minqi/DeepScaleR-1k-subset · Datasets at Fast360

problem stringlengths 10 1.99k	answer stringlengths 1 74	solution stringclasses 245 values
Inside an isosceles triangle $\mathrm{ABC}$ with equal sides $\mathrm{AB} = \mathrm{BC}$ and an angle of 80 degrees at vertex $\mathrm{B}$, a point $\mathrm{M}$ is taken such that the angle $\mathrm{MAC}$ is 10 degrees and the angle $\mathrm{MCA}$ is 30 degrees. Find the measure of the angle $\mathrm{AMB}$.	70
For how many integers $n$ between $1$ and $50$, inclusive, is $\frac{(n^2-1)!}{(n!)^n}$ an integer?	34	1. Understanding the Expression: We start by analyzing the expression \[\frac{(n^2-1)!}{(n!)^n}.\] We need to determine for how many integers $n$ between $1$ and $50$ this expression is an integer. 2. Relating to a Known Integer Expression: We know that \[\frac{(n^2)!}{(n!)^{n+1}}\] is an integer because it represents the number of ways to distribute $n^2$ objects into $n$ groups each of size $n$. This can be rewritten using the factorial definition: \[\frac{(n^2)!}{(n!)^{n+1}} = \frac{(n^2)!}{(n!)^n \cdot n!}.\] 3. Connecting the Two Expressions: We can relate the given expression to the known integer expression: \[\frac{(n^2-1)!}{(n!)^n} = \frac{(n^2)!}{n^2 \cdot (n!)^n} = \frac{(n^2)!}{(n!)^{n+1}} \cdot \frac{n!}{n^2}.\] For this to be an integer, $\frac{n!}{n^2}$ must also be an integer, which simplifies to checking if $n^2$ divides $n!$. 4. Analyzing $n^2$ Dividing $n!$: The condition $n^2 \mid n!$ is equivalent to checking if $\frac{n!}{n}$ is an integer. By Wilson's Theorem, this condition fails when $n$ is a prime number or $n=4$ (since $4! = 24$ and $4^2 = 16$ does not divide $24$). 5. Counting Prime Numbers and $n=4$: There are $15$ prime numbers between $1$ and $50$. Including $n=4$, there are $16$ values of $n$ for which the expression is not an integer. 6. Calculating the Total: Since there are $50$ integers from $1$ to $50$, and $16$ of these do not make the expression an integer, the number of integers for which the expression is an integer is $50 - 16 = 34$. 7. Conclusion: Therefore, there are $\boxed{34}$ integers $n$ between $1$ and $50$ for which \[\frac{(n^2-1)!}{(n!)^n}\] is an integer.
The height of a cone and its slant height are 4 cm and 5 cm, respectively. Find the volume of a hemisphere inscribed in the cone, whose base lies on the base of the cone.	\frac{1152}{125} \pi
Jerry and Neil have a 3-sided die that rolls the numbers 1, 2, and 3, each with probability $\frac{1}{3}$. Jerry rolls first, then Neil rolls the die repeatedly until his number is at least as large as Jerry's. Compute the probability that Neil's final number is 3.	\frac{11}{18}	If Jerry rolls $k$, then there is a $\frac{1}{4-k}$ probability that Neil's number is 3, since Neil has an equal chance of rolling any of the $4-k$ integers not less than $k$. Thus, the answer is $$\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)=\frac{11}{18}$$.
Given a triangle \( \triangle ABC \) with interior angles \( \angle A, \angle B, \angle C \) and opposite sides \( a, b, c \) respectively, where \( \angle A - \angle C = \frac{\pi}{2} \) and \( a, b, c \) are in arithmetic progression, find the value of \( \cos B \).	\frac{3}{4}
A sequence consists of $2020$ terms. Each term after the first is 1 larger than the previous term. The sum of the $2020$ terms is $5410$. When every second term is added up, starting with the first term and ending with the second last term, what is the sum?	2200
In $\triangle ABC$, the sides opposite to angles $A$, $B$, $C$ are $a$, $b$, $c$, respectively, and $C= \frac{3}{4}\pi$, $\sin A= \frac{\sqrt{5}}{5}$. (I) Find the value of $\sin B$; (II) If $c-a=5-\sqrt{10}$, find the area of $\triangle ABC$.	\frac{5}{2}
Points $M$ and $N$ are located on side $AC$ of triangle $ABC$, and points $K$ and $L$ are on side $AB$, with $AM : MN : NC = 1 : 3 : 1$ and $AK = KL = LB$. It is known that the area of triangle $ABC$ is 1. Find the area of quadrilateral $KLNM$.	7/15
A granite pedestal. When constructing a square foundation and a cubic pedestal for a monument, granite cubic blocks of size \(1 \times 1\) meter were used. The pedestal used exactly as many blocks as the square foundation upon which it stood. All the blocks were used whole and uncut. Look at the picture and try to determine the total number of blocks used. The foundation has a thickness of one block.	128
Given that $\angle A$ and $\angle B$ are within the interval $\left(0, \frac{\pi}{2}\right)$, and that $\frac{\sin A}{\sin B} = \sin (A+B)$, find the maximum value of $\tan A$.	4/3
In a factory, a total of $100$ parts were produced. Among them, A produced $0$ parts, with $35$ being qualified. B produced $60$ parts, with $50$ being qualified. Let event $A$ be "Selecting a part from the $100$ parts at random, and the part is qualified", and event $B$ be "Selecting a part from the $100$ parts at random, and the part is produced by A". Then, the probability of $A$ given $B$ is \_\_\_\_\_\_.	\dfrac{7}{8}
Given an isosceles triangle $ABC$ satisfying $AB=AC$, $\sqrt{3}BC=2AB$, and point $D$ is on side $BC$ with $AD=BD$, then the value of $\sin \angle ADB$ is ______.	\frac{2 \sqrt{2}}{3}
Evaluate the expression $3 + 2\sqrt{3} + \frac{1}{3 + 2\sqrt{3}} + \frac{1}{2\sqrt{3} - 3}$.	3 + \frac{10\sqrt{3}}{3}
Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$. Find $d$.	450	Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\sqrt{3}$, so that $AB=2\sqrt{3}$. Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$, which of course includes $P, Q$. From there, we can call $OU=h$. There are two crucial equations we can thus generate. WLOG set $PU<QU$, then we call $PU=d-h, QU=d+h$. First equation: using the Pythagorean Theorem on $\triangle UOB$, $h^2+2^2=d^2$. Next, using the tangent addition formula on angles $\angle PHU, \angle UHQ$ we see that after simplifying $-d^2+h^2=-4, 2d=3\sqrt{3}$ in the numerator, so $d=\frac{3\sqrt{3}}{2}$. Multiply back the scalar and you get $\boxed{450}$. Not that hard, was it?
How many values of $x$, $-17<x<100$, satisfy $\cos^2 x + 3\sin^2 x = \cot^2 x$? (Note: $x$ is measured in radians.)	37
Using the trapezoidal rule with an accuracy of 0.01, calculate $\int_{2}^{3} \frac{d x}{x-1}$.	0.6956
The function $f(x)=x^5-20x^4+ax^3+bx^2+cx+24$ has the interesting property that its roots can be arranged to form an arithmetic sequence. Determine $f(8)$ .	-24
How many integers $n$ are there such that $0 \le n \le 720$ and $n^2 \equiv 1$ (mod $720$ )?	16
Given $$\frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} = 2$$, find the value of $$\frac{1 + \sin4\alpha - \cos4\alpha}{1 + \sin4\alpha + \cos4\alpha}$$.	\frac{3}{4}
Define the hotel elevator cubicas the unique cubic polynomial $P$ for which $P(11) = 11$ , $P(12) = 12$ , $P(13) = 14$ , $P(14) = 15$ . What is $P(15)$ ? Proposed by Evan Chen	13
If $x = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$, then:	1 < x < 2	1. Identify the equation: Given the expression for $x$, we can write it as: \[ x = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}} \] Notice that the expression under the square root is the same as $x$ itself. Therefore, we can set up the equation: \[ x = \sqrt{1 + x} \] 2. Square both sides: To eliminate the square root, square both sides of the equation: \[ x^2 = 1 + x \] 3. Rearrange into a standard quadratic form: Bring all terms to one side of the equation: \[ x^2 - x - 1 = 0 \] 4. Apply the quadratic formula: The solutions to a quadratic equation $ax^2 + bx + c = 0$ are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plugging in $a = 1$, $b = -1$, and $c = -1$, we get: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \] 5. Determine the appropriate root: Since $x$ is defined as a series of nested square roots, it must be non-negative. Thus, we discard the negative root and keep the positive root: \[ x = \frac{1 + \sqrt{5}}{2} \] 6. Approximate the value of $x$: The value of $\sqrt{5}$ is approximately $2.236$. Therefore: \[ x \approx \frac{1 + 2.236}{2} \approx 1.618 \] 7. Compare $x$ with the options provided: The value $1.618$ lies between $1$ and $2$. Therefore, the correct choice is: \[ \boxed{\textbf{(C)}\ 1 < x < 2} \]
Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ$ , and $AB = 10$. Let $P$ be randomly chosen inside $ABC$ , and extend $\overline{BP}$ to meet $\overline{AC}$ at $D$. What is the probability that $BD > 5\sqrt2$?	\frac{3-\sqrt3}{3}	1. Identify the lengths of sides in triangle $ABC$: Given that $\angle ACB = 90^\circ$ and $\angle ABC = 60^\circ$, triangle $ABC$ is a 30-60-90 triangle. In such triangles, the sides are in the ratio $1:\sqrt{3}:2$. Since $AB = 10$ (hypotenuse), the other sides are: - $BC = \frac{1}{2} \times AB = \frac{1}{2} \times 10 = 5$ (opposite the $30^\circ$ angle), - $AC = \sqrt{3} \times BC = \sqrt{3} \times 5 = 5\sqrt{3}$ (opposite the $60^\circ$ angle). 2. Determine the condition for $BD > 5\sqrt{2}$: Extend $BP$ to meet $AC$ at $D$. We need to find the condition under which $BD > 5\sqrt{2}$. Consider a specific $D'$ on $AC$ such that $BD' = 5\sqrt{2}$. By the Pythagorean theorem in triangle $BD'C$, we have: \[ BD'^2 = BC^2 + CD'^2 \implies (5\sqrt{2})^2 = 5^2 + CD'^2 \implies 50 = 25 + CD'^2 \implies CD'^2 = 25 \implies CD' = 5. \] 3. Analyze the geometric condition: Since $CD' = 5$ and $AC = 5\sqrt{3}$, the point $D'$ divides $AC$ into segments of $5$ and $5\sqrt{3} - 5$. For $BD > 5\sqrt{2}$, $CD$ must be greater than $5$, meaning $D$ must be between $D'$ and $C$. 4. Calculate the probability: The probability that $P$ lies in the region where $BD > 5\sqrt{2}$ is equivalent to the probability that $P$ lies in triangle $ABD'$, which is a smaller triangle within triangle $ABC$. The ratio of their areas is the same as the ratio of $AD'$ to $AC$ because the height from $B$ to line $AC$ is the same for both triangles. Thus, \[ \text{Probability} = \frac{AD'}{AC} = \frac{AC - CD'}{AC} = \frac{5\sqrt{3} - 5}{5\sqrt{3}} = 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} = \frac{3 - \sqrt{3}}{3}. \] 5. Conclusion: The probability that $BD > 5\sqrt{2}$ is $\boxed{\frac{3-\sqrt{3}}{3}}$.
