcal P$ ; recall that $\mathcal P$ is the set of points $T$ such that the distance from $T$ to $\ell$ is equal to $TF$ . Let $P$ and $Q$ be the orthogonal projections of $F$ and $A$ onto $\ell$ , and further let $X$ and $Y$ be the orthogonal projections of $F$ and $V$ onto $AQ$ . Because $AF < AV$ , there are two possible configurations which may arise, and they are shown below. $[asy] import olympiad; size(230); defaultpen(linewidth(0.8)+fontsize(11pt)); real d = 1.1, edge = 2.5, Ax = 1.6; real f(real x) { return 1/(4*d) * x * x; } pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax)); pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q); draw(P--F--A--V--Y^^F--X--Q^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red); draw(graph(f,-2.5,2.5)); draw(la -- lb); dot(F^^A^^V); label("$F$",F,NW); label("$V$",V,SW); label("$A$",A,dir(F--A)); label("$P$",P,S,red); label("$Q$",Q,S,red); label("$X$",X,E,red); label("$Y$",Y,E,red); [/asy]$ $[asy] import olympiad; size(200); defaultpen(linewidth(0.8)+fontsize(11pt)); real d = 0.7, edge = 2.5, Ax = 1.9; real f(real x) { return 1/(4*d) * x * x; } pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax)); pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q); draw(Q--A--F--P^^F--X^^A--V--Y^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red); draw(la -- lb); draw(graph(f,-2.5,2.5)); dot(F^^A^^V); label("$F$",F,NW); label("$V$",V,SW); label("$A$",A,dir(F--A)); label("$P$",P,S,red); label("$Q$",Q,S,red); label("$X$",X,E,red); label("$Y$",Y,E,red); [/asy]$ Set $d = FV$ , which by the definition of a parabola also equals $VP$ . Then as $AQ = AF = 20$ , we have $AY = 20 - d$ and $AX = |20 - 2d|$ . Since $FXYV$ is a rectangle, $FX = VY$ , so by Pythagorean Theorem on triangles $AFX$ and $AVY$ , $\begin{align*} 21^2 - (20 - d)^2 &= AV^2 - AY^2 = VY^2\\ &= FX^2 = AF^2 - AX^2 = 20^2 - (20 - 2d)^2 \end{align*}$ This equation simplifies to $3d^2 - 40d + 41 = 0$ , which has solutions $d = \tfrac{20\pm\sqrt{277}}3$ . Both values of $d$ work - the smaller solution with the right configuration and the larger solution with the left configuration - and so the requested answer is $\boxed{\tfrac{40}3}$ .

