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In the final stage of a professional bowling competition, the top five players compete as follows:
- The fifth place player competes against the fourth place player.
- The loser of the match receives the 5th place award.
- The winner then competes against the third place player.
- The loser of this match receives the 4th place award.
- The winner competes against the second place player.
- The loser receives the 3rd place award.
- The winner competes against the first place player.
- The loser of this final match receives the 2nd place award, and the winner receives the 1st place award.
How many different possible sequences of award distribution are there?
|
16
|
deepscale
| 10,901
| ||
Find the number of four-digit numbers, composed of the digits 1, 2, 3, 4, 5, 6, 7 (each digit can be used no more than once), that are divisible by 15.
|
36
|
deepscale
| 13,853
| ||
An equilateral triangle is inscribed in a circle. A smaller equilateral triangle has one vertex coinciding with a vertex of the larger triangle and another vertex on the midpoint of a side of the larger triangle. What percent of the area of the larger triangle is the area of the smaller triangle?
|
25\%
|
deepscale
| 29,776
| ||
For a French class, I need to master a list of 600 vocabulary words for an upcoming test. The score on the test is based on the percentage of words I recall correctly. In this class, I have noticed that even when guessing the words I haven't studied, I have about a 10% chance of getting them right due to my prior knowledge. What is the minimum number of words I need to learn in order to guarantee at least a 90% score on this test?
|
534
|
deepscale
| 23,709
| ||
2 diagonals of a regular decagon (a 10-sided polygon) are chosen. What is the probability that their intersection lies inside the decagon?
|
\dfrac{42}{119}
|
deepscale
| 23,559
| ||
A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$, and that $f(x) = 1-|x-2|$ for $1\le x \le 3$. Find the smallest $x$ for which $f(x) = f(2001)$.
|
Iterating the condition $f(3x) = 3f(x)$, we find that $f(x) = 3^kf\left(\frac{x}{3^k}\right)$ for positive integers $k$. We know the definition of $f(x)$ from $1 \le x \le 3$, so we would like to express $f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6$. Indeed,
\[f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.\]
We now need the smallest $x$ such that $f(x) = 3^kf\left(\frac{x}{3^k}\right) = 186$. The range of $f(x),\ 1 \le x \le 3$, is $0 \le f(x) \le 1$. So when $1 \le \frac{x}{3^k} \le 3$, we have $0 \le f\left(\frac{x}{3^k}\right) = \frac{186}{3^k} \le 1$. Multiplying by $3^k$: $0 \le 186 \le 3^k$, so the smallest value of $k$ is $k = 5$. Then,
\[186 = {3^5}f\left(\frac{x}{3^5}\right).\]
Because we forced $1 \le \frac{x}{3^5} \le 3$, so
\[186 = {3^5}f\left(\frac{x}{3^5}\right) = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 \cdot 243.\]
We want the smaller value of $x = \boxed{429}$.
An alternative approach is to consider the graph of $f(x)$, which iterates every power of $3$, and resembles the section from $1 \le x \le 3$ dilated by a factor of $3$ at each iteration.
|
429
|
deepscale
| 6,723
| |
On one particular Wednesday, Jack worked \( t-2 \) hours and earned \( 3t-2 \) dollars per hour. His coworker Bob worked 1.5 times more hours than Jack but earned \( 2t-7 \) dollars per hour less than Jack. After paying a fixed tax of $10 each, they both netted the same amount of earnings. Determine the value of \( t \).
|
\frac{19}{3}
|
deepscale
| 16,563
| ||
A nine-digit integer is formed by repeating a positive three-digit integer three times. For example, 123,123,123 or 456,456,456 are integers of this form. What is the greatest common divisor of all nine-digit integers of this form?
|
1001001
|
deepscale
| 21,772
| ||
Find all natural numbers which are divisible by $30$ and which have exactly $30$ different divisors.
(M Levin)
|
To find all natural numbers divisible by 30 with exactly 30 different divisors, we first explore the properties of divisors and the structure of such numbers.
A natural number \( n \) has exactly 30 divisors if its prime factorization can be expressed to fit the divisors' formula:
\[
(n = p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k} \Rightarrow \text{number of divisors} = (a_1+1)(a_2+1)\cdots(a_k+1) = 30).
\]
Given that the number must be divisible by 30, its prime factorization must include at least the primes 2, 3, and 5. Therefore, the factorization will be of the form:
\[
n = 2^a \cdot 3^b \cdot 5^c \cdot m,
\]
where \( m \) is coprime to 30. We will solve this in steps by considering cases that result in the number of divisors being 30.
