AMC12 2019 A

The area of the smaller pizza is $\pi \cdot 3^2 = 9\pi$ in$^2$. The area of the larger pizza is $\pi \cdot 4^2 = 16\pi$ in$^2$. The difference is $7\pi$ in$^2$. Therefore the larger pizza's area is $\frac{7\pi}{9\pi} = \frac{7}{9} = 0.777 \dots \approx 78\%$ larger.

Because $a = \frac{150}{100}b = \frac{3}{2}b$, it follows that $3b = 2a$, so $3b$ is 200% of $a$.

The greatest number of balls that can be drawn without getting 15 of one color is 14 red + 14 green + 14 yellow + 13 blue + 11 white + 9 black = 75 balls. The next ball drawn must be a 15th red, green, or yellow ball. Thus, 76 balls guarantee at least 15 of one color.

Answer (D): The sum $(-44)+(-43)+\cdots+43+44+45=45$. This sum has 90 consecutive integers. There is no longer list because for the sum of consecutive integers to be positive, there must be more positive integers than negative integers. Further, if there are more than 90 consecutive integers as part of a list that sums to a positive number, then there must be a positive integer greater than 45 that is not cancelled out by its additive inverse.

Answer (C): Let $P(2,2)$ be the intersection point. The two lines have equations $y=\frac{1}{2}x+1$ and $y=2x-2$. They intersect $x+y=10$ at $A(6,4)$ and $B(4,6)$. Consider $\overline{AB}$ to be the base of the triangle; then the altitude of the triangle is the segment joining $(2,2)$ and $(5,5)$. By the Distance Formula, the area of $\triangle PAB$ is $\frac{1}{2}\cdot\sqrt{(6-4)^2+(4-6)^2}\cdot\sqrt{(5-2)^2+(5-2)^2}=\frac{1}{2}\cdot2\sqrt{2}\cdot3\sqrt{2}=6.$ Note: The area of the triangle with vertices $(2,2)$, $(6,4)$, and $(4,6)$ can be calculated in a number of other ways, such as by enclosing it in a $4\times4$ square with sides parallel to the coordinate axes and subtracting the areas of three right triangles; by splitting it into two triangles with the line $y=4$; by the shoelace formula: $\frac{1}{2}\cdot\left|(2\cdot4+6\cdot6+4\cdot2)-(6\cdot2+4\cdot4+2\cdot6)\right|=6;$ or by observing that there are $4$ lattice points in the interior of the triangle and $6$ lattice points on the boundary, and using Pick’s Formula: $4+\frac{6}{2}-1=6.$

A translation in the direction parallel to line $\ell$ by an amount equal to the distance between the left sides of successive squares above the line (or any integer multiple thereof), will take the figure to itself. The translation vector could be PQ in the figure below. In addition, a rotation of 180° around any point on line $\ell$ that is halfway between the bases on $\ell$ of a square above the line and a nearest square below the line, such as point R in the figure, will also take the figure to itself. Either of the given reflections, however, will result in a figure in which the “tails” attached to the squares above the line are on the left side of the squares instead of the right side. Therefore 2 of the listed non-identity transformations will transform this figure into itself.

Each of the values 1 through 28 is a mode, so $d = (14+15)/2 = 14.5$. There are $15\cdot12 = 180 < 365/2$ data values less than or equal to 15, and there are $16\cdot12 = 192 > 365/2$ values less than or equal to 16. Therefore more than half of the values are greater than or equal to 16 and more than half of the values are less than or equal to 16, so $M = 16$. To see the relationship between $\mu$ and 16, note that if every month had 31 days, then there would be 12 of each value from 1 to 31, and the mean would be 16; because the actual data are missing some of the larger values, $\mu < 16$. To see the relationship between $\mu$ and 14.5, note that if every month had 28 days, then there would be 12 of each value from 1 to 28, and the mean would be 14.5; because the actual data consist of all of these values together with some larger values, $\mu > 14.5$. Therefore $d = 14.5 < \mu < 16 = M$.

