AMC 12 2021 Fall Test A
Instructions
- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have 75 minutes working time to complete the test.
What is the value of $\dfrac{(2112-2021)^2}{169}$?
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$
$\textbf{C}$
$$\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\frac{(13\times7)^2}{13^2}=7^2=49$$
Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?
$\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }19\qquad\textbf{(E) }20$
$\textbf{E}$
If Menkara shortens the length of the 4-inch side by 1 inch, she will get a rectangle of $3\times6=18$ square inches. If Menkara shortens the length of the 6-inch side by 1 inch, she will get a rectangle of $4\times5=20$ square inches. Obviously Menkara shortens the 4-inch side in the first time. So in the next time she will get a rectangle of 20 square inches instead.
Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\dfrac{1}{2}$-mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A?
$\textbf{(A)}\ 2 \dfrac{3}{4} \qquad\textbf{(B)}\ 3 \dfrac{3}{4} \qquad\textbf{(C)}\ 4 \dfrac{1}{2} \qquad\textbf{(D)}\ 5 \dfrac{1}{2} \qquad\textbf{(E)}\ 6 \dfrac{3}{4}$
$\textbf{B}$
For Route A, Mr. Lopez will spend $\dfrac{6}{30}=\dfrac{1}{5}$ hour.
For Route B, Mr. Lopez will spend $\dfrac{5-0.5}{40}+\dfrac{0.5}{20}=\dfrac{11}{80}$ hour.
Route B is quicker than Route A by $\dfrac{1}{5}-\dfrac{11}{80}=\dfrac{1}{16}$ hour, or $\dfrac{1}{16}\times60=\dfrac{15}{4}=3\dfrac{3}{4}$ minutes.
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$(\textbf{A})\: 1\qquad(\textbf{B}) \: 3\qquad(\textbf{C}) \: 5 \qquad(\textbf{D}) \: 7\qquad(\textbf{E}) \: 9$
$\textbf{E}$
We need to consider all possible choices for $A$ from 0 to 9.
In case the six-digit number is divisible by 2, $A$ could be 0, 2, 4, 6, 8.
In case the six-digit number is divisible by 5, $A$ could be 0 or 5.
In case the six-digit number is divisible by 3, $2+0+2+1+0+A=A+5$ is a multiple of 3. Hence, $A$ could be 1, 4, or 7.
In case the six-digit number is divisible by 11, $A+1+0-0-2-2=A-3$ is a multiple of 11. Hence, $A$ could be 3.
The only choice we does not mention above is 9. So the answer is 9.
Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$st pole along this road is exactly one mile ($5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$
$\textbf{B}$
There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.
Each of Elmer's strides covers $132\div44=3$ feet.
Each of Oscar's leaps covers $132\div12=11$ feet.
Therefore, Oscar's leap is $11-3=8$ feet longer than Elmer's stride.
As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE$?
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$
$\textbf{D}$
First, $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ$.
By $DE=DF$ we know $\triangle DEF$ is isosceles, so $\angle DFE = \dfrac{180^\circ - \angle ADE}{2}=10^\circ$.
Finally, we get $\angle AFE = 180^\circ - \angle DFE = 170^\circ$.
A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$. Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$?
$\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$
$\textbf{B}$
Based on the formula of expected value $$E(x)=\sum x\cdot p(x)$$ We have $$t= \frac15\times50 + \frac15\times20 + \frac15\times20 + \frac15\times5 + \frac15\times5=20$$ $$s= \frac{50}{100}\times50 + \frac{20}{100}\times20 + \frac{20}{100}\times20 + \frac{5}{100}\times5 + \frac{5}{100}\times5=33.5$$ So $$t-s=-13.5$$
Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\dfrac{N}{M}?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886$
$\textbf{D}$
By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, so \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29\] Based on the same logic, we find that\[N = M \cdot 2 \cdot 37\] So the answer is $\dfrac{N}{M} = 2\cdot 37 = 74.$
A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_{2}x, \log_{3}x,$ and $\log_{4}x.$ What is $x?$
$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576$
$\textbf{E}$
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 $\log_{2}x\log_{3}x\log_{4}x.$
Now we know the surface area and the volume are numerically equal, so 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\[2\left(\frac{1}{\log_{4}x}+\frac{1}{\log_{3}x}+\frac{1}{\log_{2}x}\right)=1\] Recall that $\log_{b}a=\dfrac{1}{\log_{a}b}$, the equation above can be rewrote as \[2(\log_{x}4+\log_{x}3+\log_{x}2)=1\] By adding together we have \[\log_x{(4\cdot3\cdot2)}^2=\log_x{576}=1\] Finally we get $x=576$.
