AMC12 2020 B

A direct calculation shows $\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16} = 1 + 2 + 3 + 4 = 10$. Note: This problem illustrates the fact that the sum of the first $n$ odd integers equals $n^2$.

Because $a^2 - b^2$ factors as $(a-b)(a+b)$, the given expression simplifies to 1: $\frac{100^2 - 7^2}{70^2 - 11^2} \cdot \frac{(70 - 11)(70 + 11)}{(100 - 7)(100 + 7)} = \frac{100^2 - 7^2}{(100 - 7)(100 + 7)} \cdot \frac{(70 - 11)(70 + 11)}{70^2 - 11^2} = 1$.

Because $\frac{w}{y} = \frac{w}{x} \cdot \frac{x}{z} \cdot \frac{z}{y} = \frac{4}{3} \cdot 6 \cdot \frac{2}{3} = \frac{16}{3}$, the requested ratio is $16:3$.

Consider in order the first five prime numbers that are possible values of $b$, namely 2, 3, 5, 7, and 11. The corresponding values of $a = 90 - b$ are 88, 87, 85, 83, and 79. The first prime in this latter list is 83, so $b = 7$ is the least value for $b$ for which both $a$ and $b$ are prime and $a > b$.

Let $g$ be the number of games played by A; then B played $g + 7 + 7$ games. Focusing on the number of games won by B implies $\frac{2}{3}g + 7 = \frac{5}{8}(g + 14)$. Solving this equation gives $g = 42$.

For all nonnegative integers $n$, $$\frac{(n+2)! - (n+1)!}{n!} = (n+2)(n+1) - (n+1) = (n+1)(n+2-1) = (n+1)^2,$$ a perfect square. When $n = 10$, the value is 121, which is not a multiple of 4, a multiple of 10, a prime number, or a perfect cube.

Let $\theta_1$ and $\theta_2$ be the angles that the two lines make with the rightward-pointing $x$-axis, and let $m$ and $6m$ be their slopes. Without loss of generality, $\tan \theta_1 = m$ and $\tan \theta_2 = 6m$. It follows from the identity for the tangent of the difference of two angles that $$1 = \tan 45^\circ = \tan(\theta_1 - \theta_2) = \frac{m - 6m}{1 + 6m^2}$$ or $$1 = \tan 45^\circ = \tan(\theta_2 - \theta_1) = \frac{6m - m}{1 + 6m^2}.$$ Therefore $6m^2 \pm 5m + 1 = 0$, and $m = \pm \frac{1}{2}$ or $\pm \frac{1}{3}$. Then $6m = \pm 3$ or $\pm 2$, respectively. The product of the slopes is either $(\pm \frac{1}{2})( \pm 3) = \frac{3}{2}$ or $(\pm \frac{1}{3})( \pm 2) = \frac{2}{3}$, so the greatest possible value of the product of the slopes is $\frac{3}{2}$. For example, the lines could have equations $y = \frac{1}{2}x$ and $y = 3x$.

The given equation is equivalent to $$x^{2020} + y^2 - 2y + 1 = 1,$$ which is equivalent to $x^{2020} + (y - 1)^2 = 1$. The only way to write 1 as the sum of the squares of two integers is as $0 + 1$ or $1 + 0$, so $x$ must be $0$ or $\pm 1$. The complete set of solutions is $\{(0,0), (0,2), (1,1), (-1,1)\}$, which has 4 elements.

The length of the circular portion of the sector is $\frac{3}{4} \cdot 2\pi \cdot 4 = 6\pi$. That arc forms the circumference of the base of the cone, so the radius of the base of the cone is $r = \frac{6\pi}{2\pi} = 3$. The slant height of the cone is the radius of the sector. By the Pythagorean Theorem the cone's height is $h = \sqrt{4^2 - 3^2} = \sqrt{7}$. The volume of the cone is $$\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \cdot 3^2 \cdot \sqrt{7} = 3\pi\sqrt{7}.$$

Let Q be the midpoint of $\overline{AB}$. By the Power of a Point Theorem, $AP \cdot AM = AQ^2 = \frac{1}{4}$. By the Pythagorean Theorem, $AM = \frac{1}{2}\sqrt{5}$. Thus $$AP = \frac{\frac{1}{4}}{\frac{1}{2}\sqrt{5}} = \frac{\sqrt{5}}{10}.$$

