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Given that $\{a_n\}$ is a geometric sequence with a common ratio of $q$, and $a_m$, $a_{m+2}$, $a_{m+1}$ form an arithmetic sequence.
(Ⅰ) Find the value of $q$;
(Ⅱ) Let the sum of the first $n$ terms of the sequence $\{a_n\}$ be $S_n$. Determine whether $S_m$, $S_{m+2}$, $S_{m+1}$ form an arithmetic sequence and explain the reason.
|
-\frac{1}{2}
|
deepscale
| 11,325
| ||
Determine the maximum number of elements in the set \( S \) that satisfy the following conditions:
1. Each element in \( S \) is a positive integer not exceeding 100;
2. For any two different elements \( a \) and \( b \) in \( S \), there exists another element \( c \) in \( S \) such that the greatest common divisor of \( a + b \) and \( c \) is 1;
3. For any two different elements \( a \) and \( b \) in \( S \), there exists another element \( c \) in \( S \) such that the greatest common divisor of \( a + b \) and \( c \) is greater than 1.
|
50
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deepscale
| 28,103
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Calculate the angle \(\theta\) for the expression \[ e^{11\pi i/60} + e^{23\pi i/60} + e^{35\pi i/60} + e^{47\pi i/60} + e^{59\pi i/60} \] in the form \( r e^{i \theta} \), where \( 0 \leq \theta < 2\pi \).
|
\frac{7\pi}{12}
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deepscale
| 9,599
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Find the set of values of the parameter \(a\) for which the sum of the cubes of the roots of the equation \(x^{2}-a x+a+2=0\) is equal to -8.
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-2
|
deepscale
| 16,238
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Let $Z$ be as in problem 15. Let $X$ be the greatest integer such that $|X Z| \leq 5$. Find $X$.
|
Problems 13-15 go together. See below.
|
2
|
deepscale
| 3,550
| |
The area of rectangle PRTV is divided into four rectangles, PQXW, QRSX, XSTU, and WXUV. Given that the area of PQXW is 9, the area of QRSX is 10, and the area of XSTU is 15, find the area of rectangle WXUV.
|
\frac{27}{2}
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deepscale
| 16,220
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Define $A\star B$ as $A\star B = \frac{(A+B)}{3}$. What is the value of $(2\star 10) \star 5$?
|
3
|
deepscale
| 33,901
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Compute without using a calculator: $\dfrac{9!}{6!3!}$
|
84
|
deepscale
| 35,164
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In a bookshelf, there are four volumes of Astrid Lindgren's collected works in order, each containing 200 pages. A little worm living in these volumes burrowed a path from the first page of the first volume to the last page of the fourth volume. How many pages did the worm burrow through?
|
400
|
deepscale
| 25,351
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In what ratio does the angle bisector of an acute angle of an isosceles right triangle divide the area of the triangle?
|
1 : \sqrt{2}
|
deepscale
| 14,918
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Given triangle \( \triangle ABC \) with circumcenter \( O \) and orthocenter \( H \), and \( O \neq H \). Let \( D \) and \( E \) be the midpoints of sides \( BC \) and \( CA \) respectively. Let \( D' \) and \( E' \) be the reflections of \( D \) and \( E \) with respect to \( H \). If lines \( AD' \) and \( BE' \) intersect at point \( K \), find the value of \( \frac{|KO|}{|KH|} \).
|
3/2
|
deepscale
| 14,630
| ||
A, B, and C are guessing a two-digit number.
A says: The number has an even number of factors, and it is greater than 50.
B says: The number is odd, and it is greater than 60.
C says: The number is even, and it is greater than 70.
If each of them is only half correct, what is the number?
|
64
|
deepscale
| 11,702
| ||
The interior of a right, circular cone is 8 inches tall with a 2-inch radius at the opening. The interior of the cone is filled with ice cream, and the cone has a hemisphere of ice cream exactly covering the opening of the cone. What is the volume of ice cream? Express your answer in terms of $\pi$.
|
16\pi
|
deepscale
| 35,552
| ||
An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$. Find $N$.
|
First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$.
The probability both are green is $\frac{4}{10}\cdot\frac{16}{16+N}$, and the probability both are blue is $\frac{6}{10}\cdot\frac{N}{16+N}$, so \[\frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}\] Solving this equation, \[20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29\] Multiplying both sides by $16+N$, we get
\begin{align*} 20\cdot16+30\cdot N&=29(16+N)\\ 320+30N&=464+29N\\ N&=\boxed{144} \end{align*}
|
144
|
deepscale
| 7,092
| |
Find $x+y+z$ when $$ a_1x+a_2y+a_3z= a $$ $$ b_1x+b_2y+b_3z=b $$ $$ c_1x+c_2y+c_3z=c $$ Given that $$ a_1\left(b_2c_3-b_3c_2\right)-a_2\left(b_1c_3-b_3c_1\right)+a_3\left(b_1c_2-b_2c_1\right)=9 $$ $$ a\left(b_2c_3-b_3c_2\right)-a_2\left(bc_3-b_3c\right)+a_3\left(bc_2-b_2c\right)=17 $$ $$ a_1\left(bc_3-b_3c\right)-a\left(b_1c_3-b_3c_1\right)+a_3\left(b_1c-bc_1\right)=-8 $$ $$ a_1\left(b_2c-bc_2\right)-a_2\left(b_1c-bc_1\right)+a\left(b_1c_2-b_2c_1\right)=7. $$ *2017 CCA Math Bonanza Lightning Round #5.1*
|
16/9
|
deepscale
| 27,884
| ||
The hour and minute hands of a clock move continuously and at constant speeds. A moment of time $X$ is called interesting if there exists such a moment $Y$ (the moments $X$ and $Y$ do not necessarily have to be different), so that the hour hand at moment $Y$ will be where the minute hand is at moment $X$, and the minute hand at moment $Y$ will be where the hour hand is at moment $X$. How many interesting moments will there be from 00:01 to 12:01?
|
143
|
deepscale
| 26,362
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In $\triangle ABC$, the angles $A$, $B$, $C$ correspond to the sides $a$, $b$, $c$, and $A$, $B$, $C$ form an arithmetic sequence.
(I) If $b=7$ and $a+c=13$, find the area of $\triangle ABC$.
(II) Find the maximum value of $\sqrt{3}\sin A + \sin(C - \frac{\pi}{6})$ and the size of angle $A$ when the maximum value is reached.
|
\frac{\pi}{3}
|
deepscale
| 32,165
| ||
Given functions $y_1=\frac{k_1}{x}$ and $y_{2}=k_{2}x+b$ ($k_{1}$, $k_{2}$, $b$ are constants, $k_{1}k_{2}\neq 0$).<br/>$(1)$ If the graphs of the two functions intersect at points $A(1,4)$ and $B(a,1)$, find the expressions of functions $y_{1}$ and $y_{2}$.<br/>$(2)$ If point $C(-1,n)$ is translated $6$ units upwards and falls exactly on function $y_{1}$, and point $C(-1,n)$ is translated $2$ units to the right and falls exactly on function $y_{2}$, and $k_{1}+k_{2}=0$, find the value of $b$.
|
-6
|
deepscale
| 22,945
| ||
Find all pairs of integer solutions $(n, m)$ to $2^{3^{n}}=3^{2^{m}}-1$.
|
We find all solutions of $2^{x}=3^{y}-1$ for positive integers $x$ and $y$. If $x=1$, we obtain the solution $x=1, y=1$, which corresponds to $(n, m)=(0,0)$ in the original problem. If $x>1$, consider the equation modulo 4. The left hand side is 0, and the right hand side is $(-1)^{y}-1$, so $y$ is even. Thus we can write $y=2 z$ for some positive integer $z$, and so $2^{x}=(3^{z}-1)(3^{z}+1)$. Thus each of $3^{z}-1$ and $3^{z}+1$ is a power of 2, but they differ by 2, so they must equal 2 and 4 respectively. Therefore, the only other solution is $x=3$ and $y=2$, which corresponds to $(n, m)=(1,1)$ in the original problem.
|
(0,0) \text{ and } (1,1)
|
deepscale
| 4,100
| |
What fraction of the entire wall is painted red if Matilda paints half of her section red and Ellie paints one third of her section red?
|
Matilda and Ellie each take $\frac{1}{2}$ of the wall. Matilda paints $\frac{1}{2}$ of her half, or $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the entire wall. Ellie paints $\frac{1}{3}$ of her half, or $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the entire wall. Therefore, $\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$ of the wall is painted red.
|
\frac{5}{12}
|
deepscale
| 5,852
| |
Find an approximate value of $0.998^6$ such that the error is less than $0.001$.
|
0.988
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deepscale
| 24,577
| ||
A positive integer $N$ greater than $1$ is described as special if in its base- $8$ and base- $9$ representations, both the leading and ending digit of $N$ are equal to $1$ . What is the smallest special integer in decimal representation?
