AMC12 2019 B

Let $x$ be the volume of the first container and $y$ the volume of the second container. Then $\frac{5}{6}x = \frac{3}{4}y$, so $\frac{x}{y} = \frac{3}{4} \cdot \frac{6}{5} = \frac{9}{10}$.

An implication is false if and only if the hypothesis is true but the conclusion is false. Choice (E), $n = 27$, is a counterexample to the statement because the hypothesis is true (27 is not prime) but the conclusion is false (27 - 2 = 25 is not prime). For answer choices (A) and (C), $n$ is prime, so the hypothesis is false and these values of $n$ do not provide a counterexample. For choices (B), (C), and (D), $n - 2$ is prime, so the conclusion is true and these values of $n$ do not provide a counterexample.

Rotation around the origin by 180° can be described by the rule $(x, y) \rightarrow (-x, -y)$, so (E) is the correct answer. The observation that $\overline{AB}$ is in the second quadrant and $\overline{A'B'}$ is in the fourth quadrant excludes answer options (A), (B), and (D), because these transformations map $\overline{AB}$ into a neighboring quadrant. The translation described in (C) maps points in $\overline{AB}$ onto points in $\overline{A'B'}$, but the image is not oriented properly (for instance, $B$ is mapped to $A'$).

Dividing the given equation by $n!$, simplifying, and completing the square yields $(n+1) + (n+2)(n+1) = 440$, $n^2 + 4n + 3 = 440$, $n^2 + 4n + 4 = 441$, $(n+2)^2 = 21^2$. Thus $n+2 = 21$ and $n = 19$. The requested sum of the digits of $n$ is $1+9 = 10$.

Answer (B): Because Casper has exactly enough money to buy 12 pieces of red candy, the amount of money he has must be a multiple of 12 cents. Similarly, it must be a multiple of both 14 cents and 15 cents. Furthermore, this amount of money will buy a whole number of purple candies that cost 20 cents each, so the amount of money must also be a multiple of 20. The least common multiple of $12=2^2\cdot 3$, $14=2\cdot 7$, $15=3\cdot 5$, and $20=2^2\cdot 5$ is $2^2\cdot 3\cdot 5\cdot 7=420$. Therefore the number of purple candies that Casper can buy, $n$, must be a multiple of $420\div 20=21$. Thus the least possible value of $n$ is 21. In this case the red candies cost $420\div 12=35$ cents each, the green candies cost $420\div 14=30$ cents each, and the blue candies cost $420\div 15=28$ cents each.

Answer (A): In order for the area of $\triangle ABC$ to be 100, the altitude to the base $AB$ must be 20. Thus $C$ must lie on one of the two lines parallel to and 20 units from line $AB$. In order for the perimeter of $\triangle ABC$ to be 50, the sum of the lengths of the other two sides must be 40, which implies that point $C$ lies on an ellipse whose foci are $A$ and $B$ and whose semi-minor axis has length $\frac{1}{2}\sqrt{40^2-10^2}=\sqrt{375}$, which is less than 20. Therefore the ellipse does not intersect either of the parallel lines, and there are no such points $C$.

Answer (A): The mean of the given numbers is $\dfrac{4+6+8+17+x}{5}=\dfrac{x+35}{5}=\dfrac{x}{5}+7.$ The median depends on the value of $x$. If $x<6$, then the median is $6$. If the mean and median are equal, then $\dfrac{x}{5}+7=6$, which is equivalent to $x=-5$. If $6\le x\le 8$, then the median is $x$. If the mean and median are equal, then $\dfrac{x}{5}+7=x$, which is equivalent to $x=\dfrac{35}{4}$. But this is outside of the given range. If $x>8$, then the median is $8$. If the mean and median are equal then $\dfrac{x}{5}+7=8$, which is equivalent to $x=5$. Again this is outside of the given range. Therefore the only value of $x$ for which the mean equals the median is $-5$, so the requested sum is also $-5$.

