AMC12 2018 A

There are currently 36 red balls in the urn. In order for the 36 red balls to represent 72% of the balls in the urn after some blue balls are removed, there must be $36 \div 0.72 = 50$ balls left in the urn. This requires that $100 - 50 = 50$ blue balls be removed.

Answer (C): The 5-pound rocks have a value of \$14 ÷ 5 = \$2.80 per pound; the 4-pound rocks have a value of \$11 ÷ 4 = \$2.75 per pound; the 1-pound rocks have a value of \$2 per pound. It is not to Carl’s advantage to take 1-pound rocks when he can take the larger rocks. Therefore the only issue is how many of the more valuable 5-pound rocks to take, including as many 4-pound rocks as possible in each case. The viable choices are displayed in the following table. \[ \begin{array}{cccc} \text{5-pound rocks} & \text{4-pound rocks} & \text{1-pound rocks} & \text{value} \\ (\$14\ \text{each}) & (\$11\ \text{each}) & (\$2\ \text{each}) & \\ 3 & 0 & 3 & \$48 \\ 2 & 2 & 0 & \$50 \\ 1 & 3 & 1 & \$49 \\ 0 & 4 & 2 & \$48 \\ \end{array} \] The maximum possible value is \$50.

Answer (E): There are 4 choices for the periods in which the mathematics courses can be taken: periods 1, 3, 5; periods 1, 3, 6; periods 1, 4, 6; and periods 2, 4, 6. Each choice of periods allows $3! = 6$ ways to order the 3 mathematics courses. Therefore there are $4 \cdot 6 = 24$ ways of arranging a schedule.

Answer (D): Because the statements of Alice, Bob, and Charlie are all incorrect, the actual distance $d$ satisfies $d<6$, $d>5$, and $d>4$. Hence the actual distance lies in the interval $(5,6)$.

Answer (E): Factoring $x^2-3x+2$ as $(x-1)(x-2)$ shows that its roots are 1 and 2. If 1 is a root of $x^2-5x+k$, then $1^2-5\cdot1+k=0$ and $k=4$. If 2 is a root of $x^2-5x+k$, then $2^2-5\cdot2+k=0$ and $k=6$. The sum of all possible values of $k$ is $4+6=10$.

Answer (B): Note that the given conditions imply that the 6 values are listed in increasing order. Because the median of the these 6 values is $n$, the mean of the middle two values must be $n$, so $$\frac{(m+10)+(n+1)}{2}=n,$$ which implies $m=n-11$. Because the mean of the set is also $n$, $$\frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6}=n,$$ so $7n-16=6n$ and $n=16$. Then $m=16-11=5$, and the requested sum is $5+16=21$.

Answer (E): Because $4000=2^5\cdot 5^3$, $$4000\cdot \left(\frac{2}{5}\right)^n = 2^{5+n}\cdot 5^{3-n}.$$ This product will be an integer if and only if both of the factors $2^{5+n}$ and $5^{3-n}$ are integers, which happens if and only if both exponents are nonnegative. Therefore the given expression is an integer if and only if $5+n\ge 0$ and $3-n\ge 0$. The solutions are exactly the integers satisfying $-5\le n\le 3$. There are $3-(-5)+1=9$ such values.

The length of the base $DE$ of $\triangle ADE$ is 4 times the length of the base of a small triangle, so the area of $\triangle ADE$ is $4^2 \cdot 1 = 16$. Therefore the area of $DBCE$ is the area of $\triangle ABC$ minus the area of $\triangle ADE$, which is $40 - 16 = 24$.

Answer (E): If $0 \le x \le \pi$ and $0 \le y \le \pi$, then $\sin(x) \ge 0$, $\sin(y) \ge 0$, $\cos(x) \le 1$, and $\cos(y) \le 1$. Therefore $$ \sin(x+y)=\sin(x)\cdot\cos(y)+\cos(x)\cdot\sin(y)\le \sin(x)+\sin(y). $$ The given inequality holds for all $y$ such that $0 \le y \le \pi$.