### Case Analysis
#### Case 1: \((a_1+1)(a_2+1)(a_3+1) = 30 \)
The positives factorization of 30 can be:
1. \( 2 \times 3 \times 5 \)
2. \( 2 \times 15 \)
3. \( 3 \times 10 \)
4. \( 5 \times 6 \)
Let's consider these factorizations using the minimum primes 2, 3, and 5:
1. **\( 2 \times 3 \times 5 \) decomposition**:
- Possible distributions:
- \( (a_1+1, a_2+1, a_3+1) = (2, 3, 5) \)
- This implies:
- \( 2^{1} \cdot 3^{2} \cdot 5^{4-1} = 2 \cdot 9 \cdot 5 \)
- Reorder cases to cover permutations and solve algebraically assuring divisibility by 30.
2. **\( 2 \times 15 \) decomposition**:
- Consider:
- \( 2 = (a_1+1) \), \( 15 = (a_2+1)(a_3+1)(a_4+1) \)
- This constraint is hard to satisfy with three primes without additional minor primes.
3. **\( 3 \times 10 \) decomposition**:
- Similar issues as some combinations with three define another framework.
4. **\( 5 \times 6 \) decomposition**:
- Here potential factors would involve higher powers as needed but was explored similarly.
From the computations, potential valid numbers fitting these criteria turn out to be:
- 11250, 4050, 7500, 1620, 1200, 720
Thus, the natural numbers divisible by 30 that have exactly 30 different divisors are:
\[
\boxed{11250, 4050, 7500, 1620, 1200, 720}
\]
|
11250, 4050, 7500, 1620, 1200, 720
|
deepscale
| 6,117
| |
The line $12x+5y=60$ forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?
|
1. **Find the intercepts of the line with the axes:**
- Set $y = 0$ in the equation $12x + 5y = 60$:
\[
12x = 60 \implies x = \frac{60}{12} = 5
\]
- Set $x = 0$ in the equation $12x + 5y = 60$:
\[
5y = 60 \implies y = \frac{60}{5} = 12
\]
The line intersects the x-axis at $(5,0)$ and the y-axis at $(0,12)$.
2. **Identify the vertices of the triangle:**
The vertices of the triangle formed with the coordinate axes are $(0,0)$, $(5,0)$, and $(0,12)$.
3. **Calculate the lengths of two altitudes:**
- The altitude from $(0,0)$ to the line $x = 5$ (vertical line through $(5,0)$) is $5$.
- The altitude from $(0,0)$ to the line $y = 12$ (horizontal line through $(0,12)$) is $12$.
4. **Calculate the area of the triangle:**
The area $A$ of the triangle can be calculated using the formula for the area of a triangle with base and height:
\[
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30
\]
5. **Calculate the length of the third altitude:**
The third altitude is from the origin to the hypotenuse. The hypotenuse is the line $12x + 5y = 60$. The formula for the distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is:
\[
d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\]
Substituting the values, we get:
\[
d = \frac{|12 \cdot 0 + 5 \cdot 0 - 60|}{\sqrt{12^2 + 5^2}} = \frac{60}{\sqrt{144 + 25}} = \frac{60}{13}
\]
6. **Sum the lengths of the altitudes:**
The sum of the lengths of the altitudes is:
\[
5 + 12 + \frac{60}{13} = \frac{65}{13} + \frac{156}{13} + \frac{60}{13} = \frac{281}{13}
\]
Thus, the sum of the lengths of the altitudes of the triangle is $\boxed{\textbf{(E)} \dfrac{281}{13}}$.
|
\frac{281}{13}
|
deepscale
| 1,253
| |
What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?
|
1. **Substitute and Simplify**: We start by substituting $y = x^2 + 18x + 30$ into the given equation:
\[
x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}
\]
This becomes:
\[
y = 2 \sqrt{y + 15}
\]
2. **Square Both Sides**: To eliminate the square root, we square both sides:
\[
y^2 = 4(y + 15)
\]
Simplifying, we get:
\[
y^2 - 4y - 60 = 0
\]
3. **Solve the Quadratic Equation**: We solve the quadratic equation for $y$:
\[
y^2 - 4y - 60 = 0
\]
Factoring, we find:
\[
(y - 10)(y + 6) = 0
\]
Thus, $y = 10$ or $y = -6$.
4. **Check for Extraneous Roots**: We need to check if $y = -6$ is valid:
\[
2 \sqrt{-6 + 15} = 2 \sqrt{9} = 6
\]
But substituting $y = -6$ in the original substitution gives:
\[
-6 = 2 \sqrt{9} = 6
\]
which is false. Therefore, $y = -6$ is extraneous.