Consider the semicircles on two adjacent sides of the hexagon, where $F$ and $G$ are the centers and midpoints of the sides. Because $\angle FBG = 120^\circ$, $\angle FBD = 60^\circ$ and $\triangle FBD$ is equilateral. The segment of the circle at $F$ by arc $BD$ has area $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$. The area is hexagon area minus six semicircles plus 12 segments: $6 \left( \frac{\sqrt{3}}{4} \cdot 4 \right) - 6 \left( \frac{\pi \cdot 1^2}{2} \right) + 12 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = 3\sqrt{3} - \pi$.

Answer (E): Let $\overline{DF}$ be a chord of the circle that is perpendicular to the diameter $\overline{AB}$, and let $G$ be the intersection of $\overline{AB}$ and $\overline{DF}$. Let $O$ be the center of the circle. Because $\triangle EGD$ and $\triangle EGF$ are congruent and $\angle GED = 45^\circ$, it follows that $\triangle DEF$ is a right isosceles triangle with $EF = ED$. Then $\angle CDF = \angle EDF = 45^\circ$. Because $\angle COF$ subtends arc $CF$, it follows that $\angle COF = 2\angle CDF = 90^\circ$. Because $\triangle CEF$ and $\triangle COF$ are right triangles, $$ CE^2 + EF^2 = CF^2 = CO^2 + OF^2 = 2(5\sqrt{2})^2 = 100. $$ Thus $CE^2 + ED^2 = CE^2 + EF^2 = 100$. Notice that the position of $E$ on $\overline{AB}$ is irrelevant.

Answer (D): Let $a=\sqrt{\log_2 3}$. Note that $\frac{1}{a}=\sqrt{\log_3 2}$. Then \[ \sqrt{\log_2 6+\log_3 6} =\sqrt{\log_2 3+\log_3 2+2} =\sqrt{a^2+2+\frac{1}{a^2}} =a+\frac{1}{a} =\sqrt{\log_2 3}+\sqrt{\log_3 2}. \] It remains to show that the other choices given are not equal to $\sqrt{\log_2 6+\log_3 6}$, and to do so it is enough to show that the choices are in strictly increasing order. Because $6>5$ it follows that $1<\sqrt{\log_5 6}$. Because $6<25$ it follows that $\sqrt{\log_5 6}<2$. The Arithmetic Mean-Geometric Mean Inequality and the fact that $\log_2 3\ne \log_3 2$ imply that \[ \frac{\sqrt{\log_2 3}+\sqrt{\log_3 2}}{2}>\sqrt{\sqrt{\log_2 3}\cdot \sqrt{\log_3 2}}=1, \] from which it follows that $2<\sqrt{\log_2 3}+\sqrt{\log_3 2}$. Finally each term of choice (D) is less than the corresponding term in choice (E), so the sum is less as well.

Answer (A): Bela has a straightforward strategy that will guarantee winning. At his first turn he chooses the number $\frac{n}{2}$. This splits the playing field into two parts—the numbers less than $\frac{n}{2}$ and the numbers greater than $\frac{n}{2}$. Thereafter, whatever legal move Jenn makes, Bela makes the symmetric move in the other half of the playing field. Specifically, if Jenn chooses $x$, then Bela chooses $n-x$. By symmetry, if Jenn’s move is possible, so is Bela’s. Thus the first player to be unable to make a legal move will be Jenn, and she will lose. The game will end after a finite number of moves, because of the restriction that the chosen numbers are at least one unit apart. Note: The game is much more interesting if the chosen numbers are restricted to being integers, in which case it is known as Dawson’s Chess. See Winning Ways for Your Mathematical Plays by Berlekamp, Conway, and Guy for a mathematical theory of games like this.

