AMC12 2020 A

After Carlos took 70% of the pie, 30% remained. Maria took a third of that, which was 10% of the whole pie. Therefore 30% − 10% = 20% of the original pie was left.

There are 5 horizontal and 8 vertical segments, each of length 1 unit. There are 4 slanted segments, each corresponding to the hypotenuse of an isosceles right triangle with leg length 1. By the Pythagorean Theorem, the length of each slanted segment is $\sqrt{1^{2} + 1^{2}} = \sqrt{2}$. Therefore the sum of all the lengths is $13 + 4\sqrt{2}$ units.

The driver travels $2 \cdot 60 = 120$ miles, which requires $120 \div 30 = 4$ gallons of gasoline. Her gasoline expense is $4 \cdot \$2 = \$8$. She is paid $120 \cdot \$0.50 = \$60$, so after her gasoline expense, she makes $\$60 - \$8 = \$52$ in 2 hours, which is $\$$26 per hour.

For an integer to satisfy the given conditions, the thousands digit must lie in $\{2, 4, 6, 8\}$, the hundreds and tens digits must lie in $\{0, 2, 4, 6, 8\}$, and the units digit must be 0. By the Multiplication Principle for counting, the number of choices of digits is $4 \cdot 5 \cdot 5 \cdot 1 = 100$.

Answer (C): The sum of all 25 integers from $-10$ to $14$, inclusive, equals $11+12+13+14=50$ because the integers in that sum from $1$ to $10$ can be paired with their additive inverses to contribute $0$ to the sum. Because each of the $5$ rows in the $5$-by-$5$ square has the same sum and there are $5$ rows, that common sum is $50 \div 5 = 10$. Here is one such magic square, in which each row, each column, and each main diagonal sums to $10$. \[ \begin{array}{|c|c|c|c|c|} \hline 6 & 13 & -10 & -3 & 4\\ \hline 12 & -6 & -4 & 3 & 5\\ \hline -7 & -5 & 2 & 9 & 11\\ \hline -1 & 1 & 8 & 10 & -8\\ \hline 0 & 7 & 14 & -9 & -2\\ \hline \end{array} \]

A non-square rectangle has exactly two lines of symmetry in the plane, as shown below. The additional unit squares that must be shaded are indicated with an X in the figure. Indeed, because the square in the second column of the first (topmost) row is shaded, for the figure to have symmetry around the vertical line, the square in the fourth column of the first row must be shaded. Then for the figure to have symmetry around the horizontal line, the squares in the second and fourth columns of the fourth (bottommost) row must be shaded. Similarly, because the square in the lower right corner is shaded, so must be the squares in the other three corners; and because the square in the middle of the third row is shaded, so must be the middle square in the second row. The resulting figure will then have symmetry around both lines. Thus the least number of additional unit squares that must be shaded to give the figure two lines of symmetry is 7.

Because the volumes of the cubes are given to be $1^3, 2^3, \dots, 7^3$, the edge lengths are 1, 2, $\dots$, 7. If the cubes were not stacked, the total surface area would be $6 \cdot (1^2 + 2^2 + \dots + 7^2)$. When two cubes meet along a surface, two of the smaller face areas are lost, for a total area of $2 \cdot (1^2 + 2^2 + \dots + 6^2)$ that needs to be subtracted from the surface area of the unstacked cubes to get the total surface area of the tower. The result is $$4 \cdot (1^2 + 2^2 + \dots + 6^2) + 6 \cdot 7^2 = 4 \cdot 91 + 6 \cdot 49 = 658.$$ OR The horizontal surface area is $2 \cdot 7^2 = 98$, and the vertical surface area is $4 \cdot (1^2 + 2^2 + \dots + 7^2) = 560$, for a total of 658.