In a right triangle \(ABC\) with a right angle at \(B\) and \(\angle A = 30^\circ\), a height \(BD\) is drawn. Then, in triangle \(BDC\), a median \(DE\) is drawn, and in triangle \(DEC\), an angle bisector \(EF\) is drawn. Find the ratio \( \frac{FC}{AC} \).	1/8
Given the function $y=\cos (2x-\frac{\pi }{4})$, determine the horizontal translation of the graph of the function $y=\sin 2x$.	\frac{\pi }{8}
Find the smallest positive integer $m$ such that for all positive integers $n \geq m$, there exists a positive integer $l$ satisfying $$ n < l^2 < \left(1+\frac{1}{2009}\right)n. $$	16144325
Let $r$, $s$, and $t$ be solutions of the equation $x^3-5x^2+6x=9$. Compute $\frac{rs}t + \frac{st}r + \frac{tr}s$.	-6
A right triangle with integer leg lengths is called "cool'' if the number of square units in its area is equal to twice the number of units in the sum of the lengths of its legs. What is the sum of all the different possible areas of cool right triangles?	118
Maria ordered a certain number of televisions for the stock of a large store, paying R\$ 1994.00 per television. She noticed that in the total amount to be paid, the digits 0, 7, 8, and 9 do not appear. What is the smallest number of televisions she could have ordered?	56
Given an ellipse $C$: $\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1 (a > b > 0)$ with a focal length of $2$, and point $Q( \frac{a^{2}}{ \sqrt{a^{2}-b^{2}}},0)$ on the line $l$: $x=2$. (1) Find the standard equation of the ellipse $C$; (2) Let $O$ be the coordinate origin, $P$ a moving point on line $l$, and $l'$ a line passing through point $P$ that is tangent to the ellipse at point $A$. Find the minimum value of the area $S$ of $\triangle POA$.	\frac{ \sqrt{2}}{2}
Given that $2^x+ 2^x+ 2^x+ 2^x= 128$, what is the value of $(x + 1)(x - 1)$?	24
Given an arithmetic sequence $\{a_{n}\}$, where $a_{1}+a_{8}=2a_{5}-2$ and $a_{3}+a_{11}=26$, calculate the sum of the first 2022 terms of the sequence $\{a_{n} \cdot \cos n\pi\}$.	2022
Determine the area and the circumference of a circle with the center at the point \( R(2, -1) \) and passing through the point \( S(7, 4) \). Express your answer in terms of \( \pi \).	10\pi \sqrt{2}
Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$	33	Let $1-x=a;1-y=b;1-z=c$, rewrite those equations $\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1$; $\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}$ $\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}$ square both sides, get three equations: $2ab-1=2\sqrt{(1-a^2)(1-b^2)}$ $2bc=2\sqrt{(1-b^2)(1-c^2)}$ $2ac+1=2\sqrt{(1-c^2)(1-a^2)}$ Getting that $a^2+b^2-ab=\frac{3}{4}$ $b^2+c^2=1$ $a^2+c^2+ac=\frac{3}{4}$ Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$, $a=b-c$ Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}$, $bc=\frac{1}{4}$ Since $a^2=b^2+c^2-2bc=\frac{1}{2}$, the final answer is $\frac{1}{4}\frac{1}{4}\frac{1}{2}=\frac{1}{32}$ the final answer is $\boxed{033}$ ~bluesoul
If P and Q are points on the line y = 1 - x and the curve y = -e^x, respectively, find the minimum value of \|PQ\|.	\sqrt{2}
If $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$ are unit vectors, find the largest possible value of \[ \\|\mathbf{a} - \mathbf{b}\\|^2 + \\|\mathbf{a} - \mathbf{c}\\|^2 + \\|\mathbf{a} - \mathbf{d}\\|^2 + \\|\mathbf{b} - \mathbf{c}\\|^2 + \\|\mathbf{b} - \mathbf{d}\\|^2 + \\|\mathbf{c} - \mathbf{d}\\|^2. \]	16
Betty goes to the store to get flour and sugar. The amount of flour she buys, in pounds, is at least 6 pounds more than half the amount of sugar, and is no more than twice the amount of sugar. Find the least number of pounds of sugar that Betty could buy.	4
Compute $\cos 90^\circ$.	0
Let $\triangle PQR$ be a right triangle such that $Q$ is a right angle. A circle with diameter $QR$ intersects side $PR$ at $S$. If $PS = 3$ and $QS = 9$, what is $RS$?	27
Given an arithmetic sequence $\{a\_n\}$, where $a\_n \in \mathbb{N}^*$, and $S\_n = \frac{1}{8}(a\_n + 2)^2$. If $b\_n = \frac{1}{2}a\_n - 30$, find the minimum value of the sum of the first $\_\_\_\_\_\_$ terms of the sequence $\{b\_n\}$.	15
Find maximum value of number $a$ such that for any arrangement of numbers $1,2,\ldots ,10$ on a circle, we can find three consecutive numbers such their sum bigger or equal than $a$ .	18
Given that \( F \) is the right focus of the hyperbola \( x^{2} - y^{2} = 1 \), \( l \) is the right directrix of the hyperbola, and \( A \) and \( B \) are two moving points on the right branch of the hyperbola such that \( A F \perp B F \). The projection of the midpoint \( M \) of line segment \( AB \) onto \( l \) is \( N \). Find the maximum value of \( \frac{\|MN\|}{\|AB\|} \).	1/2
Let the function $f(x) = (\sin x + \cos x)^2 - \sqrt{3}\cos 2x$. (Ⅰ) Find the smallest positive period of $f(x)$; (Ⅱ) Find the maximum value of $f(x)$ on the interval $\left[0, \frac{\pi}{2}\right]$ and the corresponding value of $x$ when the maximum value is attained.	\frac{5\pi}{12}
Farmer Yang has a \(2015 \times 2015\) square grid of corn plants. One day, the plant in the very center of the grid becomes diseased. Every day, every plant adjacent to a diseased plant becomes diseased. After how many days will all of Yang's corn plants be diseased?	2014
A geometric sequence starts $16$, $-24$, $36$, $-54$. What is the common ratio of this sequence?	-\frac{3}{2}
The sequence \(a_{n}\) is defined as follows: \[ a_{1} = 1, \quad a_{n+1} = a_{n} + \frac{2a_{n}}{n} \text{ for } n \geq 1 \] Find \(a_{100}\).	5151
At around 8 o'clock in the morning, two cars left the fertilizer plant one after another, heading toward Happy Village. Both cars travel at a speed of 60 kilometers per hour. At 8:32, the distance the first car had traveled from the fertilizer plant was three times the distance traveled by the second car. At 8:39, the distance the first car had traveled from the fertilizer plant was twice the distance traveled by the second car. At what exact time did the first car leave the fertilizer plant?	8:11
Given that the complex number \( z \) satisfies \( \|z\|=1 \), find the maximum value of \( \left\| z^3 - 3z - 2 \right\| \).	3\sqrt{3}
If a school bus leaves school with 48 students on board, and one-half of the students get off the bus at each of the first three stops, how many students remain on the bus after the third stop?	6
Given that the power function $y=x^{m}$ is an even function and is a decreasing function when $x \in (0,+\infty)$, determine the possible value of the real number $m$.	-2
Let $p(x)$ be a monic polynomial of degree 6 such that $p(1) = 1,$ $p(2) = 2,$ $p(3) = 3,$ $p(4) = 4,$ $p(5) = 5,$ and $p(6) = 6.$ Find $p(7).$	727
Martha can make 24 cookies with 3 cups of flour. How many cookies can she make with 5 cups of flour, and how many cups of flour are needed to make 60 cookies?	7.5
Ilya takes a triplet of numbers and transforms it following the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after the 1989th application of this rule, if the initial triplet of numbers was \(\{70, 61, 20\}\)? If the question allows for multiple solutions, list them all as a set.	50
For real numbers \(x\), \(y\), and \(z\), consider the matrix \[ \begin{pmatrix} x+y & x & y \\ x & y+z & y \\ y & x & x+z \end{pmatrix} \] Determine whether this matrix is invertible. If not, list all possible values of \[ \frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y}. \]	-3
Xiaoming's home is 30 minutes away from school by subway and 50 minutes by bus. One day, due to some reasons, Xiaoming first took the subway and then transferred to the bus, taking 40 minutes to reach the school. The transfer process took 6 minutes. How many minutes did Xiaoming spend on the bus that day?	10
In a scalene triangle with integer side lengths $a, b, c$, the following relation holds. What is the smallest height of the triangle? $$ \frac{a^{2}}{c}-(a-c)^{2}=\frac{b^{2}}{c}-(b-c)^{2} $$	2.4
The angles of a convex $n$-sided polygon form an arithmetic progression whose common difference (in degrees) is a non-zero integer. Find the largest possible value of $n$ for which this is possible.	27	The exterior angles form an arithmetic sequence too (since they are each $180^{\circ}$ minus the corresponding interior angle). The sum of this sequence must be $360^{\circ}$. Let the smallest exterior angle be $x$ and the common difference be $d$. The sum of the exterior angles is then $x+(x+a)+(x+2a)+\ldots+(x+(n-1)a)=\frac{n(n-1)}{2} \cdot a+nx$. Setting this to 360, and using $nx>0$, we get $n(n-1)<720$, so $n \leq 27$.
