AMC 12 2021 Spring Test B
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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.
How many integer values of $x$ satisfy $|x|<3\pi?$
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20$
$\textbf{D}$
Here we have $-3\pi<x<3\pi$, or approximately $-3(3.14)=-9.42<x<3(3.14)=9.42$. So the answer is $9+1+9=19$.
At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
$\textbf{(A) }23 \qquad \textbf{(B) }32 \qquad \textbf{(C) }37 \qquad \textbf{(D) }41 \qquad \textbf{(E) }64$
$\textbf{B}$
$23\cdot2=46$ students in blue shirts are in pairs with each other, so $57-46=11$ students in blue shirts are in pairs with 11 students with yellow shirts. Hence, $75-11=64$ students in yellow shirts are in pairs with each other, which means 32 pairs of yellow shirts.
Suppose\[2+\dfrac{1}{1+\dfrac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.\]What is the value of $x?$
$\textbf{(A) }\dfrac34 \qquad \textbf{(B) }\dfrac78 \qquad \textbf{(C) }\dfrac{14}{15} \qquad \textbf{(D) }\dfrac{37}{38} \qquad \textbf{(E) }\dfrac{52}{53}$
$\textbf{A}$
\begin{align*}
2+\dfrac{1}{1+\dfrac{1}{2+\frac{2}{3+x}}}&=\frac{144}{53}\\
\dfrac{1}{1+\dfrac{1}{2+\frac{2}{3+x}}}&=\frac{38}{53}\\
1+\dfrac{1}{2+\frac{2}{3+x}}&=\frac{53}{38}\\
\dfrac{1}{2+\frac{2}{3+x}}&=\frac{15}{38}\\
2+\frac{2}{3+x}&=\frac{38}{15}\\
\frac{2}{3+x}&=\frac{8}{15}\\
x+3&=\dfrac{15}{4}\\
x&=\dfrac34
\end{align*}
Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$, and the afternoon class's mean score is $70$. The ratio of the number of students in the morning class to the number of students in the afternoon class is $\dfrac34$. What is the mean of the score of all the students?
$\textbf{(A) }74 \qquad \textbf{(B) }75 \qquad \textbf{(C) }76 \qquad \textbf{(D) }77 \qquad \textbf{(E) }78$
$\textbf{C}$
Assuming that there are $3n$ students in the morning class and $4n$ students in the afternoon class. The total number of students is $7n$, and the total score is $3n\cdot84+4n\cdot70=532n$. So the average is $\dfrac{532n}{7n}=76$.
The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y=-x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b-a?$
$\textbf{(A) }1 \qquad \textbf{(B) }3 \qquad \textbf{(C) }5 \qquad \textbf{(D) }7 \qquad \textbf{(E) }9$
$\textbf{D}$
The final coordinate is $(-6,3)$. Assuming the coordinate before reflecting is $(m,n)$. The slope of the line between the two points is $\dfrac{n-3}{m-(-6)}=\dfrac{n-3}{m+6}$. The line should be perpendicular to the line $y=-x$. Hence, we have $$\dfrac{n-3}{m+6}\cdot(-1)=-1$$ The middle point of $(-6.3)$ and $(m,n)$ is $\left(\dfrac{m-6}2,\dfrac{n+3}2\right)$. The middle point must be on the line $y=-x$. So we get $$\dfrac{n+3}{2}=-\dfrac{m-6}{2}$$ Finally we get $m=-3$, $n=6$.
Now consider the rotation. The line connecting $(1,5)$ and $(-3,6)$ must be perpendicular to the line connecting $(1,5)$ and $(a,b)$. So we have $$\dfrac{6-5}{-3-1}\cdot\dfrac{b-5}{a-1}=-1$$ Rotation doesn't change the distance between points, so we have $$\sqrt{(-3-1)^2+(6-5)^2}=\sqrt{(a-1)^2+(b-5)^2}$$ Solving, we find $(a,b)=(0,1)\ \text{or}\ (2,9)$. By the direction of rotation (counterclockwise), we rule out $(0,1)$.
Therefore, the answer is $b-a=9-2=7$.
An inverted cone with base radius $12 \textrm{cm}$ and height $18\textrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of $24\textrm{cm}$. What is the height in centimeters of the water in the cylinder?
