AMC12 2021 A

We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{\textbf{(C) } 49}.\]

We construct the following table: \[ \begin{array}{l|c|c|c} \text{Scenario} & \text{Length} & \text{Width} & \text{Area} \\ \hline \text{Initial} & 4 & 6 & 24 \\ \text{Menkara shortens one side.} & 3 & 6 & 18 \\ \text{Menkara shortens other side instead.} & 4 & 5 & 20 \\ \end{array} \] Therefore, the answer is (E) 20.

If Mr. Lopez chooses Route A, then he will spend $\frac{6}{30}=\frac{1}{5}$ hour, or $\frac{1}{5}\cdot60=12$ minutes. If Mr. Lopez chooses Route B, then he will spend $\frac{9/2}{40}+\frac{1/2}{20}=\frac{11}{80}$ hour, or $\frac{11}{80}\cdot60=8\frac14$ minutes. Therefore, Route B is quicker than Route A by $12-8\frac14=\boxed{\textbf{(B)}\ 3 \frac{3}{4}}$ minutes.

First, modulo $2$ or $5$, $\underline{20210A} \equiv A$. Hence, $A \neq 0, 2, 4, 5, 6, 8$. Second modulo $3$, $\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A$. Hence, $A \neq 1, 4, 7$. Third, modulo $11$, $\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3$. Hence, $A \neq 3$. Therefore, the answer is $\boxed{\textbf{(E)}\ 9}$.

There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet. Each of Oscar's leaps covers $132\div12=11$ feet, and each of Elmer's strides covers $132\div44=3$ feet. Therefore, Oscar's leap is $11-3=\boxed{\textbf{(B) }8}$ feet longer than Elmer's stride.

By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$ Note that $\triangle DEF$ is isosceles, so $\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle DFE = \boxed{\textbf{(D) }170}$ degrees.

The formula for expected values is \[\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).\] We have \begin{align*} t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\ &= (50+20+20+5+5)\cdot\frac15 \\ &= 100\cdot\frac15 \\ &= 20, \\ s &= 50\cdot\frac{50}{100} + 20\cdot\frac{20}{100} + 20\cdot\frac{20}{100} + 5\cdot\frac{5}{100} + 5\cdot\frac{5}{100} \\ &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5. \end{align*} Therefore, the answer is $t-s=\boxed{\textbf{(B)}\ {-}13.5}.$

By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, that we are taking the $\text{lcm}$ of. In this case, \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\] Now, using the same logic, we find that \[N = M \cdot 2 \cdot 37,\] because we have an extra power of $2$ and an extra power of $37.$ Thus, $\frac{N}{M} = 2\cdot 37 = \boxed{\textbf{(D)}\ 74}.$

The surface area of this right rectangular prism is $2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x).$ The volume of this right rectangular prism is $\log_{2}x\log_{3}x\log_{4}x.$ Equating the numerical values of the surface area and the volume, we have \[2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x)=\log_{2}x\log_{3}x\log_{4}x.\] Dividing both sides by $\log_{2}x\log_{3}x\log_{4}x,$ we get \[2\left(\frac{1}{\log_{4}x}+\frac{1}{\log_{3}x}+\frac{1}{\log_{2}x}\right)=1. \hspace{15mm} (\bigstar)\] Recall that $\log_{b}a=\frac{1}{\log_{a}b}$ and $\log_{b}\left(a^n\right)=n\log_{b}a,$ so we rewrite $(\bigstar)$ as \begin{align*} 2(\log_{x}4+\log_{x}3+\log_{x}2)&=1 \\ 2\log_{x}24&=1 \\ \log_{x}576&=1 \\ x&=\boxed{\textbf{(E)}\ 576}. \end{align*}

Recall that $9\equiv-1\pmod{5}.$ We expand $N$ by the definition of bases: \begin{align*} N&=27{,}006{,}000{,}052_9 \\ &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 2 &&\pmod{5} \\ &\equiv 2-7+6+2 &&\pmod{5} \\ &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. \end{align*}