How many different graphs with 9 vertices exist where each vertex is connected to 2 others?	4	It suffices to consider the complements of the graphs, so we are looking for graphs with 9 vertices, where each vertex is connected to 2 others. There are $\mathbf{4}$ different graphs.
For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?	48	1. Let $x$ be the number of acorns that both the chipmunk and the squirrel hid. According to the problem, the chipmunk hides 3 acorns per hole and the squirrel hides 4 acorns per hole. 2. Let $h_c$ be the number of holes dug by the chipmunk and $h_s$ be the number of holes dug by the squirrel. We know that: \[ 3h_c = x \quad \text{and} \quad 4h_s = x \] This implies that the total number of acorns hidden by each animal is the same. 3. We are also given that the squirrel needed 4 fewer holes than the chipmunk: \[ h_c = h_s + 4 \] 4. Substitute $h_c$ from the third equation into the first equation: \[ 3(h_s + 4) = x \] Simplifying, we get: \[ 3h_s + 12 = x \] 5. From the second equation, we know $4h_s = x$. We can set the expressions for $x$ equal to each other: \[ 3h_s + 12 = 4h_s \] Solving for $h_s$, we subtract $3h_s$ from both sides: \[ 12 = h_s \] 6. Substitute $h_s = 12$ back into the equation $4h_s = x$ to find $x$: \[ x = 4 \times 12 = 48 \] Thus, the number of acorns that the chipmunk hid is $\boxed{48}$, which corresponds to answer choice $\textbf{(D)}$.
Given that the terms of the geometric sequence $\{a_n\}$ are positive, and the common ratio is $q$, if $q^2 = 4$, then $$\frac {a_{3}+a_{4}}{a_{4}+a_{5}}$$ equals \_\_\_\_\_\_.	\frac {1}{2}
For real numbers $a$ and $b$ , define $$ f(a,b) = \sqrt{a^2+b^2+26a+86b+2018}. $$ Find the smallest possible value of the expression $$ f(a, b) + f (a,-b) + f(-a, b) + f (-a, -b). $$	4 \sqrt{2018}
A line through the points $(2, -9)$ and $(j, 17)$ is parallel to the line $2x + 3y = 21$. What is the value of $j$?	-37
A bakery sells three kinds of rolls. How many different combinations of rolls could Jack purchase if he buys a total of six rolls and includes at least one of each kind?	10
Determine the smallest positive integer $n$ such that $n^2$ is divisible by 50 and $n^3$ is divisible by 294.	210
Richard starts with the string HHMMMMTT. A move consists of replacing an instance of HM with MH , replacing an instance of MT with TM, or replacing an instance of TH with HT. Compute the number of possible strings he can end up with after performing zero or more moves.	70	The key claim is that the positions of the Ms fully determines the end configuration. Indeed, since all Hs are initially left of all Ts, the only successful swaps that can occur will involve Ms. So, picking $\binom{8}{4}=70$ spots for Ms and then filling in the remaining 4 spots with Hs first and then Ts gives all possible arrangements. It is not hard to show that all of these arrangements are also achievable; just greedily move Ms to their target positions.
Compute $$\sum_{n_{60}=0}^{2} \sum_{n_{59}=0}^{n_{60}} \cdots \sum_{n_{2}=0}^{n_{3}} \sum_{n_{1}=0}^{n_{2}} \sum_{n_{0}=0}^{n_{1}} 1$$	1953	The given sum counts the number of non-decreasing 61-tuples of integers $\left(n_{0}, \ldots, n_{60}\right)$ from the set $\{0,1,2\}$. Such 61-tuples are in one-to-one correspondence with strictly increasing 61-tuples of integers $\left(m_{0}, \ldots, m_{60}\right)$ from the set $\{0,1,2, \ldots, 62\}$: simply let $m_{k}=n_{k}+k$. But the number of such $\left(m_{0}, \ldots, m_{60}\right)$ is almost by definition $\binom{63}{61}=\binom{63}{2}=1953$.
What is the smallest positive integer $n$ such that $\frac{n}{n+101}$ is equal to a terminating decimal?	24
Find the least positive integer $n$ , such that there is a polynomial \[ P(x) = a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+\dots+a_1x+a_0 \] with real coefficients that satisfies both of the following properties: - For $i=0,1,\dots,2n$ it is $2014 \leq a_i \leq 2015$ . - There is a real number $\xi$ with $P(\xi)=0$ .	2014
A list of $3042$ positive integers has a unique mode, which occurs exactly $15$ times. Calculate the least number of distinct values that can occur in the list.	218
The average of the numbers $1, 2, 3, \dots, 44, 45, x$ is $50x$. What is $x$?	\frac{1035}{2299}
In triangle \(ABC\), the angle at vertex \(B\) is \(\frac{\pi}{3}\), and the line segments connecting the incenter to vertices \(A\) and \(C\) are 4 and 6, respectively. Find the radius of the circle inscribed in triangle \(ABC\).	\frac{6 \sqrt{3}}{\sqrt{19}}
Chords \(AB\) and \(CD\) of a circle with center \(O\) both have a length of 5. The extensions of segments \(BA\) and \(CD\) beyond points \(A\) and \(D\) intersect at point \(P\), where \(DP=13\). The line \(PO\) intersects segment \(AC\) at point \(L\). Find the ratio \(AL:LC\).	13/18
For what value of $n$ does $\|6 + ni\| = 6\sqrt{5}$?	12
Mrs. Everett recorded the performance of her students in a chemistry test. However, due to a data entry error, 5 students who scored 60% were mistakenly recorded as scoring 70%. Below is the corrected table after readjusting these students. Using the data, calculate the average percent score for these $150$ students. \begin{tabular}{\|c\|c\|} \multicolumn{2}{c}{}\\\hline \textbf{$\%$ Score}&\textbf{Number of Students}\\\hline 100&10\\\hline 95&20\\\hline 85&40\\\hline 70&40\\\hline 60&20\\\hline 55&10\\\hline 45&10\\\hline \end{tabular}	75.33
In the expression \(5 * 4 * 3 * 2 * 1 = 0\), replace the asterisks with arithmetic operators \(+, -, \times, \div\), using each operator exactly once, so that the equality holds true (note: \(2 + 2 \times 2 = 6\)).	5 - 4 \times 3 : 2 + 1
The sequence \(\{a_n\}\) is a geometric sequence with a common ratio of \(q\), where \(\|q\| > 1\). Let \(b_n = a_n + 1 (n \in \mathbb{N}^*)\), if \(\{b_n\}\) has four consecutive terms in the set \(\{-53, -23, 19, 37, 82\}\), find the value of \(q\).	-\dfrac{3}{2}
A boss plans a business meeting at Starbucks with the two engineers below him. However, he fails to set a time, and all three arrive at Starbucks at a random time between 2:00 and 4:00 p.m. When the boss shows up, if both engineers are not already there, he storms out and cancels the meeting. Each engineer is willing to stay at Starbucks alone for an hour, but if the other engineer has not arrived by that time, he will leave. What is the probability that the meeting takes place?	\frac{7}{24}
For distinct positive integers $a, b<2012$, define $f(a, b)$ to be the number of integers $k$ with $1\le k<2012$ such that the remainder when $ak$ divided by $2012$ is greater than that of $bk$ divided by $2012$. Let $S$ be the minimum value of $f(a, b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than $2012$. Determine $S$.	502	To solve for \( S \), the minimum value of \( f(a, b) \), where distinct positive integers \( a, b < 2012 \), we first need to analyze the function \( f(a, b) \). This function represents the number of integers \( k \) with \( 1 \leq k < 2012 \) such that: \[ ak \mod 2012 > bk \mod 2012 \] ### Steps to find \( S \): 1. Understanding the Remainder Condition: For each \( k \), we compare: \[ ak \equiv r_a \pmod{2012} \] and \[ bk \equiv r_b \pmod{2012} \] We need \( r_a > r_b \). 2. Expressing the Condition: The condition becomes: \[ ak - bk \equiv (a-b)k \equiv r_a - r_b \pmod{2012} \] 3. Analyzing \( f(a, b) \): Note that both \( a \) and \( b \) are distinct and less than \( 2012 \). For a specific \( k \), the behavior of \( ak \mod 2012 \) and \( bk \mod 2012 \) involves cycling through the possible remainders from \( 0 \) to \( 2011 \). 4. Distribution of Remainders: Since \( a \) and \( b \) are distinct, their multiplicative properties will result in differences in the cycle of remainders. 5. Symmetry Argument: By symmetry, as \( k \) ranges from \( 1 \) to \( 2012 - 1 \), there will be a balance in the number of \( k \) for which \( ak \mod 2012 > bk \mod 2012 \) and \( ak \mod 2012 < bk \mod 2012 \). 6. Calculating \( S \): Since for each pair \( (a, b) \) except permutations, the integer values \( k \) will be split symmetrically, Hence, we anticipate that on average, the set of \( k \) is divided equally between remainders being higher for \( a \) or \( b \). Therefore: \[ S = \frac{2012 - 1}{2} = 1005 \] However, due to rounding down because of distinct integer properties (as \( k \) values cannot be split fractionally), the precise minimum value \( S \) is: \[ \boxed{502} \] This accounts for any adjustments due to parity and nearest integer calculations rounding down for edge differences.
If three different numbers are selected from 2, 3, 4, 5, 6 to be $a$, $b$, $c$ such that $N = abc + ab + bc + a - b - c$ reaches its maximum value, then this maximum value is.	167
Place each of the digits 4, 5, 6, and 7 in exactly one square to make the smallest possible product. What is this product?	2622
For any real number $x$ , we let $\lfloor x \rfloor$ be the unique integer $n$ such that $n \leq x < n+1$ . For example. $\lfloor 31.415 \rfloor = 31$ . Compute \[2020^{2021} - \left\lfloor\frac{2020^{2021}}{2021} \right \rfloor (2021).\] 2021 CCA Math Bonanza Team Round #3	2020
If in an arithmetic sequence, the sum of the first three terms is 34, the sum of the last three terms is 146, and the sum of all terms is 390, then the number of terms in the sequence is __________.	13