Answer (D): There are several cases to consider. If all four lines are concurrent, then there is 1 intersection point. If three of the lines are concurrent and the fourth line is parallel to one of those three, then there are 3 intersection points. If three of the lines are concurrent and the fourth line is parallel to none of those three, then there are 4 intersection points. In the remaining cases no three lines are concurrent. If they are all parallel, then there are 0 intersection points. If only three of them are parallel, then there are again 3 intersection points. If two of them are parallel but no three are mutually parallel, then there are either again 4 intersection points, if the other two lines are parallel to each other; or 5 intersection points, if the other two lines intersect. In the final case, every line intersects every other line, giving 6 points of intersection. These are all the cases, so the requested sum is $1+3+4+0+5+6=19$.

Answer (E): The sequence begins $1,\ \frac{3}{7},\ \frac{3}{11},\ \frac{3}{15},\ \frac{3}{19},\ \cdots$. This pattern leads to the conjecture that $a_n=\frac{3}{4n-1}$. Checking the initial conditions $n=1$ and $n=2$, and observing that for $n\ge 3$, $$ \frac{\frac{3}{4(n-2)-1}\cdot \frac{3}{4(n-1)-1}}{2\cdot \frac{3}{4(n-2)-1}-\frac{3}{4(n-1)-1}} = \frac{\frac{3}{4n-9}\cdot \frac{3}{4n-5}}{\frac{6}{4n-9}-\frac{3}{4n-5}} = \frac{9}{6(4n-5)-3(4n-9)} = \frac{9}{12n-3} = \frac{3}{4n-1}, $$ confirms the conjecture. Therefore $a_{2019}=\frac{3}{4\cdot 2019-1}=\frac{3}{8075}$, and the requested sum is $3+8075=8078$.

Answer (A): Let $A$, $B$, $C$, and $D$ be the centers of four of the circles as shown below, and let $P$ be the intersection of the diagonals of rhombus $ABDC$. Then $PC = 1$ and $AC = 2$, so $AP = \sqrt{3}$; similarly $PD = \sqrt{3}$. The radius of the large circle is therefore $1 + 2\sqrt{3}$. The requested area is $$ \pi(1 + 2\sqrt{3})^2 - 13\pi = 4\pi\sqrt{3}. $$

Answer (D): The number $0.\overline{23}_k$ is the sum of an infinite geometric series with first term $\frac{2}{k}+\frac{3}{k^2}$ and common ratio $\frac{1}{k^2}$. Therefore the sum is \[ \frac{\frac{2}{k}+\frac{3}{k^2}}{1-\frac{1}{k^2}}=\frac{2k+3}{k^2-1}=\frac{7}{51}. \] Then $0=7k^2-102k-160=(k-16)(7k+10)$, and therefore $k=16$.

Answer (B): Let $a=\log_2 x$ and $b=\log_2 y$. Then \[ \log_y 16 = 4\log_y 2 = \frac{4}{\log_2 y}=\frac{4}{b}, \] so the equation $\log_2 x=\log_y 16$ becomes $ab=4$. Because $xy=2^a\cdot 2^b=64$, it follows that $a+b=6$. Then \[ \left(\log_2\frac{x}{y}\right)^2=(\log_2 x-\log_2 y)^2 =(a-b)^2 =a^2+b^2-2ab =(a+b)^2-4ab =6^2-4\cdot 4 =20. \] Note: Solving the system of equations $\{ab=4,\ a+b=6\}$ gives $\{a,b\}=\{3\pm \sqrt5,\ 3\mp \sqrt5\}$, and $x=2^a$ and $y=2^b$ satisfy the conditions of the problem.

Answer (E): The restriction amounts to the following constraints: 2, 4, and 8 get different colors; 2 and 6 get different colors; 3 and 6 get different colors; and 3 and 9 get different colors. There are $3!$ ways to assign the colors for 2, 4, and 8. Once this is done, there are 2 ways to assign the color for 6, then 2 ways for 3, and 2 ways for 9. Finally, there are 3 ways to assign the color for each of 5 and 7. Hence the total number of paintings is $3!\cdot 2^3 \cdot 3^2 = 432$.

Answer (E): The roots of $x^2-2x+2$ are $1\pm i$, and the roots of $x^2-4x+8$ are $2\pm 2i=2(1\pm i)$. The two roots of $x^2-cx+4$ must be chosen from among those four numbers, and their product must be $4$. The only two pairs of possible roots with a product of $4$ are $\{1+i,\,2-2i\}$ and $\{1-i,\,2+2i\}$, and in each case the value of $c$ is the sum of the roots. Thus the only possible values of $c$ are $3\pm i$, and in both cases $\lvert c\rvert=\sqrt{10}$.