*Proposed by Michael Ren*
|
793
|
deepscale
| 16,840
| ||
The cells of a $5 \times 5$ grid are each colored red, white, or blue. Sam starts at the bottom-left cell of the grid and walks to the top-right cell by taking steps one cell either up or to the right. Thus, he passes through 9 cells on his path, including the start and end cells. Compute the number of colorings for which Sam is guaranteed to pass through a total of exactly 3 red cells, exactly 3 white cells, and exactly 3 blue cells no matter which route he takes.
|
Let $c_{i, j}$ denote the cell in the $i$-th row from the bottom and the $j$-th column from the left, so Sam starts at $c_{1,1}$ and is traveling to $c_{5,5}$. The key observation (from, say, trying small cases) is that Claim. For $1 \leq i, j<5$, the cells $c_{i+1, j}$ and $c_{i, j+1}$ must be the same color. Proof. Choose a path $P$ from $c_{1,1}$ to $c_{i, j}$, and a path $Q$ from $c_{i+1, j+1}$ to $c_{5,5}$. Then consider the two paths $P \rightarrow c_{i+1, j} \rightarrow Q$ and $P \rightarrow c_{i, j+1} \rightarrow Q$. These both must have 3 cells of each color, but they only differ at cells $c_{i+1, j}$ and $c_{i, j+1}$. So these cells must be the same color. Hence, every diagonal $D_{k}=\left\{c_{a, b}: a+b=k\right\}$ must consist of cells of the same color. Moreover, any path that goes from $c_{1,1}$ to $c_{5,5}$ contains exactly one cell in $D_{k}$ for $k=2,3, \ldots, 10$. So we simply need to color the diagonals $D_{2}, \ldots, D_{10}$ such that there are 3 diagonals of each color. The number of ways to do this is $\binom{9}{3,3,3}=1680$.
|
1680
|
deepscale
| 4,300
| |
Ron has eight sticks, each having an integer length. He observes that he cannot form a triangle using any three of these sticks as side lengths. The shortest possible length of the longest of the eight sticks is:
|
21
|
deepscale
| 31,231
| ||
The rules for a race require that all runners start at $A$, touch any part of the 1500-meter wall, and stop at $B$. What is the number of meters in the minimum distance a participant must run? Express your answer to the nearest meter. Assume the distances from A to the nearest point on the wall is 400 meters, and from B to the nearest point on the wall is 600 meters.
|
1803
|
deepscale
| 30,294
| ||
Using an electric stove with a power of $P=500 \mathrm{W}$, a certain amount of water is heated. When the electric stove is turned on for $t_{1}=1$ minute, the water temperature increases by $\Delta T=2^{\circ} \mathrm{C}$, and after turning off the heater, the temperature decreases to the initial value in $t_{2}=2$ minutes. Determine the mass of the heated water, assuming the thermal power losses are constant. The specific heat capacity of water is $c_{B}=4200$ J/kg$\cdot{ }^{\circ} \mathrm{C}$.
|
2.38
|
deepscale
| 11,329
| ||
Let $a,b,c$ be positive real numbers such that $a+b+c=10$ and $ab+bc+ca=25$. Let $m=\min\{ab,bc,ca\}$. Find the largest possible value of $m$.
|
\frac{25}{9}
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deepscale
| 37,530
| ||
There are positive integers $x$ and $y$ that satisfy the system of equations\begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*}Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$, and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$. Find $3m+2n$.
|
880
|
deepscale
| 38,186
| ||
Suppose that $a$ is a multiple of 4 and $b$ is a multiple of 8. Which of the following statements are true?
A. $a+b$ must be even.
B. $a+b$ must be a multiple of 4.
C. $a+b$ must be a multiple of 8.
D. $a+b$ cannot be a multiple of 8.
Answer by listing your choices in alphabetical order, separated by commas. For example, if you think all four are true, then answer $\text{A,B,C,D}$
|
\text{A,B}
|
deepscale
| 38,434
| ||
Five packages are delivered to five houses, one to each house. If the packages are randomly delivered, what is the probability that exactly three of them are delivered to their correct houses?
|
\frac{1}{6}
|
deepscale
| 25,936
| ||
Two $4 \times 4$ squares are randomly placed on an $8 \times 8$ chessboard so that their sides lie along the grid lines of the board. What is the probability that the two squares overlap?
|
$529 / 625$. Each square has 5 horizontal $\cdot 5$ vertical $=25$ possible positions, so there are 625 possible placements of the squares. If they do not overlap, then either one square lies in the top four rows and the other square lies in the bottom four rows, or one square lies in the left four columns and the other lies in the right four columns. The first possibility can happen in $2 \cdot 5 \cdot 5=50$ ways (two choices of which square goes on top, and five horizontal positions for each square); likewise, so can the second. However, this double-counts the 4 cases in which the two squares are in opposite corners, so we have $50+50-4=96$ possible non-overlapping arrangements $\Rightarrow 25^{2}-96=529$ overlapping arrangements.
|
529/625
|
deepscale
| 3,447
| |
If 2 cards from a standard deck are selected randomly, what is the probability that either two kings or at least 1 ace occurs? (There are 4 Aces, 4 kings and 52 total cards in a standard deck.)
|
\frac{2}{13}
|
deepscale
| 35,399
| ||
A farmer contracted several acres of fruit trees. This year, he invested 13,800 yuan, and the total fruit yield was 18,000 kilograms. The fruit sells for a yuan per kilogram in the market and b yuan per kilogram when sold directly from the orchard (b < a). The farmer transports the fruit to the market for sale, selling an average of 1,000 kilograms per day, requiring the help of 2 people, paying each 100 yuan per day, and the transportation cost of the agricultural vehicle and other taxes and fees average 200 yuan per day.
(1) Use algebraic expressions involving a and b to represent the income from selling the fruit in both ways.
(2) If a = 4.5 yuan, b = 4 yuan, and all the fruit is sold out within the same period using both methods, calculate which method of selling is better.