A sphere intersects the $xy$-plane in a circle centered at $(3, 5, 0)$ with radius 2. The sphere also intersects the $yz$-plane in a circle centered at $(0, 5, -8),$ with radius $r.$ Find $r.$	\sqrt{59}
On one side of the acute angle \(A\), points \(P\) and \(Q\) are marked such that \(AP = 4\), \(AQ = 12\). On the other side, points \(M\) and \(N\) are marked at distances of 6 and 10 from the vertex. Find the ratio of the areas of triangles \(MNO\) and \(PQO\), where \(O\) is the intersection point of the lines \(MQ\) and \(NP\).	1:5
Let $a_n$ be the integer closest to $\sqrt{n}$. Find the sum $$ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{1980}}. $$	88
Given that the vertex of a parabola is at the origin and the center of the circle $(x-2)^2 + y^2 = 4$ is exactly the focus of the parabola. 1. Find the equation of the parabola. 2. A line with a slope of 2 passes through the focus of the parabola and intersects the parabola at points A and B. Find the area of triangle OAB.	4\sqrt{5}
$a,b,c$ - are sides of triangle $T$ . It is known, that if we increase any one side by $1$ , we get new a) triangle b)acute triangle Find minimal possible area of triangle $T$ in case of a) and in case b)	\frac{\sqrt{3}}{4}
A point is chosen randomly from within a circular region with radius $r$. A related concentric circle with radius $\sqrt{r}$ contains points that are closer to the center than to the boundary. Calculate the probability that a randomly chosen point lies closer to the center than to the boundary.	\frac{1}{4}
Determine all integers $k$ such that there exists infinitely many positive integers $n$ [b]not[/b] satisfying \[n+k \|\binom{2n}{n}\]	k \neq 1	Determine all integers \( k \) such that there exist infinitely many positive integers \( n \) not satisfying \[ n + k \mid \binom{2n}{n}. \] We claim that all integers \( k \neq 1 \) satisfy the desired property. First, recall that \(\frac{1}{n + 1} \binom{2n}{n}\) is the \( n \)-th Catalan number. Since the Catalan numbers are a sequence of integers, it follows that \( n + 1 \mid \binom{2n}{n} \) for all \( n \). Hence, \( k = 1 \) certainly cannot satisfy the problem statement. Now, we consider two cases: Case 1: \( k \neq 2 \). Suppose that \( p \) is a prime divisor of \( k \) and let \( n = p^\alpha \) for any \( \alpha \in \mathbb{N} \). Then, since \( p \mid n + k \), in order to prove that \( n + k \nmid \binom{2n}{n} \), it suffices to show that \[ p \nmid \binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}. \] Note that the greatest power of \( p \) that divides any term in the numerator or denominator of \(\frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)}\) is less than \( p^{\alpha} \). Since the sets \(\{1, 2, \cdots, n - 1\}\) and \(\{n + 1, n + 2, \cdots, 2n - 1\}\) are congruent modulo \( p^{\alpha} \), the numerator and denominator of the fraction \(\frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)}\) both contain the same number of factors of \( p \). Therefore, \( p \nmid \frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)} \). Now, if we can show that \( p \nmid 2 \), we will be able to conclude that \( p \nmid \binom{2n}{n} \), as desired. Indeed, if \( p \neq 2 \), then trivially \( p \nmid 2 \). Meanwhile, if \( p = 2 \), then let us take \( \alpha \geq 2 \) so that \( 2^2 \mid n + k \). Hence, we wish to show that \( 2^2 \nmid \binom{2n}{n} \). But since \( 2 \nmid \frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)} = \frac{\binom{2n}{n}}{2} \), we need only show that \( 2^2 \nmid 2 \), which is obvious. This concludes Case 1. Case 2: \( k = 2 \). Seeking a nice expression for \( n + k \), we choose to set \( n = 2^{\alpha} - 2 \) for any \( \alpha \in \mathbb{N} \) with \( \alpha \geq 2 \). Then, since \( n + k = 2^{\alpha} \), we wish to show that \[ 2^{\alpha} \nmid \binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}. \] Notice that since \( 2n < 2^{\alpha + 1} \), the greatest power of \( 2 \) that divides any term in the numerator or denominator of \(\frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}\) is \( 2^{\alpha} \). Then, because the sets \(\{1, 2, \cdots, n - 2\}\) and \(\{n + 3, n + 4, \cdots, 2n\}\) are congruent modulo \( 2^{\alpha} \), we deduce that \( 2 \nmid \frac{(n + 3)(n + 4) \cdots (2n)}{1 \cdot 2 \cdots (n - 2)} \). Removing this fraction from the fraction \(\binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}\), it suffices to show that \( 2^{\alpha} \nmid \frac{(n + 1)(n + 2)}{(n - 1)n} \). Keeping in mind that \( n + 2 = 2^{\alpha} \), we see that the largest power of \( 2 \) that divides the numerator is \( 2^{\alpha} \), while the largest power of \( 2 \) that divides the denominator is \( 2^1 \) (since \( 2 \mid n \)). Therefore, \( 2^{\alpha - 1} \) is the largest power of \( 2 \) that divides \(\frac{(n + 1)(n + 2)}{(n - 1)n}\), so \[ 2^{\alpha} \nmid \frac{(n + 1)(n + 2)}{(n - 1)n} \implies n + k \nmid \binom{2n}{n}, \] as desired. Thus, the integers \( k \) that satisfy the condition are all integers \( k \neq 1 \). The answer is: \boxed{k \neq 1}.
Let $A$ be the set $\{k^{19}-k: 1<k<20, k\in N\}$ . Let $G$ be the GCD of all elements of $A$ . Then the value of $G$ is?	798
Given that $F_{1}$ and $F_{2}$ are two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{7}=1$, $A$ is a point on the ellipse, and $\angle AF_{1}F_{2}=45^{\circ}$, calculate the area of triangle $AF_{1}F_{2}$.	\frac{7}{2}
Cube $ABCDEFGH,$ labeled as shown below, has edge length $2$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. Find the volume of the smaller of the two solids.	\frac{1}{6}
Given that $\cos{\alpha}=-\frac{4}{5}$, where $\alpha$ is an angle in the third quadrant, find the value of $\sin{\left(\alpha-\frac{\pi}{4}\right)$.	\frac{\sqrt{2}}{10}
A covered rectangular football field with a length of 90 m and a width of 60 m is being designed to be illuminated by four floodlights, each hanging from some point on the ceiling. Each floodlight illuminates a circle, with a radius equal to the height at which the floodlight is hanging. Determine the minimally possible height of the ceiling, such that the following conditions are met: every point on the football field is illuminated by at least one floodlight, and the height of the ceiling must be a multiple of 0.1 m (for example, 19.2 m, 26 m, 31.9 m, etc.).	27.1
The sequence $(a_n)$ satisfies \[a_1 + a_2 + a_3 + \dots + a_n = n^2 a_n\]for all $n \ge 2.$ If $a_{63} = 1,$ find $a_1.$	2016
Determine the largest natural number $r$ with the property that among any five subsets with $500$ elements of the set $\{1,2,\ldots,1000\}$ there exist two of them which share at least $r$ elements.	200
Five soccer teams play a match where each team plays every other team exactly once. Each match awards 3 points to the winner, 0 points to the loser, and 1 point to each team in the event of a draw. After all matches have been played, the total points of the five teams are found to be five consecutive natural numbers. Let the teams ranked 1st, 2nd, 3rd, 4th, and 5th have drawn $A$, $B$, $C$, $D$, and $E$ matches respectively. Determine the five-digit number $\overline{\mathrm{ABCDE}}$.	13213
In triangle $ABC$, $AB = 8$, $BC = 8$, and $CA = 6$. Point $P$ is randomly selected inside triangle $ABC$. What is the probability that $P$ is closer to vertex $C$ than it is to either vertex $A$ or $B$?	\frac{1}{4}
Determine all integral solutions of \[ a^2\plus{}b^2\plus{}c^2\equal{}a^2b^2.\]	(0, 0, 0)	We are tasked with finding all integral solutions to the equation: \[ a^2 + b^2 + c^2 = a^2b^2. \] First, let's rewrite the equation and rearrange the terms: \[ a^2b^2 - a^2 - b^2 = c^2. \] This suggests that \( c^2 \) must be non-negative, which means \( a^2b^2 \ge a^2 + b^2 \). ### Case Analysis: #### Case 1: \( a = 0 \) or \( b = 0 \) Without loss of generality, consider \( a = 0 \). Then the equation becomes: \[ b^2 + c^2 = 0. \] This implies that \( b^2 = 0 \) and \( c^2 = 0 \), hence \( b = 0 \) and \( c = 0 \). Similarly, if \( b = 0 \), we also get \( a = 0 \) and \( c = 0 \). Thus, one solution is \( (a, b, c) = (0, 0, 0) \). #### Case 2: \( a \neq 0 \) and \( b \neq 0 \) Suppose both \( a \) and \( b \) are non-zero. Since \( a^2b^2 \geq a^2 + b^2 \), divide both sides by positive \( a^2b^2 \): 1. Rearrange the equation to \( 1 \geq \frac{a^2}{b^2} + \frac{b^2}{a^2} \). 2. By the AM-GM inequality, we have: \[ \frac{a^2}{b^2} + \frac{b^2}{a^2} \geq 2. \] Therefore, the equation \( 1 \geq 2 \) leads to a contradiction. This contradiction implies there cannot be any non-zero integer solutions for \( a \neq 0 \) and \( b \neq 0 \). ### Conclusion: The only integral solution satisfying the equation \( a^2 + b^2 + c^2 = a^2b^2 \) is: \[ \boxed{(0, 0, 0)}. \]
The quadrilateral \(ABCD\) is circumscribed around a circle with a radius of \(1\). Find the greatest possible value of \(\left\| \frac{1}{AC^2} + \frac{1}{BD^2} \right\|\).	1/4
Given that $F$ is the focus of the parabola $y^{2}=4x$, and a perpendicular line to the directrix is drawn from a point $M$ on the parabola, with the foot of the perpendicular being $N$. If $\|MF\|= \frac{4}{3}$, then $\angle NMF=$ .	\frac{2\pi}{3}

