2021 AMC12B真题与答案 高清文字版

2021 AMC 12B 真题与答案


Problem 1

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

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

Problem 2

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

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

Problem 3

Suppose\[2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.\]What is the value of $x?$

$\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}$

Problem 4

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$, and the afternoon class’s mean score is $70$. The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac34$. What is the mean of the score of all the students?

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

Problem 5

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

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

Problem 6

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

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

Problem 7

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

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

Problem 8

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

$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$

Problem 9

What is the value of\[\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?\]

$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }\frac54 \qquad \textbf{(D) }2 \qquad \textbf{(E) }\log_2 5$

Problem 10

Two distinct numbers are selected from the set $\{1,2,3,4,\dots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?

$\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Problem 11

Triangle $ABC$ has $AB=13,BC=14$ and $AC=15$. Let $P$ be the point on $\overline{AC}$ such that $PC=10$. There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$

$\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18$

Problem 12

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

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

Problem 13

How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy\[1-3\sin\theta+5\cos3\theta = 0?\]

$\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8$

Problem 14

Let $ABCD$ be a rectangle and let $\overline{DM}$ be a segment perpendicular to the plane of $ABCD$. Suppose that $\overline{DM}$ has integer length, and the lengths of $\overline{MA},\overline{MC},$ and $\overline{MB}$ are consecutive odd positive integers (in this order). What is the volume of pyramid $MABCD?$

$\textbf{(A) }24\sqrt5 \qquad \textbf{(B) }60 \qquad \textbf{(C) }28\sqrt5\qquad \textbf{(D) }66 \qquad \textbf{(E) }8\sqrt{70}$

Problem 15

The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written is $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$[asy] /* Made by samrocksnature */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-2,2); pair G=(-3,2); draw(A--B--C--D--E--A); draw(A--F--A--G); draw(B--F--C); draw(E--G--D); label("A",A,N); label("B",B,W); label("C",C,W); label("D",D,dir(0)); label("E",E,dir(0)); dot(A^^B^^C^^D^^E^^F^^G); [/asy]

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

Problem 16

Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$

$\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}$

Problem 17

Let $ABCD$ be an isosceles trapezoid having parallel bases $\overline{AB}$ and $\overline{CD}$ with $AB>CD.$ Line segments from a point inside $ABCD$ to the vertices divide the trapezoid into four triangles whose areas are $2, 3, 4,$ and $5$ starting with the triangle with base $\overline{CD}$ and moving clockwise as shown in the diagram below. What is the ratio $\frac{AB}{CD}?$[asy] unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.3, 0.9), D=(-0.3, 0.9), P=(0.2, 0.5), E=(0.1, 0.75), F=(0.4, 0.5), G=(0.15, 0.2), H=(-0.3, 0.5); draw(A--B--C--D--cycle, black); draw(A--P, black); draw(B--P, black); draw(C--P, black); draw(D--P, black); label("$A$",A,(-1,0)); label("$B$",B,(1,0)); label("$C$",C,(1,-0)); label("$D$",D,(-1,0)); label("$2$",E,(0,0)); label("$3$",F,(0,0)); label("$4$",G,(0,0)); label("$5$",H,(0,0)); dot(A^^B^^C^^D^^P); [/asy]

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

Problem 18

Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$

$\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4$

Problem 19

Two fair dice, each with at least $6$ faces are rolled. On each face of each dice is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum if $7$ is $\frac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\frac{1}{12}$. What is the least possible number of faces on the two dice combined?

$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20$

Problem 20

Let $Q(z)$ and $R(z)$ be the unique polynomials such that\[z^{2021}+1=(z^2+z+1)Q(z)+R(z)\]and the degree of $R$ is less than $2.$ What is $R(z)?$

$\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1$

Problem 21

Let $S$ be the sum of all positive real numbers $x$ for which\[x^{2^{\sqrt2}}=\sqrt2^{2^x}.\]Which of the following statements is true?

$\textbf{(A) }S<\sqrt2 \qquad \textbf{(B) }S=\sqrt2 \qquad \textbf{(C) }\sqrt2<S<2\qquad \textbf{(D) }2\le S<6 \qquad \textbf{(E) }S\ge 6$

Problem 22

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

[asy] unitsize(4mm); real[] boxes = {0,1,2,3,5,6,13,14,15,17,18,21,22,24,26,27,30,31,32,33}; for(real i:boxes){ draw(box((i,0),(i+1,3))); } draw((8,1.5)--(12,1.5),Arrow()); defaultpen(fontsize(20pt)); label(",",(20,0)); label(",",(29,0)); label(",...",(35.5,0)); [/asy]

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

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

Problem 23

Three balls are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $i$ is $2^{-i}$ for $i=1,2,3,....$ More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is $\frac pq,$ where $p$ and $q$ are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins $3,17,$ and $10.$) What is $p+q?$

$\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59$

Problem 24

Let $ABCD$ be a parallelogram with area $15$. Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.

[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]

Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$

$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$

Problem 25

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

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



Solution 1

1、Since $3\pi$ is about $9.42$, we multiply 9 by 2 and add 1 to get $\boxed{\textbf{(D)}\ ~19}$

2、$|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$. Since $\pi$ is approximately $3.14$$3\pi$ is approximately $9.42$. We are trying to solve for $-9.42<x<9.42$, where $x\in\mathbb{Z}$. Hence, $-9.42<x<9.42$ $\implies$ $-9\leq x\leq9$, for $x\in\mathbb{Z}$. The number of integer values of $x$ is $9-(-9)+1=19$. Therefore, the answer is $\boxed{\textbf{(D)}19}$.


There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\tfrac{64}{2}=\boxed{\textbf{(B)} ~32}$


Subtracting $2$ from both sides and taking reciprocals gives $1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}$. Subtracting $1$ from both sides and taking reciprocals again gives $2+\frac{2}{3+x}=\frac{38}{15}$. Subtracting $2$ from both sides and taking reciprocals for the final time gives $\frac{x+3}{2}=\frac{15}{8}$ or $x=\frac{3}{4} \implies \boxed{\text{A}}$.

Solution 4

1、WLOG, assume there are $3$ students in the morning class and $4$ in the afternoon class. Then the average is $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}$

2、Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$. The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$. Therefore, the answer is $\boxed{\textbf{(C)}76}$.