(3) If the farmer strengthens orchard management, aiming for a net income of 72,000 yuan next year, and uses the better selling method from (2), what is the growth rate of the net income (Net income = Total income - Total expenses)?
|
20\%
|
deepscale
| 22,791
| ||
If $a(x+2)+b(x+2)=60$ and $a+b=12$, what is the value of $x$?
|
The equation $a(x+2)+b(x+2)=60$ has a common factor of $x+2$ on the left side. Thus, we can re-write the equation as $(a+b)(x+2)=60$. When $a+b=12$, we obtain $12 \cdot(x+2)=60$ and so $x+2=5$ which gives $x=3$.
|
3
|
deepscale
| 5,341
| |
By multiplying a natural number by the number that is one greater than it, the product takes the form $ABCD$, where $A, B, C, D$ are different digits. Starting with the number that is 3 less, the product takes the form $CABD$. Starting with the number that is 30 less, the product takes the form $BCAD$. Determine these numbers.
|
8372
|
deepscale
| 25,302
| ||
Simplify first, then evaluate: $(\frac{{x-3}}{{{x^2}-1}}-\frac{2}{{x+1}})\div \frac{x}{{{x^2}-2x+1}}$, where $x=(\frac{1}{2})^{-1}+\left(\pi -1\right)^{0}$.
|
-\frac{2}{3}
|
deepscale
| 23,318
| ||
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 $m + n ?$
|
1. **Identify Key Triangles and Midpoint**:
Let $M$ be the midpoint of $CD$. Given that the figure is constructed from line segments each of length $2$, and considering the symmetry and angles in the figure, we can identify that $\triangle AED$ and $\triangle ABC$ are $30^\circ-60^\circ-90^\circ$ triangles. This is deduced from the fact that they are derived from splitting equilateral triangles ($60^\circ-60^\circ-60^\circ$) along their altitudes.
2. **Calculate $AM$ Using Pythagoras' Theorem**:
Since $M$ is the midpoint of $CD$, $MD = \frac{1}{2}CD = 1$. In $\triangle AMD$, we have:
\[
AM = \sqrt{AD^2 - MD^2} = \sqrt{2^2 - 1^2} = \sqrt{4 - 1} = \sqrt{3}
\]
However, this calculation seems to have an error in the initial solution provided. Correcting this:
\[
AM = \sqrt{AD^2 - MD^2} = \sqrt{3^2 - 1^2} = \sqrt{9 - 1} = \sqrt{8}
\]
3. **Calculate Area of $\triangle ACD$**:
The area of $\triangle ACD$ can be calculated using the formula for the area of a triangle, $\frac{1}{2}ab\sin C$. Here, $a = b = 2$ (sides $AC$ and $CD$), and $C = 120^\circ$ (angle at $C$):
\[
[ACD] = \frac{1}{2} \cdot 2 \cdot 2 \cdot \sin 120^\circ = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
\]
4. **Calculate Area of $\triangle AED$**:
Similarly, for $\triangle AED$, using the same formula:
\[
[AED] = \frac{1}{2} \cdot 2 \cdot 2 \cdot \sin 120^\circ = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
\]
5. **Calculate Total Area of Pentagon $ABCDE$**:
The total area of pentagon $ABCDE$ is the sum of the areas of $\triangle ACD$ and twice the area of $\triangle AED$ (since there are two such triangles):
\[
[ABCDE] = [ACD] + 2[AED] = \sqrt{3} + 2\sqrt{3} = 3\sqrt{3}
\]
This can be rewritten as $\sqrt{27}$, which is $\sqrt{9 \times 3} = 3\sqrt{3}$.
6. **Conclusion**:
The area of pentagon $ABCDE$ is $\sqrt{27} = 3\sqrt{3}$, which is not in the form $\sqrt{m} + \sqrt{n}$. There seems to be a misunderstanding or misinterpretation in the problem statement or the initial solution. If we consider the area as $\sqrt{27}$, then $m + n = 27$ which is not an option in the given choices. Revisiting the problem, if we assume the area is $\sqrt{11} + \sqrt{12}$, then $m + n = 11 + 12 = 23$.
Therefore, the correct answer is $\boxed{\textbf{(D)} ~23}$.
|
23
|
deepscale
| 2,737
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Marcelle and Jaclyn each think of a polynomial. Each of their polynomials is monic, has degree 4, and has the same positive constant term and the same coefficient of $z$. The product of their polynomials is \[z^8 +3z^7 +z^6 +3z^5 +4z^4 +6z^3 +2z^2 +4.\]What is the constant term of Jaclyn's polynomial?
|
2
|
deepscale
| 33,860
| ||
Murtha has decided to start a pebble collection. She collects one pebble the first day and two pebbles on the second day. On each subsequent day she collects one more pebble than the previous day. How many pebbles will she have collected at the end of the twelfth day?
|
78
|
deepscale
| 33,695
| ||
Given two non-zero planar vectors $\overrightarrow{a}, \overrightarrow{b}$ that satisfy the condition: for any $λ∈R$, $| \overrightarrow{a}-λ \overrightarrow{b}|≥slant | \overrightarrow{a}- \frac {1}{2} \overrightarrow{b}|$, then:
$(①)$ If $| \overrightarrow{b}|=4$, then $\overrightarrow{a}· \overrightarrow{b}=$ _______ ;
$(②)$ If the angle between $\overrightarrow{a}, \overrightarrow{b}$ is $\frac {π}{3}$, then the minimum value of $\frac {|2 \overrightarrow{a}-t· \overrightarrow{b}|}{| \overrightarrow{b}|}$ is _______ .
|
\sqrt {3}
|
deepscale
| 22,629
| ||
Jane's quiz scores were 98, 97, 92, 85 and 93. What was her mean score?
|
93
|
deepscale
| 39,193
| ||
A convex polyhedron $P$ has $26$ vertices, $60$ edges, and $36$ faces, $24$ of which are triangular and $12$ of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $P$ have?
|
Every pair of vertices of the polyhedron determines either an edge, a face diagonal or a space diagonal. We have ${26 \choose 2} = \frac{26\cdot25}2 = 325$ total line segments determined by the vertices. Of these, $60$ are edges. Each triangular face has $0$ face diagonals and each quadrilateral face has $2$, so there are $2 \cdot 12 = 24$ face diagonals. This leaves $325 - 60 - 24 = \boxed{241}$ segments to be the space diagonals.
|
241
|
deepscale
| 6,793
| |
Assuming that the birth of a boy or a girl is equally likely, what is the probability that the three children in a family include at least one boy and one girl? Express your answer as a common fraction.
|
\frac{3}{4}
|
deepscale
| 35,402
| ||
Find the minimum of the expression \(6 t^{2} + 3 s^{2} - 4 s t - 8 t + 6 s + 5\).
|
\frac{8}{7}
|
deepscale
| 15,057
| ||
If a bag contains only green, yellow, and red marbles in the ratio $3: 4: 2$ and 63 of the marbles are not red, how many red marbles are in the bag?
|
Since the ratio of green marbles to yellow marbles to red marbles is $3: 4: 2$, then we can let the numbers of green, yellow and red marbles be $3n, 4n$ and $2n$ for some positive integer $n$. Since 63 of the marbles in the bag are not red, then $3n+4n=63$ and so $7n=63$ or $n=9$, which means that the number of red marbles in the bag is $2n=2 \times 9=18$.
|
18
|
deepscale
| 5,616
| |
Given the function $f(x)=(\sin x+\cos x)^{2}+2\cos ^{2}x$,
(1) Find the smallest positive period and the monotonically decreasing interval of the function $f(x)$;
(2) When $x\in[0, \frac{\pi}{2}]$, find the maximum and minimum values of $f(x)$.
|
2+\sqrt{2}
|
deepscale
| 29,167
| ||
Given that $\log_{10}\sin x + \log_{10}\cos x= -2$ and that $\log_{10}(\sin x+\cos x)=\frac{1}{2}(\log_{10}m-2)$, find $m$.
|
102
|
deepscale
| 20,794
| ||
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
|
1. **Label the seats and calculate unrestricted seating for Derek and Eric:**
Label the seats from left to right as $1$ through $5$. Without any restrictions, Derek and Eric can be seated in $5 \times 4 = 20$ ways, since Derek has 5 choices and Eric has 4 remaining choices.
2. **Calculate the restricted seating for Derek and Eric (not sitting next to each other):**
Consider Derek and Eric as a block to find the number of ways they can sit next to each other. This block can be placed in 4 different positions (seats 1-2, 2-3, 3-4, 4-5), and within each block, Derek and Eric can switch places, giving $4 \times 2 = 8$ ways. Thus, the number of ways they can sit without being next to each other is $20 - 8 = 12$.