Inside an isosceles triangle $\mathrm{ABC}$ with equal sides $\mathrm{AB} = \mathrm{BC}$ and an angle of 80 degrees at vertex $\mathrm{B}$, a point $\mathrm{M}$ is taken such that the angle $\mathrm{MAC}$ is 10 degrees and the angle $\mathrm{MCA}$ is 30 degrees. Find the measure of the angle $\mathrm{AMB}$.

70

For how many integers $n$ between $1$ and $50$, inclusive, is $\frac{(n^2-1)!}{(n!)^n}$ an integer?

34

1. **Understanding the Expression**: We start by analyzing the expression \[\frac{(n^2-1)!}{(n!)^n}.\] We need to determine for how many integers $n$ between $1$ and $50$ this expression is an integer. 2. **Relating to a Known Integer Expression**: We know that \[\frac{(n^2)!}{(n!)^{n+1}}\] is an integer because it represents the number of ways to distribute $n^2$ objects into $n$ groups each of size $n$. This can be rewritten using the factorial definition: \[\frac{(n^2)!}{(n!)^{n+1}} = \frac{(n^2)!}{(n!)^n \cdot n!}.\] 3. **Connecting the Two Expressions**: We can relate the given expression to the known integer expression: \[\frac{(n^2-1)!}{(n!)^n} = \frac{(n^2)!}{n^2 \cdot (n!)^n} = \frac{(n^2)!}{(n!)^{n+1}} \cdot \frac{n!}{n^2}.\] For this to be an integer, $\frac{n!}{n^2}$ must also be an integer, which simplifies to checking if $n^2$ divides $n!$. 4. **Analyzing $n^2$ Dividing $n!$**: The condition $n^2 \mid n!$ is equivalent to checking if $\frac{n!}{n}$ is an integer. By Wilson's Theorem, this condition fails when $n$ is a prime number or $n=4$ (since $4! = 24$ and $4^2 = 16$ does not divide $24$). 5. **Counting Prime Numbers and $n=4$**: There are $15$ prime numbers between $1$ and $50$. Including $n=4$, there are $16$ values of $n$ for which the expression is not an integer. 6. **Calculating the Total**: Since there are $50$ integers from $1$ to $50$, and $16$ of these do not make the expression an integer, the number of integers for which the expression is an integer is $50 - 16 = 34$. 7. **Conclusion**: Therefore, there are $\boxed{34}$ integers $n$ between $1$ and $50$ for which \[\frac{(n^2-1)!}{(n!)^n}\] is an integer.

The height of a cone and its slant height are 4 cm and 5 cm, respectively. Find the volume of a hemisphere inscribed in the cone, whose base lies on the base of the cone.

\frac{1152}{125} \pi

Jerry and Neil have a 3-sided die that rolls the numbers 1, 2, and 3, each with probability $\frac{1}{3}$. Jerry rolls first, then Neil rolls the die repeatedly until his number is at least as large as Jerry's. Compute the probability that Neil's final number is 3.

\frac{11}{18}

If Jerry rolls $k$, then there is a $\frac{1}{4-k}$ probability that Neil's number is 3, since Neil has an equal chance of rolling any of the $4-k$ integers not less than $k$. Thus, the answer is $$\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)=\frac{11}{18}$$.

Given a triangle $ \triangle ABC $ with interior angles $ \angle A, \angle B, \angle C $ and opposite sides $ a, b, c $ respectively, where $ \angle A - \angle C = \frac{\pi}{2} $ and $ a, b, c $ are in arithmetic progression, find the value of $ \cos B $.

\frac{3}{4}

A sequence consists of $2020$ terms. Each term after the first is 1 larger than the previous term. The sum of the $2020$ terms is $5410$. When every second term is added up, starting with the first term and ending with the second last term, what is the sum?

2200

In $\triangle ABC$, the sides opposite to angles $A$, $B$, $C$ are $a$, $b$, $c$, respectively, and $C= \frac{3}{4}\pi$, $\sin A= \frac{\sqrt{5}}{5}$. (I) Find the value of $\sin B$; (II) If $c-a=5-\sqrt{10}$, find the area of $\triangle ABC$.

\frac{5}{2}

Points $M$ and $N$ are located on side $AC$ of triangle $ABC$, and points $K$ and $L$ are on side $AB$, with $AM : MN : NC = 1 : 3 : 1$ and $AK = KL = LB$. It is known that the area of triangle $ABC$ is 1. Find the area of quadrilateral $KLNM$.

7/15

A granite pedestal. When constructing a square foundation and a cubic pedestal for a monument, granite cubic blocks of size $1 \times 1$ meter were used. The pedestal used exactly as many blocks as the square foundation upon which it stood. All the blocks were used whole and uncut. Look at the picture and try to determine the total number of blocks used. The foundation has a thickness of one block.

128

Given that $\angle A$ and $\angle B$ are within the interval $\left(0, \frac{\pi}{2}\right)$, and that $\frac{\sin A}{\sin B} = \sin (A+B)$, find the maximum value of $\tan A$.

4/3

In a factory, a total of $100$ parts were produced. Among them, A produced $0$ parts, with $35$ being qualified. B produced $60$ parts, with $50$ being qualified. Let event $A$ be "Selecting a part from the $100$ parts at random, and the part is qualified", and event $B$ be "Selecting a part from the $100$ parts at random, and the part is produced by A". Then, the probability of $A$ given $B$ is \_\_\_\_\_\_.

\dfrac{7}{8}

Given an isosceles triangle $ABC$ satisfying $AB=AC$, $\sqrt{3}BC=2AB$, and point $D$ is on side $BC$ with $AD=BD$, then the value of $\sin \angle ADB$ is ______.

\frac{2 \sqrt{2}}{3}

Evaluate the expression $3 + 2\sqrt{3} + \frac{1}{3 + 2\sqrt{3}} + \frac{1}{2\sqrt{3} - 3}$.

3 + \frac{10\sqrt{3}}{3}

Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$. Find $d$.

450

Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\sqrt{3}$, so that $AB=2\sqrt{3}$. Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$, which of course includes $P, Q$. From there, we can call $OU=h$. There are two crucial equations we can thus generate. WLOG set $PU<QU$, then we call $PU=d-h, QU=d+h$. First equation: using the Pythagorean Theorem on $\triangle UOB$, $h^2+2^2=d^2$. Next, using the tangent addition formula on angles $\angle PHU, \angle UHQ$ we see that after simplifying $-d^2+h^2=-4, 2d=3\sqrt{3}$ in the numerator, so $d=\frac{3\sqrt{3}}{2}$. Multiply back the scalar and you get $\boxed{450}$. Not that hard, was it?

How many values of $x$, $-17<x<100$, satisfy $\cos^2 x + 3\sin^2 x = \cot^2 x$? (Note: $x$ is measured in radians.)

37

Using the trapezoidal rule with an accuracy of 0.01, calculate $\int_{2}^{3} \frac{d x}{x-1}$.

0.6956

The function $f(x)=x^5-20x^4+ax^3+bx^2+cx+24$ has the interesting property that its roots can be arranged to form an arithmetic sequence. Determine $f(8)$ .

-24

How many integers $n$ are there such that $0 \le n \le 720$ and $n^2 \equiv 1$ (mod $720$ )?

16

Given $$\frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} = 2$$, find the value of $$\frac{1 + \sin4\alpha - \cos4\alpha}{1 + \sin4\alpha + \cos4\alpha}$$.

\frac{3}{4}

Define the *hotel elevator cubic*as the unique cubic polynomial $P$ for which $P(11) = 11$ , $P(12) = 12$ , $P(13) = 14$ , $P(14) = 15$ . What is $P(15)$ ? *Proposed by Evan Chen*

13

If $x = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$, then:

1 < x < 2

1. **Identify the equation**: Given the expression for $x$, we can write it as: \[ x = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}} \] Notice that the expression under the square root is the same as $x$ itself. Therefore, we can set up the equation: \[ x = \sqrt{1 + x} \] 2. **Square both sides**: To eliminate the square root, square both sides of the equation: \[ x^2 = 1 + x \] 3. **Rearrange into a standard quadratic form**: Bring all terms to one side of the equation: \[ x^2 - x - 1 = 0 \] 4. **Apply the quadratic formula**: The solutions to a quadratic equation $ax^2 + bx + c = 0$ are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plugging in $a = 1$, $b = -1$, and $c = -1$, we get: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \] 5. **Determine the appropriate root**: Since $x$ is defined as a series of nested square roots, it must be non-negative. Thus, we discard the negative root and keep the positive root: \[ x = \frac{1 + \sqrt{5}}{2} \] 6. **Approximate the value of $x$**: The value of $\sqrt{5}$ is approximately $2.236$. Therefore: \[ x \approx \frac{1 + 2.236}{2} \approx 1.618 \] 7. **Compare $x$ with the options provided**: The value $1.618$ lies between $1$ and $2$. Therefore, the correct choice is: \[ \boxed{\textbf{(C)}\ 1 < x < 2} \]

Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ$ , and $AB = 10$. Let $P$ be randomly chosen inside $ABC$ , and extend $\overline{BP}$ to meet $\overline{AC}$ at $D$. What is the probability that $BD > 5\sqrt2$?