3、Suppose the morning class has $m$ students and the afternoon class has $a$ students. We have the following chart:\[\begin{array}{c|c|c|c} & \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\ \hline \textbf{Morning} & m & 84 & 84m \\ \hline \textbf{Afternoon} & a & 70 & 70a \end{array}\]

We are also given that $\frac ma=\frac34,$ which rearranges as $m=\frac34a.$

The mean of the scores of all the students is\[\frac{84m+70a}{m+a}=\frac{84\left(\frac34a\right)+70a}{\frac34a+a}=\frac{133a}{\frac74a}=133\cdot\frac47=\boxed{\textbf{(C)} ~76}.\]


The final image of $P$ is $(-6,3)$. We know the reflection rule for reflecting over $y=-x$ is $(x,y) --> (-y, -x)$. So before the reflection and after rotation the point is $(-3,6)$.

By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$. The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$. This means the slope of $P$ and $(1,5)$ is $4$.

Rotations also preserve distance to the center of rotation, and since we only “travelled” up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$.

Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$. The answer is $9-2 = 7 = \boxed{\textbf{(D)}}$.


Solution 6

1、The volume of a cone is $\frac{1}{3} \cdot\pi \cdot r^2 \cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi$.

The volume of a cylinder is $\pi \cdot r^2 \cdot h$ so the volume of the water in the cylinder would be $24\cdot24\cdot\pi\cdot h$.

We can equate these two expressions because the water volume stays the same like this $24\cdot24\cdot\pi\cdot h = 6\cdot144\pi$. We get $4h = 6$ and $h=\frac{6}{4}$.

So the answer is $1.5 = \boxed{\textbf{(A)}}.$

2、The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has $\frac{1}{3}$ of the volume of the cylinder, and so the height is divided by $3$. Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since $2^2 = 4$).

Therefore, the height is divided by $3$ and divided by $4$, which is $18 \div 3 \div 4 = 1.5 = \boxed{\textbf{(A)}}.$

Solution 7

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

2、Prime factorizing $N$, we see $N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}$. The sum of $N$‘s odd divisors are the sum of the factors of $N$ without $2$, and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors. The sum of odd divisors is given by\[a = (1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2)\]and the total sum of divisors is\[(1+2+4+8)(1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2) = 15a\]. Thus, our ratio is\[\frac{a}{15a-a} = \frac{a}{14a} = \frac{1}{14}\]$\boxed{C}$

Solution 8

1、[asy] size(6cm); pair O = (0, 4), A = (0, 5), B = (0, 7), R = (3.873, 5), L = (2.645, 7); draw(O--A--B); draw(O--R); draw(O--L); label("$A$", A, NW); label("$B$", B, N); label("$R$", R, NE); label("$L$", L, N); label("$O$", O, S); label("$d$", O--A, W); label("$2d$", A--B, W*2+0.5*N); label("$r$", O--R, S); label("$r$", O--L, S*0.5 + 1.5 * E); dot(O); dot(A); dot(B); dot(R); dot(L); draw(circle((0, 4), 4)); draw((-3.873, 3) -- (3.873, 3)); draw((-3.873, 5) -- (3.873, 5)); draw((-2.645, 7) -- (2.645, 7)); [/asy]

Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is

\[2d + d = 3d\]

and the distance between each of the chords is just $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we create two different right triangles:

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

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

By the Pythagorean theorem, we can create the following system of equations:

\[19^2 + d^2 = r^2\]

\[17^2 + (3d)^2 = r^2\]

Solving, we find $d = 3$, so $2d = \boxed{\textbf{(B)}\ 6}$.

2、Because we know that the equation of a circle is $(x-a)^2 + (y-b)^2 = r^2$ where the center of the circle is $(a, b)$ and the radius is $r$, we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is $x^2 + y^2 = r^2$. Now, we can set the distance between the chords as $2d$ so the distance from the chord with length 38 to the diameter is $d$.

Therefore, the following points are on the circle as the y-axis splits the chord in half, that is where we get our x value:

$(19, d)$

$(19, -d)$

$(17, -3d)$

Now, we can plug one of the first two value in as well as the last one to get the following equations:

\[19^2 + d^2 = r^2\]

\[17^2 + (3d)^2 = r^2\]

Subtracting these two equations, we get $19^2 - 17^2 = 8d^2$ – therefore, we get $72 = 8d^2 \rightarrow d^2 = 9 \rightarrow d = 3$. We want to find $2d = 6$ because that’s the distance between two chords. So, our answer is $\boxed{B}$.



Note that $\log_{40}{2}=\frac{1}{\log_{2}{40}}$, and similarly $\log_{20}{2}=\frac{1}{\log_{2}{20}}$

\[= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}\]




All the log terms cancel, so the answer is $2\implies\boxed{\text{(D)}}$.


The sum of the first $37$ integers is given by $n(n+1)/2$, so $37(37+1)/2=703$.

Therefore, $703-x-y=xy$

Rearranging, $xy+x+y=703$


Looking at the possible divisors of $704 = 2^6*11$$22$ and $32$ are within the constraints of $0 < x \leq y \leq 37$ so we try those:

$(x+1)(y+1) = 22 * 32$

$x+1=22, y+1 = 32$

$x = 21, y = 31$

Therefore, the difference $y-x=31-21=10$, choice E).


Solution 11 

[asy] size(8cm); pair A = (5,12); pair B = (0,0); pair C = (14,0); pair P = 2/3*A+1/3*C; pair D = 3/2*P; pair E = 3*P; draw(A--B--C--A); draw(A--D); draw(C--E--B); dot("$A$",A,N); dot("$B$",B,W); dot("$C$",C,ESE); dot("$D$",D,N); dot("$P$",P,W); dot("$E$",E,N); defaultpen(fontsize(9pt)); label("$13$", (A+B)/2, NW); label("$14$", (B+C)/2, S); label("$5$",(A+P)/2, NE); label("$10$", (C+P)/2, NE); [/asy]

1、Toss on the Cartesian plane with $A=(5, 12), B=(0, 0),$ and $C=(14, 0)$. Then $\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}$ by the trapezoid condition, where $D, E\in\overline{BP}$. Since $PC=10$, point $P$ is $\tfrac{10}{15}=\tfrac{2}{3}$ of the way from $C$ to $A$ and is located at $(8, 8)$. Thus line $BP$ has equation $y=x$. Since $\overline{AD}\parallel\overline{BC}$ and $\overline{BC}$ is parallel to the ground, we know $D$ has the same $y$-coordinate as $A$, except it’ll also lie on the line $y=x$. Therefore, $D=(12, 12). \, \blacksquare$