Solution 21

方法1、The solutions to this equation are $z = 1$ , $z = -1 \pm i\sqrt 3$ , and $z = -2\pm i\sqrt 2$ . Consider the five points $(1,0)$ , $(-1,\pm\sqrt 3)$ , and $(-2,\pm\sqrt 2)$ ; these are the five points which lie on $\mathcal E$ . Note that since these five points are symmetric about the $x$ -axis, so must $\mathcal E$ .
Now let $r:= b/a$ denote the ratio of the length of the minor axis of $\mathcal E$ to the length of its major axis. Remark that if we perform a transformation of the plane which scales every $x$ -coordinate by a factor of $r$ , $\mathcal E$ is sent to a circle $\mathcal E'$ . Thus, the problem is equivalent to finding the value of $r$ such that $(r,0)$ , $(-r,\pm\sqrt 3)$ , and $(-2r,\pm\sqrt 2)$ all lie on a common circle; equivalently, it suffices to determine the value of $r$ such that the circumcenter of the triangle formed by the points $P_1 = (r,0)$ , $P_2 = (-r,\sqrt 3)$ , and $P_3 = (-2r,\sqrt 2)$ lies on the $x$ -axis.
Recall that the circumcenter of a triangle $ABC$ is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments $\overline{P_1P_2}$ and $\overline{P_1P_3}$ are $y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{and}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}(x + \tfrac r2)$ respectively. These two lines have different slopes for $r\neq 0$ , so indeed they will intersect at some point $(x_0,y_0)$ ; we want $y_0 = 0$ . Plugging $y = 0$ into the first equation yields $x = -\tfrac{3}{4r}$ , and so plugging $y=0$ into the second equation and simplifying yields $-\tfrac{1}{3r} = x + \tfrac r2 = -\tfrac{3}{4r} + \tfrac{r}{2}.$ Solving yields $r=\sqrt{\tfrac 56}$ .
Finally, recall that the lengths $a$ , $b$ , and $c$ (where $c$ is the distance between the foci of $\mathcal E$ ) satisfy $c = \sqrt{a^2 - b^2}$ . Thus the eccentricity of $\mathcal E$ is $\tfrac ca = \sqrt{1 - (\tfrac ba)^2} = \sqrt{\tfrac 16}$ and the requested answer is $\boxed{7\textbf{ (A)}}$ .
方法2、Completing the square in the original equation, we get $(z-1)\left((z+1)^2+3\right)\left((z+2)^2+2\right)=0,$ from which $z=1,-1\pm\sqrt{3}i,-2\pm\sqrt{2}i.$ Now, we will find the equation of an ellipse $\mathbb{E}$ that passes through $(1,0),(-1,\pm\sqrt3),$ and $(-2,\pm\sqrt2)$ in the $xy$ -plane. By symmetry, the center of $\mathbb{E}$ must be on the $x$ -axis. The formula of $\mathbb{E}$ is $\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1,$ with the center at $(h,0)$ and the axes' lengths $2a$ and $2b.$ Plugging the points $(1,0),(-1,\sqrt3),$ and $(-2,\sqrt2)$ in, respectively, we get the following system of three equations: $\begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*}$ Clearing fractions gives $\begin{align*} (1-h)^2&=a^2, \\ b^2(-1-h)^2 + 3a^2 &= a^2b^2, \\ b^2(-2-h)^2 + 2a^2 &= a^2b^2. \end{align*}$ Since $t^2=(-t)^2$ for all real numbers $t,$ we rewrite the system as $\begin{align*} (1-h)^2&=a^2, \ \ \ \ \ \ \ \ \ \ \ \ &\text{(1)}\\ b^2(1+h)^2 + 3a^2 &= a^2b^2, &\text{(2)}\\ b^2(2+h)^2 + 2a^2 &= a^2b^2. &\text{(3)} \end{align*}$ Applying the Transitive Property in $\text{(2)}$ and $\text{(3)},$ we get $\begin{align*} b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\ a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\ a^2 &= b^2(2h+3). \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(4)} \end{align*}$ Applying the results of $\text{(1)}$ and $\text{(4)}$ on $\text{(2)},$ we get $\begin{align*} b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_\text{by (4)} &= \underbrace{(1-h)^2}_\text{by (1)}b^2 \\ (1+h)^2+3(2h+3)&=(1-h)^2 \\ 1+2h+h^2+6h+9&=1-2h+h^2 \\ 10h&=-9 \\ h&=-\frac{9}{10}. \end{align*}$ Substituting this into $\text{(1)},$ we get $a^2=\frac{361}{100}.$ Substituting the current results into $\text{(4)},$ we get $b^2=\frac{361}{120}.$ Finally, we have $c^2 = a^2-b^2 = 361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},$ and $\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.$ Our answer is $1+6=\boxed{\textbf{(A) } 7}.$

Solution 22

方法1、Part 1: solving for a $a$ is the negation of the sum of roots $a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right)$ The real values of the 7th roots of unity are: $1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7$ and they sum to $0$ .
If we subtract 1, and condense identical terms, we get: $2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1$ Therefore, we have $a = -\left(-\frac{1}2\right) = \frac{1}2$ Part 2: solving for b $b$ is the sum of roots two at a time by Vieta's $b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7$ We know that $\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}$ By plugging all the parts in we get: $\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2$ Which ends up being: $\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7$ Which was shown in the first part to equal $-\frac{1}2$ , so $b = -\frac{1}2$ Part 3: solving for c
Notice that $\cos \frac{6\pi}7 = \cos \frac{8\pi}7$ $c$ is the negation of the product of roots by Vieta's formulas $c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$ Multiply by $8 \sin{\frac{2\pi}{7}}$ $c 8 \sin{2\pi}7 = -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$ Then use sine addition formula backwards: $2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7$ $c \cdot 8 \sin{\frac{2\pi}{7}} = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$ $c \cdot 8 \sin{\frac{2\pi}{7}} = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7$ $c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{16\pi}7$ $c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{2\pi}7$ $c = -\frac{1}8$ Finally multiply $abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}{32}$ or $\boxed{D) \frac{1}{32}}$ .