\frac{3-\sqrt3}{3}

1. **Identify the lengths of sides in triangle $ABC$**: Given that $\angle ACB = 90^\circ$ and $\angle ABC = 60^\circ$, triangle $ABC$ is a 30-60-90 triangle. In such triangles, the sides are in the ratio $1:\sqrt{3}:2$. Since $AB = 10$ (hypotenuse), the other sides are: - $BC = \frac{1}{2} \times AB = \frac{1}{2} \times 10 = 5$ (opposite the $30^\circ$ angle), - $AC = \sqrt{3} \times BC = \sqrt{3} \times 5 = 5\sqrt{3}$ (opposite the $60^\circ$ angle). 2. **Determine the condition for $BD > 5\sqrt{2}$**: Extend $BP$ to meet $AC$ at $D$. We need to find the condition under which $BD > 5\sqrt{2}$. Consider a specific $D'$ on $AC$ such that $BD' = 5\sqrt{2}$. By the Pythagorean theorem in triangle $BD'C$, we have: \[ BD'^2 = BC^2 + CD'^2 \implies (5\sqrt{2})^2 = 5^2 + CD'^2 \implies 50 = 25 + CD'^2 \implies CD'^2 = 25 \implies CD' = 5. \] 3. **Analyze the geometric condition**: Since $CD' = 5$ and $AC = 5\sqrt{3}$, the point $D'$ divides $AC$ into segments of $5$ and $5\sqrt{3} - 5$. For $BD > 5\sqrt{2}$, $CD$ must be greater than $5$, meaning $D$ must be between $D'$ and $C$. 4. **Calculate the probability**: The probability that $P$ lies in the region where $BD > 5\sqrt{2}$ is equivalent to the probability that $P$ lies in triangle $ABD'$, which is a smaller triangle within triangle $ABC$. The ratio of their areas is the same as the ratio of $AD'$ to $AC$ because the height from $B$ to line $AC$ is the same for both triangles. Thus, \[ \text{Probability} = \frac{AD'}{AC} = \frac{AC - CD'}{AC} = \frac{5\sqrt{3} - 5}{5\sqrt{3}} = 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} = \frac{3 - \sqrt{3}}{3}. \] 5. **Conclusion**: The probability that $BD > 5\sqrt{2}$ is $\boxed{\frac{3-\sqrt{3}}{3}}$.

In a right triangle $ABC$ with a right angle at $B$ and $\angle A = 30^\circ$, a height $BD$ is drawn. Then, in triangle $BDC$, a median $DE$ is drawn, and in triangle $DEC$, an angle bisector $EF$ is drawn. Find the ratio $ \frac{FC}{AC} $.

1/8

Given the function $y=\cos (2x-\frac{\pi }{4})$, determine the horizontal translation of the graph of the function $y=\sin 2x$.

\frac{\pi }{8}

Find the smallest positive integer $m$ such that for all positive integers $n \geq m$, there exists a positive integer $l$ satisfying $$ n < l^2 < \left(1+\frac{1}{2009}\right)n. $$

16144325

Let $r$, $s$, and $t$ be solutions of the equation $x^3-5x^2+6x=9$. Compute $\frac{rs}t + \frac{st}r + \frac{tr}s$.

-6

A right triangle with integer leg lengths is called "cool'' if the number of square units in its area is equal to twice the number of units in the sum of the lengths of its legs. What is the sum of all the different possible areas of cool right triangles?

118

Maria ordered a certain number of televisions for the stock of a large store, paying R\$ 1994.00 per television. She noticed that in the total amount to be paid, the digits 0, 7, 8, and 9 do not appear. What is the smallest number of televisions she could have ordered?

56

Given an ellipse $C$: $\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1 (a > b > 0)$ with a focal length of $2$, and point $Q( \frac{a^{2}}{ \sqrt{a^{2}-b^{2}}},0)$ on the line $l$: $x=2$. (1) Find the standard equation of the ellipse $C$; (2) Let $O$ be the coordinate origin, $P$ a moving point on line $l$, and $l'$ a line passing through point $P$ that is tangent to the ellipse at point $A$. Find the minimum value of the area $S$ of $\triangle POA$.

\frac{ \sqrt{2}}{2}

Given that $2^x+ 2^x+ 2^x+ 2^x= 128$, what is the value of $(x + 1)(x - 1)$?

24

Given an arithmetic sequence $\{a_{n}\}$, where $a_{1}+a_{8}=2a_{5}-2$ and $a_{3}+a_{11}=26$, calculate the sum of the first 2022 terms of the sequence $\{a_{n} \cdot \cos n\pi\}$.

2022

Determine the area and the circumference of a circle with the center at the point $ R(2, -1) $ and passing through the point $ S(7, 4) $. Express your answer in terms of $ \pi $.

10\pi \sqrt{2}

Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

33

Let $1-x=a;1-y=b;1-z=c$, rewrite those equations $\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1$; $\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}$ $\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}$ square both sides, get three equations: $2ab-1=2\sqrt{(1-a^2)(1-b^2)}$ $2bc=2\sqrt{(1-b^2)(1-c^2)}$ $2ac+1=2\sqrt{(1-c^2)(1-a^2)}$ Getting that $a^2+b^2-ab=\frac{3}{4}$ $b^2+c^2=1$ $a^2+c^2+ac=\frac{3}{4}$ Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$, $a=b-c$ Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}$, $bc=\frac{1}{4}$ Since $a^2=b^2+c^2-2bc=\frac{1}{2}$, the final answer is $\frac{1}{4}*\frac{1}{4}*\frac{1}{2}=\frac{1}{32}$ the final answer is $\boxed{033}$ ~bluesoul

If P and Q are points on the line y = 1 - x and the curve y = -e^x, respectively, find the minimum value of |PQ|.

\sqrt{2}

If $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$ are unit vectors, find the largest possible value of \[ \|\mathbf{a} - \mathbf{b}\|^2 + \|\mathbf{a} - \mathbf{c}\|^2 + \|\mathbf{a} - \mathbf{d}\|^2 + \|\mathbf{b} - \mathbf{c}\|^2 + \|\mathbf{b} - \mathbf{d}\|^2 + \|\mathbf{c} - \mathbf{d}\|^2. \]

16

Betty goes to the store to get flour and sugar. The amount of flour she buys, in pounds, is at least 6 pounds more than half the amount of sugar, and is no more than twice the amount of sugar. Find the least number of pounds of sugar that Betty could buy.

4

Compute $\cos 90^\circ$.

0

Let $\triangle PQR$ be a right triangle such that $Q$ is a right angle. A circle with diameter $QR$ intersects side $PR$ at $S$. If $PS = 3$ and $QS = 9$, what is $RS$?

27

Given an arithmetic sequence $\{a\_n\}$, where $a\_n \in \mathbb{N}^*$, and $S\_n = \frac{1}{8}(a\_n + 2)^2$. If $b\_n = \frac{1}{2}a\_n - 30$, find the minimum value of the sum of the first $\_\_\_\_\_\_$ terms of the sequence $\{b\_n\}$.

15

Find maximum value of number $a$ such that for any arrangement of numbers $1,2,\ldots ,10$ on a circle, we can find three consecutive numbers such their sum bigger or equal than $a$ .

18

Given that $ F $ is the right focus of the hyperbola $ x^{2} - y^{2} = 1 $, $ l $ is the right directrix of the hyperbola, and $ A $ and $ B $ are two moving points on the right branch of the hyperbola such that $ A F \perp B F $. The projection of the midpoint $ M $ of line segment $ AB $ onto $ l $ is $ N $. Find the maximum value of $ \frac{|MN|}{|AB|} $.

1/2

Let the function $f(x) = (\sin x + \cos x)^2 - \sqrt{3}\cos 2x$. (Ⅰ) Find the smallest positive period of $f(x)$; (Ⅱ) Find the maximum value of $f(x)$ on the interval $\left[0, \frac{\pi}{2}\right]$ and the corresponding value of $x$ when the maximum value is attained.

\frac{5\pi}{12}

Farmer Yang has a $2015 \times 2015$ square grid of corn plants. One day, the plant in the very center of the grid becomes diseased. Every day, every plant adjacent to a diseased plant becomes diseased. After how many days will all of Yang's corn plants be diseased?

2014

A geometric sequence starts $16$, $-24$, $36$, $-54$. What is the common ratio of this sequence?

-\frac{3}{2}

The sequence $a_{n}$ is defined as follows: \[ a_{1} = 1, \quad a_{n+1} = a_{n} + \frac{2a_{n}}{n} \text{ for } n \geq 1 \] Find $a_{100}$.

5151

At around 8 o'clock in the morning, two cars left the fertilizer plant one after another, heading toward Happy Village. Both cars travel at a speed of 60 kilometers per hour. At 8:32, the distance the first car had traveled from the fertilizer plant was three times the distance traveled by the second car. At 8:39, the distance the first car had traveled from the fertilizer plant was twice the distance traveled by the second car. At what exact time did the first car leave the fertilizer plant?

8:11

Given that the complex number $ z $ satisfies $ |z|=1 $, find the maximum value of $ \left| z^3 - 3z - 2 \right| $.

3\sqrt{3}

If a school bus leaves school with 48 students on board, and one-half of the students get off the bus at each of the first three stops, how many students remain on the bus after the third stop?

6

Given that the power function $y=x^{m}$ is an even function and is a decreasing function when $x \in (0,+\infty)$, determine the possible value of the real number $m$.

-2

Let $p(x)$ be a monic polynomial of degree 6 such that $p(1) = 1,$ $p(2) = 2,$ $p(3) = 3,$ $p(4) = 4,$ $p(5) = 5,$ and $p(6) = 6.$ Find $p(7).$

727

Martha can make 24 cookies with 3 cups of flour. How many cookies can she make with 5 cups of flour, and how many cups of flour are needed to make 60 cookies?

7.5

Ilya takes a triplet of numbers and transforms it following the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after the 1989th application of this rule, if the initial triplet of numbers was $\{70, 61, 20\}$? If the question allows for multiple solutions, list them all as a set.

50

For real numbers $x$, $y$, and $z$, consider the matrix \[ \begin{pmatrix} x+y & x & y \\ x & y+z & y \\ y & x & x+z \end{pmatrix} \] Determine whether this matrix is invertible. If not, list all possible values of \[ \frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y}. \]

-3

Xiaoming's home is 30 minutes away from school by subway and 50 minutes by bus. One day, due to some reasons, Xiaoming first took the subway and then transferred to the bus, taking 40 minutes to reach the school. The transfer process took 6 minutes. How many minutes did Xiaoming spend on the bus that day?

10

In a scalene triangle with integer side lengths $a, b, c$, the following relation holds. What is the smallest height of the triangle? $$ \frac{a^{2}}{c}-(a-c)^{2}=\frac{b^{2}}{c}-(b-c)^{2} $$

2.4

The angles of a convex $n$-sided polygon form an arithmetic progression whose common difference (in degrees) is a non-zero integer. Find the largest possible value of $n$ for which this is possible.