Because there are 4040 numbers in the list, the median is the mean of the numbers in positions 2020 and 2021 when the list is put into nondecreasing order. All of the squared entries are greater than all of the non-squared entries except for $1^2, 2^2, \dots, 44^2 = 1936$. When the numbers are put into nondecreasing order, those 44 entries will be in the first half, and the greatest 44 non-squared entries, namely 1977, 1978, $\dots$, 2020, will be in the second half. Therefore the entry in position 2020 will be 1976, the entry in position 2021 will be 1977, and the median will be 1976.5.

Let $f(x) = \tan(2x)$ and $g(x) = \cos\left(\frac{x}{2}\right)$, so that the given equation is $f(x) = g(x)$. On the interval $[0, 2\pi]$, the values of $g(x)$ decrease from 1 to -1. On each of the intervals $$\left( \frac{(2k-1)\pi}{4}, \frac{(2k+1)\pi}{4} \right), \quad k = 1, 2, 3,$$ the values of $f(x)$ increase and range over all real numbers, so there is a unique solution in each of those intervals. On the interval $[0, \frac{\pi}{4})$, the values of $f(x)$ increase with range $[0, \infty)$ and $g(x) > 0$, so there is a unique solution in that interval. On the interval $(\frac{7\pi}{4}, 2\pi]$, the values of $f(x)$ increase with range $(-\infty, 0]$ and $g(x) < 0$, so there is a unique solution in that interval. The total number of solutions is 5, as displayed in the following graph.

Converting all the logarithms to base 2 gives $$\log_2\left(\frac{1}{4} \log_2 n\right) = \frac{1}{2} \log_2\left(\frac{1}{2} \log_2 n\right).$$ Then $$\log_2\left(\frac{1}{4}\right) + \log_2(\log_2 n) = \frac{1}{2} \log_2\left(\frac{1}{2}\right) + \frac{1}{2} \log_2(\log_2 n).$$ This simplifies to $\log_2(\log_2 n) = 3$, so $n = 2^{(2^3)} = 256$. The requested sum of digits is $2 + 5 + 6 = 13$. OR Let the values of $\log_2(\log_{16} n)$ and $\log_4(\log_4 n)$ both be equal to the real number $x$. Then $2^x = \log_{16} n$ and hence $$n = 16^{(2^x)} = (4^2)^{(2^x)} = 4^{(2 \cdot 2^x)}.$$ Similarly, $4^x = \log_4 n$ and hence $n = 4^{(4^x)} = 4^{(2^x \cdot 2^x)}$. It must therefore be the case that $2 \cdot 2^x = 2^x \cdot 2^x$ and hence $x = 1$. Therefore $n = 4^4 = 256$, and the requested sum is $2 + 5 + 6 = 13$.

After the first jump, the frog is at one of $(0,2)$, $(1,1)$, $(1,3)$, $(2,2)$, each with probability $\frac{1}{4}$. If at $(0,2)$, it ends on a vertical side. In the other cases, it is on a diagonal, so by symmetry, equally likely to end on horizontal or vertical side. Thus, $\frac{1}{4}\cdot1 + \frac{3}{4}\cdot\frac{1}{2}=\frac{5}{8}$.

Answer (B): Note that the point (20, 20) lies on both line $\ell$ and line $k$. The slope of a nonhorizontal, nonvertical line is the tangent of the angle it makes in the upper half-plane with the rightward-pointing $x$-axis. Because line $\ell$ has slope $\frac{3}{5}$, it follows that the slope of line $k$ will be $$ \tan\left(\arctan\frac{3}{5}+45^\circ\right)=\frac{\frac{3}{5}+1}{1-\frac{3}{5}\cdot 1}=4. $$ Therefore an equation of line $k$ is $y-20=4(x-20)$, and this line has $x$-intercept (15, 0).