方法2、Letting the roots be $p$ , $q$ , and $r$ , Vietas gives $p + q + r = a$ $pq + qr + pq = -b$ $pqr = c$ We use the Taylor series for $\cos x$ , $\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}$ to approximate the roots. Taking the sum up to $k = 3$ yields a close approximation, so we have $\cos\left(\frac{2\pi}{7}\right) \simeq 1-\frac{\left(\frac{2\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{2\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{2\pi}{7}\right)^{6}}{720} \simeq 0.623$ $\cos\left(\frac{4\pi}{7}\right) \simeq 1-\frac{\left(\frac{4\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{4\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{4\pi}{7}\right)^{6}}{720} \simeq -0.225$ $\cos\left(\frac{6\pi}{7}\right) \simeq 1-\frac{\left(\frac{6\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{6\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{6\pi}{7}\right)^{6}}{720} \simeq -0.964.$ Note that these approximations get worse as $x$ gets larger, but they will be fine for the purposes of this problem. We then have $p + q + r = a \simeq -0.56$ $pq + qr + pr = -b \simeq -0.524$ $pqr = c \simeq 0.135$ We further approximate these values to $a \simeq -0.5$ , $b \simeq 0.5$ , and $c \simeq 0.125$ (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have $abc \simeq \boxed{\textbf{(D) } \frac{1}{32}}$ . ~ciceronii
Remark: In order to be more confident in your answer, you can go a few terms further in the Taylor series.

方法3、Note sum of roots of unity equal zero, sum of real parts equal zero, and $\text{Re} \omega^{m} = \text{Re} \omega^{-m},$ thus $\cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} = 1/2(0 - \cos 0) = -1/2$ which means $A = \frac{1}{2}.$ By product to sum, $\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} + \cos \frac{2 \pi}{7} \cos \frac{6 \pi}{7} + \cos \frac{4 \pi}{7} \cos \frac{6 \pi}{7} = \frac{1}{2} (2 \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{10 \pi}{7})$ $= \frac{1}{2}(2 \cos \frac{2 \pi}{7} + 2 \cos \frac{4 \pi}{7} + 2 \cos \frac{6 \pi}{7}) = -1/2,$ so $B = - \frac{1}{2}.$ By product to sum, $\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{6 \pi}{7} = \frac{1}{2}(\cos \frac{2 \pi}{7} + \cos \frac{6 \pi}{7}) \cos \frac{6 \pi}{7} = \frac{1}{4}(\cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7}) + \frac{1}{4}(1 + \cos \frac{12 \pi}{7})$ $= \frac{1}{4}(1 + \cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}) = 1/8,$ so $C = -1/8.$ $ABC =\boxed{ \frac{1}{32}}.$

方法4、Using geometric series, we can show that $\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=0:$ $\begin{align*} \sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}&=1+e^{\frac{2\pi i}{7}}+e^{\frac{4\pi i}{7}}+\cdots+e^{\frac{12\pi i}{7}} \\ &=\frac{1-1}{1-e^{\frac{2\pi i}{7}}} \\ &=0. \end{align*}$ Desmos graph of $e^{\frac{2k\pi i}{7}},$ where $k=0,1,2,\cdots,6$ (the $7$ th roots of unity): https://www.desmos.com/calculator/kjnnkhgq6u

Graphically, the imaginary parts of these complex numbers sum to $0.$ Using the above result, the real parts of these complex numbers sum to $0$ too. It follows that $\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}=\left(\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}\right)-1=-1,$ from which $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac12,$ as it contributes half the real part of $\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}.$ Two solutions follow from here:

4.1 We know that $\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}$ are solutions of $\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \ (*)$ as they can be verified graphically. Now, let $x=\cos\theta.$ It follows that $\begin{align*} \cos(2\theta)&=2\cos^2\theta-1 \\ &=2x^2-1, \\ \cos(3\theta)&=\cos(2\theta+\theta) \\ &=\cos(2\theta)\cos\theta-\sin(2\theta)\sin\theta \\ &=(2x^2-1)x-\sin(2\theta)\sin\theta \\ &=(2x^2-1)x-2\sin^2\theta\cos\theta \\ &=(2x^2-1)x-2(1-\cos^2\theta)\cos\theta \\ &=(2x^2-1)x-2(1-x^2)x \\ &=2x^3-x-2x+2x^3 \\ &=4x^3-3x. \end{align*}$ Rewriting $(*)$ from above in terms of $x,$ we have $\begin{align*} x+(2x^2-1)+(4x^3-3x)&=-\frac12 \\ 4x^3+2x^2-2x-\frac12&=0 \\ x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0. \end{align*}$ It follows that $(a,b,c)=\left(\frac12,-\frac12,-\frac18\right),$ and $abc=\boxed{\textbf{(D) }\frac{1}{32}}.$