To find the location of point $E$, we need to find the intersection of $y=x$ with a line parallel to $\overline{AB}$ passing through $C$. The slope of this line is the same as the slope of $\overline{AB}$, or $\tfrac{12}{5}$, and has equation $y=\tfrac{12}{5}x-\tfrac{168}{5}$. The intersection of this line with $y=x$ is $(24, 24)$. Therefore point $E$ is located at $(24, 24). \, \blacksquare$

The distance $DE$ is equal to the distance between $(12, 12)$ and $(24, 24)$, which is $\boxed{\textbf{(D)} ~12\sqrt{2}}$

2、Using Stewart’s Theorem we find $BP = 8\sqrt{2}$. From the similar triangles $BPA\sim DPC$ and $BPC\sim EPA$ we have\[DP = BP\cdot\frac{PC}{PA} = 2BP\]\[EP = BP\cdot\frac{PA}{PC} = \frac12 BP\]So\[DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}\]

3、Let $x$ be the length $PE$. From the similar triangles $BPA\sim DPC$ and $BPC\sim EPA$ we have\[BP = \frac{PA}{PC}x = \frac12 x\]\[PD = \frac{PA}{PC}BP = \frac14 x\]Therefore $BD = DE = \frac{3}{4}x$. Now extend line $CD$ to the point $Z$ on $AE$, forming parallelogram $ZABC$. As $BD = DE$ we also have $EZ = ZC = 13$ so $EC = 26$.

We now use the Law of Cosines to find $x$ (the length of $PE$):\[x^2 = EC^2 + PC^2 - 2(EC)(PC)\cos{(PCE)} = 26^2 + 10^2 - 2\cdot 26\cdot 10\cos(\angle PCE)\]As $\angle PCE = \angle BAC$, we have (by Law of Cosines on triangle $BAC$)\[\cos(\angle PCE) = \frac{13^2 + 15^2 - 14^2}{2\cdot 13\cdot 15}.\]Therefore\begin{align*} x^2 &= 26^2 + 10^2 - 2\cdot 26\cdot 10\cdot\frac{198}{2\cdot 13\cdot 15}\\ &= 776 - 264\\ &= 512 \end{align*}And $x = 16\sqrt2$. The answer is then $\frac34x = \boxed{\textbf{(D) }12\sqrt2}$

4、Let the brackets denote areas. By Heron’s Formula, we have\begin{align*} [ABC]&=\sqrt{\frac{13+14+15}{2}\left(\frac{13+14+15}{2}-13\right)\left(\frac{13+14+15}{2}-14\right)\left(\frac{13+14+15}{2}-15\right)} \\ &=\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)} \\ &=\sqrt{21\left(8\right)\left(7\right)\left(6\right)} \\ &=\sqrt{\left(3\cdot7\right)\left(2^3\right)\left(7\right)\left(2\cdot3\right)} \\ &=2^2\cdot3\cdot7 \\ &=84. \end{align*}It follows that the height of $ABCD$ is $\frac{2[ABC]}{14}=12.$

Next, we drop the altitudes $\overline{AF}$ and $\overline{DG}$ of $ABCD.$ By the Pythagorean Theorem on $\triangle AFB,$ we get $BF=5.$ By the AA Similarity, $\triangle ADP\sim\triangle CBP,$ with the ratio of similitude $1:2.$ It follows that $AD=7.$ Since $ADGF$ is a rectangle, $FG=AD=7.$ By the Pythagorean Theorem on $\triangle DGB,$ we get $BD=12\sqrt2.$

By $\triangle ADP\sim\triangle CBP$ again, we have $BP=8\sqrt2$ and $DP=4\sqrt2.$ Also, by the AA Similarity, $\triangle ABP\sim\triangle CEP,$ with the ratio of similitude $1:2.$ It follows that $EP=16\sqrt2.$

Finally, $DE=EP-DP=\boxed{\textbf{(D) }12\sqrt2}.$

Solution 12

1、Let $x$ be the greatest integer, $y$ be the smallest, $z$ be the sum of the numbers in S excluding $x$ and $y$, and $k$ be the number of elements in S.

Then, $S=x+y+z$

Firstly, when the greatest integer is removed, $\frac{S-x}{k-1}=32$

When the smallest integer is also removed, $\frac{S-x-y}{k-2}=35$

When the greatest integer is added back, $\frac{S-y}{k-1}=40$

We are given that $x=y+72$

After you substitute $x=y+72$, you have 3 equations with 3 unknowns $S,$$y$ and $k$.




This can be easily solved to yield $k=10$$y=8$$S=368$.

$\therefore$ average value of all integers in the set $=S/k = 368/10 = 36.8$, D)

2、We should plug in $36.2$ and assume everything is true except the $35$ part. We then calculate that part and end up with $35.75$. We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$. The answer is then $\boxed{(D) 36.8}$.


Solution 13

1、First, move terms to get $1+5cos3x=3sinx$. After graphing, we find that there are $\boxed{6}$ solutions (two in each period of $5cos3x$). -dstanz5

2、We can graph two functions in this case: $5\cos{3x}$ and $3\sin{x} -1$.\[\newline\]Using transformation of functions, we know that $5\cos{3x}$ is just a cos function with amplitude 5 and period $\frac{2\pi}{3}$. Similarly, $3\sin{x} -1$ is just a sin function with amplitude 3 and shifted 1 unit downwards. So:[asy] import graph; size(400,200,IgnoreAspect); real Sin(real t) {return 3*sin(t) - 1;} real Cos(real t) {return 5*cos(3*t);} draw(graph(Sin,0, 2pi),red,"$3\sin{x} -1 $"); draw(graph(Cos,0, 2pi),blue,"$5\cos{3x}$"); xaxis("$x$",BottomTop,LeftTicks); yaxis("$y$",LeftRight,RightTicks(trailingzero)); add(legend(),point(E),20E,UnFill); [/asy]We have $\boxed{(A) 6}$


Solution 14

1、This question is just about pythagorean theorem\[a^2+(a+2)^2-b^2 = (a+4)^2\]\[2a^2+4a+4-b^2 = a^2+8a+16\]\[a^2-4a+4-b^2 = 16\]\[(a-2+b)(a-2-b) = 16\]\[a=3, b=7\]With these calculation, we find out answer to be $\boxed{\textbf{(A) }24\sqrt5}$ ~Lopkiloinm

2、Let $\overline{AD}$ be $b$$\overline{CD}$ be $a$$\overline{MD}$ be $x$$\overline{MC}$$\overline{MA}$$\overline{MB}$ be $t$$t-2$$t+2$ respectively.