The base-nine representation of the number $N$ is $27{,}006{,}000{,}052_{nine}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$
$\textbf{D}$
\[N= 2\times9^{10} + 7\times9^9 + 6\times9^6 + 5\times9 + 2\] Since $9\equiv-1\pmod{5}$, we have \[N\equiv 2\times(-1)^{10} + 7\times(-1)^9 + 6\times(-1)^6 + 5\times(-1) + 2\pmod{5}\] So we get \[N\equiv-2\pmod{5}\] or \[N\equiv 3\pmod{5}\] Another way to solve the problem is to think about the last digit of $N$. We can find the pattern of the units digit of the power of 9. When the power is odd, the units digit is 9. When the power is even, the units digit is 1. So the expression of $N$ can be simplified as $$N=2\times1+7\times9+6\times1+5\times9+2=118$$ Hence, the reminder when $N$ is divided by 5 is 3.
Consider two concentric circles of radius $17$ and $19.$ The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?
$\textbf{(A)}\ 12\sqrt{2} \qquad\textbf{(B)}\ 10\sqrt{3} \qquad\textbf{(C)}\ \sqrt{17 \cdot 19} \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 8\sqrt{6}$
$\textbf{E}$
Let the length of the chord be $x$. By the Pythagorean Theorem, the distance from the center of circle to the chord can be expressed as $\sqrt{17^2-\left(\dfrac{x}{4}\right)^2}$ or $\sqrt{19^2-\left(\dfrac{x}{2}\right)^2}$. Hence, we have $$\sqrt{17^2-\left(\dfrac{x}{4}\right)^2}=\sqrt{19^2-\left(\dfrac{x}{2}\right)^2}$$ So the length of the chord is $x=8\sqrt6$.
What is the number of terms with rational coefficients among the $1001$ terms in the expansion of $\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 166 \qquad\textbf{(C)}\ 167 \qquad\textbf{(D)}\ 500 \qquad\textbf{(E)}\ 501$
$\textbf{C}$
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 number of $k$ such that both $\dfrac k3$ and $\dfrac{1000-k}{2}$ 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 167 such values of $k$, namely $6\times0, 6\times1, 6\times2, \ldots, 6\times166$. So the answer is C.
The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y=3x$ has equation $y=kx.$ What is $k?$
$\textbf{(A)} \ \dfrac{1+\sqrt{5}}{2} \qquad \textbf{(B)} \ \dfrac{1+\sqrt{7}}{2} \qquad \textbf{(C)} \ \dfrac{2+\sqrt{3}}{2} \qquad \textbf{(D)} \ 2\qquad \textbf{(E)} \ \dfrac{2+\sqrt{5}}{2}$
$\textbf{A}$
Let the angle made by $y=x$ and the $x$-axis be $\theta_{1}$, and the angle made by $y=3x$ and the $x$-axis be $\theta_{3}$. Then we have $\tan(\theta_{1})=1$ and $\tan(\theta_{3})=3$.
Let the angle made by $y=kx$ and the $x$-axis be $\theta_{k}$. Thus, we have $\tan(\theta_{k})=k$. Since $y=kx$ bisects the two lines, we have\[\theta_k-\theta_1=\theta_3-\theta_k\] Now, we can take the tangent and apply the tangent subtraction formula:
\begin{align*} \tan(\theta_k-\theta_1)&=\tan(\theta_3-\theta_k)\\ \frac{\tan\theta_k-\tan\theta_1}{1+\tan\theta_k\tan\theta_1}&=\frac{\tan\theta_3-\tan\theta_k}{1+\tan\theta_3\tan\theta_k}\\ \frac{k-1}{1+k}&=\frac{3-k}{1+3k}\\ (k-1)(1+3k)&=(1+k)(3-k)\\ 3k^2-2k-1&=-k^2+2k+3\\ 4k^2-4k-4&=0\\ k^2-k-1&=0\\ k&=\frac{1\pm \sqrt{5}}{2}
\end{align*} The slope of $y=kx$ must be positive, which means $k>0$. Thus, we get $k=\dfrac{1+\sqrt{5}}{2}$.