3. **Case analysis based on the positions of Derek and Eric:**
- **Case 1: Derek and Eric at each end (seats 1 and 5 or seats 5 and 1):**
This case is impossible as it forces Alice, Bob, and Carla into the middle three seats, where Alice would have to sit next to either Bob or Carla. Thus, this case contributes $0$ ways.
- **Case 2: Derek and Eric in seats 2 and 4 (or vice versa):**
There are $2$ ways to arrange Derek and Eric in these seats. The remaining seats (1, 3, 5) are non-consecutive, allowing Alice, Bob, and Carla to sit in any order without Alice being next to Bob or Carla. There are $3! = 6$ ways to arrange Alice, Bob, and Carla. Thus, this case contributes $2 \times 6 = 12$ ways.
- **Case 3: Derek and Eric such that exactly one pair of consecutive seats are available:**
From the total $12$ ways Derek and Eric can sit without being next to each other, subtract the $2$ ways from Case 2 and the $2$ ways from Case 1, leaving $12 - 2 - 2 = 8$ ways. In these configurations, Alice must avoid the consecutive seats to not sit next to Bob or Carla. Alice has one choice, and Bob and Carla can be arranged in $2! = 2$ ways in the remaining seats. Thus, this case contributes $8 \times 2 = 16$ ways.
4. **Summing all cases:**
The total number of valid arrangements is $0 + 12 + 16 = 28$.
Thus, the number of ways for the five of them to sit in a row under the given conditions is $\boxed{\textbf{(C)}\ 28}$.
|
28
|
deepscale
| 555
| |
Given a tesseract (4-dimensional hypercube), calculate the sum of the number of edges, vertices, and faces.
|
72
|
deepscale
| 19,553
| ||
The graph of the line $x+y=b$ intersects the line segment from $(2,5)$ to $(4,9)$ at its midpoint. What is the value of $b$?
|
10
|
deepscale
| 33,477
| ||
For how many four-digit whole numbers does the sum of the digits equal $30$?
|
20
|
deepscale
| 28,880
| ||
An integer is called snakelike if its decimal representation $a_1a_2a_3\cdots a_k$ satisfies $a_i<a_{i+1}$ if $i$ is odd and $a_i>a_{i+1}$ if $i$ is even. How many snakelike integers between 1000 and 9999 have four distinct digits?
|
Let's create the snakelike number from digits $a < b < c < d$, and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of $5\cdot{10 \choose 4}$ But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits $a < b < c$. There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is $2\cdot{9 \choose 3}$. Thus our answer is $5\cdot{10 \choose 4} - 2\cdot{9 \choose 3} = \boxed{882}$
|
882
|
deepscale
| 6,796
| |
Let $\Omega$ be a sphere of radius 4 and $\Gamma$ be a sphere of radius 2 . Suppose that the center of $\Gamma$ lies on the surface of $\Omega$. The intersection of the surfaces of $\Omega$ and $\Gamma$ is a circle. Compute this circle's circumference.
|
Take a cross-section of a plane through the centers of $\Omega$ and $\Gamma$, call them $O_{1}$ and $O_{2}$, respectively. The resulting figure is two circles, one of radius 4 and center $O_{1}$, and the other with radius 2 and center $O_{2}$ on the circle of radius 4 . Let these two circles intersect at $A$ and $B$. Note that $\overline{A B}$ is a diameter of the desired circle, so we will find $A B$. Focus on triangle $O_{1} O_{2} A$. The sides of this triangle are $O_{1} O_{2}=O_{1} A=4$ and $O_{2} A=2$. The height from $O_{1}$ to $A O_{2}$ is $\sqrt{4^{2}-1^{2}}=\sqrt{15}$, and because $O_{1} O_{2}=2 \cdot A O_{2}$, the height from $A$ to $O_{1} O_{2}$ is $\frac{\sqrt{15}}{2}$. Then the distance $A B$ is two times this, or $\sqrt{15}$. Thus, the circumference of the desired circle is $\pi \sqrt{15}$.
|
\pi \sqrt{15}
|
deepscale
| 4,209
| |
We color some cells in $10000 \times 10000$ square, such that every $10 \times 10$ square and every $1 \times 100$ line have at least one coloring cell. What minimum number of cells we should color ?
|
10000
|
deepscale
| 19,861
| ||
Find the smallest positive integer whose cube ends in $888$.
|
Let $x^3 = 1000a + 888$. We factor an $8$ out of the right hand side, and we note that $x$ must be of the form $x = 2y$, where $y$ is a positive integer. Then, this becomes $y^3 = 125a + 111$. Taking mod $5$, $25$, and $125$, we get $y^3 \equiv 1\pmod 5$, $y^3 \equiv 11\pmod{25}$, and $y^3 \equiv 111\pmod{125}$.
We can work our way up, and find that $y\equiv 1\pmod 5$, $y\equiv 21\pmod{25}$, and finally $y\equiv 96\pmod{125}$. This gives us our smallest value, $y = 96$, so $x = \boxed{192}$, as desired. - Spacesam
|
192
|
deepscale
| 6,499
| |
If the function
$$
f(x) = |a \sin x + b \cos x - 1| + |b \sin x - a \cos x| \quad (a, b \in \mathbf{R})
$$
attains a maximum value of 11, then $a^{2} + b^{2} = \, \, \, $ .
|
50
|
deepscale
| 13,038
| ||
Cyclic quadrilateral $A B C D$ has side lengths $A B=1, B C=2, C D=3$ and $D A=4$. Points $P$ and $Q$ are the midpoints of $\overline{B C}$ and $\overline{D A}$. Compute $P Q^{2}$.
|
Construct $\overline{A C}, \overline{A Q}, \overline{B Q}, \overline{B D}$, and let $R$ denote the intersection of $\overline{A C}$ and $\overline{B D}$. Because $A B C D$ is cyclic, we have that $\triangle A B R \sim \triangle D C R$ and $\triangle A D R \sim \triangle B C R$. Thus, we may write $A R=4 x, B R=2 x, C R=6 x, D R=12 x$. Now, Ptolemy applied to $A B C D$ yields $140 x^{2}=1 \cdot 3+2 \cdot 4=11$. Now $\overline{B Q}$ is a median in triangle $A B D$. Hence, $B Q^{2}=\frac{2 B A^{2}+2 B D^{2}-A D^{2}}{4}$. Likewise, $C Q^{2}=\frac{2 C A^{2}+2 C D^{2}-D A^{2}}{4}$. But $P Q$ is a median in triangle $B Q C$, so $P Q^{2}=\frac{2 B Q^{2}+2 C Q^{2}-B C^{2}}{4}=\frac{A B^{2}+B D^{2}+C D^{2}+C A^{2}-B C^{2}-A D^{2}}{4}=$ $\frac{(196+100) x^{2}+1^{2}+3^{2}-2^{2}-4^{2}}{4}=\frac{148 x^{2}-5}{2}=\frac{148 \cdot \frac{11}{140}-5}{2}=\frac{116}{35}$.
|
\frac{116}{35}
|
deepscale
| 3,477
| |
A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
|
Without loss of generality, assume that the $8$ numbers on Sharon's cards are $1$, $1$, $2$, $3$, $4$, $5$, $6$, and $7$, in that order, and assume the $8$ colors are red, red, and six different arbitrary colors. There are ${8\choose2}-1$ ways of assigning the two red cards to the $8$ numbers; we subtract $1$ because we cannot assign the two reds to the two $1$'s. In order for Sharon to be able to remove at least one card and still have at least one card of each color, one of the reds have to be assigned with one of the $1$s. The number of ways for this to happen is $2 \cdot 6 = 12$. Each of these assignments is equally likely, so desired probability is $\frac{12}{{8\choose2}-1}=\frac{4}{9} \implies 4 + 9 = 13 = \boxed{013}$.
|
13
|
deepscale
| 7,204
| |
Let $a,$ $b,$ and $c$ be nonnegative real numbers such that $a + b + c = 8.$ Find the maximum value of
\[\sqrt{3a + 2} + \sqrt{3b + 2} + \sqrt{3c + 2}.\]
|
3\sqrt{10}
|
deepscale
| 20,673
| ||
Samuel's birthday cake is in the form of a $4 \times 4 \times 4$ inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into $64$ smaller cubes, each measuring $1 \times 1 \times 1$ inch, as shown below. How many of the small pieces will have icing on exactly two sides?
|
To solve this problem, we need to determine how many of the smaller $1 \times 1 \times 1$ inch cubes have icing on exactly two sides. We will analyze the positions of these cubes on the cake.