4.2 Let $z=e^{\frac{2\pi i}{7}}.$ Since $z$ is a $7$ th root of unity, $z^7=1.$ Graphically, it follows that $\begin{array}{ccccc} \cos{\frac{2\pi}{7}} &=& \frac{z+z^6}{2} &=& \frac{z+z^{-1}}{2} \\ [2ex] \cos{\frac{4\pi}{7}} &=& \frac{z^2+z^5}{2} &=& \frac{z^2+z^{-2}}{2} \\ [2ex] \cos{\frac{7\pi}{7}} &=& \frac{z^3+z^4}{2} &=& \frac{z^3+z^{-3}}{2} \end{array}$ Recall that $\sum_{k=0}^{6}z^k=0$ (so that $\sum_{k=1}^{6}z^k=-1$ ), and let $(r,s,t)=\left(\cos{\frac{2\pi}{7}},\cos{\frac{4\pi}{7}},\cos{\frac{6\pi}{7}}\right).$ By Vieta's Formulas and the results above, the answer is $\begin{align*} abc&=[-(r+s+t)](rs+st+tr)(-rst) \\ &=(r+s+t)(rs+st+tr)(rst) \\ &=\left(\frac{\sum_{k=1}^{6}z^k}{2}\right)\left(\frac{2\sum_{k=1}^{6}z^k}{4}\right)\left(\frac{1+\sum_{k=0}^{6}z^k}{8}\right) \\ &=\frac{1}{32}\left(\sum_{k=1}^{6}z^k\right)\left(\sum_{k=1}^{6}z^k\right)\left(1+\sum_{k=0}^{6}z^k\right) \\ &=\frac{1}{32}(-1)(-1)(1) \\ &=\boxed{\textbf{(D) }\frac{1}{32}}. \end{align*}$

Solution 23(complementary counting)

We will use complementary counting. First, the frog can go left with probability $\frac14$ . We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4\cdot \frac14=1$ , we will ignore the leading probability.
From the left, she either goes left to another edge ( $\frac14$ ) or back to the center ( $\frac14$ ). Time for some casework. $\textbf{Case 1:}$ She goes back to the center.
Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac12$ . --End case 1 $\textbf{Case 2:}$ She goes to another edge (rightmost).
Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac12$ Subcase 2: She goes to the center. Now any move works. $\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38$ for this case. --End case 2
She goes back to the center in Case 1 with probability $\frac14$ , and to the right edge with probability $\frac14$ So, our answer is $\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}$ But, don't forget complementary counting. So, we get $1-\frac7{32}=\frac{25}{32} \implies \boxed{D}$ .

Solution24

Let $O=\Gamma$ be the center of the semicircle, $X=\Omega$ be the center of the circle, and $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Theorem Converse, we have $\overline{XM}\perp\overline{QR}$ and $\overline{OM}\perp\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear.
Applying the Extended Law of Sines on $\triangle PQR,$ we have $XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3,$ in which the radius of $\odot \Omega$ is $3.$ By the SAS Congruence, we have $\triangle QXM\cong\triangle RXM,$ both of which are $30^\circ$ - $60^\circ$ - $90^\circ$ triangles. By the side-length ratios, $RM=\frac{3\sqrt3}{2}, RX=3,$ and $MX=\frac{3}{2}.$ By the Pythagorean Theorem in $\triangle ORM,$ we get $OM=\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on $\triangle OXP,$ we obtain $OP=4.$