In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ$. The enclosed area of the hexagon is $6\sqrt{3}$. What is the perimeter of the hexagon?
$\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3$
$\textbf{E}$
Divide the equilateral hexagon $ABCDEF$ into isosceles triangles $ABF$, $CBD$, $EDF$ and equilateral triangle $BDF$.
Let the side length of the hexagon be $s$. The area of each isosceles triangle (for example, $\triangle ABF$) is\[\frac{1}{2} AB\cdot AF \sin\angle A = \frac{1}{2} \cdot s \cdot s \cdot \sin{30^{\circ}} = \frac{1}{4}s^2\] By the Law of Cosines on $\triangle ABF$,\[BF^2=s^2+s^2-2s^2\cos{30^{\circ}}=\left(2-\sqrt3\right)s^2\] Hence, the area of equilateral triangle $BDF$ is\[\frac{\sqrt{3}}{4} BF^2 = \frac{\sqrt{3}}{4}\cdot\left(2-\sqrt3\right)s^2=\frac{2\sqrt3-3}{4}s^2\] The total area of the hexagon is three times the area of each isosceles triangle plus the area of the equilateral triangle: \[3\left(\frac{1}{4}s^2\right)+ \frac{2\sqrt3-3}{4}s^2=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}\]Hence, $s=2\sqrt{3}$, and the perimeter of the hexagon is $6s= 12\sqrt3$.
Recall that the conjugate of the complex number $w = a + bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$, is the complex number $\overline{w} = a - bi$. For any complex number $z$, let $f(z) = 4i\hspace{1pt}\overline{z}$. The polynomial\[P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1\]has four complex roots: $z_1$, $z_2$, $z_3$, and $z_4$. Let\[Q(z) = z^4 + Az^3 + Bz^2 + Cz + D\]be the polynomial whose roots are $f(z_1)$, $f(z_2)$, $f(z_3)$, and $f(z_4)$, where the coefficients $A,$ $B,$ $C,$ and $D$ are complex numbers. What is $B + D?$
$(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304$
$\textbf{D}$
By comparing the coefficients of $$(z-z_1)(z-z_2)(z-z_3)(z-z_4)=z^4+4z^3+3z^2+2z+1$$ we get
\begin{align*}
-(z_1+z_2+z_3+z_4) &=4\\
z_1z_2+z_1z_3+z_1z_4+z_2z_3+z_2z_4+z_3z_4 &=3\\
-(z_1z_2z_3+z_1z_2z_4+z_1z_3z_4+z_2z_3z_4) &=2\\
z_1z_2z_3z_4 &=1
\end{align*}
Similarly, by comparing the coefficients of $$\left(z-f(z_1)\right)\left(z-f(z_2)\right)\left(z-f(z_3)\right)\left(z-f(z_4)\right)=z^4 + Az^3 + Bz^2 + Cz + D$$ we get
\begin{align*}
B &=f(z_1)f(z_2)+f(z_1)f(z_3)+f(z_1)f(z_4)+f(z_2)f(z_3)+f(z_2)f(z_4)+f(z_3)f(z_4) \\
D &=f(z_1)f(z_2)f(z_3)f(z_4)
\end{align*}
Since $f(z) = 4i\overline{z}$, we get $$B=(4i)^2\left(\overline{z}_1\overline{z}_2+\overline{z}_1\overline{z}_3+\overline{z}_1\overline{z}_4+\overline{z}_2\overline{z}_3+\overline{z}_2\overline{z}_4+\overline{z}_3\overline{z}_4\right)$$ Considering that $\overline{a}\cdot\overline{b}=\overline{ab},$ we get \[B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3}+\overline{z_2z_4}+\overline{z_3z_4}\right)\] Also we have $\overline{a}+\overline{b}=\overline{a+b},$ so \[B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48\] Now think about $D$ \[D=(4i)^4\left(\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4\right)=256\left(\overline{z_1z_2z_3z_4}\right)=256(\overline{1})=256.\] Our answer is $B+D=-48+256=208$.