27

The exterior angles form an arithmetic sequence too (since they are each $180^{\circ}$ minus the corresponding interior angle). The sum of this sequence must be $360^{\circ}$. Let the smallest exterior angle be $x$ and the common difference be $d$. The sum of the exterior angles is then $x+(x+a)+(x+2a)+\ldots+(x+(n-1)a)=\frac{n(n-1)}{2} \cdot a+nx$. Setting this to 360, and using $nx>0$, we get $n(n-1)<720$, so $n \leq 27$.

How many different graphs with 9 vertices exist where each vertex is connected to 2 others?

4

It suffices to consider the complements of the graphs, so we are looking for graphs with 9 vertices, where each vertex is connected to 2 others. There are $\mathbf{4}$ different graphs.

For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?

48

1. Let $x$ be the number of acorns that both the chipmunk and the squirrel hid. According to the problem, the chipmunk hides 3 acorns per hole and the squirrel hides 4 acorns per hole. 2. Let $h_c$ be the number of holes dug by the chipmunk and $h_s$ be the number of holes dug by the squirrel. We know that: \[ 3h_c = x \quad \text{and} \quad 4h_s = x \] This implies that the total number of acorns hidden by each animal is the same. 3. We are also given that the squirrel needed 4 fewer holes than the chipmunk: \[ h_c = h_s + 4 \] 4. Substitute $h_c$ from the third equation into the first equation: \[ 3(h_s + 4) = x \] Simplifying, we get: \[ 3h_s + 12 = x \] 5. From the second equation, we know $4h_s = x$. We can set the expressions for $x$ equal to each other: \[ 3h_s + 12 = 4h_s \] Solving for $h_s$, we subtract $3h_s$ from both sides: \[ 12 = h_s \] 6. Substitute $h_s = 12$ back into the equation $4h_s = x$ to find $x$: \[ x = 4 \times 12 = 48 \] Thus, the number of acorns that the chipmunk hid is $\boxed{48}$, which corresponds to answer choice $\textbf{(D)}$.

Given that the terms of the geometric sequence $\{a_n\}$ are positive, and the common ratio is $q$, if $q^2 = 4$, then $$\frac {a_{3}+a_{4}}{a_{4}+a_{5}}$$ equals \_\_\_\_\_\_.

\frac {1}{2}

For real numbers $a$ and $b$ , define $$ f(a,b) = \sqrt{a^2+b^2+26a+86b+2018}. $$ Find the smallest possible value of the expression $$ f(a, b) + f (a,-b) + f(-a, b) + f (-a, -b). $$

4 \sqrt{2018}

A line through the points $(2, -9)$ and $(j, 17)$ is parallel to the line $2x + 3y = 21$. What is the value of $j$?

-37

A bakery sells three kinds of rolls. How many different combinations of rolls could Jack purchase if he buys a total of six rolls and includes at least one of each kind?

10

Determine the smallest positive integer $n$ such that $n^2$ is divisible by 50 and $n^3$ is divisible by 294.

210

Richard starts with the string HHMMMMTT. A move consists of replacing an instance of HM with MH , replacing an instance of MT with TM, or replacing an instance of TH with HT. Compute the number of possible strings he can end up with after performing zero or more moves.

70

The key claim is that the positions of the Ms fully determines the end configuration. Indeed, since all Hs are initially left of all Ts, the only successful swaps that can occur will involve Ms. So, picking $\binom{8}{4}=70$ spots for Ms and then filling in the remaining 4 spots with Hs first and then Ts gives all possible arrangements. It is not hard to show that all of these arrangements are also achievable; just greedily move Ms to their target positions.

Compute $$\sum_{n_{60}=0}^{2} \sum_{n_{59}=0}^{n_{60}} \cdots \sum_{n_{2}=0}^{n_{3}} \sum_{n_{1}=0}^{n_{2}} \sum_{n_{0}=0}^{n_{1}} 1$$

1953

The given sum counts the number of non-decreasing 61-tuples of integers $\left(n_{0}, \ldots, n_{60}\right)$ from the set $\{0,1,2\}$. Such 61-tuples are in one-to-one correspondence with strictly increasing 61-tuples of integers $\left(m_{0}, \ldots, m_{60}\right)$ from the set $\{0,1,2, \ldots, 62\}$: simply let $m_{k}=n_{k}+k$. But the number of such $\left(m_{0}, \ldots, m_{60}\right)$ is almost by definition $\binom{63}{61}=\binom{63}{2}=1953$.

What is the smallest positive integer $n$ such that $\frac{n}{n+101}$ is equal to a terminating decimal?

24

Find the least positive integer $n$ , such that there is a polynomial \[ P(x) = a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+\dots+a_1x+a_0 \] with real coefficients that satisfies both of the following properties: - For $i=0,1,\dots,2n$ it is $2014 \leq a_i \leq 2015$ . - There is a real number $\xi$ with $P(\xi)=0$ .

2014

A list of $3042$ positive integers has a unique mode, which occurs exactly $15$ times. Calculate the least number of distinct values that can occur in the list.

218

The average of the numbers $1, 2, 3, \dots, 44, 45, x$ is $50x$. What is $x$?

\frac{1035}{2299}

In triangle $ABC$, the angle at vertex $B$ is $\frac{\pi}{3}$, and the line segments connecting the incenter to vertices $A$ and $C$ are 4 and 6, respectively. Find the radius of the circle inscribed in triangle $ABC$.

\frac{6 \sqrt{3}}{\sqrt{19}}

Chords $AB$ and $CD$ of a circle with center $O$ both have a length of 5. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ intersect at point $P$, where $DP=13$. The line $PO$ intersects segment $AC$ at point $L$. Find the ratio $AL:LC$.

13/18

For what value of $n$ does $|6 + ni| = 6\sqrt{5}$?

12

Mrs. Everett recorded the performance of her students in a chemistry test. However, due to a data entry error, 5 students who scored 60% were mistakenly recorded as scoring 70%. Below is the corrected table after readjusting these students. Using the data, calculate the average percent score for these $150$ students. \begin{tabular}{|c|c|} \multicolumn{2}{c}{}\\\hline \textbf{$\%$ Score}&\textbf{Number of Students}\\\hline 100&10\\\hline 95&20\\\hline 85&40\\\hline 70&40\\\hline 60&20\\\hline 55&10\\\hline 45&10\\\hline \end{tabular}

75.33

In the expression $5 * 4 * 3 * 2 * 1 = 0$, replace the asterisks with arithmetic operators $+, -, \times, \div$, using each operator exactly once, so that the equality holds true (note: $2 + 2 \times 2 = 6$).

5 - 4 \times 3 : 2 + 1

The sequence $\{a_n\}$ is a geometric sequence with a common ratio of $q$, where $|q| > 1$. Let $b_n = a_n + 1 (n \in \mathbb{N}^*)$, if $\{b_n\}$ has four consecutive terms in the set $\{-53, -23, 19, 37, 82\}$, find the value of $q$.

-\dfrac{3}{2}

A boss plans a business meeting at Starbucks with the two engineers below him. However, he fails to set a time, and all three arrive at Starbucks at a random time between 2:00 and 4:00 p.m. When the boss shows up, if both engineers are not already there, he storms out and cancels the meeting. Each engineer is willing to stay at Starbucks alone for an hour, but if the other engineer has not arrived by that time, he will leave. What is the probability that the meeting takes place?

\frac{7}{24}

For distinct positive integers $a, b<2012$, define $f(a, b)$ to be the number of integers $k$ with $1\le k<2012$ such that the remainder when $ak$ divided by $2012$ is greater than that of $bk$ divided by $2012$. Let $S$ be the minimum value of $f(a, b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than $2012$. Determine $S$.

502

To solve for $ S $, the minimum value of $ f(a, b) $, where distinct positive integers $ a, b < 2012 $, we first need to analyze the function $ f(a, b) $. This function represents the number of integers $ k $ with $ 1 \leq k < 2012 $ such that: \[ ak \mod 2012 > bk \mod 2012 \] ### Steps to find $ S $: 1. **Understanding the Remainder Condition**: For each $ k $, we compare: \[ ak \equiv r_a \pmod{2012} \] and \[ bk \equiv r_b \pmod{2012} \] We need $ r_a > r_b $. 2. **Expressing the Condition**: The condition becomes: \[ ak - bk \equiv (a-b)k \equiv r_a - r_b \pmod{2012} \] 3. **Analyzing $ f(a, b) $**: Note that both $ a $ and $ b $ are distinct and less than $ 2012 $. For a specific $ k $, the behavior of $ ak \mod 2012 $ and $ bk \mod 2012 $ involves cycling through the possible remainders from $ 0 $ to $ 2011 $. 4. **Distribution of Remainders**: Since $ a $ and $ b $ are distinct, their multiplicative properties will result in differences in the cycle of remainders. 5. **Symmetry Argument**: By symmetry, as $ k $ ranges from $ 1 $ to $ 2012 - 1 $, there will be a balance in the number of $ k $ for which $ ak \mod 2012 > bk \mod 2012 $ and $ ak \mod 2012 < bk \mod 2012 $. 6. **Calculating $ S $**: Since for each pair $ (a, b) $ except permutations, the integer values $ k $ will be split symmetrically, Hence, we anticipate that on average, the set of $ k $ is divided equally between remainders being higher for $ a $ or $ b $. Therefore: \[ S = \frac{2012 - 1}{2} = 1005 \] However, due to rounding down because of distinct integer properties (as $ k $ values cannot be split fractionally), the precise minimum value $ S $ is: \[ \boxed{502} \] This accounts for any adjustments due to parity and nearest integer calculations rounding down for edge differences.

If three different numbers are selected from 2, 3, 4, 5, 6 to be $a$, $b$, $c$ such that $N = abc + ab + bc + a - b - c$ reaches its maximum value, then this maximum value is.

167

Place each of the digits 4, 5, 6, and 7 in exactly one square to make the smallest possible product. What is this product?

2622

For any real number $x$ , we let $\lfloor x \rfloor$ be the unique integer $n$ such that $n \leq x < n+1$ . For example. $\lfloor 31.415 \rfloor = 31$ . Compute \[2020^{2021} - \left\lfloor\frac{2020^{2021}}{2021} \right \rfloor (2021).\] *2021 CCA Math Bonanza Team Round #3*

2020

If in an arithmetic sequence, the sum of the first three terms is 34, the sum of the last three terms is 146, and the sum of all terms is 390, then the number of terms in the sequence is __________.