$\textbf{(A) }1.5 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }4.5 \qquad \textbf{(E) }6$
$\textbf{A}$
The volume of water in the cone is $\dfrac12\cdot\pi(12)^2\cdot18=864\pi$. Let the height of water in the cylinder be $h$. Then the volume of water in the cylinder is $\pi(24)^2h=864\pi\rightarrow h=1.5$.
Let $N=34\cdot34\cdot63\cdot270.$ What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N?$
$\textbf{(A) }1:16 \qquad \textbf{(B) }1:15 \qquad \textbf{(C) }1:14 \qquad \textbf{(D) }1:8 \qquad \textbf{(E) }1:3$
$\textbf{C}$
Prime factorize $N$ to get $N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}$. For each odd divisor $n$ of $N$, there exist even divisors $2n, 4n, 8n$ of $N$. Therefore, the ratio is $1:(2+4+8)=1 : 14.$
Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$. What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\dfrac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\dfrac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\dfrac12$
$\textbf{B}$
Since two parallel chords have the same length $38$, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is $2d + d = 3d$, and the distance between each of the chords is just $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we create two different right triangles:
- One with base $\dfrac{38}{2}= 19$, height $d$, and hypotenuse $r$ ($\triangle RAO$ on the diagram)
- Another with base $\dfrac{34}{2} = 17$, height $3d$, and hypotenuse $r$ ($\triangle LBO$ on the diagram)
By the Pythagorean theorem, we can create the following system of equations: \[19^2 + d^2 = r^2\] \[17^2 + (2d + d)^2 = r^2\] Solving, we find $d = 3$, so $2d = 6$.
What is the value of\[\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?\] $\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }\dfrac54 \qquad \textbf{(D) }2 \qquad \textbf{(E) }\log_2 5$
$\textbf{D}$
\begin{align*}
\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}&= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}\\
&=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}\\
&=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}\\
&=2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{20})^2-3\log_{2}{20}-(\log_{2}{20})^2\\
&=2
\end{align*}
Two distinct numbers are selected from the set $\{1,2,3,4,\dots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?
$\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10$
$\textbf{E}$
The sum of all numbers in the set is $1+2+\cdots+37=\dfrac12\cdot37\cdot38=703$. Let the two selected numbers be $x$ and $y$. We have $703-x-y=xy\rightarrow (x+1)(y+1)=704$.
Prime factorization of $740$ gives $704=2^6\cdot11$. Supposing $x+1$ is the divisor who is a multiple of 11. If $x+1=11$, $y+1=2^6=64$, where $y$ is not the set $\{1,2,3,4,\dots,36,37\}$. If $x+1=22$, $y+1=2^5=32$, then we get $(x,y)=(21,31)$. So the answer is $31-21=10$.
Triangle $ABC$ has $AB=13,BC=14$ and $AC=15$. Let $P$ be the point on $\overline{AC}$ such that $PC=10$. There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$
$\textbf{(A) }\dfrac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\dfrac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18$
$\textbf{D}$
We have $AP\parallel CD$ and $AD\parallel BC$ in the diagram. Apparently $\triangle ADP\sim\triangle CBP$, so $\dfrac{AD}{CB}=\dfrac{DP}{BP}=\dfrac{AP}{CP}=\dfrac{5}{10}=\dfrac12$ To find $DP$, we need to find $BP$ first. To find $BP$ in $\triangle ABP$, we need to find $\angle BAP$. By the law of cosines in $\triangle ABC$, we have $$\cos\angle BAC=\dfrac{AB^2+AC^2-BC^2}{2AB\cdot AC}=\dfrac{13^2+15^2-14^2}{2\cdot13\cdot15}=\dfrac{33}{65}$$ Hence, in $\triangle ABP$, we have $$BP=\sqrt{AB^2+AP^2-2AB\cdot AP\cos\angle BAP}=\sqrt{13^2+5^2-2\cdot13\cdot5\cdot\dfrac{33}{65}}=8\sqrt2$$ Then we get $DP=\dfrac12BP=4\sqrt2$, $BD=BP+PD=12\sqrt2$. Since $\triangle ABD\sim\triangle CEB$, we have $\dfrac{BD}{EB}=\dfrac{AD}{CB}=\dfrac12$. So $DE=BD=12\sqrt2$.
Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$, then the average value (arithmetic mean) of the integers remaining is $32$. If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$. If the greatest integer is then returned to the set, the average value of the integers rises to $40$. The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$. What is the average value of all the integers in the set $S$?
$\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37$
$\textbf{D}$
Supposing that there are $n$ numbers in the set $S$. The smallest integer is $x$, and the largest integer is $x+72$. Let $s$ be the sum of all numbers in the set $S$. Now we have
\begin{align*}
s&=32(n-1)+x+72\\
s&=35(n-2)+x+x+72\\
s&=40(n-1)+x
\end{align*} Solving, we find $x=8$, $n=10$, $s=368$. So the average of all integers in the set $S$ is $\dfrac{s}{n}=\dfrac{368}{10}=36.8$.
How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy\[1-3\sin\theta+5\cos3\theta = 0?\]
$\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8$
$\textbf{D}$
We rearrange to get\[5\cos3\theta = 3\sin\theta-1\]We can graph two functions in this case: $y=5\cos{3x}$ and $y=3\sin{x} -1$. Using transformation of functions, we know that $5\cos{3x}$ is just a cosine function with amplitude $5$ and period $\dfrac{2\pi}{3}$. Similarly, $3\sin{x} -1$ is just a sine function with amplitude $3$ and shifted $1$ unit downward:
So, we have $6$ solutions.
Let $ABCD$ be a rectangle and let $\overline{DM}$ be a segment perpendicular to the plane of $ABCD$. Suppose that $\overline{DM}$ has integer length, and the lengths of $\overline{MA},\overline{MC},$ and $\overline{MB}$ are consecutive odd positive integers (in this order). What is the volume of pyramid $MABCD?$
$\textbf{(A) }24\sqrt5 \qquad \textbf{(B) }60 \qquad \textbf{(C) }28\sqrt5\qquad \textbf{(D) }66 \qquad \textbf{(E) }8\sqrt{70}$
$\textbf{A}$
Let $MA=a$ and $MD=d.$ It follows that $MC=a+2$ and $MB=a+4.$ As shown below, note that $\triangle MAD$ and $\triangle MBC$ are both right triangles.
By the Pythagorean Theorem, we have\begin{alignat*}{6} AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2. \end{alignat*}Since $AD=BC$ in rectangle $ABCD,$ we equate the expressions for $AD^2$ and $BC^2,$ then rearrange and factor:\begin{align*} a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\ a^2 - d^2 &= 4a + 12 \\ a^2 - 4a - d^2 &= 12 \\ (a-2)^2 - d^2 &= 16 \\ (a+d-2)(a-d-2) &= 16. \end{align*}As $a+d-2$ and $a-d-2$ have the same parity, we get $a+d-2=8$ and $a-d-2=2,$ from which $(a,d)=(7,3).$
Applying the Pythagorean Theorem to right $\triangle MAD$ and right $\triangle MCD,$ we obtain $AD=2\sqrt{10}$ and $CD=6\sqrt2,$ respectively.
Let the brackets denote areas. Together, the volume of pyramid $MABCD$ is\[\frac13\cdot [ABCD]\cdot MD = \frac13\cdot (AD\cdot CD)\cdot MD = 24\sqrt5\]
The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$
$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$
$\textbf{D}$
Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end.
Note that $\triangle AED$ and $\triangle ABC$ are $120$ - $30$ - $30$ triangles. The area of each of them is $$\dfrac12AB\cdot AC\sin\angle ABC=\dfrac12\cdot2\cdot2\cdot\sin120^\circ=\sqrt3$$ For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$.
Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11}$. So the answer is $m+n=12+11=23$.
Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$
$\textbf{(A) }\dfrac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\dfrac{1+a+b+c}{c^2}\qquad \textbf{(D) }\dfrac{a+b+c}{c^2} \qquad \textbf{(E) }\dfrac{1+a+b+c}{a+b+c}$
$\textbf{A}$
Note that $f(1/x)$ has the same roots as $g(x)$, if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have\[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\]so we can see that\[g(x) = \frac{x^3}{c}f(1/x)\]Therefore\[g(1) = \frac{1}{c}f(1) = \frac{1+a+b+c}c\]
Let $ABCD$ be an isosceles trapezoid having parallel bases $\overline{AB}$ and $\overline{CD}$ with $AB>CD.$ Line segments from a point inside $ABCD$ to the vertices divide the trapezoid into four triangles whose areas are $2, 3, 4,$ and $5$ starting with the triangle with base $\overline{CD}$ and moving clockwise as shown in the diagram below. What is the ratio $\dfrac{AB}{CD}?$
$\textbf{(A)}\: 3\qquad\textbf{(B)}\: 2+\sqrt{2}\qquad\textbf{(C)}\: 1+\sqrt{6}\qquad\textbf{(D)}\: 2\sqrt{3}\qquad\textbf{(E)}\: 3\sqrt{2}$
$\textbf{B}$
Supposing $AB=x$, $CD=y$. Then the distance from the interior point to $AB$ is $\dfrac{8}{x}$, the distance to $CD$ is $\dfrac{4}{y}$. The area of the trapezoid $ABCD$ is $$\dfrac12(x+y)\left(\dfrac{8}{x}+\dfrac{4}{y}\right)=6+\dfrac{2x}{y}+\dfrac{4y}{x}=14$$ Let $\dfrac{AB}{CD}=\dfrac{x}{y}=z$. Then we have $$6+2z+\dfrac{4}{z}=14$$ Solving, we get the answer $z=2+\sqrt2$.
Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\dfrac 6z?$
$\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\dfrac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4$
$\textbf{A}$
Using the fact $z\bar{z}=|z|^2$, the equation rewrites itself as\begin{align*} 12z\bar{z}&=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31 \\ 0&=-12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32 \\ 0&=\left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right) \\ 0&=(z+\bar{z}+2)^2+(z\bar{z}-6)^2 \end{align*}As the two quantities in the parentheses are real, both quantities must equal $0$. So \begin{align*}
z+\bar{z}+2&=0\\
z\bar{z}-6&=0
\end{align*}The answer is \[z+\frac6z=z+\bar{z}=-2\]
Two fair dice, each with at least $6$ faces are rolled. On each face of each dice is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum of $7$ is $\dfrac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\dfrac{1}{12}$. What is the least possible number of faces on the two dice combined?
$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20$
$\textbf{B}$
Suppose the dice have $a$ and $b$ faces, and WLOG $a\geq{b}$. Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$. As a result, there must be $\dfrac{4}{3}\cdot6=8$ ways to sum to $10$. There are at most nine distinct ways to get a sum of $10$, which are possible whenever $a,b\geq{9}$. To achieve exactly eight ways, $b$ must have $8$ faces, and $a\geq9$. Let $n$ be the number of ways to obtain a sum of $12$. The total number of arrangements of $(a,b)$ is $ab=8a$. So $$\dfrac{n}{8a}=\dfrac{1}{12}\implies n=\dfrac{2}{3}a$$ Since $b=8$, $n\leq8\implies a\leq{12}$. Considering that $a$ is divisible by 3 as $n=\dfrac23a$, we only have to test $a=9,12$, of which both work. Taking the smaller one, our answer becomes $a+b=9+8=17$.
Let $Q(z)$ and $R(z)$ be the unique polynomials such that\[z^{2021}+1=(z^2+z+1)Q(z)+R(z)\]and the degree of $R$ is less than $2.$ What is $R(z)?$
$\textbf{(A) }{-}z \qquad \textbf{(B) }{-}1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1$
$\textbf{A}$
Let $z=s$ be a root of $z^2+z+1$ so that $s^2+s+1=0.$ It follows that\[(s-1)\left(s^2+s+1\right)=s^3-1=0,\]from which $s^3=1,$ but $s\neq1.$
Note that\begin{align*} s^{2021}+1 &= s^{3\cdot673+2}+1 \\ &= (s^3)^{673}\cdot s^2+1 \\ &= s^2+1 \\ &= \left(s^2+s+1\right)-s \\ &= -s. \end{align*}Since $z^{2021}+1=-z$ for each root $z=s$ of $z^2+z+1,$ the remainder when $z^{2021}+1$ is divided by $z^2+z+1$ is $R(z)=-z.$
Let $S$ be the sum of all positive real numbers $x$ for which\[x^{2^{\sqrt2}}=\sqrt2^{2^x}.\]Which of the following statements is true?