An organization has $30$ employees, $20$ of whom have a brand A computer while the other $10$ have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with each other. What is the maximum possible number of cables used?
$\textbf{(A)}\ 190 \qquad\textbf{(B)}\ 191 \qquad\textbf{(C)}\ 192 \qquad\textbf{(D)}\ 195 \qquad\textbf{(E)}\ 196$
$\textbf{B}$
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 $191.$
For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
$\textbf{B}$
A quadratic equation does not have distinct real solutions if and only if the discriminant is nonpositive. We conclude that:
$\quad\bullet\quad$Since $x^2+bx+c=0$ does not have distinct real solutions, we have $b^2\leq 4c.$
$\quad\bullet\quad$Since $x^2+cx+b=0$ does not have distinct 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\[b^4\leq 16c^2\leq 64b\]We apply casework to the value of $b:$
$\quad\bullet\quad$ If $b=1,$ then $1\leq 16c^2\leq 64,$ from which $c=1,2.$
$\quad\bullet\quad$ If $b=2,$ then $16\leq 16c^2\leq 128,$ from which $c=1,2.$
$\quad\bullet\quad$ If $b=3,$ then $81\leq 16c^2\leq 192,$ from which $c=3.$
$\quad\bullet\quad$ If $b=4,$ then $256\leq 16c^2\leq 256,$ from which $c=4.$
Together, there are $6$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$
Each of $20$ balls is tossed independently and at random into one of $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\dfrac{p}{q}$?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
$\textbf{E}$
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
For each ball, we must decide which of the 5 bins should it go into. So we has 5 choices. And 20 balls have $5^{20}$ arrangements.
In case $q$, we need to decide which 4 ball go into the first bin. So there are $\dbinom{20}{4}$ choices. For the second bin, we have $\dbinom{16}{4}$ choices, etc. Hence, the probability is $$q=\dfrac{\dbinom{20}{4}\dbinom{16}{4}\dbinom{12}{4}\dbinom{8}{4}\dbinom{4}{4}}{5^{20}}=\dfrac{\dbinom{20}{4,4,4,4,4}}{5^{20}}$$ In case $p$, first we need to decide which of the 5 bins should be the one with 5 balls, and which 5 balls they are. This gives us $5\times\dbinom{20}{5}$ choices. Next, we need to decide which of the rest 4 bins should be the one with 3 balls, and which 3 balls they are. This gives us $4\times\dbinom{15}{3}$ choices. The rest 12 balls are evenly distributed, so we have $\dbinom{12}{4,4,4}$ arrangements. Therefore, the probability is $$p=\dfrac{5\times\dbinom{20}{5}\times4\times\dbinom{15}{3}\times\dbinom{12}{4,4,4}}{5^{20}}=\dfrac{5\times4\times\dbinom{20}{5,3,4,4,4}}{5^{20}}$$ The ratio is \[\dfrac pq=\dfrac{5\times4\times\dbinom{20}{5,3,4,4,4}}{\dbinom{20}{4,4,4,4,4}}=\dfrac{5\times4\times\dfrac{20!}{5!\times3!\times4!\times4!\times4!}}{\dfrac{20!}{4!\times4!\times4!\times4!\times4!}}=\dfrac{5\times4\times(4!)^5}{5!\times3!\times(4!)^3}=\dfrac{5\times4\times4}{5}=16\]
Let $x$ be the least real number greater than $1$ such that $\sin(x) = \sin(x^2)$, where the arguments are in degrees. What is $x$ rounded up to the closest integer?
$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$
$\textbf{B}$
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 $x^2+x=180$. Hence, we get $$x=\dfrac{-1\pm\sqrt{1+4\cdot180}}2\approx\dfrac{-1\pm\sqrt{4\cdot180}}2=\dfrac{-1\pm12\sqrt5}2$$ Since $x$ is greater than 1, we have $x\approx\dfrac{-1+12\sqrt5}2\approx6\sqrt5\approx6\times2.2\approx13$.