Answer (C): This problem can be modeled by a regular decagon with the 5 diagonals connecting each point to the point opposite to it. A line segment between two vertices indicates that the people represented by those vertices know each other. See the figure below. The required pairings are then sets of 5 of the 15 line segments in this figure such that no 2 line segments in a set share an endpoint. (This is known as a perfect matching in graph theory.) The number of pairings can be counted by focusing on the number of diagonals used in the pairing. • There is 1 pairing using all 5 diagonals, and any pairing that includes 4 of the diagonals must use all of them. • Suppose a pairing uses exactly 3 diagonals. If 3 of the endpoints of these diagonals are consecutive points on the decagon, then a pairing can be formed by including sides of the decagon as the other 2 segments. There are 5 different pairings of this form. On the other hand, any pairing that includes 3 of the diagonals whose endpoints are not consecutive points on the decagon must use all the diagonals. • There are no pairings using exactly 2 of the diagonals, because any pairing that includes 2 of the diagonals will force at least 3 of the diagonals to be used. • If a pairing uses exactly 1 diagonal, then the rest of the pairing is uniquely determined. There are 5 different pairings of this form. • If a pairing uses no diagonals, then it must use nonadjacent sides of the decagon, and there are 2 different pairings of this form. As shown in the figure, the requested number of pairings is $1+5+5+2=13$.

Answer (B): By symmetry it may be assumed without loss of generality that George’s first draw is red and therefore the urn contains two red balls and one blue ball before the second draw. With probability $\frac{2}{3}$, George will draw a red ball next. In that case the only way for the urn to end up with three balls of each color is for the next two draws to be blue, which will happen with probability $\frac{1}{4}\cdot\frac{2}{5}=\frac{1}{10}$. On the other hand, George will get a blue ball on his second draw with probability $\frac{1}{3}$, and the urn will then have two balls of each color. In that case no matter what color he gets on his third draw, for the urn to end up with three balls of each color the fourth draw must be different from the third draw, which will happen with probability $\frac{2}{5}$. Therefore, the probability that the urn will end up with three balls of each color is $$ \frac{2}{3}\cdot\frac{1}{4}\cdot\frac{2}{5}+\frac{1}{3}\cdot\frac{2}{5}=\frac{1}{5}. $$

Answer (C): Because any such polynomial has constant term $2020\neq 0$, the polynomial must have five nonzero roots whose product is $-2020$. Because the polynomial has real coefficients, all nonreal roots appear in complex-conjugate pairs. Because the polynomial has exactly five roots, there must be a real root $s\neq 0$ such that each root of the polynomial is one of $$ s,\qquad \alpha=\frac{-1+i\sqrt3}{2}\cdot s,\qquad \text{or}\qquad \overline{\alpha}=\frac{-1-i\sqrt3}{2}\cdot s, $$ for otherwise there would be at least six distinct roots. Because the product of $\alpha$ and $\overline{\alpha}$ is $s^2$, it follows that $s=-\sqrt[5]{2020}$. Either $\alpha$ and $\overline{\alpha}$ are both roots of multiplicity $2$, or the complex roots each have multiplicity $1$ and $s$ has multiplicity $3$. These are the only possibilities, so there are $2$ such polynomials. Explicitly, they are, respectively, $$ x^5-\sqrt[5]{2020}\,x^4+\sqrt[5]{2020^2}\,x^3+\sqrt[5]{2020^3}\,x^2-\sqrt[5]{2020^4}\,x+2020 $$ and $$ x^5+2\sqrt[5]{2020}\,x^4+\sqrt[5]{2020^2}\,x^3+\sqrt[5]{2020^3}\,x^2+2\sqrt[5]{2020^4}\,x+2020. $$

Answer (B): Square $ABCD$ has area $4$, so its side length is $AB=2$. Isosceles right triangle $HAE$ has leg length $AE=\sqrt{2}$, so $EB=2-\sqrt{2}$. Extend $EH$ and $BC$ to meet at $K$. Besides the right angles, all the other angles in the diagram below are one of a single supplementary pair. Isosceles right triangle $KBE$ has area $$ \frac{1}{2}(2-\sqrt{2})^2=3-2\sqrt{2}, $$ so isosceles right triangle $FIK$ has area $4-2\sqrt{2}=\frac{1}{2}FI^2$. Hence $FI^2=8-4\sqrt{2}$.