Label the center of both circles $O$. Label the chord in the larger circle as $\overline{ABCD}$, where $A$ and $D$ are on the larger circle and $B$ and $C$ are on the smaller circle. Construct the radius perpendicular to the chord and label their intersection as $M$. Because a radius that is perpendicular to a chord bisects the chord, $M$ is the midpoint of the chord. Construct segments $\overline{AO}$ and $\overline{BO}$. These are radii with lengths 17 and 19 respectively. Then, use the Pythagorean Theorem. In $\triangle OMA$, we have \begin{align*} OM^2 & = OA^2 - AM^2 \\ & = OA^2 - \left( \frac{AD}{2} \right)^2 \\ & = 19^2 - \frac{AD^2}{4} . \end{align*} In $\triangle OMB$, we have \begin{align*} OM^2 & = OB^2 - BM^2 \\ & = OB^2 - \left( \frac{BC}{2} \right)^2 \\ & = OB^2 - \left( \frac{AD}{4} \right)^2 \\ & = 17^2 - \frac{AD^2}{16} . \end{align*} Equating these two expressions, we get \[19^2 - \frac{AD^2}{4} = 17^2 - \frac{AD^2}{16}\] and $AD=\boxed{\textbf{(E) }8 \sqrt{6}}$.

By the Binomial Theorem, each term in the expansion is of the form \[\binom{1000}{k}\left(x\sqrt[3]{2}\right)^k\left(y\sqrt{3}\right)^{1000-k}=\binom{1000}{k}2^{\frac k3}3^{\frac{1000-k}{2}}x^k y^{1000-k},\] where $k\in\{0,1,2,\ldots,1000\}.$ This problem is equivalent to counting the values of $k$ such that both $\frac k3$ and $\frac{1000-k}{2}$ are integers. Note that $k$ must be a multiple of $3$ and a multiple of $2,$ so $k$ must be a multiple of $6.$ There are $\boxed{\textbf{(C)}\ 167}$ such values of $k:$ \[6(0), 6(1), 6(2), \ldots, 6(166).\]

This solution refers to the Diagram section. Let $O=(0,0), A=(3,3), B=(1,3),$ and $C=\left(\frac3k,3\right).$ As shown below, note that $\overline{OA}, \overline{OB},$ and $\overline{OC}$ are on the lines $y=x, y=3x,$ and $y=kx,$ respectively. By the Distance Formula, we have $OA=3\sqrt2, OB=\sqrt{10}, AC=3-\frac3k,$ and $BC=\frac3k-1.$ By the Angle Bisector Theorem, we get $\frac{OA}{OB}=\frac{AC}{BC},$ or \begin{align*} \frac{3\sqrt2}{\sqrt{10}}&=\frac{3-\frac3k}{\frac3k-1} \\ \frac{3\sqrt2}{\sqrt{10}}&=\frac{3k-3}{3-k} \\ \frac{\sqrt2}{\sqrt{10}}&=\frac{k-1}{3-k} \\ \frac15&=\frac{(k-1)^2}{(3-k)^2} \\ 5(k-1)^2&=(3-k)^2 \\ 4k^2-4k-4&=0 \\ k^2-k-1&=0 \\ k&=\frac{1\pm\sqrt5}{2}. \end{align*} Since $k>0,$ the answer is $k=\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.$ Remark The value of $k$ is known as the Golden Ratio: $\phi=\frac{1+\sqrt{5}}{2}\approx 1.61803398875.$