1. **Understanding the Cake Structure**:
- The cake is a $4 \times 4 \times 4$ cube.
- Icing is on the top and all four vertical sides, but not on the bottom.
- The cake is divided into $64$ smaller cubes, each $1 \times 1 \times 1$ inch.
2. **Identifying Cubes with Icing on Two Sides**:
- **Edge Cubes**: These are the cubes along the edges of the cake, excluding the corners. They have two sides exposed (one vertical and the top, except for the bottom layer).
- **Corner Cubes**: These are at the vertices of the cake. They have three sides exposed (two vertical and the top, except for the bottom layer).
3. **Counting Edge Cubes with Icing on Two Sides**:
- Each face of the cake has 12 edge positions (excluding the 4 corners), but only the top three layers contribute to cubes with icing on two sides.
- For each of the four vertical faces, the top three layers have 3 edge cubes each (excluding corners), totaling $3 \times 3 = 9$ per face.
- There are four such faces, so $4 \times 9 = 36$ edge cubes.
- However, each edge cube is shared between two faces, so we must divide by 2 to avoid double-counting: $36 / 2 = 18$ edge cubes with icing on two sides.
4. **Counting Corner Cubes with Icing on Two Sides**:
- The bottom layer corner cubes are the only corner cubes with exactly two sides iced (one vertical side and the top).
- There are 4 corners on the bottom layer, each contributing one such cube.
5. **Total Cubes with Icing on Exactly Two Sides**:
- Adding the edge cubes and the bottom layer corner cubes: $18 + 4 = 22$.
However, upon reviewing the solution, it appears there was an error in the initial count of edge cubes. Let's correct this:
- **Correct Count for Edge Cubes**:
- Each of the four vertical faces has 4 edge positions per layer (excluding corners), and only the top three layers contribute.
- For each face, $4 \times 3 = 12$ edge cubes per face.
- Four faces contribute, but each edge cube is shared between two faces, so $4 \times 12 / 2 = 24$ edge cubes with icing on two sides.
- **Revised Total**:
- Adding the corrected edge cubes and the bottom layer corner cubes: $24 + 4 = 28$.
However, this count exceeds any of the provided options, indicating a need to re-evaluate the shared edges and corners. The correct approach is to count only the edge cubes that are not on the bottom layer and are not corners, which gives us:
- **Final Count for Edge Cubes**:
- Each face has 4 edge positions per layer (excluding corners), and only the top three layers contribute.
- For each face, $4 \times 3 = 12$ edge cubes per face.
- Four faces contribute, but each edge cube is shared between two faces, so $4 \times 12 / 2 = 24$ edge cubes with icing on two sides.
- Subtract the 8 bottom edge cubes (which have only one frosted face), and add the 4 bottom corner cubes (which have two frosted faces).
Thus, the final count is $24 - 8 + 4 = \boxed{\textbf{(D) }20}$.
|
20
|
deepscale
| 615
| |
Suppose that $a$ varies inversely with $b^2$. If $a=9$ when $b=2$, find the value of $a$ when $b=3$.
|
4
|
deepscale
| 33,954
| ||
The solution to the equation \(\arcsin x + \arcsin 2x = \arccos x + \arccos 2x\) is
|
\frac{\sqrt{5}}{5}
|
deepscale
| 9,194
| ||
Let
\[f(x) = \frac{x^2 - 6x + 6}{2x - 4}\]and
\[g(x) = \frac{ax^2 + bx + c}{x - d}.\]You are given the following properties:
$\bullet$ The graphs of $f(x)$ and $g(x)$ have the same vertical asymptote.
$\bullet$ The oblique asymptotes of $f(x)$ and $g(x)$ are perpendicular, and they intersect on the $y$-axis.
$\bullet$ The graphs of $f(x)$ and $g(x)$ have two intersection points, one of which is on the line $x = -2.$
Find the point of intersection of the graphs of $f(x)$ and $g(x)$ that does not lie on the line $x = -2.$
|
\left( 4, -\frac{1}{2} \right)
|
deepscale
| 36,629
| ||
The sequence $b_1, b_2, b_3, \dots$ satisfies $b_1 = 25,$ $b_{12} = 125,$ and for all $n \ge 3,$ $b_n$ is the arithmetic mean of the first $n - 1$ terms. Find $b_2.$
|
225
|
deepscale
| 22,135
| ||
Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
|
To solve this problem, we need to consider the constraints given by the diagonals of the pentagon. Each diagonal connects two vertices, and the vertices at the ends of each diagonal must have different colors. We will analyze the possible colorings by considering different cases based on the color assignments of the vertices.
#### Step 1: Assign colors to vertices $A$ and $B$
- Vertex $A$ can be colored in any of the 6 available colors.
- Vertex $B$, which is connected to $A$ by an edge, must have a different color from $A$. Therefore, $B$ has 5 choices of colors.
#### Step 2: Consider vertex $C$
- Vertex $C$ is connected by diagonals to both $A$ and $B$. We consider two subcases:
- **Subcase 1:** $C$ has the same color as $A$.
- **Subcase 2:** $C$ has a different color from both $A$ and $B$.
#### Subcase 1: $C$ has the same color as $A$
- $C$ has 1 choice (the same color as $A$).
- $D$, connected to $A$, $B$, and $C$, must have a different color from $A$ (and hence $C$). Thus, $D$ has 5 choices.
- $E$, connected to $A$, $B$, $C$, and $D$, must have a different color from $A$ and $D$. Thus, $E$ has 4 choices.
- Total combinations for this subcase: $6 \cdot 5 \cdot 1 \cdot 5 \cdot 4 = 600$.
#### Subcase 2: $C$ has a different color from both $A$ and $B$
- $C$ has 4 choices (excluding the colors of $A$ and $B$).
- We further split this into two scenarios based on the color of $D$:
- **Scenario 1:** $D$ has the same color as $A$.
- **Scenario 2:** $D$ has a different color from $A$.
##### Scenario 1: $D$ has the same color as $A$
- $D$ has 1 choice (the same color as $A$).
- $E$, connected to $A$, $B$, $C$, and $D$, must have a different color from $A$ and $D$. Thus, $E$ has 5 choices.
- Total combinations for this scenario: $6 \cdot 5 \cdot 4 \cdot 1 \cdot 5 = 600$.
##### Scenario 2: $D$ has a different color from $A$
- $D$ has 4 choices (excluding the colors of $A$, $B$, and $C$).
- $E$, connected to $A$, $B$, $C$, and $D$, must have a different color from $A$ and $D$. Thus, $E$ has 4 choices.
- Total combinations for this scenario: $6 \cdot 5 \cdot 4 \cdot 4 \cdot 4 = 1920$.
#### Step 3: Summing all combinations
- Total combinations = Combinations from Subcase 1 + Combinations from Scenario 1 + Combinations from Scenario 2
- Total combinations = $600 + 600 + 1920 = 3120$.
Thus, the total number of different colorings possible is $\boxed{3120 \ \textbf{(C)}}$.
|
3120
|
deepscale
| 369
| |
Given that $a$, $b$, $c$ are the opposite sides of angles $A$, $B$, $C$ in triangle $ABC$, and $\frac{a-c}{b-\sqrt{2}c}=\frac{sin(A+C)}{sinA+sinC}$.