Answer (C): Label the sides of the square $p, q, r,$ and $s$, with $p$ being the side that starts as the upper horizontal side, $q$ being the side that starts as the left vertical side, $r$ being the side that starts as the bottom horizontal side, $s$ being the side that starts as the right vertical side. The position of the square after a sequence of transformations can be represented by a list of these four letters, starting with the side that is in the top horizontal position and moving counterclockwise. Thus $L$ takes $wxyz$ to $zwxy$, $R$ takes $wxyz$ to $xyzw$, $H$ takes $wxyz$ to $yxwz$, and $V$ takes $wxyz$ to $wzyx$. Let $E=\{pqrs, qpsr, rspq, srqp\}$ and let $O=\{psrq, qrsp, rqps, spqr\}$. Each of the given transformations $L, R, H,$ and $V$ will map each element of $E$ to a different element of $O$, and vice versa. Because 19 is an odd number, any sequence of 19 transformations will map the square to a position in set $O$, and exactly one of the given transformations will map this position back to position $pqrs$. Therefore there are $4^{19}=2^{38}$ possible sequences that will return the square to its original position.

Answer (D): There are $2^6=64$ ways to color each cube, for a total of $2^{12}=4096$ colorings of the two cubes. The 64 colorings may be grouped as follows: • There is 1 coloring with no white faces, and the same number with no black faces. • There are 6 colorings with exactly one white face, and the same number with exactly one black face. • There are 3 colorings with only two white faces that are opposite each other, and the same number with only two black faces that are opposite each other. • There are 12 colorings with only two white faces that share an edge, and the same number with only two black faces that share an edge. • There are 8 colorings with exactly three white faces that share a vertex. • There are 12 colorings with exactly three white faces that include a pair of opposite faces. The two cubes can be rotated to be identical in appearance if and only if their colorings are in the same group. Therefore the requested probability is $\dfrac{2\cdot(1^2+6^2+3^2+12^2)+8^2+12^2}{4096}=\dfrac{588}{4096}=\dfrac{147}{1024}.$

Answer (C): Each positive integer $n$ can be uniquely expressed as $k^2+j$, where $j$ and $k$ are integers with $k\ge 1$ and $0\le j\le 2k$. When $n$ is so expressed, $\lfloor\sqrt{n}\rfloor=k$, and the given equation can be written as $$k^2-(70k-j)+1000=0.$$ For a fixed value of $k$, there is a solution for $j$ if and only if $$k^2-70k+1000\le 0\qquad\text{and}\qquad k^2-68k+1000\ge 0.$$ Solving the first inequality gives $20\le k\le 50$, and solving the second gives $k\le 34-\sqrt{156}\approx 21\frac12$ or $k\ge 34+\sqrt{156}\approx 46\frac12$. Thus the suitable values of $k$ are $20,21,47,48,49,$ and $50$, and, for each of these, the unique value of $j$ is $(k-20)(50-k)$. Hence the given equation has 6 solutions for $n$, namely $400,470,2290,2360,2430,$ and $2500$.

Answer (C): Let $x=\dfrac{3t}{2^t}$. Then the given expression is $y=\dfrac{1}{3}x-\dfrac{1}{3}x^2$, whose graph is a downward-opening parabola with vertex $\left(\dfrac{1}{2},\dfrac{1}{12}\right)$. Because $x=0$ when $t=0$ and $x=\dfrac{3}{2}$ when $t=1$, there must be some value of $t$ between 0 and 1 for which $x=\dfrac{1}{2}$, and therefore the maximum $y=\dfrac{1}{12}$ is attained. (There is also a value of $t>1$ for which $x=\dfrac{1}{2}$.)