Answer (A): Note that $f(1-x)=(1-x)^2(1-(1-x))^2=(1-x)^2x^2=f(x).$ Therefore $f\left(\frac{1}{2019}\right)=f\left(\frac{2018}{2019}\right)$, so the first and last terms of the given sum combine to give 0. Similarly, $f\left(\frac{2}{2019}\right)-f\left(\frac{2017}{2019}\right)=0$, $f\left(\frac{3}{2019}\right)-f\left(\frac{2016}{2019}\right)=0$, and so on. Because there are an even number of terms, all the terms cancel out in pairs, and the sum is 0.

Answer (B): A triangle of positive area can be formed if and only if all three Triangle Inequalities are satisfied: $\log_2 x + \log_4 x > 3,$ $3 + \log_4 x > \log_2 x,$ and $3 + \log_2 x > \log_4 x.$ By the Change of Base Formula, $\log_4 x=\dfrac{\log_2 x}{\log_2 4}=\dfrac{1}{2}\log_2 x.$ Therefore the inequalities are equivalent, respectively, to $\log_2 x>2,$ $6>\log_2 x,$ and $\log_2 x>-6.$ These are in turn equivalent to $x>4$, $x<64$, and $x>\dfrac{1}{64}$. Because $x$ must be an integer, it must lie between $5$ and $63$, inclusive, and there are $63-4=59$ such numbers.

Answer (E): Label the remaining cities as shown below. Because 3 roads reach cities $B$, $C$, $E$, $H$, $J$, and $K$, at least one of these roads must be excluded from Paula’s route at each of these 6 cities. Furthermore, because the route must begin at $A$ and conclude at $L$, one of the two roads reaching $A$ and one reaching $L$ must be excluded. There are 17 roads and only 4 can be excluded from a 13-road route. Because 8 cities must involve excluded roads and no road can reach more than 2 cities, the 4 excluded roads must each involve 2 of these 8 cities. Because $C$ can be paired only with $B$, and $J$ can be paired only with $K$, $\overline{BC}$ and $\overline{JK}$ must be excluded. For similar reasons, $\overline{AE}$ and $\overline{HL}$ must be the other two excluded roads and Paula’s route must be as shown below. There are exactly 4 routes, depending on the choice of path (clockwise or counterclockwise) around the loops at $F$ and $G$, namely $ABFJIEFGCDHGKL$, $ABFEIJFGHDCGKL$, $ABFJIEFGCDHGKL$, and $ABFEIJFGHDCGKL$.

Each of the 12 edges of the cube shares a face with each of 6 other edges and is parallel to 1 additional edge. The other 4 edges are skew to the given edge. Thus each of the 12 edges can be paired with any of 7 others to determine a plane. After taking into account the double-counting of each pair, it follows that the number of unordered pairs of edges that determine a plane is $\frac{1}{2} \cdot 12 \cdot 7 = 42$.

Answer (D): Let $\theta=\angle CAD$, $x=CD$, and $y=DA$. It follows from the given information that $AC=\sqrt{2}$ and $\angle BAC=45^\circ$. Because the perimeters of the two triangles are equal, $x+y=2$. By the Pythagorean Theorem, $2+x^2=y^2$. Solving these two equations simultaneously gives $x=\frac{1}{2}$ and $y=\frac{3}{2}$. It follows that $\sin\theta=\frac{1}{3}$ and $\cos\theta=\frac{2}{3}\sqrt{2}$. Then $$ \begin{array}{rcl} \sin(2\angle BAD) & = & 2\sin(\angle BAD)\cdot \cos(\angle BAD)\\ & = & 2\sin(\theta+45^\circ)\cdot \cos(\theta+45^\circ)\\ & = & 2(\sin\theta\cos45^\circ+\cos\theta\sin45^\circ)\cdot(\cos\theta\cos45^\circ-\sin\theta\sin45^\circ)\\ & = & 2\left(\frac{4+\sqrt{2}}{6}\right)\left(\frac{4-\sqrt{2}}{6}\right)\\ & = & \frac{7}{9}. \end{array} $$