$(Ⅰ)$ Find the measure of angle $A$;
$(Ⅱ)$ If $a=\sqrt{2}$, $O$ is the circumcenter of triangle $ABC$, find the minimum value of $|3\overrightarrow{OA}+2\overrightarrow{OB}+\overrightarrow{OC}|$;
$(Ⅲ)$ Under the condition of $(Ⅱ)$, $P$ is a moving point on the circumcircle of triangle $ABC$, find the maximum value of $\overrightarrow{PB}•\overrightarrow{PC}$.
|
\sqrt{2} + 1
|
deepscale
| 31,566
| ||
Given a sequence ${a_n}$ that satisfies $a_1=1$ and $a_n=a_1+ \frac {1}{2}a_2+ \frac {1}{3}a_3+…+ \frac {1}{n-1}a_{n-1}$ for $n\geqslant 2, n\in\mathbb{N}^*$, if $a_k=2017$, then $k=$ \_\_\_\_\_\_.
|
4034
|
deepscale
| 23,809
| ||
A circle of radius 2 is centered at $A$. An equilateral triangle with side 4 has a vertex at $A$. What is the difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?
|
1. **Calculate the area of the circle:**
The formula for the area of a circle is $\pi r^2$. Given the radius $r = 2$, the area of the circle is:
\[
\text{Area of the circle} = \pi \times 2^2 = 4\pi
\]
2. **Calculate the area of the equilateral triangle:**
The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4} s^2$, where $s$ is the side length. Given $s = 4$, the area of the triangle is:
\[
\text{Area of the triangle} = \frac{\sqrt{3}}{4} \times 4^2 = 4\sqrt{3}
\]
3. **Determine the regions of interest:**
We need to find the difference between two areas:
- The area inside the circle but outside the triangle.
- The area inside the triangle but outside the circle.
4. **Understand the geometric relationship:**
Since the vertex of the triangle is at the center of the circle and the side length of the triangle is twice the radius of the circle, all vertices of the triangle lie on the circle. This implies that the triangle is inscribed in the circle.
5. **Calculate the difference in areas:**
The difference between the area of the circle and the area of the triangle represents the total area outside the triangle but inside the circle and the area inside the triangle but outside the circle. Since the triangle is inscribed, the area outside the triangle but inside the circle is the area of the circle minus the area of the triangle:
\[
\text{Difference in areas} = \text{Area of the circle} - \text{Area of the triangle} = 4\pi - 4\sqrt{3}
\]
6. **Simplify the expression:**
Factor out the common term:
\[
4\pi - 4\sqrt{3} = 4(\pi - \sqrt{3})
\]
7. **Conclude with the final answer:**
The difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle is:
\[
\boxed{\textbf{(D)}\; 4(\pi - \sqrt{3})}
\]
|
$4(\pi - \sqrt{3})$
|
deepscale
| 610
| |
I have two 12-sided dice, each with 3 maroon sides, 4 teal sides, 4 cyan sides, and one sparkly side. If I roll both dice simultaneously, what is the probability that they will display the same color?
|
\frac{7}{24}
|
deepscale
| 16,331
| ||
There are 456 natives on an island, each of whom is either a knight who always tells the truth or a liar who always lies. All residents have different heights. Once, each native said, "All other residents are shorter than me!" What is the maximum number of natives who could have then said one minute later, "All other residents are taller than me?"
|
454
|
deepscale
| 15,182
| ||
An equilateral triangle ABC has a side length of 4. A right isosceles triangle DBE, where $DB=EB=1$ and angle $D\hat{B}E = 90^\circ$, is cut from triangle ABC. Calculate the perimeter of the remaining quadrilateral.
|
10 + \sqrt{2}
|
deepscale
| 16,541
| ||
Given eight students, including Abby and Bridget, are randomly assigned to the 12 spots arranged in three rows of four as shown, calculate the probability that Abby and Bridget are seated directly adjacent to each other (in the same row or same column).
|
\frac{17}{66}
|
deepscale
| 14,555
| ||
Let $Q$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$, and $(0,1)$. The probability that the slope of the line determined by $Q$ and the point $\left(\frac{1}{2}, \frac{1}{4} \right)$ is greater than or equal to $1$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
|
41
|
deepscale
| 32,069
| ||
Before the soccer match between the "North" and "South" teams, five predictions were made:
a) There will be no draw;
b) "South" will concede goals;
c) "North" will win;
d) "North" will not lose;
e) Exactly 3 goals will be scored in the match.
After the match, it was found that exactly three predictions were correct. What was the final score of the match?
|
2-1
|
deepscale
| 25,894
| ||
An equilateral triangle $ABC$ shares a common side $BC$ with a square $BCDE,$ as pictured. What is the number of degrees in $\angle DAE$ (not pictured)? [asy]
pair pA, pB, pC, pD, pE;
pA = (0, 0);
pB = pA + dir(300);
pC = pA + dir(240);
pD = pC + dir(270);
pE = pB + dir(270);
draw(pA--pB--pC--pA);
draw(pB--pC--pD--pE--pB);
label("$A$", pA, N);
label("$B$", pB, E);
label("$C$", pC, W);
label("$D$", pD, SW);
label("$E$", pE, SE);
[/asy]
|
30^\circ.
|
deepscale
| 35,822
| ||
In the expansion of $(x^2+ \frac{4}{x^2}-4)^3(x+3)$, find the constant term.
|
-240
|
deepscale
| 32,669
| ||
Triangle $ABC$, with sides of length $5$, $6$, and $7$, has one vertex on the positive $x$-axis, one on the positive $y$-axis, and one on the positive $z$-axis. Let $O$ be the origin. What is the volume of tetrahedron $OABC$?
|
1. **Assigning Coordinates**: Assume without loss of generality that vertex $A$ is on the $x$-axis, vertex $B$ is on the $y$-axis, and vertex $C$ is on the $z$-axis. Let the coordinates of $A$, $B$, and $C$ be $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$ respectively.
2. **Using Triangle Side Lengths**: Given the side lengths of triangle $ABC$ are $5$, $6$, and $7$, we can use the distance formula to set up equations for the sides:
- $AB = 5$ gives $\sqrt{a^2 + b^2} = 5$
- $BC = 6$ gives $\sqrt{b^2 + c^2} = 6$
- $CA = 7$ gives $\sqrt{c^2 + a^2} = 7$
3. **Squaring the Equations**: Square each equation to eliminate the square roots:
- $a^2 + b^2 = 25$
- $b^2 + c^2 = 36$
- $c^2 + a^2 = 49$
4. **Solving for $a^2$, $b^2$, $c^2$**:
- Adding all three equations: $2(a^2 + b^2 + c^2) = 110 \Rightarrow a^2 + b^2 + c^2 = 55$
- Solving for individual squares:
- $a^2 = \frac{(a^2 + b^2) + (c^2 + a^2) - (b^2 + c^2)}{2} = \frac{25 + 49 - 36}{2} = 19$
- $b^2 = \frac{(a^2 + b^2) + (b^2 + c^2) - (c^2 + a^2)}{2} = \frac{25 + 36 - 49}{2} = 6$
- $c^2 = \frac{(b^2 + c^2) + (c^2 + a^2) - (a^2 + b^2)}{2} = \frac{36 + 49 - 25}{2} = 30$
5. **Volume of Tetrahedron $OABC$**:
- The volume $V$ of a tetrahedron with mutually perpendicular adjacent edges of lengths $a$, $b$, and $c$ originating from the same vertex (the origin in this case) is given by:
\[
V = \frac{1}{6}abc
\]
- Substituting $a = \sqrt{19}$, $b = \sqrt{6}$, and $c = \sqrt{30}$:
\[
V = \frac{1}{6} \sqrt{19} \cdot \sqrt{6} \cdot \sqrt{30} = \frac{1}{6} \sqrt{19 \cdot 6 \cdot 30} = \frac{1}{6} \sqrt{3420}
\]
- Simplifying $\sqrt{3420}$:
\[
\sqrt{3420} = \sqrt{114 \cdot 30} = \sqrt{114 \cdot 6 \cdot 5} = \sqrt{570 \cdot 6} = \sqrt{3420} = 6\sqrt{95}
\]
- Therefore, $V = \frac{1}{6} \cdot 6\sqrt{95} = \sqrt{95}$
6. **Conclusion**:
- The volume of tetrahedron $OABC$ is $\boxed{\sqrt{95}}$, which corresponds to choice $\boxed{\text{C}}$. $\blacksquare$
|
\sqrt{95}
|
deepscale
| 2,591
| |
What is the remainder when $2^{87} +3$ is divided by $7$?