As shown above, we construct an altitude $\overline{PC}$ of $\triangle PQR.$ Since $\overline{PC}\perp\overline{RQ}$ and $\overline{OM}\perp\overline{RQ},$ we know that $\overline{PC}\parallel\overline{OM}.$ We construct $D$ on $\overline{PC}$ such that $\overline{XD}\perp\overline{PC}.$ Clearly, $MXDC$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $\triangle XPD\sim\triangle OXP$ by the AA Similarity, with ratio of similitude $\frac{XP}{OX}=\frac 35.$ Therefore, we get that $PD=\frac 95$ and $PC=PD+DC=PD+MX=\frac 95 + \frac 32 = \frac{33}{10}.$ The area of $\triangle PQR$ is $\frac12(RQ)(PC)=\frac12\left(3\sqrt3\right)\left(\frac{33}{10}\right)=\frac{99\sqrt3}{20},$ and the answer is $99+3+20=\boxed{\textbf{(D) } 122}.$

Solution 25

方法1、Consider the prime factorization $n=\prod_{i=1}^{k}p_i^{e_i}.$ By the Multiplication Principle, $d(n)=\prod_{i=1}^{k}(e_i+1).$ Now, we rewrite $f(n)$ as $f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_1+1}{p_i^{{e_i}/3}}.$ As $f(n)>0$ for all positive integers $n,$ it follows that for all positive integers $a$ and $b$ , $f(a)>f(b)$ if and only if $f(a)^3>f(b)^3.$ So, $f(n)$ is maximized if and only if $f(n)^3=\prod_{i=1}^{k}\frac{(e_1+1)^3}{p_i^{{e_i}}}$ is maximized.
For every factor $\frac{(e_i+1)^3}{p_i^{e_i}}$ with a fixed $p_i,$ where $1\leq i\leq k,$ the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime $p_i=2,3,5,7,\cdots,$ we look for the $e_i$ for which $\frac{(e_i+1)^3}{p_i^{e_i}}$ is a relative maximum: $\begin{array}{c|c|c|c} \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{(e_i+1)^3/\left(p_i^{e_i}\right)} & \textbf{max?} \\ \hline\hline 2 & 0 & 1 & \\ 2 & 1 & 4 & \\ 2 & 2 & 27/4 &\\ 2 & 3 & 8 & \text{yes}\\ 2 & 4 & 125/16 & \\ \hline 3 & 0 & 1 &\\ 3 & 1 & 8/3 & \\ 3 & 2 & 3 & \text{yes}\\ 3 & 3 & 64/27 & \\ \hline 5 & 0 & 1 & \\ 5 & 1 & 8/5 & \text{yes}\\ 5 & 2 & 27/25 & \\ \hline 7 & 0 & 1 & \\ 7 & 1 & 8/7 & \text{yes}\\ 7 & 2 & 27/49 & \\ \hline 11 & 0 & 1 & \text{yes} \\ 11 & 1 & 8/11 & \\ \hline \cdots & \cdots & \cdots & \end{array}$ Finally, the number we seek is $N=2^3 3^2 5^1 7^1 = 2520.$ The sum of its digits is $2+5+2+0=\boxed{\textbf{(E) }9}.$ Actually, once we get that $3^2$ is a factor of $N,$ we know that the sum of the digits of $N$ must be a multiple of $9.$ Only choice $\textbf{(E)}$ is possible.

方法2、Using the answer choices to our advantage, we can show that $N$ must be divisible by 9 without explicitly computing $N$ , by exploiting the following fact: Claim: If $n$ is not divisible by 3, then $f(9n) > f(3n) > f(n)$ . Proof: Since $d(\cdot)$ is a multiplicative function, we have $2021 AMC12A真题与答案$ and $d(9n) = 3d(n)$ . Then $\begin{align*} f(3n) &= \frac{2d(n)}{\sqrt[3]{3n}} \approx 1.38 f(n)\\ f(9n) &= \frac{3d(n)}{\sqrt[3]{9n}} \approx 1.44 f(n) \end{align*}$ Note that the values $\frac{2}{\sqrt[3]{3}}$ and $\frac{3}{\sqrt[3]{9}}$ do not have to be explicitly computed; we only need the fact that $\frac{3}{\sqrt[3]{9}} > \frac{2}{\sqrt[3]{3}} > 1$ which is easy to show by hand.
The above claim automatically implies $N$ is a multiple of 9: if $N$ was not divisible by 9, then $f(9N) > f(N)$ which is a contradiction, and if $N$ was divisible by 3 and not 9, then $f(3N) > f(N) > f\left(\frac{N}{3}\right)$ , also a contradiction. Then the sum of digits of $N$ must be a multiple of 9, so only choice $\boxed{\textbf{(E) } 9}$ works.