The graph of the first equation is a line with $x$-intercept $(3,0)$ and $y$-intercept $(0,1)$. To draw the graph of the second equation, consider the equation quadrant by quadrant. In the first quadrant $x>0$ and $y>0$, and thus the second equation is equivalent to $\lvert x-y\rvert=1$, which in turn is equivalent to $y=x\pm1$. Its graph consists of the rays with endpoints $(0,1)$ and $(1,0)$, as shown. In the second quadrant $x<0$ and $y>0$. The corresponding graph is the reflection of the first quadrant graph across the $y$-axis. The rest of the graph can be sketched by further reflections of the first-quadrant graph across the coordinate axes, resulting in the figure shown. There are 3 intersection points: $(-3,2)$, $(0,1)$, and $\left(\frac{3}{2},\frac{1}{2}\right)$, as shown.

Answer (D): The paper’s long edge $\overline{AB}$ is the hypotenuse of right triangle $ACB$, and the crease lies along the perpendicular bisector of $\overline{AB}$. Because $AC>BC$, the crease hits $\overline{AC}$ rather than $\overline{BC}$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the intersection of $\overline{AC}$ and the line through $D$ perpendicular to $\overline{AB}$. Then the crease in the paper is $\overline{DE}$. Because $\triangle ADE \sim \triangle ACB$, it follows that $\dfrac{DE}{AD}=\dfrac{CB}{AC}=\dfrac{3}{4}$. Thus $$ DE=AD\cdot \frac{CB}{AC}=\frac{5}{2}\cdot \frac{3}{4}=\frac{15}{8}. $$

If 1 $\in S$, then S can have only 1 element, not 6 elements. If 2 is the least element of S, then 2, 3, 5, 7, 9, and 11 are available to be in S, but 3 and 9 cannot both be in S, so the largest possible size of S is 5. If 3 is the least element, then 3, 4, 5, 7, 8, 10, and 11 are available, but at most one of 4 and 8 can be in S and at most one of 5 and 10 can be in S, so again S has size at most 5. The set S = ${4, 6, 7, 9, 10, 11}$ has the required property, so 4 is the least possible element of S.

Answer (D): Let $S$ be the set of integers, both negative and non-negative, having the given form. Increasing the value of $a_i$ by 1 for $0 \le i \le 7$ creates a one-to-one correspondence between $S$ and the ternary (base 3) representation of the integers from 0 through $3^8-1$, so $S$ contains $3^8=6561$ elements. One of those is 0, and by symmetry, half of the others are positive, so $S$ contains $1+\frac{1}{2}\cdot(6561-1)=3281$ elements.

Answer (D): By the change-of-base formula, the given equation is equivalent to $$\frac{\log 4}{\log 3x}=\frac{\log 8}{\log 2x}$$ $$\frac{2\log 2}{\log 3+\log x}=\frac{3\log 2}{\log 2+\log x}$$ $$2\log 2+2\log x=3\log 3+3\log x$$ $$\log x=2\log 2-3\log 3$$ $$\log x=\log \frac{4}{27}.$$ Therefore $x=\frac{4}{27}$, and the requested sum is $4+27=31$.

Answer (B): None of the squares that are marked with dots in the sample scanning code shown below can be mapped to any other marked square by reflections or non-identity rotations. Therefore these 10 squares can be arbitrarily colored black or white in a symmetric scanning code, with the exception of “all black” and “all white”. On the other hand, reflections or rotations will map these squares to all the other squares in the scanning code, so once these 10 colors are specified, the symmetric scanning code is completely determined. Thus there are $2^{10}-2=1022$ symmetric scanning codes.

Answer (E): Solving the second equation for $x^2$ gives $x^2=y+a$, and substituting into the first equation gives $y^2+y+(a-a^2)=0$. The polynomial in $y$ can be factored as $(y+(1-a))(y+a)$, so the solutions are $y=a-1$ and $y=-a$. (Alternatively, the solutions can be obtained using the quadratic formula.) The corresponding equations for $x$ are $x^2=2a-1$ and $x^2=0$. The second equation always has the solution $x=0$, corresponding to the point of tangency at the vertex of the parabola $y=x^2-a$. The first equation has 2 solutions if and only if $a>\frac12$, corresponding to the 2 symmetric intersection points of the parabola with the circle. Thus the two curves intersect at 3 points if and only if $a>\frac12$.

