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【同步首發(fā)】2018AMC12A官方真題及答案

Category: AMC美國數(shù)學競賽, 國際競賽, 福利干貨, 競賽資料 Date: 2018年2月9日上午10:49

2018 AMC 12A真題

答案解析請參考文末

Problem 1

A large urn contains $100$ balls, of which $36 \%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72 \%$ ? (No red balls are to be removed.) $\textbf{(A)}\ 28 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 64$

Problem 2

While exploring a cave, Carl comes across a collection of $5$ -pound rocks worth $$14$ each, $4$ -pound rocks worth $$11$ each, and $1$ -pound rocks worth $$2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave? $\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52$

Problem 3

How many ways can a student schedule 3 mathematics courses -- algebra, geometry, and number theory -- in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.) $\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

Problem 4

Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least 6 miles away," Bob replied, "We are at most 5 miles away." Charlie then remarked, "Actually the nearest town is at most 4 miles away." It turned out that none of the three statements were true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d$ ? $\textbf{(A) } (0,4) \qquad \textbf{(B) } (4,5) \qquad \textbf{(C) } (4,6) \qquad \textbf{(D) } (5,6) \qquad \textbf{(E) } (5,\infty)$

Problem 5

What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common? $\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10 \qquad$

Problem 6

For positive integers $m$ and $n$ such that $m+10<n+1$ , both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$ . What is $m+n$ ? $\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$

Problem 7

For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer? $\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

Problem 8

All of the triangles in the diagram below are similar to iscoceles triangle $ABC$ , in which $AB=AC$ . Each of the 7 smallest triangles has area 1, and $\triangle ABC$ has area 40. What is the area of trapezoid $DBCE$ ? $[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]$ $\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$

Problem 9

Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which $\sin(x+y)\leq \sin(x)+\sin(y)$ for every $x$ between $0$ and $\pi$ , inclusive? $\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi$

Problem 10

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? $x+3y=3$ $\big||x|-|y|\big|=1$ $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$

Problem 11

A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point $A$ falls on point $B$ . What is the length in inches of the crease? $[asy] draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$4$", (2,0), S); label("$3$", (4,1.5), E); label("$5$", (2,1.5), NW); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); [/asy]$ $\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$

Problem 12

Let $S$ be a set of 6 integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S?$ $\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$

Problem 13

How many nonnegative integers can be written in the form $a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,$ where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$ ? $\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$

Problem 14

The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$ ? $\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$

Problem 15

A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called [i]symmetric[/i] if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes? $\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$

Problem 16

Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points? $\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Problem 17

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is 2 units. What fraction of the field is planted? $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); label("$4$", (2,0), N); label("$3$", (0,1.5), E); label("$2$", (.8,1), E); label("$S$", (0,0), NE); draw((0.3,0.3)--(1.4,1.9), dashed); [/asy]$ $\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$

Problem 18

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$ ? $\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Problem 19

Let $A$ be the set of positive integers that have no prime factors other than $2$ , $3$ , or $5$ . The infinite sum $\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots$ of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? $\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}$

Problem 20

Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ , $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$ ? $\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Problem 21

Which of the following polynomials has the greatest real root? $\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018$

Problem 22

The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$ $\textbf{(A)} 20 \qquad \textbf{(B)} 21 \qquad \textbf{(C)} 22 \qquad \textbf{(D)} 23 \qquad \textbf{(E)} 24$

Problem 23

In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$ $\textbf{(A)} 76 \qquad \textbf{(B)} 77 \qquad \textbf{(C)} 78 \qquad \textbf{(D)} 79 \qquad \textbf{(E)} 80$

Problem 24

Alice, Bob, and Carol play a game in which each of them chooses a real number between 0 and 1. The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between 0 and 1, and Bob announces that he will choose his number uniformly at random from all the numbers between $\tfrac{1}{2}$ and $\tfrac{2}{3}.$ Armed with this information, what number should Carol choose to maximize her chance of winning? $\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{13}{24} \qquad \textbf{(C) }\frac{7}{12} \qquad \textbf{(D) }\frac{5}{8} \qquad \textbf{(E) }\frac{2}{3}\qquad$

Problem 25

For a positive integer $n$ and nonzero digits $a$ , $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$ ? $\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}$