Divide the equilateral hexagon $ABCDEF$ into isosceles triangles $ABF$, $CBD$, and $EDF$ and triangle $BDF$. The three isosceles triangles are congruent by SAS congruence. By CPCTC, $BF=BD=DF$, so triangle $BDF$ is equilateral. Let the side length of the hexagon be $s$. The area of each isosceles triangle is \[\frac{1}{2} a b \sin\angle C = \frac{1}{2} \cdot s \cdot s \cdot \sin{30^{\circ}} = \frac{1}{4}s^2.\] By the Law of Cosines on triangle $ABF$, \[BF^2=s^2+s^2-2s^2\cos{30^{\circ}}=2s^2-\sqrt{3}s^2.\] Hence, the area of the equilateral triangle $BDF$ is \[\frac{\sqrt{3}}{4} BF^2 = \frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2.\] The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or \[3\left(\frac{1}{4}s^2\right)+ \left( \frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2 \right)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}.\] Hence, $s=2\sqrt{3}$, and the perimeter of the hexagon is $6s=\boxed{\textbf{(E)} \: 12\sqrt3}$.

By Vieta's formulas, $z_1z_2z_3z_4=1$, and $D= (4i)^4\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4.$ Since $\overline{a}\cdot\overline{b}=\overline{ab},$ \[D=(4i)^4\overline{z_1z_2z_3z_4} = 256(\overline{1}) = 256\] By Vieta's formulas, $z_1z_2+z_1z_3+\dots+z_3z_4=3$, and $B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).$ Since $\overline{a}\cdot\overline{b}=\overline{ab},$ \[B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3}+\overline{z_2z_4}+\overline{z_3z_4}\right).\] Since $\overline{a}+\overline{b}=\overline{a+b},$ \[B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48\] Our answer is $B+D=256-48=\boxed{(\textbf{D}) \: 208}.$

We claim that to maximize the number of cables used, we isolate one computer and connect all cables for the remaining $29$ computers, then connect one more cable for the isolated computer. If a brand A computer is isolated, then the technician can use at most $19\cdot10+1=191$ cables. If a brand B computer is isolated instead, then the technician can use at most $20\cdot9+1=181$ cables. Therefore, the answer is $\boxed{\textbf{(B)}\ 191}.$ Remark Suppose that the computers are labeled $A_1, A_2, \ldots, A_{20}$ and $B_1, B_2, \ldots, B_{10}.$ We will prove the claim in the first paragraph of this solution: 1. With exactly $\boldsymbol{190}$ cables, it is possible that some computers cannot communicate with each other. 2. With exactly $\boldsymbol{191}$ cables, all computers must be able to communicate with each other.

A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that: 1. Since $x^2+bx+c=0$ does not have real solutions, we have $b^2\leq 4c.$ 2. Since $x^2+cx+b=0$ does not have real solutions, we have $c^2\leq 4b.$ Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get \[b^4\leq 16c^2\leq 64b.\] Since $b^4\leq 64b,$ it follows that $b\leq4.$ We apply casework to the value of $b:$ - If $b=1,$ then $1\leq 16c^2\leq 64,$ from which $c=1,2.$ - If $b=2,$ then $16\leq 16c^2\leq 128,$ from which $c=1,2.$ - If $b=3,$ then $81\leq 16c^2\leq 192,$ from which $c=3.$ - If $b=4,$ then $256\leq 16c^2\leq 256,$ from which $c=4.$ Together, there are $\boxed{\textbf{(B) } 6}$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$

For simplicity purposes, we assume that the balls and the bins are both distinguishable. Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. For $p,$ we choose one of the $5$ bins to have $3$ balls and another one of the $4$ bins to have $5$ balls. We get \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.\]

The smallest $x$ to make $\sin(x) = \sin(x^2)$ would require $x=x^2$, but since $x$ needs to be greater than $1$, these solutions are not valid. The next smallest $x$ would require $x=180-x^2$, or $x^2+x=180$. After a bit of guessing and checking, we find that $12^2+12=156$, and $13^2+13=182$, so the solution lies between $12{ }$ and $13$, making our answer $\boxed{\textbf{(B) } 13}.$ Note: One can also solve the quadratic and estimate the radical.