Answer (C): The probability that the two balls are tossed into the same bin is $$ \frac{1}{2}\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{8}\cdot\frac{1}{8}+\cdots =\sum_{n=1}^{\infty}\frac{1}{4^n} =\frac{\frac{1}{4}}{1-\frac{1}{4}} =\frac{1}{3}. $$ Therefore the probability that the balls are tossed into different bins is $\frac{2}{3}$. By symmetry the probability that the red ball is tossed into a higher-numbered bin than the green ball is $\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$.

Answer (C): Note that $100,000 = 2^5 \cdot 5^5$. This implies that for a number to be a product of two elements in $S$ it must be of the form $2^a \cdot 5^b$ with $0 \le a \le 10$ and $0 \le b \le 10$. The corresponding product for the remainder of this solution will be denoted $(a, b)$. Note that the pairs $(0,0)$, $(0,10)$, $(10,0)$, and $(10,10)$ cannot be obtained as the product of two distinct elements of $S$; these products can be obtained only as $1 \cdot 1 = 1$, $5^5 \cdot 5^5 = 5^{10}$, $2^5 \cdot 2^5 = 2^{10}$, and $10^5 \cdot 10^5 = 10^{10}$, respectively. This gives at most $11 \cdot 11 - 4 = 117$ possible products. To see that all these pairs can be achieved, consider four cases: If $0 \le a \le 5$ and $0 \le b \le 5$, other than $(0,0)$, then $(a,b)$ can be achieved with the divisors $1$ and $2^a \cdot 5^b$. If $6 \le a \le 10$ and $0 \le b \le 5$, other than $(10,0)$, then $(a,b)$ can be achieved with the divisors $2^5$ and $2^{a-5} \cdot 5^b$. If $0 \le a \le 5$ and $6 \le b \le 10$, other than $(0,10)$, then $(a,b)$ can be achieved with the divisors $5^5$ and $2^a \cdot 5^{b-5}$. Finally, if $6 \le a \le 10$ and $6 \le b \le 10$, other than $(10,10)$, then $(a,b)$ can be achieved with the divisors $2^5 \cdot 5^5$ and $2^{a-5} \cdot 5^{b-5}$.

Answer (E): Let $H$ and $I$ be the intersections of $\overline{AD}$ with the circle centered at $F$, where $H$ lies between $A$ and $B$, and $I$ lies between $C$ and $D$; and let $K$ be the foot of the perpendicular line segment from $F$ to $\overline{AD}$. The specified region consists of three subregions: a semicircle of radius 2, a $4\times 1$ rectangle with 4 quarter circles of radius 1 removed, and the segment of the circle cut off by chord $\overline{HI}$, as shown in the figure below. The semicircle of radius 2 has area $2\pi$. The rectangle minus the 4 quarter circles has area $4-\pi$. Because $FK=1$ and $FI=2$, it follows that $\angle KFI$ has measure $60^\circ$, and therefore the segment of the circle is a third of the circle with $\triangle HFI$ removed. The area of the segment is $$ \frac{4}{3}\pi-\frac12\cdot 2\sqrt3\cdot 1=\frac{4}{3}\pi-\sqrt3. $$ Adding the areas of the three subregions gives $\frac{7}{3}\pi-\sqrt3+4$, and the requested sum is $7+3+3+4=17$.

Answer (A): In order to clear the first hurdle, the sequence of jump lengths must be 1-1-2 or 2-2. These occur with probabilities $(\frac{1}{2})^3$ and $(\frac{1}{2})^2$, respectively, so the probability of landing safely on pad 4 is $\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$. In order for Fiona to make it safely to pad 7 once she has arrived at pad 4, she must jump 1-2, which has probability $\frac{1}{4}$. Once on pad 7, in order to reach the food the frog must jump 1-2 or 2-1 or 1-1-1, with probability $\frac{1}{4}+\frac{1}{4}+\frac{1}{8}=\frac{5}{8}$; otherwise she skips over the food. Therefore the requested probability is $\frac{3}{8}\cdot\frac{1}{4}\cdot\frac{5}{8}=\frac{15}{256}$.