Answer (D): If $n$ is a positive integer such that $n=\sqrt{\log a}$, then $a=10^{n^2}$; and if $m$ is a positive integer such that $m=\log\sqrt{a}$, then $a=10^{2m}$. Thus if both expressions are positive integers, then $a=10^{4k^2}$ for some positive integer $k$, and $$ \sqrt{\log a}+\log\sqrt{a}=\sqrt{4k^2}+\log 10^{2k^2}=2k+2k^2. $$ A similar conclusion holds for $b$. The first few possible values of $a$ and $b$ are $10^4$, $10^{16}$, $10^{36}$, $10^{64}$, $10^{100}$, $10^{144}$, and $10^{196}$, and the corresponding values of $\sqrt{\log a}+\log\sqrt{a}$ and $\sqrt{\log b}+\log\sqrt{b}$ less than 100 are 4, 12, 24, 40, 60, and 84. Because the sum of all four terms is 100, it follows that $\{a,b\}=\{10^{64},10^{100}\}$, and $ab=10^{64}\cdot 10^{100}=10^{164}$.

Answer (B): The sum of three integers is odd exactly when either all of the integers are odd, or one is odd and two are even. Five of the numbers 1, 2, ..., 9 are odd, so at least one row must contain two or more odd numbers. Thus one row must contain three odd numbers and no even numbers, and the other two rows must contain one odd number and two even numbers. The same is true of the three columns. There are $3\times 3=9$ ways to choose which row and which column contain all odd numbers, and then the remaining four squares must have even numbers. There are $\binom{9}{4}=126$ ways in total to choose which squares have odd numbers and which have even numbers, so the desired probability is $\frac{9}{126}=\frac{1}{14}$.

Answer (D): Let $r$ denote a root of the polynomial. Then $r^3 = 5r^2 - 8r + 13$. For $k \ge 2$ multiplying both sides by $r^{k-2}$ gives $r^{k+1} = 5r^k - 8r^{k-1} + 13r^{k-2}$. Summing this identity over all roots $r$ yields $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}$. Thus $a + b + c = 5 - 8 + 13 = 10$.

Answer (D): Let $\triangle ABC$ be the given triangle, with $AB=24$ and $AC=BC=15$, and let $D$ be the midpoint of $AB$. The length of the altitude to $AB$ is $CD=\sqrt{15^2-12^2}=9$. The area of $\triangle ABC$ is $\frac12\cdot24\cdot9=108$. The plane of the triangle intersects the sphere in a circle, which is the inscribed circle for $\triangle ABC$. Let $r$ and $I$ be the radius and the center of the inscribed circle, respectively. The semiperimeter of the triangle is $\frac12(AB+BC+AC)=27$, so $r=\frac{108}{27}=4$. In right triangle $DIO$ the hypotenuse $OD$ has length $6$ (the radius of the sphere) and $DI=r=4$, so $OI=\sqrt{36-16}=2\sqrt5$, the requested distance between the center of the sphere and the plane determined by $\triangle ABC$.

Answer (A): The sines of the angles can be calculated using the Pythagorean Identity: $\sin A=\frac{3\sqrt{15}}{16},\quad \sin B=\frac{\sqrt{15}}{8}=\frac{2\sqrt{15}}{16},\quad \sin C=\frac{\sqrt{15}}{4}=\frac{4\sqrt{15}}{16}.$ The ratio of these values is $3:2:4$. By the Law of Sines, the side lengths are in the same ratio. The smallest triangle having this ratio has sides of lengths $3,2,$ and $4$; the requested least possible perimeter is $9$.