Answer (D): Let the triangle’s vertices in the coordinate plane be $(4,0)$, $(0,3)$, and $(0,0)$, with $[0,s]\times[0,s]$ representing the unplanted portion of the field. The equation of the hypotenuse is $3x+4y-12=0$, so the distance from $(s,s)$, the corner of $S$ closest to the hypotenuse, to this line is given by \[ \frac{|3s+4s-12|}{\sqrt{3^2+4^2}}. \] Setting this equal to $2$ and solving for $s$ gives $s=\frac{22}{7}$ and $s=\frac{2}{7}$, and the former is rejected because the square must lie within the triangle. The unplanted area is thus $\left(\frac{2}{7}\right)^2=\frac{4}{49}$, and the requested fraction is \[ 1-\frac{\frac{4}{49}}{\frac{1}{2}\cdot 4\cdot 3}=\frac{145}{147}. \]

Answer (D): Because $AB$ is $\frac{5}{6}$ of $AB+AC$, it follows from the Angle Bisector Theorem that $DF$ is $\frac{5}{6}$ of $DE$, and $BG$ is $\frac{5}{6}$ of $BC$. Because trapezoids $FDBG$ and $EDBC$ have the same height, the area of $FDBG$ is $\frac{5}{6}$ of the area of $EDBC$. Furthermore, the area of $\triangle ADE$ is $\frac{1}{4}$ of the area of $\triangle ABC$, so its area is $30$, and the area of trapezoid $EDBC$ is $120-30=90$. Therefore the area of quadrilateral $FDBG$ is $\frac{5}{6}\cdot 90=75$.

Answer (C): Elements of set $A$ are of the form $2^i\cdot 3^j\cdot 5^k$ for nonnegative integers $i$, $j$, and $k$. Note that the product $\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots\right)\left(1+\frac{1}{5}+\frac{1}{5^2}+\cdots\right)$ will produce the desired sum. By the formula for infinite geometric series, this product evaluates to $\frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{1}{3}}\cdot\frac{1}{1-\frac{1}{5}}=2\cdot\frac{3}{2}\cdot\frac{5}{4}=\frac{15}{4}.$ The requested sum is $15+4=19$.

Answer (D): It follows from the Pythagorean Theorem that $CM = MB = \frac{3}{2}\sqrt{2}$. Because quadrilateral $AIME$ is cyclic, opposite angles are supplementary and thus $\angle IME$ is a right angle. Let $x = CI$ and $y = BE$; then $AI = 3 - x$ and $AE = 3 - y$. By the Law of Cosines in $\triangle MCI$, $$ IM^2 = x^2 + \left(\frac{3}{2}\sqrt{2}\right)^2 - 2\cdot x \cdot \frac{3}{2}\sqrt{2}\cdot \cos 45^\circ = x^2 - 3x + \frac{9}{2}. $$ Similarly, $ME^2 = y^2 - 3y + \frac{9}{2}$. By the Pythagorean Theorem in right triangles $EMI$ and $IAE$, $$ \left(x^2 - 3x + \frac{9}{2}\right) + \left(y^2 - 3y + \frac{9}{2}\right) = (3 - x)^2 + (3 - y)^2, $$ which simplifies to $x + y = 3$. Because the area of $\triangle EMI$ is $2$, it follows that $IM^2\cdot ME^2 = 16$. Therefore $$ \left(x^2 - 3x + \frac{9}{2}\right)\left((3 - x)^2 - 3(3 - x) + \frac{9}{2}\right)=16, $$ which simplifies to $$ \left(x^2 - 3x + \frac{9}{2}\right)^2 = 16. $$ Because $y > x$, the only real solution is $x = \frac{3-\sqrt{7}}{2}$. The requested sum is $3 + 7 + 2 = 12$.