Answer (D): For $0$, $z$, and $z^3$ to be vertices of an equilateral triangle, the distances that $z$ and $z^3$ are from $0$ must be equal, which implies that $|z|=1$. Let $z=\cos\theta+i\sin\theta$, where $0\le\theta<360^\circ$. Then by De Moivre’s Theorem, $z^3=\cos(3\theta)+i\sin(3\theta)$. For the three points to form an equilateral triangle, the angle from $z$ to $0$ to $z^3$ must be $60^\circ$, which implies that $2\theta=\pm60^\circ+k\cdot360^\circ$ for some integer $k$. It follows that $\theta$ must be one of $30^\circ$, $150^\circ$, $210^\circ$, or $330^\circ$. In each of these 4 cases, $0$, $z$, and $z^3$ are vertices of a triangle with two congruent sides and an included angle of $60^\circ$, so they are vertices of an equilateral triangle.

Answer (C): Place the pyramid in coordinate space with point $A$ at $(0,0,0)$, point $B$ at $(3,0,0)$, point $C$ at $(3,3,0)$, point $D$ at $(0,3,0)$, and point $E$ at $(0,0,6)$. Then $P$ has coordinates $(2,0,2)$, $Q$ has coordinates $(0,2,2)$, and $R$ has coordinates $(1,1,4)$. It follows from the distance formula that $PR=QR=\sqrt{6}$, and $PQ=2\sqrt{2}$. Because $\triangle PQR$ is isosceles, the length of its altitude to $PQ$ can be found from the Pythagorean Theorem to be $2$, and thus the area of $\triangle PQR$ is $2\sqrt{2}$.

Answer (B): No player can ever end up with \$3 at the end of a round, because that player had to give away one of the dollars in play. Therefore the only two possible distributions of the money are 1-1-1 and 2-1-0. Suppose that a round of the game starts at 1-1-1. Without loss of generality, assume that Raashan gives his dollar to Sylvia. Then the only way for the round to end at 1-1-1 is for Ted to give his dollar to Raashan (otherwise Sylvia would end up with \$2) and for Sylvia to give her dollar to Ted; the probability of this is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. Next suppose that a round starts at 2-1-0; without loss of generality, assume that Raashan has \$2 and Sylvia has \$1. Then the only way for the round to end at 1-1-1 is for Sylvia to give her dollar to Ted (otherwise Raashan would end up with \$2) and for Raashan to give his dollar to Sylvia; the probability of this is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. Thus no matter how the round starts, the probability that the round will end at 1-1-1 is $\frac{1}{4}$. In particular, the probability is $\frac{1}{4}$ that at the end of the 2019th round each player will have \$1.

Answer (C): Let $T$ be the point where the tangents at $A$ and $B$ intersect. By symmetry $T$ lies on the perpendicular bisector $\ell$ of $AB$, so in fact $T$ is the (unique) intersection point of line $\ell$ with the $x$-axis. Computing the midpoint $M$ of $AB$ gives $(9,12)$, and computing the slope of $AB$ gives $\frac{13-11}{6-12}=-\frac{1}{3}$. This means that the slope of $\ell$ is $3$, so the equation of $\ell$ is given by $y-12=3(x-9)$. Setting $y=0$ yields that $T=(5,0)$.