Answer (B): Written with exponential notation, the left-hand side of the given equation is $((N^{\frac{1}{c}}\cdot N^1)^{\frac{1}{b}}\cdot N^1)^{\frac{1}{a}}=((N^{\frac{1+c}{c}})^{\frac{1}{b}}\cdot N^1)^{\frac{1}{a}}$ $=(N^{\frac{1+c+bc}{bc}})^{\frac{1}{a}}$ $=N^{\frac{1+c+bc}{abc}}.$ Equating exponents and multiplying by $a$ gives $\frac{1+c+bc}{bc}=a\cdot\frac{25}{36}.$ Because $b$ and $c$ are given to be greater than 1, the left-hand side is less than 2, so $a$ cannot be as large as 3. Because $a>1$ as well, $a$ must equal 2, and this equation is equivalent to $18(bc+c+1)=25bc$, from which it follows that $18(c+1)=7bc$. Therefore $c$ divides 18 and 7 divides $c+1$, and thus $c=6$ and $b=3$.

Answer (B): Without loss of generality, let the side length of the regular octagon be 2. Extend sides $AB$ and $DC$ to meet at point $P$. Then $BP=CP=\sqrt{2}$. By symmetry, $ACEG$ is a square. Let its side length be $s$. Then by the Pythagorean Theorem applied to $\triangle APC$, the area of $ACEG$ is \[ m=s^2=(2+\sqrt{2})^2+(\sqrt{2})^2=8+4\sqrt{2}. \] The area of the octagon can be found by subtracting the areas of 4 triangles congruent to $\triangle BPC$ from the area of the square circumscribing the octagon shown below. Thus \[ n=(2+2\sqrt{2})^2-4\cdot\frac12\cdot(\sqrt{2})^2=8+8\sqrt{2}. \] Then \[ \frac{m}{n}=\frac{8+4\sqrt{2}}{8+8\sqrt{2}}=\frac{2+\sqrt{2}}{2+2\sqrt{2}}=\frac{\sqrt{2}}{2}. \]

Answer (D): The solutions to the first equation are 2 times the third roots of unity, namely, 2, $-1 + i\sqrt{3}$, and $-1 - i\sqrt{3}$. The polynomial in the second equation can be rewritten as $(z - 8)(z^2 - 8) = (z - 8)(z - 2\sqrt{2})(z + 2\sqrt{2}),$ so the roots are 8, $2\sqrt{2}$, and $-2\sqrt{2}$. The maximum distance between solutions of the respective equations is the distance between $-1 \pm i\sqrt{3}$ and 8. This is $\sqrt{(-1 - 8)^2 + (\sqrt{3} - 0)^2} = \sqrt{84} = 2\sqrt{21}.$

Answer (B): Without loss of generality, it can be assumed that the point falls within the unit square with vertices $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$. The probability that the point is within $d$ units of a lattice point is the total area of the four quarter-circles of radius $d$ centered at the corners of the square, provided $d<\frac{1}{2}$. This is the shaded region in the figure. Thus $\pi d^2=\frac{1}{2}$, so $d=\frac{1}{\sqrt{2\pi}}<\frac{1}{2}$. Estimating gives $$ d=\frac{1}{\sqrt{2\pi}}\approx\frac{1}{\sqrt{6.28}}\approx\frac{10}{\sqrt{625}}=\frac{10}{25}=0.4. $$ For a rigorous justification of this estimate, it must be shown that $\frac{1}{\sqrt{2\pi}}$ satisfies $$ \frac{7}{20}=0.35<\frac{1}{\sqrt{2\pi}}<0.45=\frac{9}{20}, $$ which is equivalent to $$ \frac{49\pi}{400}<\frac{1}{2}<\frac{81\pi}{400}. $$ Indeed, $$ \frac{49\pi}{400}<\frac{50\pi}{400}=\frac{\pi}{8}<\frac{1}{2} $$ and $$ \frac{81\pi}{400}>\frac{80\pi}{400}=\frac{\pi}{5}>\frac{1}{2}. $$