Answer (B): By Descartes’ Rule of Signs, none of these polynomials has a positive root, and each one has exactly one negative root. Because each polynomial is positive at $x=0$ and negative at $x=-1$, it follows that each has exactly one root between $-1$ and $0$. Note also that each polynomial is increasing throughout the interval $(-1,0)$. Because $x^{19} > x^{17}$ for all $x$ in the interval $(-1,0)$, it follows that the polynomial in choice A is greater than the polynomial in choice B on that interval, which implies that the root of the polynomial in choice A is less than the root of the polynomial in choice B. Because $x^{13} > x^{11}$ for all $x$ in the interval $(-1,0)$, it follows that the polynomial in choice C is greater than the polynomial in choice A on that interval, which implies that the root of the polynomial in choice C is less than the root of the polynomial in choice A and therefore less than the root of the polynomial in choice B. The same reasoning shows that the root of the polynomial in choice D is less than the root of the polynomial in choice B. Furthermore, $2018 > 2018x^6$ on the interval $(-1,0)$, so $x^6+2018 > 2019x^6$, from which it follows that $x^{11}(x^6+2018) < 2019x^{17}$. Therefore the polynomial in choice B is less than $2019x^{17}+1$ on the interval $(-1,0)$. The polynomial in choice E has root $-\left(1-\frac{1}{2019}\right)$. Bernoulli’s Inequality shows that $(1+x)^{17} > 1+17x$ for all $x>-1$, which implies that \[ -2019\left(1-\frac{1}{2019}\right)^{17}+1 < -2019\left(1-\frac{17}{2019}\right)+1 = -2001 < 0, \] so the polynomial in choice B is negative at the root of the polynomial in choice E. This shows that the root of the polynomial in choice B is greater than the root in choice E. Because the unique real root of the polynomial in choice B is greater than the unique root of the polynomial in each of the other choices, that polynomial has the greatest real root.

Answer (A): Let $z=a+bi$ be a solution of the first equation, where $a$ and $b$ are real numbers. Then $(a+bi)^2=4+4\sqrt{15}i$. Expanding the left-hand side and equating real and imaginary parts yields $$a^2-b^2=4 \quad \text{and} \quad 2ab=4\sqrt{15}.$$ AMC 12A Solutions 13 From the second equation, $b=\dfrac{2\sqrt{15}}{a}$, and substituting this into the first equation and simplifying gives $(a^2)^2-4a^2-60=0$, which factors as $(a^2-10)(a^2+6)=0$. Because $a$ is real, it follows that $a=\pm\sqrt{10}$, from which it then follows that $b=\pm\sqrt{6}$. Thus two vertices of the parallelogram are $\sqrt{10}+\sqrt{6}i$ and $-\sqrt{10}-\sqrt{6}i$. A similar calculation with the other given equation shows that the other two vertices of the parallelogram are $\sqrt{3}+i$ and $-\sqrt{3}-i$. The area of this parallelogram can be computed using the shoelace formula, which gives the area of a polygon in terms of the coordinates of its vertices $(x_1,y_1)$, $(x_2,y_2)$, $\ldots$, $(x_n,y_n)$ in clockwise or counter-clockwise order: $$\frac12\cdot\left|\,(x_1y_2+x_2y_3+\cdots+x_{n-1}y_n+x_ny_1)-(y_1x_2+y_2x_3+\cdots+y_{n-1}x_n+y_nx_1)\,\right|.$$ In this case $x_1=\sqrt{10}$, $y_1=\sqrt{6}$, $x_2=\sqrt{3}$, $y_2=1$, $x_3=-\sqrt{10}$, $y_3=-\sqrt{6}$, $x_4=-\sqrt{3}$, and $y_4=-1$. The area is $6\sqrt{2}-2\sqrt{10}$, and the requested sum of the four positive integers in this expression is $20$.