We have three equations:\[a^2 + x^2 = t^2\]\[a^2 + b^2 + x^2 = t^2 + 4t + 4\]\[b^2 + x^2 = t^2 - 4t + 4\]

Subbing in the first and third equation into the second equation, we get:\[t^2 - 8t - x^2 = 0\]\[(t-4)^2 - x^2 = 16\]\[(t-4-x)(t-4+x) = 16\]Therefore,\[t = 9\],\[x = 3\]Solving for other values, we get $b = 2\sqrt{10}$$a = 6\sqrt{2}$. The volume is then\[\frac{1}{3} abx = \boxed{\textbf{(A)}24\sqrt{5}}\]

Solution 15

1、Let $M$ be the midpoint of $CD$. Noting that $AED$ and $ABC$ are $120-30-30$ triangles because of the equilateral triangles, $AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}$. Also, $[AED]=2\cdot2\cdot\frac{1}{2}\cdot\sin{120^o}=\sqrt{3}$ and so $[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}$.

2、[asy] /* Made by samrocksnature, adapted by Tucker */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-2,2); pair G=(-3,2); draw(A--B--C--D--E--A); draw(A--C--A--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E); dot(F^^G, gray); draw(A--G--A--F, gray); draw(B--F--C, gray); draw(E--G--D, gray); [/asy]

Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$, they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$. For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$. Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}$.


Solution 16

1、Note that $f(1/x)$ has the same roots as $g(x)$, if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have\[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\]so we can see that\[g(x) = \frac{x^3}{c}f(1/x)\]Therefore\[g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}\]

2、Let the three roots of $f(x)$ be $d$$e$, and $f$. (Here e does NOT mean 2.7182818…) We know that $a=-(d+e+f)$$b=de+ef+df$, and $c=-def$, and that $g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}$ (Vieta’s). This is equal to $\frac{def-de-df-ef+d+e+f-1}{def}$, which equals $\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}$. -dstanz5

3、Because the problem doesn’t specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let’s take $f(x) = (x+5)^3 = x^3+15x^2+75x+125$. Then $f(x)$ has a triple root of $x = -5$. Then $g(x)$ has a triple root of $-\frac{1}{5}$, and it’s monic, so $g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}$. We can see that this is $\frac{1+a+b+c}{c}$, which is answer choice $\boxed{(A)}$.

-Darren Yao

4、If we let $p, q,$ and $r$ be the roots of $f(x)$$f(x) = (x-p)(x-q)(x-r)$ and $g(x) = (x-\frac{1}{p})(x-\frac{1}{q})(x-\frac{1}{r})$. The requested value, $g(1)$, is then\[(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \frac{(p-1)(q-1)(r-1)}{pqr}\]The numerator is $-f(1)$ (using the product form of $f(x)$ ) and the denominator is $-c$, so the answer is\[\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}\]


1、Without loss let $\mathcal T$ have vertices $A$$B$$C$, and $D$, with $AB = r$ and $CD = s$. Also denote by $P$ the point in the interior of $\mathcal T$.

Let $X$ and $Y$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$, respectively. Observe that $PX = \tfrac 8r$ and $PY = \tfrac 4s$. Now using the formula for the area of a trapezoid yields\[14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 4\cdot\frac rs + 2\cdot\frac sr.\]Thus, the ratio $\rho := \tfrac rs$ satisfies $2\rho + \rho^{-1} = 4$; solving yields $\rho = \boxed{2+\sqrt 2\textbf{ (B)}}$.

2、Let $b_1$ be the bottom base, $b_2$ be the top base, $h_1$ be the height of the bottom triangle, $h_2$ be the height of the top triangle. Thus, $b_1h_1 = 8, b_2h_2 = 4, (b_1+b_2)(h_1+h_2) = 28,$ so $b_1h_2 + b_2h_1 = 16.$ Let $b_2 = 1, h_2 = 4,$ so we get $b_1h_1 = 8, 4b_1+h_1 = 16.$ This gives us a quadratic in $b_1,$ ie. $4b_1^2+8=16b_1,$ so $b_1 = \boxed{2+\sqrt{2}}.$

Solution 18

1、Using the fact $z\bar{z}=|z|^2$, the equation rewrites itself as

\[12z\bar{z}=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31\]\[-12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32=0\]\[\left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)=0\]\[(z+\bar{z}+2)^2+(z\bar{z}-6)^2=0.\]As the two quantities in the parentheses are real, both quantities must equal $0$ so\[z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.\]

2、The answer being in the form $z+\frac 6z$ means that there are two solutions, some complex number and its complex conjugate.\[a+bi = \frac{6}{a-bi}\]\[a^2+b^2=6\]We should then be able to test out some ordered pairs of $(a, b)$. After testing it out, we get the ordered pairs of $(-1, \sqrt{5})$ and its conjugate $(-1, -\sqrt{5})$. Plugging this into answer format gives us $\boxed{\textbf{(A) }-2}$ ~Lopkiloinm

3、Let $x = z + \frac{6}{z}$. Then $z = \frac{x \pm \sqrt{x^2-24}}{2}$. From the answer choices, we know that $x$ is real and $x^2<24$, so $z = \frac{x \pm i\sqrt{24-x^2}}{2}$. Then we have\[|z|^2 = 6\]\[|z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10\]\[|z^2+1|^2 = |xz -6 +1|^2 = (\frac{x^2}{2}-5)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25\]Plugging the above back to the original equation, we have\[12*6 = 2(2x+10) + x^2 + 25 + 31\]\[(x+2)^2 = 0\]So $x = \boxed{\textbf{(A) }-2}$.


4、There are actually several ways to see that $|z|^2 = 6.$ I present two troll ways of seeing it, and a legitimate way of checking.