Answer (D): Let the $x$-coordinates of the four vertices be $a$, $a+1$, $a+2$, and $a+3$. The area of the quadrilateral can be computed as the sum of the areas of three trapezoids with horizontal height $1$, minus the area of a trapezoid with horizontal height $3$. Thus the area is $$ \frac12(\ln a+\ln(a+1))+\frac12(\ln(a+1)+\ln(a+2))+\frac12(\ln(a+2)+\ln(a+3))-\frac32(\ln a+\ln(a+3)), $$ which simplifies to $$ \ln(a+1)+\ln(a+2)-\ln a-\ln(a+3)=\ln\frac{(a+1)(a+2)}{a(a+3)}. $$ The requested value of $a$, the $x$-coordinate of the leftmost vertex of the quadrilateral, can be found by solving the quadratic equation $$ \frac{(a+1)(a+2)}{a(a+3)}=\frac{91}{90} $$ or by noting that the area is a decreasing function of $a$ and $$ \frac{91}{90}=\frac{13\cdot7}{6\cdot15}=\frac{13\cdot14}{12\cdot15}. $$ Hence $a=12$. Note: The area of the quadrilateral can also be calculated using the Shoelace Formula.

Answer (D): Because the diagonals intersect, quadrilateral $ABCD$ is convex. Let $F$ be the foot of the perpendicular from $B$ to $AC$. Because $\triangle ECD$ is a right triangle, $\angle AEB=\angle DEC$ is acute, so $F$ lies between $A$ and $E$. Let $h=BF$ and $x=EF$. See the figure below. Because $\triangle BFE\sim\triangle DCE$, $$\frac{h}{x}=\frac{30}{20-5}=2.$$ Because $\triangle ABF\sim\triangle BCF$, $$\frac{h}{5-x}=\frac{15+x}{h}.$$ Together these imply that $x^2+2x-15=0$. Solving for $x$ yields $x=3$ and $h=6$. The area of $ABCD$ is the area of $\triangle ABC$ plus the area of $\triangle ACD$, and this equals $\frac12\cdot20\cdot6+\frac12\cdot20\cdot30=360$.

Answer (C): Note that the problem is equivalent to finding the number of nonzero digits when the left-hand side is written in binary. More generally, consider the fraction \[ \frac{2^{n^2}+1}{2^n+1} \] for odd $n$. By the formula for the sum of a geometric series, \[ \frac{2^{n^2}+1}{2^n+1}=2^{(n-1)n}-2^{(n-2)n}+2^{(n-3)n}-\cdots+2^{2n}-2^n+1. \] This expression is not in the form of the problem statement because of the negative signs, but it is possible to transform it into such a form by considering pairs of adjacent terms. For $i=1,3,5,\ldots,n-2$, \[ 2^{(i+1)n}-2^{in}=2^{in}(2^n-1) \] \[ =2^{in}(1+2+\cdots+2^{n-1}) \] \[ =2^{in}+2^{in+1}+\cdots+2^{(i+1)n-1}. \] Thus each pair $2^{(i+1)n}-2^{in}$ contributes $((i+1)n-1)-in+1=n$ nonzero digits to the binary expansion of the left-hand side. The total number of expressions of this form is equal to the number of integers in the set $\{1,3,5,\ldots,n-2\}$, which is $\frac12(n-1)$. Thus, including the last term of $1$, the total number of summands is $n\cdot\frac12(n-1)+1$. For $n=17$, this value is $17\cdot 8+1=137$.

Answer (A): If two rotations that are not inverses of each other are applied in order, then there is a unique third rotation that will return $T$ to its original position. There are $3\cdot2=6$ of these. If two different reflections across the coordinate axes are applied in order, then the $180^\circ$ rotation is the unique transformation that will return $T$ to its original position. There are $2$ of these. If the first two transformations are a rotation of $180^\circ$ and a reflection, in either order, then the other reflection is the unique transformation that will return $T$ to its original position. There are $4$ such sequences. It remains to show that for all other choices of the first two transformations, none of the five choices for the third transformation will return $T$ to its original position. Two transformations that are inverses of each other already return $T$ to its original position, so applying a third transformation will move $T$ out of its original position. The other cases are a rotation of $90^\circ$ or $270^\circ$ followed by a horizontal or vertical reflection, or vice versa. In each case, a reflection would be needed to restore the correct orientation of the triangle, but neither a vertical reflection nor a horizontal reflection will return $T$ to its original position. In all, there are $6+2+4=12$ sequences of three of the given transformations that will return $T$ to its original position.