Answer (E): Extend $\overline{PN}$ through $N$ to $Q$ so that $PN = NQ$. Segments $\overline{UG}$ and $\overline{PQ}$ bisect each other, implying that $UPGQ$ is a parallelogram. Therefore $GQ \parallel PT$, so $\angle QGA = 180^\circ - \angle T = \angle TPA + \angle TAP = 36^\circ + 56^\circ = 92^\circ$. Furthermore $GQ = PU = AG$, so $\triangle QGA$ is isosceles, and $\angle QAG = \frac{1}{2}(180^\circ - 92^\circ) = 44^\circ$. Because $\overline{MN}$ is a midline of $\triangle QPA$, it follows that $\overline{MN} \parallel \overline{AQ}$ and \[ \angle NMP = \angle QAP = \angle QAG + \angle GAP = 44^\circ + 56^\circ = 100^\circ, \] so acute $\angle NMA = 80^\circ$. (Note that the value of the common length $PU = AG$ is immaterial.)

Answer (B): Because Alice and Bob are choosing their numbers uniformly at random, the cases in which two or three of the chosen numbers are equal have probability $0$ and can be ignored. Suppose Carol chooses the number $c$. She will win if her number is greater than Alice’s number and less than Bob’s, and she will win if her number is less than Alice’s number and greater than Bob’s. There are three cases. • If $c \le \frac{1}{2}$, then Carol’s number is automatically less than Bob’s, so her chance of winning is the probability that Alice’s number is less than $c$, which is just $c$. The best that Carol can do in this case is to choose $c=\frac{1}{2}$, in which case her chance of winning is $\frac{1}{2}$. • If $c \ge \frac{2}{3}$, then Carol’s number is automatically greater than Bob’s, so her chance of winning is the probability that Alice’s number is greater than $c$, which is just $1-c$. The best that Carol can do in this case is to choose $c=\frac{2}{3}$, in which case her chance of winning is $\frac{1}{3}$. • Finally suppose that $\frac{1}{2}<c<\frac{2}{3}$. The probability that Carol’s number is less than Bob’s is $$ \frac{\frac{2}{3}-c}{\frac{2}{3}-\frac{1}{2}}=4-6c, $$ so the probability that her number is greater than Alice’s and less than Bob’s is $c(4-6c)$. Similarly, the probability that her number is less than Alice’s and greater than Bob’s is $(1-c)(6c-3)$. Carol’s probability of winning in this case is therefore $$ c(4-6c)+(1-c)(6c-3)=-12c^2+13c-3. $$ The value of a quadratic polynomial with a negative coefficient on its quadratic term is maximized at $-\frac{b}{2a}$, where $a$ is the coefficient on its quadratic term and $b$ is the coefficient on its linear term; here that is when $c=\frac{13}{24}$, which is indeed between $\frac{1}{2}$ and $\frac{2}{3}$. Her probability of winning is then $$ -12\cdot\left(\frac{13}{24}\right)^2+13\cdot\frac{13}{24}-3=\frac{25}{48}>\frac{24}{48}=\frac{1}{2}. $$ Because the probability of winning in the third case exceeds the probabilities obtained in the first two cases, Carol should choose $\frac{13}{24}$.

Answer (D): The equation $C_n - B_n = A_n^2$ is equivalent to $c \cdot \dfrac{10^{2n}-1}{9} - b \cdot \dfrac{10^n-1}{9} = a^2 \left(\dfrac{10^n-1}{9}\right)^2.$ Dividing by $10^n-1$ and clearing fractions yields $(9c-a^2)\cdot 10^n = 9b-9c-a^2.$ As this must hold for two different values $n_1$ and $n_2$, there are two such equations, and subtracting them gives $(9c-a^2)\left(10^{n_1}-10^{n_2}\right)=0.$ The second factor is non-zero, so $9c-a^2=0$ and thus $9b-9c-a^2=0.$ From this it follows that $c=\left(\dfrac{a}{3}\right)^2$ and $b=2c.$ Hence digit $a$ must be $3$, $6$, or $9$, with corresponding values $1$, $4$, or $9$ for $c$, and $2$, $8$, or $18$ for $b$. The case $b=18$ is invalid, so there are just two triples of possible values for $a$, $b$, and $c$, namely $(3,2,1)$ and $(6,8,4)$. In fact, in these cases, $C_n-B_n=A_n^2$ for all positive integers $n$; for example, $4444-88=4356=66^2$. The second triple has the greater coordinate sum, $6+8+4=18$.