Answer (B): If $|z_1|=|z_2|=1$ and $z_1+z_2=0$, then $0$ is the midpoint between $z_1$ and $z_2$, which is to say that they are opposite each other on the unit circle. Thus $n=2$ has the property in question. If $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$, then let $w_k=\overline{z_1}z_k$. The angle between $w_j$ and $w_k$ is the argument of $\frac{w_j}{w_k}$. Because $\frac{w_j}{w_k}=\frac{z_j}{z_k}$, the angles between the $w$'s are the same as those between the $z$'s. Also, $w_1+w_2+w_3=\overline{z_1}(z_1+z_2+z_3)=0$. (The advantage of looking at the $w$'s rather than the $z$'s is that $w_1=1$.) Let $u_k$ and $v_k$ be the real and imaginary parts of $w_k$. Then $1+u_2+u_3=0$ and $v_2+v_3=0$. Now $u_3\ge -1$, which is to say that $-u_3\le 1$, so $u_2=-1-u_3\le 0$. Similarly, $u_3\le 0$. From $v_3=-v_2$ it follows that $v_2^2=v_3^2$, so $u_2^2=1-v_2^2=1-v_3^2=u_3^2$ and hence $u_2=u_3$, because they are both in the interval $[-1,0]$. But $1+u_2+u_3=0$, so $u_2=u_3=-\frac12$. Hence $w_1,w_2,$ and $w_3$ are the cube roots of unity, with equal angular spacing. Thus $n=3$ has the property in question. It remains to show that no $n>3$ has the property. Suppose that $n=2m\ge 4$. Take $m$ pairs of numbers on the unit circle of the form $z,-z$. Such a set of $n$ numbers sum to $0$, but in general will not be equally spaced on the unit circle. Suppose that $n=2m+3\ge 5$. Take the three cube roots of unity and $m$ pairs of numbers of the form $z,-z$. They will sum to $0$, but in general will not be equally spaced. There are just $2$ such values of $n$.

Answer (A): Clearly $D(1)=0$. Suppose $n>1$. The first factor $f_1$ in a representation is a divisor of $n$ with $f_1>1$. If $f_1=n$, then there is a representation with just one factor. Suppose that $f_1$ is a divisor of $n$ and $1<f_1<n$. Then the further factors after $f_1$ are exactly the representations of $\frac{n}{f_1}$. This gives a recurrence relation: If $n>1$, then \[ D(n)=1+\sum_{\substack{f\mid n\\1<f<n}} D\!\left(\frac{n}{f}\right). \] Thus a table of values of $D(2^j3^k)$ can be constructed for $j=0,1,\ldots,5$, $k=0,1$. \[ \begin{array}{c|cc} j\backslash k & 0 & 1\\ \hline 0 & 0 & 1\\ 1 & 1 & 3\\ 2 & 2 & 8\\ 3 & 4 & 20\\ 4 & 8 & 48\\ 5 & 16 & 112 \end{array} \] From the table, it can be seen that $D(96)=112$.

Answer (B): First consider the graph of the equation $\sin^2(\pi x)+\sin^2(\pi y)=1$ in the unit square of the $xy$-coordinate plane. The equation implies that $\sin^2(\pi y)=\cos^2(\pi x)=\sin^2\!\left(\pi\left(\frac12-x\right)\right),$ so for $0\le x\le\frac12$ and $0\le y\le\frac12$, the equation is equivalent to $y=\frac12-x$. Because $\sin(\pi x)=\sin(\pi(1-x))$ and $\sin(\pi y)=\sin(\pi(1-y))$, the graph is symmetric about the lines $x=\frac12$ and $y=\frac12$. Thus within the unit square, the graph is a square with vertices $(0,\frac12)$, $(\frac12,0)$, $(1,\frac12)$, and $(\frac12,1)$. The solutions of the given inequality correspond to the points inside this square. For a fixed value of $a$, the set of all possible ordered pairs $(x,y)$ corresponds to an $a\times1$ rectangle, which has area $a$. If $a\le\frac12$, then the solutions of the inequality lie inside a triangle with altitude $a$ and base $2a$, which has area $a^2$, so $P(a)=\frac{a^2}{a}=a\le\frac12.$ If $a>\frac12$, then the solutions of the inequality lie inside the union of a triangle with area $\frac14$ and a trapezoid with altitude $a-\frac12$ and base lengths $1$ and $2(1-a)$. The total area of this region is $\frac14+\left(a-\frac12\right)\left(\frac{1+2(1-a)}{2}\right)=-a^2+2a-\frac12,$ so $P(a)=-a+2-\frac{1}{2a}.$ The value of $P(a)$ is maximized when the value of $a+\frac{1}{2a}$ is minimized. By the Arithmetic Mean–Geometric Mean Inequality, $a+\frac{1}{2a}\ge 2\sqrt{a\cdot\frac{1}{2a}}=\sqrt2,$ and the minimum value occurs when $a=\frac{1}{2a}$. This means $a=\frac12\sqrt2$ (which is indeed greater than $\frac12$), and the corresponding value of $P(a)$ is $2-\sqrt2$ (which is greater than the maximum in the case $a\le\frac12$).