13

A sphere intersects the $xy$-plane in a circle centered at $(3, 5, 0)$ with radius 2. The sphere also intersects the $yz$-plane in a circle centered at $(0, 5, -8),$ with radius $r.$ Find $r.$

\sqrt{59}

On one side of the acute angle $A$, points $P$ and $Q$ are marked such that $AP = 4$, $AQ = 12$. On the other side, points $M$ and $N$ are marked at distances of 6 and 10 from the vertex. Find the ratio of the areas of triangles $MNO$ and $PQO$, where $O$ is the intersection point of the lines $MQ$ and $NP$.

1:5

Let $a_n$ be the integer closest to $\sqrt{n}$. Find the sum $$ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{1980}}. $$

88

Given that the vertex of a parabola is at the origin and the center of the circle $(x-2)^2 + y^2 = 4$ is exactly the focus of the parabola. 1. Find the equation of the parabola. 2. A line with a slope of 2 passes through the focus of the parabola and intersects the parabola at points A and B. Find the area of triangle OAB.

4\sqrt{5}

$a,b,c$ - are sides of triangle $T$ . It is known, that if we increase any one side by $1$ , we get new a) triangle b)acute triangle Find minimal possible area of triangle $T$ in case of a) and in case b)

\frac{\sqrt{3}}{4}

A point is chosen randomly from within a circular region with radius $r$. A related concentric circle with radius $\sqrt{r}$ contains points that are closer to the center than to the boundary. Calculate the probability that a randomly chosen point lies closer to the center than to the boundary.

\frac{1}{4}

Determine all integers $k$ such that there exists infinitely many positive integers $n$ [b]not[/b] satisfying \[n+k |\binom{2n}{n}\]

k \neq 1

Determine all integers $ k $ such that there exist infinitely many positive integers $ n $ not satisfying \[ n + k \mid \binom{2n}{n}. \] We claim that all integers $ k \neq 1 $ satisfy the desired property. First, recall that $\frac{1}{n + 1} \binom{2n}{n}$ is the $ n $-th Catalan number. Since the Catalan numbers are a sequence of integers, it follows that $ n + 1 \mid \binom{2n}{n} $ for all $ n $. Hence, $ k = 1 $ certainly cannot satisfy the problem statement. Now, we consider two cases: **Case 1: $ k \neq 2 $.** Suppose that $ p $ is a prime divisor of $ k $ and let $ n = p^\alpha $ for any $ \alpha \in \mathbb{N} $. Then, since $ p \mid n + k $, in order to prove that $ n + k \nmid \binom{2n}{n} $, it suffices to show that \[ p \nmid \binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}. \] Note that the greatest power of $ p $ that divides any term in the numerator or denominator of $\frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)}$ is less than $ p^{\alpha} $. Since the sets $\{1, 2, \cdots, n - 1\}$ and $\{n + 1, n + 2, \cdots, 2n - 1\}$ are congruent modulo $ p^{\alpha} $, the numerator and denominator of the fraction $\frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)}$ both contain the same number of factors of $ p $. Therefore, $ p \nmid \frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)} $. Now, if we can show that $ p \nmid 2 $, we will be able to conclude that $ p \nmid \binom{2n}{n} $, as desired. Indeed, if $ p \neq 2 $, then trivially $ p \nmid 2 $. Meanwhile, if $ p = 2 $, then let us take $ \alpha \geq 2 $ so that $ 2^2 \mid n + k $. Hence, we wish to show that $ 2^2 \nmid \binom{2n}{n} $. But since $ 2 \nmid \frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)} = \frac{\binom{2n}{n}}{2} $, we need only show that $ 2^2 \nmid 2 $, which is obvious. This concludes Case 1. **Case 2: $ k = 2 $.** Seeking a nice expression for $ n + k $, we choose to set $ n = 2^{\alpha} - 2 $ for any $ \alpha \in \mathbb{N} $ with $ \alpha \geq 2 $. Then, since $ n + k = 2^{\alpha} $, we wish to show that \[ 2^{\alpha} \nmid \binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}. \] Notice that since $ 2n < 2^{\alpha + 1} $, the greatest power of $ 2 $ that divides any term in the numerator or denominator of $\frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}$ is $ 2^{\alpha} $. Then, because the sets $\{1, 2, \cdots, n - 2\}$ and $\{n + 3, n + 4, \cdots, 2n\}$ are congruent modulo $ 2^{\alpha} $, we deduce that $ 2 \nmid \frac{(n + 3)(n + 4) \cdots (2n)}{1 \cdot 2 \cdots (n - 2)} $. Removing this fraction from the fraction $\binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}$, it suffices to show that $ 2^{\alpha} \nmid \frac{(n + 1)(n + 2)}{(n - 1)n} $. Keeping in mind that $ n + 2 = 2^{\alpha} $, we see that the largest power of $ 2 $ that divides the numerator is $ 2^{\alpha} $, while the largest power of $ 2 $ that divides the denominator is $ 2^1 $ (since $ 2 \mid n $). Therefore, $ 2^{\alpha - 1} $ is the largest power of $ 2 $ that divides $\frac{(n + 1)(n + 2)}{(n - 1)n}$, so \[ 2^{\alpha} \nmid \frac{(n + 1)(n + 2)}{(n - 1)n} \implies n + k \nmid \binom{2n}{n}, \] as desired. Thus, the integers $ k $ that satisfy the condition are all integers $ k \neq 1 $. The answer is: \boxed{k \neq 1}.

Let $A$ be the set $\{k^{19}-k: 1<k<20, k\in N\}$ . Let $G$ be the GCD of all elements of $A$ . Then the value of $G$ is?

798

Given that $F_{1}$ and $F_{2}$ are two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{7}=1$, $A$ is a point on the ellipse, and $\angle AF_{1}F_{2}=45^{\circ}$, calculate the area of triangle $AF_{1}F_{2}$.

\frac{7}{2}

Cube $ABCDEFGH,$ labeled as shown below, has edge length $2$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. Find the volume of the smaller of the two solids.

\frac{1}{6}

Given that $\cos{\alpha}=-\frac{4}{5}$, where $\alpha$ is an angle in the third quadrant, find the value of $\sin{\left(\alpha-\frac{\pi}{4}\right)$.

\frac{\sqrt{2}}{10}

A covered rectangular football field with a length of 90 m and a width of 60 m is being designed to be illuminated by four floodlights, each hanging from some point on the ceiling. Each floodlight illuminates a circle, with a radius equal to the height at which the floodlight is hanging. Determine the minimally possible height of the ceiling, such that the following conditions are met: every point on the football field is illuminated by at least one floodlight, and the height of the ceiling must be a multiple of 0.1 m (for example, 19.2 m, 26 m, 31.9 m, etc.).

27.1

The sequence $(a_n)$ satisfies \[a_1 + a_2 + a_3 + \dots + a_n = n^2 a_n\]for all $n \ge 2.$ If $a_{63} = 1,$ find $a_1.$

2016

Determine the largest natural number $r$ with the property that among any five subsets with $500$ elements of the set $\{1,2,\ldots,1000\}$ there exist two of them which share at least $r$ elements.

200

Five soccer teams play a match where each team plays every other team exactly once. Each match awards 3 points to the winner, 0 points to the loser, and 1 point to each team in the event of a draw. After all matches have been played, the total points of the five teams are found to be five consecutive natural numbers. Let the teams ranked 1st, 2nd, 3rd, 4th, and 5th have drawn $A$, $B$, $C$, $D$, and $E$ matches respectively. Determine the five-digit number $\overline{\mathrm{ABCDE}}$.

13213

In triangle $ABC$, $AB = 8$, $BC = 8$, and $CA = 6$. Point $P$ is randomly selected inside triangle $ABC$. What is the probability that $P$ is closer to vertex $C$ than it is to either vertex $A$ or $B$?

\frac{1}{4}

Determine all integral solutions of \[ a^2\plus{}b^2\plus{}c^2\equal{}a^2b^2.\]

(0, 0, 0)

We are tasked with finding all integral solutions to the equation: \[ a^2 + b^2 + c^2 = a^2b^2. \] First, let's rewrite the equation and rearrange the terms: \[ a^2b^2 - a^2 - b^2 = c^2. \] This suggests that $ c^2 $ must be non-negative, which means $ a^2b^2 \ge a^2 + b^2 $. ### Case Analysis: #### Case 1: $ a = 0 $ or $ b = 0 $ Without loss of generality, consider $ a = 0 $. Then the equation becomes: \[ b^2 + c^2 = 0. \] This implies that $ b^2 = 0 $ and $ c^2 = 0 $, hence $ b = 0 $ and $ c = 0 $. Similarly, if $ b = 0 $, we also get $ a = 0 $ and $ c = 0 $. Thus, one solution is $ (a, b, c) = (0, 0, 0) $. #### Case 2: $ a \neq 0 $ and $ b \neq 0 $ Suppose both $ a $ and $ b $ are non-zero. Since $ a^2b^2 \geq a^2 + b^2 $, divide both sides by positive $ a^2b^2 $: 1. Rearrange the equation to $ 1 \geq \frac{a^2}{b^2} + \frac{b^2}{a^2} $. 2. By the AM-GM inequality, we have: \[ \frac{a^2}{b^2} + \frac{b^2}{a^2} \geq 2. \] Therefore, the equation $ 1 \geq 2 $ leads to a contradiction. This contradiction implies there cannot be any non-zero integer solutions for $ a \neq 0 $ and $ b \neq 0 $. ### Conclusion: The only integral solution satisfying the equation $ a^2 + b^2 + c^2 = a^2b^2 $ is: \[ \boxed{(0, 0, 0)}. \]

The quadrilateral $ABCD$ is circumscribed around a circle with a radius of $1$. Find the greatest possible value of $\left| \frac{1}{AC^2} + \frac{1}{BD^2} \right|$.

1/4

Given that $F$ is the focus of the parabola $y^{2}=4x$, and a perpendicular line to the directrix is drawn from a point $M$ on the parabola, with the foot of the perpendicular being $N$. If $|MF|= \frac{4}{3}$, then $\angle NMF=$ .

\frac{2\pi}{3}