Answer (B): Because the coefficients are real, so must be the roots. Let the two real roots be $r$ and $s$. Note that $r\neq s$ because $p(x)=rx^2+rx+r$ has no real roots. The given condition implies that, without loss of generality, $p(x)=r(x-r)(x-s)$, with $r\neq 0$. There are three cases to consider. Case 1: $r(x-r)(x-s)=rx^2+rx+s$. Then equating coefficients yields $r+s=-1$ and $r^2s=s$. The solutions to this system of equations are $(r,s)=(1,-2)$, in which case $p(x)=x^2+x-2$, and $(r,s)=(-1,0)$, in which case $p(x)=-x^2-x$. Case 2: $r(x-r)(x-s)=rx^2+sx+r$. Then $r+s=-\frac{s}{r}$ and $rs=1$. Multiplying the first equation by $r^2$ and substituting for $rs$ yields $r^3+r+1=0$. Descartes’ Rule of Signs shows that this equation has no positive real solution and one negative real soution (which is approximately $-0.682$, so $p(x)\approx -0.682x^2-1.466x-0.682$). Case 3: $r(x-r)(x-s)=rx^2+sx+s$. Then $r+s=-\frac{s}{r}$ and $rs=\frac{s}{r}$. It follows from $r\neq 0$ that $s\neq 0$. Then from the second equation, $r^2=1$. If $r=-1$, then the first equation becomes $-1+s=s$, which has no solution. If $r=1$, then the first equation becomes $1+s=-s$, so $s=-\frac{1}{2}$, in which case $p(x)=x^2-\frac{1}{2}x-\frac{1}{2}$. In summary, Case 1 yielded $2$ polynomials, Case 2 yielded $1$ more, and Case 3 yielded $1$ more, for a total of $2+1+1=4$ polynomials.

It suffices to study $y_n = x_n - 4$ and find the least positive integer $m$ such that $y_m \le 1/250$. We have $y_0 = 1$ and $$ y_{n+1} = \frac{y_n(y_n + 9)}{y_n + 10}. $$ The sequence $(y_n)$ is strictly decreasing and positive. Because $$ \frac{y_{n+1}}{y_n} = 1 - \frac{1}{y_n + 10}, $$ it follows that $$ \frac{9}{10} \le \frac{y_{n+1}}{y_n} \le \frac{10}{11}, $$ and because $y_0 = 1$, $$ \left(\frac{9}{10}\right)^k \le y_k \le \left(\frac{10}{11}\right)^k \quad \text{for all integers } k \ge 2. $$ Now note that $$ \left(\frac{1}{2}\right)^{1/4} < \frac{9}{10} $$ because this is equivalent to $0.5 < (0.9)^4 = (0.81)^2$. Therefore $$ \left(\frac{1}{2}\right)^{m/4} < y_m \le \frac{1}{220} \implies m > 80. $$ Also, $$ \left(\frac{11}{10}\right)^{10} > 2 \implies \frac{10}{11} < \left(\frac{1}{2}\right)^{1/10}. $$ Thus $$ \frac{1}{220} < y_{m-1} < \left(\frac{10}{11}\right)^{m-1} < \left(\frac{1}{2}\right)^{(m-1)/10} \implies m < 201. $$ Hence $80 < m < 201$, so the correct choice is **(C)**. (Numerical computation shows $m = 133$.

Answer (C): For $n \ge 2$, let $a_n$ be the number of sequences of length $n$ that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s. In order for the sequence to end with a 0 and satisfy the conditions, it must end either 010 or 0110. Thus $a_n = a_{n-2} + a_{n-3}$. The initial conditions for this recurrence relation are $a_2 = 0$, $a_3 = 1$ (the sequence 010), and $a_4 = 1$ (the sequence 0110). Then $a_5 = a_3 + a_2 = 1 + 0 = 1$, $a_6 = a_4 + a_3 = 1 + 1 = 2$, $a_7 = a_5 + a_4 = 1 + 1 = 2$, and so on. A bit of calculation produces the display below; $a_{19} = 65$. \[ \begin{array}{c|cccccccccc} n & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline a_n & 0 & 1 & 1 & 1 & 2 & 2 & 3 & 4 & 5 & 7 \end{array} \] \[ \begin{array}{c|cccccccc} n & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\ \hline a_n & 9 & 12 & 16 & 21 & 28 & 37 & 49 & 65 \end{array} \]