$\textbf{(A) }S<\sqrt2 \qquad \textbf{(B) }S=\sqrt2 \qquad \textbf{(C) }\sqrt2<S<2\qquad \textbf{(D) }2\le S<6 \qquad \textbf{(E) }S\ge 6$
$\textbf{D}$
We rewrite the right side without using square roots, then take the base-$2$ logarithm for both sides:
\begin{align*} x^{2^{\sqrt2}}&=\left(2^\frac12\right)^{2^x} \\ x^{2^{\sqrt2}}&=2^{\frac12\cdot2^x} \\ x^{2^{\sqrt2}}&=2^{2^{x-1}} \\ \log_2{\left(x^{2^{\sqrt2}}\right)}&=\log_2{\left(2^{2^{x-1}}\right)} \\ 2^{\sqrt2}\log_2{x}&=2^{x-1} \hspace{20mm} (*) \end{align*}By observations, $x=\sqrt2$ is one solution. Graphing $f(x)=2^{\sqrt2}\log_2{x}$ and $g(x)=2^{x-1},$ we conclude that $(*)$ has two solutions, with the smaller solution $x=\sqrt2.$ We construct the following table of values:
Let $x=t$ be the larger solution. Since exponential functions outgrow logarithmic functions, we have $f(x)<g(x)$ for all $x>t.$ By the Intermediate Value Theorem, we get $t\in(2,4),$ from which\[S=\sqrt2+t\in\left(\sqrt2+2,\sqrt2+4\right).\]Finally, approximating with $\sqrt2\approx1.414$ results in $2\le S<6.$
The graphs of $y=f(x)$ and $y=g(x)$ are shown below:
Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$
Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?
$\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)$
$\textbf{B}$
Supposing that there are several walls, where each wall is composed of 1 or 2 bricks, and the numbers of both 1-brick walls and 2-brick walls are even. This is called a symmetrical configuration. For example, $(2,2,2,2,1,1)$ is a symmetrical configuration. The second player will win for any symmetrical configuration if he repeats the movements of the first player and keeps a symmetrical configuration. Come back to the example of $(2,2,2,2,1,1)$. If the first player removes a 1-brick wall, the second player should remove another 1-brick wall to keep symmetry. If the first play removes a 2-brick wall, the second player should remove another 2-brick wall. If the first player removes 1 brick from a 2-brick wall, the second player also removes 1 brick from another 2-brick wall. Hence, the second player can guarantee his success.
Now we turn to the choices:
$(6,1,1)$ can be turned into $(2,2,1,1)$ by Arjun, which is symmetric, so Beth will lose.
$(6,3,1)$ can be turned into $(3,1,3,1)$ by Arjun, which is symmetric, so Beth will lose.
$(6,2,2)$ can be turned into $(2,2,2,2)$ by Arjun, which is symmetric, so Beth will lose.
$(6,3,2)$ can be turned into $(3,2,3,2)$ by Arjun, which is symmetric, so Beth will lose.
So the only possible answer is $(6,2,1)$.
Three balls are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $i$ is $2^{-i}$ for $i=1,2,3,....$ More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is $\dfrac pq,$ where $p$ and $q$ are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins $3,17,$ and $10.$) What is $p+q?$
$\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59$
$\textbf{A}$
Note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be $a$ and the common difference be $d$. Further note that each $(a, d)$ pair uniquely determines a set of $3$ bins.
We have $a\geq1$ because the leftmost bin in the sequence can be any bin, and $d\geq1$, because the bins must be distinct.
This gives us the following sum for the probability:\begin{align*} 3!\cdot\sum_{a=1}^{\infty} \sum_{d=1}^{\infty} \left(2^{-a}\cdot2^{-(a+d)} \cdot2^{-(a+2d)}\right)&= 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a} \cdot 2^{-3d} \\ &= 6 \left( \sum_{a=1}^{\infty} 2^{-3a} \right) \left( \sum_{d=1}^{\infty} 2^{-3d} \right) \\ &= 6 \left( \sum_{a=1}^{\infty} 8^{-a} \right) \left( \sum_{d=1}^{\infty} 8^{-d} \right) \\ &= 6 \left( \frac{1}{7} \right) \left( \frac{1}{7} \right) \\ &= \frac{6}{49} .\end{align*}Therefore the answer is $6 + 49 =55$.