Answer (B): The probability that the first coin flip for both $x$ and $y$ is heads is $\frac{1}{4}$, and in half of these cases $|x-y|$ will be $0$ and in the other half of these cases $|x-y|$ will be $1$. This contributes $\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}$ to the probability that $|x-y|>\frac{1}{2}$. The probability that the first coin flip for $x$ is heads and the first coin flip for $y$ is tails or vice versa is $\frac{1}{2}$. In such cases, one of the variables is $0$ or $1$, and the probability that $|x-y|>\frac{1}{2}$ is $\frac{1}{2}$. This contributes $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$ to the probability that $|x-y|>\frac{1}{2}$. Finally, $\frac{1}{4}$ of the time both $x$ and $y$ will be chosen uniformly from $[0,1]$. In this case, the situation can be modeled by the following diagram, in which the area of the shaded region gives the probability that $|x-y|>\frac{1}{2}$. This contributes $\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}$ to the probability that $|x-y|>\frac{1}{2}$. The requested probability is $\frac{1}{8}+\frac{1}{4}+\frac{1}{16}=\frac{7}{16}$.

Answer (C): Note that $z^2=i$, so $z^4=-1$. For $k=1,2,\ldots,6$, $$z^{(2k)^2}=(-1)^{k^2} \quad \text{and} \quad z^{(2k-1)^2}=z^{4k^2-4k+1}=(-1)^{k(k-1)}z=z.$$ Because $$\sum_{k=1}^{6}(-1)^{k^2}=0,$$ the first factor is $6z=3\sqrt{2}+3\sqrt{2}i$. Because $|z|=1$, the second factor is the conjugate, $3\sqrt{2}-3\sqrt{2}i$. The requested product is therefore the square of the modulus of $3\sqrt{2}+3\sqrt{2}i$, $$\left(3\sqrt{2}\right)^2+\left(3\sqrt{2}\right)^2=18+18=36.$$

Answer (E): Let $D$ and $Q$ be the projections of points $A$ and $O$ onto line $BC$, respectively, and let $s=AB$. It follows that $AD=\frac{s\sqrt3}{2}$. Because $\frac{BP}{CP}=3$ and $D$ is the midpoint of $BC$, it follows that $P$ is the midpoint of $CD$ and $DP=\frac{s}{4}$. It then follows from the Pythagorean Theorem that $$ AP=\sqrt{AD^2+DP^2}=\frac{s\sqrt{13}}{4}. $$ Note that $\triangle QOP\sim\triangle DAP$; therefore $\frac{OP}{OQ}=\frac{AP}{AD}$, so $$ OP=\frac{\frac{s\sqrt{13}}{4}}{\frac{s\sqrt3}{2}}\cdot 17=\frac{17\sqrt{13}}{2\sqrt3}. $$ Then $$ AP+OP=\frac{s\sqrt{13}}{4}+\frac{17\sqrt{13}}{2\sqrt3}=20, $$ which yields $$ s=\frac{80}{\sqrt{13}}-\frac{34}{\sqrt3}. $$ Therefore the requested answer is $80+13+34+3=130$.

Answer (D): Note that $a \diamond b = a^{\log_7(b)} = 7^{\log_7(a)\log_7(b)}.$ It follows that $\diamond$ operates with closure on the interval $(1,\infty)$ and is commutative and associative. Furthermore, the identity element for $\diamond$ is $7$, and the inverse of $a$ is $7^{\frac{1}{\log_7(a)}}$ for all $a>1$; denote the inverse of $a$ under $\diamond$ by $a^{-1}$. Note further that $a \heartsuit b = a^{\frac{1}{\log_7(b)}} = 7^{\log_7(a)\log_7\!\left(7^{\frac{1}{\log_7(b)}}\right)} = a \diamond b^{-1}.$ $s=\dfrac{80}{\sqrt{13}}-\dfrac{34}{\sqrt{3}}.$ Therefore the requested answer is $80+13+34+3=130$. Thus $\begin{aligned} \log_7(a_{2019}) &= \log_7((2019 \heartsuit 2018)\diamond\cdots\diamond(4\heartsuit 3)\diamond(3\heartsuit 2))\\ &= \log_7((2019\diamond 2018^{-1})\diamond\cdots\diamond(4\diamond 3^{-1})\diamond(3\diamond 2^{-1}))\\ &= \log_7(2019\diamond(2018^{-1}\diamond 2018)\diamond\cdots\diamond(3^{-1}\diamond 3)\diamond 2^{-1})\\ &= \log_7(2019\diamond 7\diamond\cdots\diamond 7\diamond 2^{-1})\\ &= \log_7(2019\diamond 2^{-1})\\ &= \log_7(2019^{\frac{1}{\log_7(2)}})\\ &= \dfrac{\log_7(2019)}{\log_7(2)}\\ &= \log_2(2019). \end{aligned}$ Because $2^{11}=2048$, the closest integer to $\log_2(2019)$ is $11$. (More precisely, $\log_2(2019)<\log_2(2048)=11$, and $2^{10.5}=2^{10}\cdot\sqrt{2}<1200\cdot 1.5=1800,$ so $\log_2(2019)>\log_2(1800)>10.5.)