First note that $n$ cannot have any prime factors greater than 7 because otherwise $\operatorname{lcm}(n, 5!)$ would contain such a factor but $5 \cdot \gcd(n, 10!)$ would not. Write $n = 2^a \cdot 3^b \cdot 5^c \cdot 7^d$. Observe that $5! = 2^3 \cdot 3 \cdot 5$ and $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$. In order that the power of 2 is the same on both sides of the given equation $\operatorname{lcm}(n, 5!) = 5 \cdot \gcd(n, 10!)$, it must be that $\max(a, 3) = \min(a, 8)$. This is true if and only if $3 \le a \le 8$. For powers of 3, it must be that $\max(b, 1) = \min(b, 4)$, which holds precisely when $1 \le b \le 4$. For powers of 5, it must be that $\max(c, 1) = 1 + \min(c, 2)$. This holds for $c = 0$, but $n$ is given to be a multiple of 5, so $c > 0$ and the only value that works is $c = 3$. For powers of 7, it must be that $\max(d, 0) = \min(d, 1)$, so $0 \le d \le 1$. There are 6 possible values for $a$, 4 for $b$, 1 for $c$, and 2 for $d$, so there are $6 \cdot 4 \cdot 1 \cdot 2 = 48$ such $n$.

Answer (B): It follows from the original equality that $$\left(\frac{2+i}{\sqrt{7}}\right)^{2n}=\frac{(a_n+b_ni)^2}{7^n}=\left(\frac{a_n^2-b_n^2}{7^n}\right)+\left(\frac{2a_nb_n}{7^n}\right)i.$$ Therefore $$\sum_{n=0}^{\infty}\frac{a_nb_n}{7^n}=\frac12\,\mathrm{Im}\left(\sum_{n=0}^{\infty}\left(\frac{2+i}{\sqrt{7}}\right)^{2n}\right)$$ $$=\frac12\,\mathrm{Im}\left(\sum_{n=0}^{\infty}\left(\frac{3+4i}{7}\right)^n\right)$$ $$=\frac12\,\mathrm{Im}\left(\frac{1}{1-\frac{3+4i}{7}}\right).$$ The preceding geometric series converges as shown because the modulus of the common ratio is less than 1. From $$\frac{1}{1-\frac{3+4i}{7}}=\frac{7}{4-4i}=\frac14\left(\frac{7}{1-i}\cdot\frac{1+i}{1+i}\right)=\frac78(1+i),$$ it follows that $$\sum_{n=0}^{\infty}\frac{a_nb_n}{7^n}=\frac{7}{16}.$$

Answer (A): There are 15 ways to roll a 7 with three dice (5-1-1, 4-2-1, 3-3-1, and 3-2-2 and their permutations) out of the $6^3$ possible rolls, so the probability of rolling 7 with three dice is $p_3=\frac{5}{72}$. Also, for $1\le a\le 7$, the probability of rolling $a$ with two dice is $p_2(a)=\frac{a-1}{36}$, and the probability of rolling any number on a single die is $p_1=\frac{1}{6}$. Because $p_2(a)<p_1$ for all $a<7$, Jason will always choose to reroll one die instead of two if possible; that is, if any two of the initial rolls sum to at most 6, then Jason is better off keeping those two rolls and rerolling the third, rather than keeping just one roll. Therefore Jason will choose to reroll two of the dice only if every pair of the initial rolls sums to at least 7. Furthermore, because $p_2(2)<p_3$ and $p_2(3)<p_3$, if all the rolls are at least 4, then Jason will choose to reroll all three dice. Therefore Jason will reroll two dice if and only if each pair of the initial rolls sums to at least 7, but at least one of the rolls is at most 3. These possibilities are given by 1-6-6, 2-$a$-$b$ where $a,b\in\{5,6\}$, and 3-$c$-$d$ where $c,d\in\{4,5,6\}$. Including permutations, this gives $3+12+27=42$ possibilities, so the requested probability is $\frac{42}{216}=\frac{7}{36}$.