Rewrite using $w \overline{w} = |w|^2$

$12z \overline{z} + 2(z+2)(\overline{z} + 2) + (z^2+1)(\overline{z}^2+1)+31$ $12 z \overline{z} = 2z \overline{z} + 4z + 4 \overline{z} + 8 + z^2 \overline{z}^2+z^2+\overline{z}^2 + 1 + 31.$ $12 z \overline{z} = 4(z + \overline{z}) + (z \overline{z})^2 + (z + \overline{z})^2 + 40.$

Symmetric in $z$ and $\overline{z},$ so if $w$ is a sol, then so is $\overline{w}$

TROLL OBSERVATION #1: ALL THE ANSWERS ARE REAL. THUS, $z + \frac{6}{z} \in \mathbb{R},$ which means they must be conjugates and so $|z|^2 = 6.$

TROLL OBSERVATION #2: Note that $z+\frac{6}{z} = \overline{z} + \frac{6}{\overline{z}}$ because either solution must give the same answer! which means that $|z|^2 = 6.$

Alternatively, you can check: Let $a = w + \overline{w} \in \mathbb{R},$ and $r = |w|^2 \in \mathbb{R}.$ Thus, we have $a^2+4a+40+r^2-12r=0,$ and the discriminant of this must be nonnegative as $a$ is real. Thus, $16-4(40+r^2-12r) \geq 0$ or $(r-6)^2 \leq 0,$ which forces $r = 6,$ as claimed.

Thus, we plug in $z \overline{z} = 6,$ and get: $72 = 4(z + \overline{z}) + 76 + (z + \overline{z})^2,$ ie. $(z+\overline{z})^2 + 4(z + \overline{z}) + 4 = 0,$ or $(z+\overline{z} + 2)^2 = 0,$ which means $z + \overline{z} = \boxed{-2}$ and that’s our answer since we know $\overline{z} = 6 / z$

– ccx09

5、Observe that all the answer choices are real. Therefore, $z$ and $\frac{6}{z}$ must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product ($6$) to be real. Thus $|z|=|\tfrac{6}{z}|=\sqrt{6}$. We will test all the answer choices, starting with $\textbf{(A)}$. Suppose the answer is $\textbf{(A)}$. If $z+\tfrac{6}{z}=-2$ then $z^{2}+2z+6=0$ and $z=\frac{-2\pm\sqrt{4-24}}{2}=-1\pm\sqrt{5}i$. Note that if $z=-1+\sqrt{5}i$ works, then so does $-1-\sqrt{5}i$. It is relatively easy to see that if $z=-1+\sqrt{5}i$, then $12|z|^{2}=12\cdot 6=72, 2|z+2|^{2}=2|1+\sqrt{5}i|=2\cdot 6=12, |z^{2}+1|^{2}=|-3-2\sqrt{5}i|^{2}=29,$ and $72=12+29+31$. Thus the condition\[12|z|^{2}=2|z+2|^{2}+|z^{2}+1|^{2}+31\]is satisfied for $z+\tfrac{6}{z}=-2$, and the answer is $\boxed{\textbf{(A)} ~-2}$.


Suppose the dice have $a$ and $b$ faces, and WLOG $a\geq{b}$. Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$. As a result, there must be $\tfrac{4}{3}\cdot6=8$ ways to sum to $10$. There are at most nine distinct ways to get a sum of $10$, which are possible whenever $a,b\geq{9}$. To achieve exactly eight ways, $b$ must have $8$ faces, and $a\geq9$. Let $n$ be the number of ways to obtain a sum of $12$, then $\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a$. Since $b=8$$n\leq8\implies a\leq{12}$. In addition to $3\mid{a}$, we only have to test $a=9,12$, of which both work. Taking the smaller one, our answer becomes $a+b=9+8=\boxed{\textbf{(B)17}}$.

Solution 20

1、Note that\[z^3-1\equiv 0\pmod{z^2+z+1}\]so if $F(z)$ is the remainder when dividing by $z^3-1$,\[F(z)\equiv R(z)\pmod{z^2+z+1}.\]Now,\[z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1\]So $F(z) = z^2+1$, and\[R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}\]The answer is $\boxed{\textbf{(A) }-z}.$

2、Instead of dealing with a nasty $z^2+z+1$, we can instead deal with the nice $z^3 - 1$, as $z^2+z+1$ is a factor of $z^3-1$. Then, we try to see what $\frac{z^{2021} + 1}{z^3 - 1}$ is. Of course, we will need a $z^{2018}$, getting $z^{2021} - z^{2018}$. Then, we’ve gotta get rid of the $z^{2018}$ term, so we add a $z^{2015}$, to get $z^{2021} - z^{2015}$. This pattern continues, until we add a $z^2$ to get rid of $z^5$, and end up with $z^{2021} - z^2$. We can’t add anything more to get rid of the $z^2$, so our factor is $z^{2018} + z^{2015} + z^{2012} + \cdots + z^2$. Then, to get rid of the $z^2$, we must have a remainder of $+z^2$, and to get the $+1$ we have to also have a $+1$ in the remainder. So, our product is\[z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1.\]Then, our remainder is $z^2+1$. The remainder when dividing by $z^3-1$ must be the same when dividing by $z^2+z+1$, modulo $z^2+z+1$. So, we have that $z^2 + 1 \equiv R(z) \pmod{z^2+z+1}$, or $R(z) \equiv -z\pmod{z^2+z+1}$. This corresponds to answer choice $\boxed{\textbf{(A)} ~ -z}$. ~rocketsri

3、One thing to note is that $R(z)$ takes the form of $Az + B$ for some constants A and B. Note that the roots of $z^2 + z + 1$ are part of the solutions of $z^3 -1 = 0$ They can be easily solved with roots of unity:\[z^3 = 1\]\[z^3 = e^{i 0}\]\[z = e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i -\frac{2\pi}{3}}\]\[\newline\]Obviously the right two solutions are the roots of $z^2 + z + 1 = 0$ We substitute $e^{i \frac{2\pi}{3}}$ into the original equation, and $z^2 + z + 1$ becomes 0. Using De Moivre’s theorem, we get:\[e^{i\frac{4042\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B\]\[e^{i\frac{4\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B\]Expanding into rectangular complex number form:\[\frac{1}{2} - \frac{\sqrt{3}}{2} i = (-\frac{1}{2}A + B) + \frac{\sqrt{3}}{2} i A\]Comparing the real and imaginary parts, we get:\[A = -1, B = 0\]The answer is $\boxed{\textbf{(A) }-z}$. ~Jamess2022(burntTacos;-;)