|
4
|
deepscale
| 38,031
| ||
Let set $A=\{x \mid |x-2| \leq 2\}$, and $B=\{y \mid y=-x^2, -1 \leq x \leq 2\}$, then $A \cap B=$ ?
|
\{0\}
|
deepscale
| 17,955
| ||
Dean is playing a game with calculators. The 42 participants (including Dean) sit in a circle, and Dean holds 3 calculators. One calculator reads 1, another 0, and the last one -1. Dean starts by pressing the cube button on the calculator that shows 1, pressing the square button on the one that shows 0, and on the calculator that shows -1, he presses the negation button. After this, he passes all of the calculators to the next person in the circle. Each person presses the same buttons on the same calculators that Dean pressed and then passes them to the next person. Once the calculators have all gone around the circle and return to Dean so that everyone has had one turn, Dean adds up the numbers showing on the calculators. What is the sum he ends up with?
|
0
|
deepscale
| 38,907
| ||
In triangle \( ABC \), \( AC = 3 AB \). Let \( AD \) bisect angle \( A \) with \( D \) lying on \( BC \), and let \( E \) be the foot of the perpendicular from \( C \) to \( AD \). Find \( \frac{[ABD]}{[CDE]} \). (Here, \([XYZ]\) denotes the area of triangle \( XYZ \)).
|
1/3
|
deepscale
| 22,853
| ||
Values for $A, B, C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$)
|
1. **Understanding the Problem**: We need to find the number of ways to choose values for $A, B, C, D$ from the set $\{1, 2, 3, 4, 5, 6\}$ without replacement such that the curves $y = Ax^2 + B$ and $y = Cx^2 + D$ intersect.
2. **Condition for Intersection**: The curves intersect if there exists an $x$ such that $Ax^2 + B = Cx^2 + D$. Simplifying, we get $(A - C)x^2 = D - B$. For this equation to have a solution for $x$, we need $A \neq C$ (to avoid the trivial case of $0x^2 = 0$ which does not depend on $x$) and $\frac{D - B}{A - C} \geq 0$.
3. **Analysis of $\frac{D - B}{A - C}$**:
- Since $A \neq C$ and $B \neq D$ (values are chosen without replacement), the fraction $\frac{D - B}{A - C}$ is well-defined and non-zero.
- The fraction is positive if $D > B$ and $A > C$, or $D < B$ and $A < C$. It is negative otherwise.
4. **Counting the Total Choices**:
- There are $6$ choices for $A$, $5$ remaining choices for $C$ (since $C \neq A$), $4$ choices for $B$ (since $B \neq A, C$), and $3$ choices for $D$ (since $D \neq A, B, C$). This gives a total of $6 \times 5 \times 4 \times 3 = 360$ ways to assign values to $A, B, C, D$.
5. **Considering the Order of Curves**:
- Since the order of curves does not matter, each set of values $(A, B, C, D)$ is equivalent to $(C, D, A, B)$. Thus, each configuration is counted twice in the 360 total choices. Therefore, we need to divide by 2, giving $\frac{360}{2} = 180$ distinct configurations.
6. **Bijection Between Positive and Negative Cases**:
- For every configuration where $\frac{D - B}{A - C} > 0$, there is a corresponding configuration where $\frac{D - B}{A - C} < 0$ by swapping $B$ and $D$. This bijection ensures that exactly half of the configurations result in intersecting curves (positive case).
- Therefore, the number of configurations that lead to intersecting curves is $\frac{180}{2} = 90$.
7. **Conclusion**:
- The number of ways to choose $A, B, C, D$ such that the curves intersect is $\boxed{\textbf{(C) }90}$.
|
90
|
deepscale
| 599
| |
A $\frac 1p$ -array is a structured, infinite, collection of numbers. For example, a $\frac 13$ -array is constructed as follows:
\begin{align*} 1 \qquad \frac 13\,\ \qquad \frac 19\,\ \qquad \frac 1{27} \qquad &\cdots\\ \frac 16 \qquad \frac 1{18}\,\ \qquad \frac{1}{54} \qquad &\cdots\\ \frac 1{36} \qquad \frac 1{108} \qquad &\cdots\\ \frac 1{216} \qquad &\cdots\\ &\ddots \end{align*}
In general, the first entry of each row is $\frac{1}{2p}$ times the first entry of the previous row. Then, each succeeding term in a row is $\frac 1p$ times the previous term in the same row. If the sum of all the terms in a $\frac{1}{2008}$ -array can be written in the form $\frac mn$, where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by $2008$.
|
1
|
deepscale
| 37,473
| ||
Let $\phi$ be the smallest acute angle for which $\cos \phi,$ $\cos 2 \phi,$ $\cos 3 \phi$ form an arithmetic progression, in some order. Find $\sin \phi.$
|
\frac{\sqrt{3}}{2}
|
deepscale
| 29,524
| ||
Given vectors $\overrightarrow{a}=(\cos \alpha,\sin \alpha)$, $\overrightarrow{b}=(\cos x,\sin x)$, $\overrightarrow{c}=(\sin x+2\sin \alpha,\cos x+2\cos \alpha)$, where $(0 < \alpha < x < \pi)$.
$(1)$ If $\alpha= \frac {\pi}{4}$, find the minimum value of the function $f(x)= \overrightarrow{b} \cdot \overrightarrow{c}$ and the corresponding value of $x$;
$(2)$ If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac {\pi}{3}$, and $\overrightarrow{a} \perp \overrightarrow{c}$, find the value of $\tan 2\alpha$.
|
- \frac { \sqrt {3}}{5}
|
deepscale
| 10,003
| ||
Given that point $C$ is the midpoint between points $A$ and $B$. At 7:00 AM, Car 1 departs from $A$ towards $B$, Car 2 departs from $B$ towards $A$, and Car 3 departs from $C$ towards $A$. When Car 1 and Car 3 meet, Car 2 has traveled exactly $\frac{3}{8}$ of the total distance. At 10:00 AM, Car 3 reaches point $A$, and at 10:30 AM, Car 2 arrives at point $A$. At this moment, Car 1 is 84 km away from point $B$. What is the distance between points $A$ and $B$ in kilometers?
|
336
|
deepscale
| 11,051
| ||
If $a-b=1$ and $ab=-2$, then $\left(a+1\right)\left(b-1\right)=$____.
|
-4
|
deepscale
| 11,925
| ||
Add $123.4567$ to $98.764$ and round your answer to the nearest hundredth. Then, subtract $0.02$ from the rounded result.
|
222.20
|
deepscale
| 32,266
| ||
Among the scalene triangles with natural number side lengths, a perimeter not exceeding 30, and the sum of the longest and shortest sides exactly equal to twice the third side, there are ____ distinct triangles.
|
20
|
deepscale
| 26,755
| ||
The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$. If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segments $\overline{CP}$ and $\overline{DP}$, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.