Answer (B): Place the points in the coordinate plane so that $A=(0,0),\quad B=(s,0),\quad C=\left(\frac{s}{2},\frac{s}{2}\sqrt{3}\right),\quad \text{and}\quad P=(x,y).$ It is given that $AP^2=1=x^2+y^2,$ (1) $BP^2=3=(x-s)^2+y^2,$ and (2) $CP^2=4=\left(x-\frac{s}{2}\right)^2+\left(y-\frac{\sqrt{3}s}{2}\right)^2.$ (3) Subtracting the first equation from the other two equations gives $-2xs+s^2=2$ and (4) $-xs-\sqrt{3}ys+s^2=3.$ (5) From equation (4) it follows that $-xs=1-\frac{1}{2}s^2.$ (6) Substituting this into equation (5) leads to $y=\frac{s^2-4}{2\sqrt{3}s}.$ Equation (6) can be used to express $x$ in terms of $s$. Inserting these values for $x$ and $y$ into equation (1) produces an equation that $s$ satisfies $AP^2=1=x^2+y^2=\left(\frac{s^2-2}{2s}\right)^2+\left(\frac{s^2-4}{2\sqrt{3}s}\right)^2.$ This simplifies to $s^4-8s^2+7=0$, so $s^2=1$ or $7$. If $s=1$, then the point $P$ exists, but is outside the triangle. Hence $s=\sqrt{7}$, and indeed $P$ is inside the triangle.

Answer (C): Note that $x>0$, and the only solution with $[x]=0$ is $x=0$, so it follows that $x\ge 1$. The graph of $y=[x]\{x\}$, for $x\ge 1$, consists of line segments $\ell_k$ from $(k,0)$ to $(k+1,k)$ for $k=1,2,\ldots$ with the right endpoint of each segment excluded. If $[x]\{x\}=ax^2$ has no solutions, then the graph of $y=ax^2$ must intersect all the segments $\ell_1,\ell_2,\ldots,\ell_n$, but not $\ell_{n+1}$. Then $$1+2+\cdots+n\le 420<2+3+\cdots+(n+1),$$ which forces $n=28$. Also, note that it must be that $a<\frac14$ for there to be solutions other than $x=0$. The following figure gives an idea of what is happening. Writing $x=[x]+\{x\}$, it follows that $[x]\{x\}=a([x]+\{x\})^2$, so $$0=a\{x\}^2+(2a-1)[x]\{x\}+a[x]^2.$$ Then by the quadratic formula, $$\{x\}=\frac{-(2a-1)[x]\pm\sqrt{(2a-1)^2[x]^2-4a^2[x]^2}}{2a}$$ $$=\left(\frac{(1-2a)\pm\sqrt{1-4a}}{2a}\right)[x].$$ The minus sign must be chosen, because for $x\ge 1$, $$\{x\}=\left(\frac{(1-2a)+\sqrt{1-4a}}{2a}\right)[x]$$ $$\ge \frac{1-2a}{2a}$$ $$=\frac{1}{2a}-1$$ $$>\frac{1}{2\cdot\frac14}-1$$ $$=1,$$ which is impossible. Therefore the solutions are $$x=[x]+\{x\}=\left(\frac{1-\sqrt{1-4a}}{2a}\right)[x]$$ for $1\le [x]\le 28$. Note that this equation implies that for each positive integer $k$, there is at most one solution with $k\le x<k+1$. It follows that the sum of the solutions is $$420=\left(\frac{1-\sqrt{1-4a}}{2a}\right)(1+2+\cdots+28)=\left(\frac{1-\sqrt{1-4a}}{2a}\right)\cdot 14\cdot 29.$$ Thus $$\frac{1-\sqrt{1-4a}}{2a}=\frac{30}{29}.$$ Writing this in the form $29\sqrt{1-4a}=29-60a$, then squaring and simplifying gives $3600a^2-116a=0,$ so