4、Note that the equation above is in the form of polynomial division, with $z^{2021}+1$ being the dividend, $z^2+z+1$ being the divisor, and $Q(x)$ and $R(x)$ being the quotient and remainder respectively. Since the degree of the dividend is $2021$ and the degree of the divisor is $2$, that means the degree of the quotient is $2021-2 = 2019$. Note that R(x) can’t influence the degree of the right hand side of this equation since its degree is either $1$ or $0$. Since the coefficients of the leading term in the dividend and the divisor are both $1$, that means the coefficient of the leading term of the quotient is also $1$. Thus, the leading term of the quotient is $z^{2019}$. Multiplying $z^{2019}$ by the divisor gives $z^{2021}+z^{2020}+z^{2019}$. We have our $z^{2021}$ term but we have these unnecessary terms like $z^{2020}$. We can get rid of these terms by adding $-z^{2018}$ to the quotient to cancel out these terms, but this then gives us $z^{2021}-z^{2018}$. Our first instinct will probably be to add $z^{2017}$, but we can’t do this as although this will eliminate the $-z^{2018}$ term, it will produce a $z^{2019}$ term. Since no other term of the form $z^n$ where $n$ is an integer less than $2017$ will produce a $z^{2019}$ term when multiplied by the divisor, we can’t add $z^{2017}$ to the quotient. Instead, we can add $z^{2016}$ to the coefficient to get rid of the $-z^{2018}$ term. Continuing this pattern, we get the quotient as\[z^{2019}-z^{2018}+z^{2016}-z^{2015}+....-z^2+1.\]The last term when multiplied with the divisor gives $z^2+z+1$. This will get rid of the $-z^2$ term but will produce the expression $z+1$, giving us the dividend as $z^{2021}+z+1$. Note that the dividend we want is of the form $z^{2021}+1$. Therefore, our remainder will have to be $-z$ in order to get rid of the $z$ term in the expression and give us $z^{2021}+1$, which is what we want. Therefore, the remainder is $\boxed{\textbf{(A) }-z \qquad}$

Solution 21

1、Note that this solution is not recommended unless you’re running out of time.

Upon pure observation, it is obvious that one solution to this equality is $x=\sqrt{2}$. From this, we can deduce that this equality has two solutions, since $\sqrt{2}^{2^{x}}$ grows faster than $x^{2^{\sqrt{2}}}$ (for greater values of $x$) and $\sqrt{2}^{2^{x}}$ is greater than $x^{2^{\sqrt{2}}}$ for $x<\sqrt{2}$ and less than $x^{2^{\sqrt{2}}}$ for $\sqrt{2}<x<n$, where $n$ is the second solution. Thus, the answer cannot be $\text{A}$ or $\text{B}$. We then start plugging in numbers to roughly approximate the answer. When $x=2$$x^{2^{\sqrt{2}}}>\sqrt{2}^{2^{x}}$, thus the answer cannot be $\text{C}$. Then, when $x=4$$x^{2^{\sqrt{2}}}=4^{2^{\sqrt{2}}}<64<\sqrt{2}^{2^{x}}=256$. Therefore, $S<4+\sqrt{2}<6$, so the answer is $\boxed{\textbf{(D) } 2 \le S < 6}$. ~Baolan

2、$x^{2^{\sqrt{2}}} = {\sqrt{2}}^{2^x}$

$2^{\sqrt{2}} \log x = 2^{x} \log \sqrt{2}$ (At this point we see by inspection that $x=\sqrt{2}$ is a solution.)

$\sqrt{2} \log 2 + \log \log x = x \log 2 + \log \log \sqrt{2}$

$\sqrt{2} + \log_2 \log_2 x = x + \log_2 \log_2 \sqrt{2} = x -1.$

$\log_2 \log_2 x = x - 1 - \sqrt{2}.$

LHS is a line. RHS is a concave curve that looks like a logarithm and has $x$ intercept at $(2,0).$ There are at most two solutions, one of which is $\sqrt{2}.$ But note that at $x=2,$ we have $\log_2 \log_2 (2) = 0 > 2 - 1 - \sqrt{2},$ meaning that the log log curve is above the line, so it must intersect the line again at a point $x > 2.$ Now we check $x=4$ and see that $\log_2 \log_2 (4) = 1 < 4 - 1 - \sqrt{2},$ which means at $x=4$ the line is already above the log log curve. Thus, the second solution lies in the interval $(2,4).$ The answer is $\boxed{2 \leq S < 6}.$


1、First we note that symmetrical positions are losing for the player to move. Then we start checking small positions. $(n)$ is always winning for the first player. Furthermore, $(3, 2, 1)$ is losing and so is $(4, 1).$ We look at all the positions created from $(6, 2, 1),$ as $(6, 1, 1)$ is obviously winning by playing $(2, 2, 1, 1).$ There are several different positions that can be played by the first player from $(6, 2, 1).$ They are $(2, 2, 2, 1), (1, 3, 2, 1), (4, 2, 1), (6, 1), (5, 2, 1), (4, 1, 2, 1), (3, 2, 2, 1).$ Now we list refutations for each of these moves:

$(2, 2, 2, 1) - (2, 1, 2, 1)$

$(1, 3, 2, 1) - (3, 2, 1)$

$(4, 2, 1) - (4, 1)$

$(6, 1) - (4, 1)$

$(5, 2, 1) - (3, 2, 1)$

$(4, 1, 2, 1) - (2, 1, 2, 1)$

$(3, 2, 2, 1) - (1, 2, 2, 1)$

This proves that $(6, 2, 1)$ is losing for the first player.

-Note: In general, this game is very complicated. For example $(8, 7, 5, 3, 2)$ is winning for the first player but good luck showing that.

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

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

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

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

That leaves $(6,2,1)$ or $\boxed{\textbf{(B)}}$.

3、Let the nim-value of the ending game state, where someone has just removed the final brick, be $0$. Then, any game state with a nim-value of $0$ is losing. It is well-known that the nim-value of a supergame (a combination of two or more individual games) is the binary xor function on the nim-values of the individual games that compose the supergame. Therefore, we calculate the nim-values of the states with a single wall up to $6$ bricks long (since the answer choices only go up to $6$).