[asy] pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6); draw(B--C--P--D--C^^D--A); filldraw(A--P--B--cycle, gray, black); label("$A$", A, SE); label("$B$", B, NE); label("$C$", C, NW); label("$D$", D, SW); label("$P$", P, N); label("$13\sqrt{3}$", A--D, S); label("$12\sqrt{3}$", A--B, E);[/asy]
|
Let $X$ be the apex of the pyramid and $M$ be the midpoint of $\overline{CD}$. We find the side lengths of $\triangle XMP$.
$MP = \frac{13\sqrt3}{2}$. $PX$ is half of $AC$, which is $\frac{\sqrt{3\cdot13^2+3\cdot12^2}}{2} = \frac{\sqrt{939}}{2}$. To find $MX$, consider right triangle $XMD$; since $XD=13\sqrt3$ and $MD=6\sqrt3$, we have $MX=\sqrt{3\cdot13^2-3\cdot6^2} = \sqrt{399}$.
Let $\theta=\angle XPM$. For calculating trig, let us double all sides of $\triangle XMP$. By Law of Cosines, $\cos\theta = \frac{3\cdot13^2+939-4\cdot399}{2\cdot13\cdot3\sqrt{313}}=\frac{-150}{6\cdot13\sqrt{313}}=-\frac{25}{13\sqrt{313}}$.
Hence, \[\sin\theta = \sqrt{1-\cos^2\theta}\] \[=\sqrt{1-\frac{625}{13^2\cdot313}}\] \[=\frac{\sqrt{13^2\cdot313-625}}{13\sqrt{313}}\] \[=\frac{\sqrt{3\cdot132^2}}{13\sqrt{313}}\] \[=\frac{132\sqrt3}{13\sqrt{313}}\] Thus, the height of the pyramid is $PX\sin\theta=\frac{\sqrt{939}}{2}\cdot\frac{132\sqrt3}{13\sqrt{313}}=\frac{66\cdot3}{13}$. Since $[CPD]=3\sqrt{3}\cdot13\sqrt{3}=9\cdot13$, the volume of the pyramid is $\frac13 \cdot 9\cdot13\cdot \frac{66\cdot3}{13}=\boxed{594}$.
|
594
|
deepscale
| 6,534
| |
In this Number Wall, you add the numbers next to each other and write the sum in the block directly above the two numbers. Which number will be the block labeled '$n$'? [asy]
draw((0,0)--(8,0)--(8,2)--(0,2)--cycle);
draw((2,0)--(2,2));
draw((4,0)--(4,2));
draw((6,0)--(6,2));
draw((1,2)--(7,2)--(7,4)--(1,4)--cycle);
draw((3,2)--(3,4));
draw((5,2)--(5,4));
draw((2,4)--(2,6)--(6,6)--(6,4)--cycle);
draw((4,4)--(4,6));
draw((3,6)--(3,8)--(5,8)--(5,6));
label("$n$",(1,1));
label("4",(3,1));
label("8",(5,1));
label("7",(7,1));
label("15",(6,3));
label("46",(4,7));
[/asy]
|
3
|
deepscale
| 39,134
| ||
Given the function $f(x)=3\sin x+4\cos x$, if for any $x\in R$ we have $f(x)\geqslant f(α)$, then the value of $\tan α$ is equal to ___.
|
\frac {3}{4}
|
deepscale
| 22,461
| ||
The instructor of a summer math camp brought several shirts, several pairs of trousers, several pairs of shoes, and two jackets for the entire summer. In each lesson, he wore trousers, a shirt, and shoes, and he wore a jacket only on some lessons. On any two lessons, at least one piece of his clothing or shoes was different. It is known that if he had brought one more shirt, he could have conducted 18 more lessons; if he had brought one more pair of trousers, he could have conducted 63 more lessons; if he had brought one more pair of shoes, he could have conducted 42 more lessons. What is the maximum number of lessons he could conduct under these conditions?
|
126
|
deepscale
| 11,477
| ||
Find the maximum value of the function $y=\frac{x}{{{e}^{x}}}$ on the interval $[0,2]$.
A) When $x=1$, $y=\frac{1}{e}$
B) When $x=2$, $y=\frac{2}{{{e}^{2}}}$
C) When $x=0$, $y=0$
D) When $x=\frac{1}{2}$, $y=\frac{1}{2\sqrt{e}}$
|
\frac{1}{e}
|
deepscale
| 32,918
| ||
Given triangle $\triangle ABC$ with sides $a$, $b$, $c$ opposite to angles $A$, $B$, $C$ respectively. If $\overrightarrow{BC} \cdot \overrightarrow{BA} + 2\overrightarrow{AC} \cdot \overrightarrow{AB} = \overrightarrow{CA} \cdot \overrightarrow{CB}$. <br/>$(1)$ Find the value of $\frac{{\sin A}}{{\sin C}}$; <br/>$(2)$ If $2a \cdot \cos C = 2b - c$, find the value of $\cos B$.
|
\frac{3\sqrt{2} - \sqrt{10}}{8}
|
deepscale
| 9,454
| ||
Given rectangle ABCD where E is the midpoint of diagonal BD, point E is connected to point F on segment DA such that DF = 1/4 DA. Find the ratio of the area of triangle DFE to the area of quadrilateral ABEF.
|
\frac{1}{7}
|
deepscale
| 20,196
| ||
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}
|
deepscale
| 24,285
| ||
Given that events A and B are independent, and both are mutually exclusive with event C. It is known that $P(A) = 0.2$, $P(B) = 0.6$, and $P(C) = 0.14$. Find the probability that at least one of A, B, or C occurs, denoted as $P(A+B+C)$.
|
0.82
|
deepscale
| 29,736
| ||
All two-digit numbers divisible by 5, where the number of tens is greater than the number of units, were written on the board. There were \( A \) such numbers. Then, all two-digit numbers divisible by 5, where the number of tens is less than the number of units, were written on the board. There were \( B \) such numbers. What is the value of \( 100B + A \)?
|
413
|
deepscale
| 15,597
| ||
Anita attends a baseball game in Atlanta and estimates that there are 50,000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games note that:
i. The actual attendance in Atlanta is within $10 \%$ of Anita's estimate.
ii. Bob's estimate is within $10 \%$ of the actual attendance in Boston.
To the nearest 1,000, the largest possible difference between the numbers attending the two games is
|
To find the largest possible difference between the numbers attending the two games, we need to consider the maximum and minimum possible attendances based on the given estimates and the conditions provided.
1. **Estimate for Atlanta:**
- Anita estimates 50,000 fans.
- The actual attendance in Atlanta is within $10\%$ of Anita's estimate.
- Therefore, the actual attendance in Atlanta, $A$, can range from $50,000 \times 0.9$ to $50,000 \times 1.1$:
\[
45,000 \leq A \leq 55,000.
\]
2. **Estimate for Boston:**
- Bob estimates 60,000 fans.
- Bob's estimate is within $10\%$ of the actual attendance in Boston.
- Let the actual attendance in Boston be $B$. Then, $0.9B \leq 60,000 \leq 1.1B$.
- Solving for $B$, we get:
\[
B \geq \frac{60,000}{1.1} \approx 54,545.45 \quad \text{and} \quad B \leq \frac{60,000}{0.9} \approx 66,666.67.
\]
- Therefore, the actual attendance in Boston can range from approximately $54,545$ to $66,667$.
3. **Calculating the largest possible difference:**
- To find the largest possible difference, we consider the maximum attendance in Boston and the minimum attendance in Atlanta:
\[
\text{Difference} = \max(B) - \min(A) = 66,667 - 45,000 = 21,667.
\]
- Rounding to the nearest 1,000, we get $22,000$.
Thus, the largest possible difference between the numbers attending the two games, to the nearest 1,000, is $\boxed{22000}$. This corresponds to choice $\mathrm{(E)}$.
|
22000
|
deepscale
| 1,891
|
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