Answer (D): Let $A_n=\dfrac{(n^2-1)!}{(n!)^n}.$ AMC 12A Solutions 15 First, note that $A_n$ is an integer when $n=1$. Next, observe that if $n$ is prime, then $A_n$ is not an integer because the numerator has $n-1$ factors of $n$ but the denominator has $n$ such factors. Note also that $A_4$ is not an integer, because the numerator, $15!$, has $7+3+1=11$ factors of $2$, whereas the denominator, $(4!)^4$, has $12$ factors of $2$. Therefore for $n\ge 2$, in order for $A_n$ to be an integer, a necessary condition is that $n$ be composite and greater than $4$. The following argument shows that this condition is also sufficient. First note that $\dfrac{n!}{n^2}=\dfrac{(n-1)!}{n}.$ If $n=ab$, where $a$ and $b$ are distinct positive integers greater than $1$, then $\dfrac{(n-1)!}{n}$ is an integer because both $a$ and $b$ appear as factors in $(n-1)!$. Otherwise $n=p^2$ for some odd prime $p$. In this case $p^2-1\ge 2p$, so $(n-1)!$ has at least two factors of $p$ and again $\dfrac{(n-1)!}{n}$ is an integer. Now the number $\dfrac{(n^2)!}{(n!)^{n+1}}$ is an integer because this expression counts the number of ways to separate $n^2$ objects into $n$ groups of size $n$ without regard to the ordering of the groups (which accounts for the extra factor of $n!$ in the denominator). By combining the previous two paragraphs, it follows that $A_n=\dfrac{(n^2-1)!}{(n!)^n}=\dfrac{(n^2)!}{(n!)^{n+1}}\cdot\dfrac{n!}{n^2}$ is an integer if and only if $n=1$ or $n$ is composite and greater than $4$. Thus the answer is $50$ minus $1$ minus the number of primes less than or equal to $50$, which is $49-15=34$.

Answer (E): For any acute triangle $DEF$, the triangle $D'E'F'$ whose vertices are the feet of the respective altitudes is called its orthic triangle. Its angles are $180^\circ-2\angle D$, $180^\circ-2\angle E$, and $180^\circ-2\angle F$. To see this, note that $\angle FF'D$ and $\angle FD'D$ are right angles, so $D'$ and $F'$ lie on the circle with diameter $DF$. Therefore $DFD'F'$ is a cyclic quadrilateral and thus \[ \angle F'D'E = 180^\circ-\angle F'D'F = \angle D. \] Similarly, $\angle E'D'F = 180^\circ-\angle E'D'E = \angle D$. It follows that \[ \angle E'D'F' = 180^\circ-\angle F'D'E-\angle E'D'F = 180^\circ-2\angle D. \] In this problem $\triangle A_nB_nC_n$ is the orthic triangle of $\triangle A_{n-1}B_{n-1}C_{n-1}$. Thus if $\triangle A_iB_iC_i$ is acute for $0\le i\le n-1$, then $\triangle A_nB_nC_n$ has angles $180^\circ-2\angle A_{n-1}$, $180^\circ-2\angle B_{n-1}$, and $180^\circ-2\angle C_{n-1}$. It follows by mathematical induction that if $\triangle A_iB_iC_i$ is acute for $0\le i\le n-1$, the angles of $\triangle A_nB_nC_n$ are \[ \left(60+\frac{2^n}{1000}\right)^\circ,\quad 60^\circ,\quad \text{and}\quad \left(60-\frac{2^n}{1000}\right)^\circ. \] Thus it suffices to find the least positive integer $n$ such that \[ 60+\frac{2^n}{1000}>90, \] which is equivalent to $2^n>30{,}000$. Because $2^{14}=16{,}384$ and $2^{15}=32{,}768$, the requested value is $15$.