Let $ABCD$ be a parallelogram with area $15$. Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$
$\textbf{A}$
Let $X$ denote the intersection point of the diagonals $AC$ and $BD$. Remark that by symmetry $X$ is the midpoint of both $\overline{PQ}$ and $\overline{RS}$, so $XP = XQ = 3$ and $XR = XS = 4$. Since $\triangle XAP\sin\triangle XBR$, we have $\dfrac{XA}{XB} = \dfrac{XP}{XR} = \dfrac34$.
Thus let $x> 0$ be such that $XA = 3x$ and $XB = 4x$. Then Pythagorean Theorem on $\triangle APX$ yields $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$, and so\[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}=15\]Solving this for $x^2$ yields $x^2 = \dfrac12 + \dfrac{\sqrt{41}}8$, and so\[(8x)^2 = 64x^2 = 64\left(\dfrac12 + \dfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}\]The requested answer is $32 + 8 + 41 = 81$.
Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\dfrac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A) }31 \qquad \textbf{(B) }47 \qquad \textbf{(C) }62\qquad \textbf{(D) }72 \qquad \textbf{(E) }85$
$\textbf{E}$
First, we find a numerical representation for the number of lattice points in $S$ that are under the line $y=mx.$ For any value of $x,$ the highest lattice point under $y=mx$ is $\lfloor mx \rfloor.$ Because every lattice point from $(x, 1)$ to $(x, \lfloor mx \rfloor)$ is under the line, the total number of lattice points under the line is $\sum_{x=1}^{30}(\lfloor mx \rfloor).$
Now, we proceed by finding lower and upper bounds for $m.$ To find the lower bound, we start with an approximation. If $300$ lattice points are below the line, then around $\dfrac{1}{3}$ of the area formed by $S$ is under the line. By using the formula for a triangle's area, we find that when $x=30, y \approx 20.$ Solving for $m$ assuming that $(30, 20)$ is a point on the line, we get $m = \dfrac{2}{3}.$ Plugging in $m$ to $\sum_{x=1}^{30}(\lfloor mx \rfloor),$ we get:
\[\sum_{x=1}^{30}(\lfloor \frac{2}{3}x \rfloor) = 0 + 1 + 2 + 2 + 3 + \cdots + 18 + 18 + 19 + 20\]
We have a repeat every $3$ values (every time $y=\dfrac{2}{3}x$ goes through a lattice point). Thus, we can use arithmetic sequences to calculate the value above:
\begin{align*}
\sum_{x=1}^{30}(\lfloor \frac{2}{3}x \rfloor)& = 0 + 1 + 2 + 2 + 3 + \cdots + 18 + 18 + 19 + 20\\
&=\frac{20(21)}{2} + 2 + 4 + 6 + \cdots + 18\\
&=210 + \frac{2+18}{2}\cdot 9\\
&=300
\end{align*} This means that $\dfrac{2}{3}$ is a possible value of $m.$ Furthermore, it is the lower bound for $m.$ This is because $y=\dfrac{2}{3}x$ goes through many points (such as $(21, 14)$). If $m$ was lower, $y=mx$ would no longer go through some of these points, and there would be less than $300$ lattice points under it.
Now, we find an upper bound for $m.$ Imagine increasing $m$ slowly and rotating the line $y=mx,$ starting from the lower bound of $m=\dfrac{2}{3}.$The upper bound for $m$ occurs when $y=mx$ intersects a lattice point again
In other words, we are looking for the first $m > \dfrac{2}{3}$ that is expressible as a ratio of positive integers $\dfrac{p}{q}$ with $q \le 30.$ For each $q=1,\dots,30$, the smallest multiple of $\dfrac{1}{q}$ which exceeds $\dfrac{2}{3}$ is $$1, \frac{2}{2}, \frac{3}{3}, \frac{3}{4}, \frac{4}{5}, \cdots , \frac{19}{27}, \frac{19}{28}, \frac{20}{29}, \frac{21}{30}$$ respectively, and the smallest of these is $\dfrac{19}{28}.$
The lower bound is $\dfrac{2}{3}$ and the upper bound is $\dfrac{19}{28}.$ Their difference is $\dfrac{1}{84},$ so the answer is $1 + 84 =85.$