First, we can test values that would make $f(x)=12$ true. For this to happen $x$ must have $6$ divisors, which means its prime factorization is in the form $pq^2$ or $p^5$, where $p$ and $q$ are prime numbers. Listing out values less than $50$ which have these prime factorizations, we find $12,18,20,28,44,45,50$ for $pq^2$, and just $32$ for $p^5$. Here $12$ especially catches our eyes, as this means if one of $f_i(n)=12$, each of $f_{i+1}(n), f_{i+2}(n), ...$ will all be $12$. This is because $f_{i+1}(n)=f(f_i(n))$ (as given in the problem statement), so were $f_i(n)=12$, plugging this in we get $f_{i+1}(n)=f(12)=12$, and thus the pattern repeats. Hence, as long as for a $i$, such that $i\leq 50$ and $f_{i}(n)=12$, $f_{50}(n)=12$ must be true, which also immediately makes all our previously listed numbers, where $f_1(x)=12$, possible values of $n$. We also know that if $f_1(x)$ were to be any of these numbers, $x$ would satisfy $f_{50}(n)$ as well. Looking through each of the possibilities aside from $12$, we see that $f_1(x)$ could only possibly be equal to $20$ and $18$, and still have $x$ less than or equal to $50$. This is because if $f_1(x)=20$ or $18$, $f_2 (x) = 12$. This would mean $x$ must have $10$, or $9$ divisors, and testing out, we see that $x$ will then be of the form $p^4q$, or $p^2q^2$. The only two values less than or equal to $50$ would be $48$ and $36$ respectively. From here there are no more possible values, so tallying our possibilities we count $\boxed{\textbf{(D) }10}$ values (Namely $12,18,20,28,32,36,44,45,48,50$).

First realize that $\triangle BCY \sim \triangle DAX.$ Thus, because $CY: XA = 2:3,$ we can say that $BY = 2s$ and $DX = 3s.$ From the Pythagorean Theorem, we have $AB^2 =(2s)^2 + 4^2 = 4s^2 + 16$ and $CD^2 = (3s)^2 + 3^2 = 9s^2 + 9.$ Because $AB = CD,$ from the problem statement, we have that \[4s^2 + 16 = 9s^2 + 9.\] Solving, gives $s = \frac{\sqrt{7}}{\sqrt{5}}.$ To find the area of the trapezoid, we can compute the area of $\triangle ABC$ and add it to the area of $\triangle ACD.$ Thus, the area of the trapezoid is $\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \left(\frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 \right) = \frac{15\sqrt{7}}{\sqrt{5}} = 3\sqrt{35}.$ Thus, the answer is $\boxed{\textbf{(C)} \: 3\sqrt{35}}.$

We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row. $\textbf{Case 1}$: 3 $O$ are in a horizontal row or a vertical row. Step 1: We determine the row that 3 $O$ occupy. The number of ways is 6. Step 2: We determine the configuration of 3 $X$. The number of ways is $\binom{6}{3} - 2 = 18$. (6 open spots to put the $X$ and subtracting cases where they fill row/column) In this case, following from the rule of product, the number of ways is $6 \cdot 18 = 108$. $\textbf{Case 2}$: 3 $O$ are in a diagonal row. Step 1: We determine the row that 3 $O$ occupy. The number of ways is 2. Step 2: We determine the configuration of 3 $X$. The number of ways is $\binom{6}{3} = 20$. In this case, following from the rule of product, the number of ways is $2 \cdot 20 = 40$. Putting all cases together, the total number of ways is $108 + 40 = 148$. Therefore, the answer is $\boxed{\textbf{(D) }148}$.