Answer (C): Note that $\omega^2=\bar{\omega}$ and $\omega^3=1$. Then $1+\omega=\frac{1}{2}+\frac{1}{2}i\sqrt{3}\in S$, $\omega+\bar{\omega}=-1\in S$, and $1+\bar{\omega}=\frac{1}{2}-\frac{1}{2}i\sqrt{3}\in S$. These three points, together with the original points $1$, $\omega$, and $\omega^2$, form the vertices of a regular hexagon inscribed in the unit circle $|z|=1$. The segment from $0$ to $1$ is a convex set in the complex plane. The segment from $0$ to $\omega$ is also a convex set, as is the segment from $0$ to $\bar{\omega}$. The sum of convex sets (that is, the set of all numbers of the form $s+t$, where $s$ is in one set and $t$ is in the other) is convex. Thus the set in question contains not just the hexagon whose vertices are the sixth roots of unity, but also its interior. Now consider the largest imaginary part of any point $a+b\omega+c\omega^2$ in $S$. Because \[ \operatorname{Im}(a)=0,\qquad \operatorname{Im}(b\omega)\le \frac{1}{2}\sqrt{3},\qquad \text{and}\qquad \operatorname{Im}(c\bar{\omega})\le 0, \] all points in $S$ have imaginary part less than or equal to $\frac{1}{2}\sqrt{3}$. Similarly, the imaginary parts of points in $S$ are all greater than or equal to $-\frac{1}{2}\sqrt{3}$. Thus $S$ lies in a strip of width $\sqrt{3}$. Furthermore, $\omega S=\omega^2 S=S$. Hence $S$ lies in the intersection of three strips, and this intersection is the hexagon already obtained. The area of the hexagon is $6$ times the area of an equilateral triangle with side length $1$, namely $\frac{3}{2}\sqrt{3}$.

Answer (C): Let $G_1$, $G_2$, and $G_3$ be the centroids of $\triangle ABC$, $\triangle BCD$, and $\triangle ACD$, respectively. Then with points viewed as position vectors represented by ordered pairs, \[ G_1=\frac13A+\frac13B+\frac13C, \] \[ G_2=\frac13B+\frac13C+\frac13D, \] \[ G_3=\frac13A+\frac13C+\frac13D. \] The directed edges of $\triangle G_1G_2G_3$ are $\overrightarrow{G_1G_2}=\frac13(D-A)$, $\overrightarrow{G_2G_3}=\frac13(A-B)$, and $\overrightarrow{G_3G_1}=\frac13(B-D)$. This triangle is equilateral if and only if $|D-A|=|A-B|=|B-D|$, which in turn is equivalent to the assertion that $\triangle ABD$ is equilateral. Let $\theta$ denote the measure of $\angle BCD$. Because $BC=2$ and $CD=6$, it follows that the area of $\triangle BCD$ is $\frac12\cdot BC\cdot CD\cdot\sin\theta=6\sin\theta$. By the Law of Cosines, \[ BD^2=BC^2+CD^2-2\cdot BC\cdot CD\cdot\cos\theta=40-24\cos\theta. \] The area of the equilateral triangle $\triangle ABD$ is $\frac{\sqrt3}{4}BD^2=10\sqrt3-6\sqrt3\cos\theta$. Adding the areas of $\triangle ABD$ and $\triangle BCD$ gives the area of the quadrilateral $ABCD$: \[ 10\sqrt3-6\sqrt3\cos\theta+6\sin\theta =10\sqrt3+12\left(-\frac{\sqrt3}{2}\cos\theta+\frac12\sin\theta\right) =10\sqrt3+12\sin(\theta-60^\circ). \] This is maximized by taking $\theta=150^\circ$, in which case the area is $10\sqrt3+12$.