For each positive integer $n$, let $f_1(n)$ be twice the number of positive integer divisors of $n$, and for $j \ge 2$, let $f_j(n) = f_1(f_{j-1}(n))$. For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
$\textbf{D}$
If $f_1(n)=12$, then $n$ 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 forms of prime factorization, we find $12,18,20,28,44,45,50$ for $pq^2$, and just $32$ for $p^5$. Here $12$ is a special number. If $n=12$, then we have\begin{align*}
f_1(n)&=12\\
f_2(n)&=f_1(f_1(n))=f_1(12)=12\\
&\vdots\\
f_{50}(n)&=12
\end{align*}
If $n=18,20,28,32,44,45,50$, we also have \begin{align*}
f_1(n)&=12\\
f_2(n)&=f_1(f_1(n))=f_1(12)=12\\
&\vdots\\
f_{50}(n)&=12
\end{align*}
Furthermore, if $f_1(n)=18,20,28,32,44,45,50$, it also works. Now $n$ has $9,10,14,16,22,22.5$, or $15$ divisors. Looking through the forms of prime factorization like what we did for $f_1(n)=12$ above, we see that $f_1(n)$ could only possibly be equal to 18 and 20, and still have $n$ less than or equal to $50$. This means $n$ must have 9 or 10 divisors, then $n$ will be in the form $p^2q^2$ or $p^4q$. The only two values less than or equal to $50$ would be 36 and 48 respectively.
Together, we find 10 possible values for $n$: $12,18,20,28,32,36,44,45,48,50.$
Let $ABCD$ be an isosceles trapezoid with $\overline{BC} \parallel \overline{AD}$ and $AB=CD$. Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y$, as shown in the figure. Suppose $\angle AXD = \angle BYC = 90^\circ$, $AX = 3$, $XY = 1$, and $YC = 2$. What is the area of $ABCD$?
$\textbf{(A)}\: 15\qquad\textbf{(B)} \: 5\sqrt{11}\qquad\textbf{(C)} \: 3\sqrt{35}\qquad\textbf{(D)} \: 18\qquad\textbf{(E)} \: 7\sqrt{7}$
$\textbf{C}$
Note 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 $CD^2 = (3s)^2 + 3^2 = 9s^2 + 9.$ Since $AB = CD,$ we have \[4s^2 + 16 = 9s^2 + 9\]Solving the equation gives $s = \dfrac{\sqrt{7}}{\sqrt{5}}.$
The area of trapezoid $ABCD$ is the sum of the areas of $\triangle ABC$ and $\triangle ACD.$ Thus, the area of the trapezoid is $$\dfrac12AC\cdot BY+\dfrac12AC\cdot DX=\dfrac12AC\cdot(BY+DX)=\dfrac12\cdot6\cdot\frac{5\sqrt{7}}{\sqrt{5}}= 3\sqrt{35}$$ We have another way to find the attitudes of both triangles. Since both $\overline{BY}$ and $\overline{DX}$ are perpendicular to $\overline{AC}$, we have $$BD^2=XY^2+(BY+DX)^2$$ We also know that the trapezoid is equilateral, so $BD=AC=6$. Hence, $$BY+DX=\sqrt{6^2-1^2}=\sqrt{35}$$ Notice that both triangles $ABC$ and $ADC$ share a common base $\overline{AC}$. Therefore,\[[ABCD]=[ABC]+[ADC]=\frac{1}{2}AC\cdot(BY+DX)=\frac{1}{2}\cdot6\cdot\sqrt{35}=3\sqrt{35}\]
Azar and Carl play a game of tic-tac-toe. Azar places an $X$ in one of the boxes in a $3$-by-$3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has $3$ of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O$. How many ways can the board look after the game is over?
$\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160$
$\textbf{D}$
When the game is over, there will be 3 $X$s and 3 $O$s on the board, where 3 $O$s are in a row and 3 $X$ are not in a row.
$\textbf{Case 1}$: 3 $O$s are in a horizontal row or in a vertical row. This gives us 6 choices. Then we need to choose 3 from the rest 6 positions for 3 $X$s. This gives us $\dbinom63=20$ choices. However, the 3 $X$s can not be in a row. So the total number of arrangements is $6\times(20-2)=108$.
$\textbf{Case 2}$: 3 $O$s are in a diagonal row. This gives us 2 choices. Again there are $\dbinom63=20$ ways for 3 $X$s. So the total number of arrangements is $2\times20=40$.