First, the game with $1$ brick has a nim-value of $1$.

Similarly, the game with $2$ bricks has a nim-value of $2$.

Next, we consider a $3$ brick wall. After the next move, the possible resulting game states are $1$ brick, a $2$ brick wall, or $2$ separate bricks. The first two options have nim-values of $1$ and $2$. The final option has a nim-value of $1\oplus 1 = 0$, so the nim-value of this game state is $3$.

Next, the $4$ brick wall. The possible states are a $2$ brick wall, a $3$ brick wall, a $2$ brick wall and a $1$ brick wall, or a $1$ brick wall and a $1$ brick wall. The nim-values of these states are $2$$3$$3$, and $0$, respectively, and hence the nim-value of this game state is $1$. (Wait why is the nim-value of it $1$? – awesomediabrine)

The possible game states after the $5$ brick wall are the following: a $3$ brick wall, a $4$ brick wall, a $3$ brick wall and a $1$ brick wall, a two $2$ brick walls, and a $2$ brick wall plus a $1$ brick wall. The nim-values of these are $3$$1$$2$$0$, and $3$, respectively, meaning the nim-value of a $5$ brick wall is $4$.

Finally, we find the nim-value of a $6$ brick wall. The possible states are a $5$ brick wall, a $4$ brick wall and a $1$ brick wall, a $3$ brick wall and a $2$ brick wall, a $4$ brick wall, a $3$ brick wall and a $1$ brick wall, and finally two $2$ brick walls. The nim-values of these game states are $4$$0$$1$$1$$2$, and $0$, respectively. This means the $6$ brick wall has a nim-value of $3$.

The problem is asking which of the answer choices is losing, or has a nim-value of $0$. We see that option $\textbf{(A)}$ has a nim-value of $3\oplus1\oplus1=3$, option $\textbf{(B)}$ has a nim-value of $3\oplus2\oplus1=0$, option $\textbf{(C)}$ has a nim-value of $3\oplus2\oplus2=3$, option $\textbf{(D)}$ has a nim-value of $3\oplus3\oplus1=1$, and option $\textbf{(E)}$ has a nim-value of $3\oplus3\oplus2=2$, so the answer is $\boxed{\textbf{(B) }(6, 2, 1)}$.

This method can also be extended to solve the note after the first solution. The nim-values of the $7$ brick wall and the $8$ brick wall are $2$ and $1$, using the same method as above. The nim-value of $(8, 7, 5, 3, 2)$ is therefore $1\oplus2\oplus4\oplus3\oplus2 = 5$, which is winning.


1、”Evenly spaced” just means the bins form an arithmetic sequence.

Suppose the middle bin in the sequence is $x$. There are $x-1$ different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these $3$ bins are chosen is $6\cdot 2^{-3x} = 6\cdot \frac{1}{8^x}$, so the probability $x$ is the middle bin is $6\cdot\frac{x-1}{8^x}$. Then, we want the sum\begin{align*} 6\sum_{x=2}^{\infty}\frac{x-1}{8^x} &= \frac{6}{8}\left[\frac{1}{8} + \frac{2}{8^2} + \frac{3}{8^3}\cdots\right]\\ &= \frac34\left[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}\right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4}\right) + \cdots\right]\\ &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}\right)\right]\\ &= \frac34\cdot \frac{8}{49}\\ &= \frac{6}{49} \end{align*}The answer is $6+49=\boxed{\textbf{(A) }55}.$

2、As in solution 1, note that “evenly spaced” means the bins are in arithmetic sequence. We let the first bin be $a$ and the common difference be $d$. Further note that each $(a, d)$ pair uniquely determines a set of 3 bins.

We have $1 \leq a \leq \infty$ because the leftmost bin in the sequence can be any bin, and $1 \leq d \leq \infty$, because the bins must be distinct.

This gives us the following sum for the probability:\begin{align*} 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a-3d} &= 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a} \cdot 2^{-3d} \\ &= 6 \left( \sum_{a=1}^{\infty} 2^{-3a} \right) \left( \sum_{d=1}^{\infty} 2^{-3d} \right) \\ &= 6 \left( \sum_{a=1}^{\infty} 8^{-a} \right) \left( \sum_{d=1}^{\infty} 8^{-d} \right) \\ &= 6 \left( \frac{1}{7} \right) \left( \frac{1}{7} \right) \\ &= \frac{6}{49} .\end{align*}Therefore the answer is $6 + 49 = 55$, which is choice (A).

-Darren Yao

3、This is a slightly messier variant of solution 2. If the first ball is in bin $i$ and the second ball is in bin $j>i$, then the third ball is in bin $2j-i$. Thus the probability is\begin{align*} 6\sum_{i=1}^{\infty}\sum_{j=i+1}^\infty2^{-i}2^{-j}2^{-2j+i}&=6\sum_{i=1}^{\infty}\sum_{j=i+1}^\infty2^{-3j}\\ &=6\sum_{i=1}^{\infty}\left(\frac{2^{-3(i+1)}}{1-\tfrac{1}{8}}\right)\\ &=6\sum_{i=1}^\infty\frac{8}{7}\cdot2^{-3}\cdot2^{-3i}\\ &=\frac{6}{7}\sum_{i=1}^\infty2^{-3i}\\ &=\frac{6}{7}\frac{2^{-3}}{1-\tfrac18} = \frac{6}{49}. \end{align*}


1、Let $X$ denote the intersection point of the diagonals $AC$ and $BD$. Remark that by symmetry $X$ is the midpoint of both $\overline{PQ}$ and $\overline{RS}$, so $XP = XQ = 3$ and $XR = XS = 4$. Now note that since $\angle APB = \angle ARB = 90^\circ$, quadrilateral $ARPB$ is cyclic, and so\[XR\cdot XA = XP\cdot XB,\]which implies $\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34$.

Thus let $x> 0$ be such that $XA = 3x$ and $XB = 4x$. Then Pythagorean Theorem on $\triangle APX$ yields $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$, and so\[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.\]Solving this for $x^2$ yields $x^2 = \tfrac12 + \tfrac{\sqrt{41}}8$, and so\[(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.\]The requested answer is $32 + 8 + 41 = \boxed{81}$.