Let $r_1$ and $r_2$ be the roots of $\tilde{p}(x)$. Then, $\tilde{p}(x)=(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$. The solutions to $\tilde{p}(\tilde{p}(x))=0$ is the union of the solutions to \[\tilde{p}(x)-r_1=x^2-(r_1+r_2)x+(r_1r_2-r_1)=0\] and \[\tilde{p}(x)-r_2=x^2-(r_1+r_2)x+(r_1r_2-r_2)=0.\] Note that one of these two quadratics has one solution (a double root) and the other has two as there are exactly three solutions. WLOG, assume that the quadratic with one root is $x^2-(r_1+r_2)x+(r_1r_2-r_1)=0$. Then, the discriminant is $0$, so $(r_1+r_2)^2 = 4r_1r_2 - 4r_1$. Thus, $r_1-r_2=\pm 2\sqrt{-r_1}$, but for $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ to have two solutions, it must be the case that $r_1-r_2=- 2\sqrt{-r_1} (*)$. It follows that the sum of the roots of $\tilde{p}(x)$ is $2r_1 + 2\sqrt{-r_1}$, whose maximum value occurs when $r_1 = - \frac{1}{4} (\star)$. Solving for $r_2$ yields $r_2 = \frac{3}{4}$. Therefore, $\tilde{p}(x)=x^2 - \frac{1}{2} x - \frac{3}{16}$, so $\tilde{p}(1)= \boxed{\textbf{(A) } \frac{5}{16}}$. Remarks - For $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ to have two solutions, the discriminant $(r_1+r_2)^2-4r_1r_2+4r_2$ must be positive. From here, we get that $(r_1-r_2)^2>-4r_2$, so $-4r_1>-4r_2 \implies r_1<r_2$. Hence, $r_1-r_2$ is negative, so $r_1-r_2=-2\sqrt{-r_1}$. - Set $\sqrt{-r_1}=x$. Now $r_1+\sqrt{-r_1}=-x^2+x$, for which the maximum occurs when $x=\frac{1}{2} \rightarrow r_1=-\frac{1}{4}$.

Let $E$ be a point on $\overline{AB}$ such that $BCDE$ is a parallelogram. Suppose that $BC=ED=b, CD=BE=c,$ and $DA=d,$ so $AE=18-c,$ as shown below. We apply the Law of Cosines to $\triangle ADE:$ \begin{align*} AD^2 + AE^2 - 2\cdot AD\cdot AE\cdot\cos 60^\circ &= DE^2 \\ d^2 + (18-c)^2 - d(18-c) &= b^2 \\ (18-c)^2 - d(18-c) &= b^2 - d^2 \\ (18-c)(18-c-d) &= (b+d)(b-d). \hspace{15mm}(\bigstar) \end{align*} Let $k$ be the common difference of the arithmetic progression of the side-lengths. It follows that $b,c,$ and $d$ are $18-k, 18-2k,$ and $18-3k,$ in some order. It is clear that $0\leq k<6.$ If $k=0,$ then $ABCD$ is a rhombus with side-length $18,$ which is valid. If $k\neq0,$ then we have six cases: 1. $(b,c,d)=(18-k,18-2k,18-3k)$ 2. $(b,c,d)=(18-k,18-3k,18-2k)$ 3. $(b,c,d)=(18-2k,18-k,18-3k)$ 4. $(b,c,d)=(18-2k,18-3k,18-k)$ 5. $(b,c,d)=(18-3k,18-k,18-2k)$ 6. $(b,c,d)=(18-3k,18-2k,18-k)$ Together, the sum of all possible values of $a$ is $18+(13+3+8)+(14+12+16)=\boxed{\textbf{(E) } 84}.$

For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$ Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$ Since $\gcd(m,4)=1,$ we conclude that \[\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}\] is the complete residue system modulo $m$ for all integers $k.$ Given any ordered quadruple $(a'_1, a'_2, a'_3, a'_4)$ in modulo $m,$ it follows that exactly one of these $m$ ordered quadruples has sum $0$ modulo $m:$ We conclude that $D(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),$ so \[q(x)=(x-1)(x-2)(x-3)=c_3x^3+c_2x^2+c_1x+c_0.\] By Vieta's Formulas, we get $c_1=1\cdot2+1\cdot3+2\cdot3=\boxed{\textbf{(E)}\ 11}.$