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There are $36$ red balls; for these red balls to comprise $72 \%$ of the urn, there must be only $14$ blue balls. Since there are currently $64$ blue balls, this means we must remove $50 = \boxed{ \textbf{(D)}.}$
The answer is just $3\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds, or $54-4=\boxed{\textbf{(C)} 50.}$
We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:Periods $1, 3, 5$ Periods $1, 3, 6$ Periods $1, 4, 6$ Periods $2, 4, 6$ There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.Therefore, there are $4 \cdot 6 = \boxed{\textbf{(E) } 24}$ ways to choose the classes.
From Alice and Bob, we know that $5 < d < 6.$ From Charlie, we know that $4 < d.$ We take the union of these two intervals to yield $\boxed{\textbf{(D) } (5,6)}$ , because the nearest town is between 5 and 6 miles away.
We factor $x^2-3x+2$ into $(x-1)(x-2)$ . Thus, either $1$ or $2$ is a root of $x^2-5x+k$ . If $1$ is a root, then $1^2-5\cdot1+k=0$ , so $k=4$ . If $2$ is a root, then $2^2-5\cdot2+k=0$ , so $k=6$ . The sum of all possible values of $k$ is $\boxed{\textbf{(E)}10}$ .
The mean and median are
$\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,$

so $3m+17=2n$ and $m+11=n$ . Solving this gives $\left(m,n\right)=\left(5,16\right)$ for $m+n=\boxed{21}$ . (trumpeter)
The prime factorization of $4000$ is $2^{5}\cdot5^{3}$ . Therefore, the maximum number for $n$ is $3$ , and the minimum number for $n$ is $-5$ . Then we must find the range from $-5$ to $3$ , which is $3-(-5) + 1 = 8 + 1 = \fbox{\textbf{(E) }9}$ .
Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have $x=7+\dfrac{9}{16}x$ This gives $x=16$ , so the area of $DBCE=40-x=\boxed{24}$ .
On the interval $[0, \pi]$ sine is nonnegative; thus $\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y$ for all $x, y \in [0, \pi]$ . The answer is $\boxed{\textbf{(E) } 0\le y\le \pi}$ .
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.The graph looks something like this: $[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy]$ Now, it becomes clear that there are $\boxed{\textbf{(C) } 3}$ intersection points. (pinetree1)
First, we need to realize that the crease line is just the perpendicular bisector of side $AB$ , the hypotenuse of right triangle $\triangle ABC$ . Call the midpoint of $AB$ point $D$ . Draw this line and call the intersection point with $AC$ as $E$ . Now, $\triangle ABC$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that $\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.$ Thus, our answer is $\boxed{\textbf{D) } \frac{15}{8}}$ .
If we start with $1$ , we can include nothing else, so that won't work.If we start with $2$ , we would have to include every odd number except $1$ to fill out the set, but then $3$ and $9$ would violate the rule, so that won't work.Experimentation with $3$ shows it's likewise impossible. You can include $7$ , $11$ , and either $5$ or $10$ (which are always safe). But after adding either $4$ or $8$ we have nowhere else to go.Finally, starting with $4$ , we find that the sequence $4,5,6,7,9,11$ works, giving us $\boxed{\textbf{(C)} \text{ 4}}$ .
This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$ , which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are $3280+1=\boxed{3281}$ integers or $\boxed{D}$ .
Base switch to log 2 and you have $\frac{\log_2 4}{\log_2 3x} = \frac{\log_2 8}{\log_2 2x}$ . $\frac{2}{\log_2 3x} = \frac{3}{\log_2 2x}$ $2*\log_2 2x = 3*\log_2 3x$ Then $\log_2 (2x)^2 = \log_2 (3x)^3$ . so $4x^2=27x^3$ and we have $x=\frac{4}{27}$ leading to $\boxed{ (D) 31}$
Draw a $7 \times 7$ square. $\begin{tabular}{|c|c|c|c|c|c|c|} \hline K & J & H & G & H & J & K \\ \hline J & F & E & D & E & F & J \\ \hline H & E & C & B & C & E & H \\ \hline G & D & B & A & B & D & G \\ \hline H & E & C & B & C & E & H \\ \hline J & F & E & D & E & F & J \\ \hline K & J & H & G & H & J & K \\ \hline \end{tabular}$ Start from the center and label all protruding cells symmetrically.More specifically, since there are $4$ given lines of symmetry ( $2$ diagonals, $1$ vertical, $1$ horizontal) and they split the plot into $8$ equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has $10$ distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose $2^{10}=1024$ but then subtract the $2$ cases where all are white or all are black. That leaves us with $\boxed{(B)}, 1022$ . ?There are only ten squares we get to actually choose, and two independent choices for each, for a total of $2^{10} = 1024$ codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of $\fbox{\textbf{(B)} \text{ 1022}}$ .
Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get $x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0$ Since this is a quartic, there are 4 total roots (counting multiplicity). We see that $x=0$ always at least one intersection at $(0,-a)$ (and is in fact a double root).The other two intersection points have $x$ coordinates $\sqrt{2a-1}$ . We must have $2a-1> 0,$ otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see $\boxed{\textbf{(E) }a>\frac12}$ .
Let the square have side length $x$ . Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$ .Square $S$ has area $x^2$ , and the two thin triangle regions have area $\dfrac{x(3-x)}{2}$ and $\dfrac{x(4-x)}{2}$ . The final triangular region with the hypotenuse as its base and height $2$ has area $5$ . Thus, we have $x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6$ Solving gives $x=\dfrac{2}{7}$ . The area of $S$ is $\dfrac{4}{49}$ and the desired ratio is $\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}$ .Alternatively, once you get $x=\frac{2}{7}$ , you can avoid computation by noticing that there is a denominator of $7$ , so the answer must have a factor of $7$ in the denominator, which only $\boxed{\dfrac{145}{147}}$ does.
Let $BC = a$ , $BG = x$ , $GC = y$ , and the length of the perpendicular to $BC$ through $A$ be $h$ . By angle bisector theorem, we have that $\frac{50}{x} = \frac{10}{y},$ where $y = -x+a$ . Therefore substituting we have that $BG=\frac{5a}{6}$ . By similar triangles, we have that $DF=\frac{5a}{12}$ , and the height of this trapezoid is $\frac{h}{2}$ . Then, we have that $\frac{ah}{2}=120$ . We wish to compute $\frac{5a}{8}\cdot\frac{h}{2}$ , and we have that it is $\boxed{75}$ by substituting. (rachanamadhu)I may have read this solution incorrectly, but it seems to me that the author mistakenly assumed that the angle bisector is a perpendicular bisector, which is false since the triangle is not isosceles.
It's just $\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.$ since this represents all the numbers in the denominator.
Observe that $\triangle{EMI}$ is isosceles right ( $M$ is the midpoint of diameter arc $EI$ ), so $MI=2,MC=\frac{3}{\sqrt{2}}$ . With $\angle{MCI}=45^\circ$ , we can use Law of Cosines to determine that $CI=\frac{3\pm\sqrt{7}}{2}$ . The same calculations hold for $BE$ also, and since $CI<BE$ , we deduce that $CI$ is the smaller root, giving the answer of $\boxed{12}$ .
We can see that our real solution has to lie in the open interval $(-1,0)$ . From there, note that $x^a < x^b$ if $a$ , $b$ are odd positive integers so $a<b$ , so hence it can only either be B or E(as all of the other polynomials will be larger than the polynomial B). Finally, we can see that plugging in the root of $2019x+2018$ into B gives a negative, and so the answer is $\fbox{B}$ .
The roots are $\pm\left(\sqrt{10}+i\sqrt{6}\right),\pm\left(\sqrt{3}+i\right)$ (easily derivable by using DeMoivre and half-angle). From there, shoelace on $\left(0,0\right),\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{3},1\right)$ and multiplying by $4$ gives the area of $6\sqrt{2}-2\sqrt{10}$ , so the answer is $\boxed{20}$ .
Let $P$ be the origin, and $PA$ lie on the x axis.We can find $U=\left(\cos(36), \sin(36)\right)$ and $G=\left(10-\cos(56), \sin(56)\right)$ Then, we have $M=(5, 0)$ and $N=\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)$ Notice that the tangent of our desired points is the the absolute difference between the y coordinates of the two points divided by the absolute difference between the x coordinates of the two points.This evaluates to $\frac{\sin(36)+\sin(56)}{\cos(36)-\cos(56)}$ Now, using sum to product identities, we have this equal to $\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)$ so the answer is $\boxed{\textbf{(E)}.}$
Plug in all the answer choices to get $\boxed{\textbf{(B)}}$ .
Observe $A_n = a(1 + 10 + \dots + 10^{n - 1}) = a \cdot \tfrac{10^n - 1}{9}$ ; similarly $B_n = b \cdot \tfrac{10^n - 1}{9}$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9}$ . The relation $C_n - B_n = A_n^2$ rewrites as $c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.$ Since $n > 0$ , $10^n > 1$ and we may cancel out a factor of $\tfrac{10^n - 1}{9}$ to obtain $c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.$ This is a linear equation in $10^n$ . Thus, if two distinct values of $n$ satisfy it, then all values of $n$ will. Matching coefficients, we need $c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.$ To maximize $a + b + c = a + \tfrac{a^2}{3}$ , we need to maximize $a$ . Since $b$ and $c$ must be integers, $a$ must be a multiple of 3. If $a = 9$ then $b$ exceeds 9. However, if $a = 6$ then $b = 8$ and $c = 4$ for an answer of $\boxed{\textbf{(D)} \text{ 18}}$ .