2、Let $X$ denote the intersection point of the diagonals $AC$ and $BD,$ and let $\theta = \angle{COB}$. Then, by the given conditions, $XR = 4,$ $XQ = 3,$ $[XCB] = \frac{15}{4}$. So,\[XC = \frac{3}{\cos \theta}\]\[XB \cos \theta = 4\]\[\frac{1}{2} XB XC \sin \theta = \frac{15}{4}\]Combining the above 3 equations, we get\[\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}.\]Since we want to find $d^2 = 4XB^2 = \frac{64}{\cos^2 \theta},$ we let $x = \frac{1}{\cos^2 \theta}.$ Then\[\frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.\]Solving this, we get $x = \frac{4 + \sqrt{41}}{8},$ so $d^2 = 64x = 32 + 8\sqrt{41}$$\boxed{81}$

3、Let $X$ be the intersection of diagonals $AC$ and $BD$. By symmetry $[\triangle XCB] = \frac{15}{4}$$XQ = 3$ and $XR = 4$, so now we have reduced all of the conditions one quadrant. Let $CQ = x$$XC = \sqrt{x^2+9}$$RB = \frac{4x}{3}$ by similar triangles and using the area condition we get $\frac{4}{3} \cdot x \cdot \sqrt{x^2+9} = \frac{15}{2}$. Note that it suffices to find $OB = \frac{4}{3}\sqrt{x^2+9}$ because we can double and square it to get $d^2$. Solving for $a = x^2$ in the above equation, and then using $d^2 = \frac{64}{9}(x^2+9) = 8\sqrt{41} + 32 \Rightarrow \boxed{81}$.

4、Again, Let $X$ be the intersection of diagonals $AC$ and $BD$. Note that triangles $\triangle QXC$ and $\triangle BXR$ are similar because they are right triangles and share $\angle CXQ$. First, call the length of $XB = \frac{d}{2}$. By the definition of an area of a parallelogram, $CQ \cdot 2XB = 15$, so $CQ = \frac{15}{d}$. Using similar triangles on $\triangle QXC$ and $\triangle BXR$$\frac{CQ}{XQ} = \frac{BR}{XR}$. Therefore, finding $BR$$BR = \frac{XR}{XQ} \cdot CQ = \frac{4}{3} \cdot \frac{15}{d} = \frac{20}{d}$. Now, applying the Pythagorean theorem once, we find $(\frac{20}{d}) ^2$ + $(4)^2$ = $(\frac{d}{2}) ^2$. Solving this equation for $d^2$, we find $d^2=\frac{64+\sqrt{4096+6400}}{2}=32+8\sqrt{41} \Rightarrow \boxed{81}$.

Solution 25

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

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

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

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

\[\sum_{x=1}^{30}(\lfloor \frac{2}{3}x \rfloor) = 0 + 1 + 2 + 2 + 3 + \cdots + 18 + 18 + 19 + 20\]\[=\frac{20(21)}{2} + 2 + 4 + 6 + \cdots + 18\]\[=210 + \frac{20}{2}\cdot 9\]\[=300\]

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

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

In other words, we are looking for the first $m > \frac{2}{3}$ that is expressible as a ratio of positive integers $\frac{p}{q}$ with $q \le 30.$ For each $q=1,\dots,30$, the smallest multiple of $\frac{1}{q}$ which exceeds $\frac{2}{3}$ is $1, \frac{2}{2}, \frac{3}{3}, \frac{3}{4}, \frac{4}{5}, \cdots , \frac{19}{27}, \frac{19}{28}, \frac{20}{29}, \frac{21}{30}$ respectively, and the smallest of these is $\frac{19}{28}.$ Note: start listing the multiples of $\frac{1}{q}$ from $\frac{21}{30}$ and observe that they get further and further away from $\frac{2}{3}.$ Alternatively, see the method of finding upper bounds in solution 2.

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


2、I know that I want about $\frac{2}{3}$ of the box of integer coordinates above my line. There are a total of 30 integer coordinates in the desired range for each axis which gives a total of 900 lattice points. I estimate that the slope, m, is $\frac{2}{3}$. Now, although there is probably an easier solution, I would try to count the number of points above the line to see if there are 600 points above the line. The line $y=\frac{2}{3}x$ separates the area inside the box so that $\frac{2}{3}$ of the are is above the line.

I find that the number of coordinates with $x=1$ above the line is 30, and the number of coordinates with $x=2$ above the line is 29. Every time the line $y=\frac{2}{3}x$ hits a y-value with an integer coordinate, the number of points above the line decreases by one. I wrote out the sum of 30 terms in hopes of finding a pattern. I graphed the first couple positive integer x-coordinates, and found that the sum of the integers above the line is $30+29+28+28+27+26+26 \ldots+ 10$. The even integer repeats itself every third term in the sum. I found that the average of each of the terms is 20, and there are 30 of them which means that exactly 600 above the line as desired. This give a lower bound because if the slope decreases a little bit, then the points that the line goes through will be above the line.

To find the upper bound, notice that each point with an integer-valued x-coordinate is either $\frac{1}{3}$ or $\frac{2}{3}$ above the line. Since the slope through a point is the y-coordinate divided by the x-coordinate, a shift in the slope will increase the y-value of the higher x-coordinates. We turn our attention to $x=28, 29, 30$ which the line $y=\frac{2}{3}x$ intersects at $y= \frac{56}{3}, \frac{58}{3}, 20$. The point (30,20) is already counted below the line, and we can clearly see that if we slowly increase the slope of the line, we will hit the point (28,19) since (28, $\frac{56}{3}$) is closer to the lattice point. The slope of the line which goes through both the origin and (28,19) is $y=\frac{19}{28}x$. This gives an upper bound of $m=\frac{19}{28}$.

Taking the upper bound of m and subtracting the lower bound yields $\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$. This is answer $1+84=$ $\boxed{\textbf{(E)} ~85}$.





[asy] /* Created by Brendanb4321 */ import graph; size(16cm); defaultpen(fontsize(9pt)); xaxis(0,30,Ticks(1.0)); yaxis(0,25,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30/28*19), dotted); int c = 0; for (int i = 1; i<=30; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); label("$m=2/3$", (32,20)); label("$m=19/28$", (32.3,20.8)); [/asy]