By adding them together, we get the answer $108+40=148$.
A quadratic polynomial with real coefficients and leading coefficient $1$ is called $\textit{disrespectful}$ if the equation $p(p(x))=0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)$?
$\textbf{(A) } \dfrac{5}{16} \qquad\textbf{(B) } \dfrac{1}{2} \qquad\textbf{(C) } \dfrac{5}{8} \qquad\textbf{(D) } 1 \qquad\textbf{(E) } \dfrac{9}{8}$
$\textbf{A}$
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 solution 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, $$r_1-r_2=\pm 2\sqrt{-r_1}$$ Since $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ has 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 $$r_1-r_2=-2\sqrt{-r_1}$$ It follows that the sum of the roots of $\tilde{p}(x)$ is $r_1+r_2=2r_1 + 2\sqrt{-r_1}$, whose maximum value occurs when $r_1 = - \dfrac{1}{4}$. Solving for $r_2$ yields $r_2 = \dfrac{3}{4}$. Therefore, $$\tilde{p}(x)=x^2 - \dfrac{1}{2} x - \dfrac{3}{16}$$ so $\tilde{p}(1)=\dfrac{5}{16}$.
Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$?
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$
$\textbf{E}$
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 above.
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}(\star) \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:
- $(b,c,d)=(18-k,18-2k,18-3k)\qquad$Note that $(\star)$ becomes $2k(5k-18)=(36-4k)(2k),$ from which $k=6.$ So, this case generates no valid solutions $(b,c,d).$
- $(b,c,d)=(18-k,18-3k,18-2k)\qquad$Note that $(\star)$ becomes $3k(5k-18)=(36-3k)k,$ from which $k=5.$ So, this case generates $(b,c,d)=(13,3,8).$
- $(b,c,d)=(18-2k,18-k,18-3k)\qquad$Note that $(\star)$ becomes $k(4k-18)=(36-5k)k,$ from which $k=6.$ So, this case generates no valid solutions $(b,c,d).$
- $(b,c,d)=(18-2k,18-3k,18-k)\qquad$Note that $(\star)$ becomes $3k(4k-18)=(36-3k)(-k),$ from which $k=2.$ So, this case generates $(b,c,d)=(14,12,16).$
- $(b,c,d)=(18-3k,18-k,18-2k)\qquad$Note that $(\star)$ becomes $k(3k-18)=(36-5k)(-k),$ from which $k=9.$ So, this case generates no valid solutions $(b,c,d).$
- $(b,c,d)=(18-3k,18-2k,18-k)\qquad$Note that $(\star)$ becomes $2k(3k-18)=(36-4k)(-2k),$ from which $k=18.$ So, this case generates no valid solutions $(b,c,d).$
Together, the sum of all possible values of $a$ is $18+(13+3+8)+(14+12+16)=84.$
Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$. There is a polynomial\[q(x) = c_3x^3+c_2x^2+c_1x+c_0\]such that $D(m) = q(m)$ for all odd integers $m\ge 5$. What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$
$\textbf{E}$
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 $\dfrac1m$ 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 satisfy that $m$ divides $S:$
\[\begin{array}{c|c} & \\ [-2.5ex] \textbf{Ordered Quadruple} & \textbf{Sum Modulo }\boldsymbol{m} \\ [0.5ex] \hline & \\ [-2ex] (a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex] (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex] (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S'+4(2) \\ [0.5ex] \cdots & \cdots \\ [0.5ex] (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S'+4(m-1) \\ [0.5ex] \end{array}\]
Note that for any integer $a^\prime_i+j$ ($1\leq i\leq4, 0\leq j\leq m-1$) in the list, if $a^\prime_i+j>m$, we can replace it with $a^\prime_i+j\mod m$. Hence, each ordered quadruple in the list corresponds to one of the $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4)$, and exactly $\dfrac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S.$ So $$D(m)=\dfrac1m\cdot m(m-1)(m-2)(m-3)=(m-1)(m-2)(m-3)$$ By $D(m)=q(m)$, we have \[q(x)=(x-1)(x-2)(x-3)=x^3-6x^2+11x-6=c_3x^3+c_2x^2+c_1x+c_0\]By comparing the coefficients, we get $c_1=11$.