5. **Valid Solution for $y$**: We have $y = 10$ as the valid solution. Substituting back for $y$:
\[
x^2 + 18x + 30 = 10
\]
Simplifying, we get:
\[
x^2 + 18x + 20 = 0
\]
6. **Calculate the Discriminant**: To ensure the roots are real, we calculate the discriminant:
\[
\Delta = 18^2 - 4 \cdot 1 \cdot 20 = 324 - 80 = 244
\]
Since $\Delta > 0$, the roots are real.
7. **Apply Vieta's Formulas**: By Vieta's formulas, the product of the roots of $x^2 + 18x + 20 = 0$ is given by the constant term divided by the leading coefficient:
\[
\text{Product of roots} = \frac{20}{1} = 20
\]
Thus, the product of the real roots of the original equation is $\boxed{20}$.
|
20
|
deepscale
| 2,464
| |
In the polar coordinate system, the equation of circle C is given by ρ = 2sinθ, and the equation of line l is given by $ρsin(θ+ \frac {π}{3})=a$. If line l is tangent to circle C, find the value of the real number a.
|
- \frac {1}{2}
|
deepscale
| 30,449
| ||
Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive $x$-axis. For any line $l$, the transformation $R(l)$ produces another line as follows: $l$ is reflected in $l_1$, and the resulting line is reflected in $l_2$. Let $R^{(1)}(l)=R(l)$ and $R^{(n)}(l)=R\left(R^{(n-1)}(l)\right)$. Given that $l$ is the line $y=\frac{19}{92}x$, find the smallest positive integer $m$ for which $R^{(m)}(l)=l$.
|
945
|
deepscale
| 40,067
| ||
Three of the edges of a cube are $\overline{AB}, \overline{BC},$ and $\overline{CD},$ and $\overline{AD}$ is an interior diagonal. Points $P, Q,$ and $R$ are on $\overline{AB}, \overline{BC},$ and $\overline{CD},$ respectively, so that $AP = 5, PB = 15, BQ = 15,$ and $CR = 10.$ What is the area of the polygon that is the intersection of plane $PQR$ and the cube?
|
525
|
deepscale
| 35,905
| ||
Given the positive sequence $\{a_n\}$, where $a_1=2$, $a_2=1$, and $\frac {a_{n-1}-a_{n}}{a_{n}a_{n-1}}= \frac {a_{n}-a_{n+1}}{a_{n}a_{n+1}}(n\geqslant 2)$, find the value of the 2016th term of this sequence.
|
\frac{1}{1008}
|
deepscale
| 10,284
| ||
What is the number of square meters in the area of a circle with diameter 4 meters? Express your answer in terms of $\pi$.
|
4\pi
|
deepscale
| 39,132
| ||
Given that $2+\sqrt{3}$ is a root of the equation \[x^3 + ax^2 + bx + 10 = 0\]and that $a$ and $b$ are rational numbers, compute $b.$
|
-39
|
deepscale
| 36,832
| ||
On a $12$-hour clock, an elapsed time of four hours looks the same as an elapsed time of $16$ hours. Because of this, we can say that four hours is "clock equivalent'' to its square number of hours. What is the least whole number of hours that is greater than $4$ hours and is "clock equivalent'' to its square number of hours?
|
9
|
deepscale
| 37,900
| ||
Find the minimum value, for \(a, b > 0\), of the expression
\[
\frac{|a + 3b - b(a + 9b)| + |3b - a + 3b(a - b)|}{\sqrt{a^{2} + 9b^{2}}}
\]
|
\frac{\sqrt{10}}{5}
|
deepscale
| 11,419
| ||
Let $S$ denote the value of the sum
\[\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}\]
$S$ can be expressed as $p + q \sqrt{r}$, where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime. Determine $p + q + r$.
|
121
|
deepscale
| 37,483
| ||
The ratio $\frac{2^{2001} \cdot 3^{2003}}{6^{2002}}$ is:
|
1. **Rewrite the given expression using the properties of exponents:**
\[
\frac{2^{2001} \cdot 3^{2003}}{6^{2002}}
\]
2. **Express $6^{2002}$ in terms of $2$ and $3$:**
\[
6^{2002} = (2 \cdot 3)^{2002} = 2^{2002} \cdot 3^{2002}
\]
3. **Substitute this expression back into the original ratio:**
\[
\frac{2^{2001} \cdot 3^{2003}}{2^{2002} \cdot 3^{2002}}
\]
4. **Simplify the expression by canceling out common terms:**
- For the powers of $2$, we have $2^{2001}$ in the numerator and $2^{2002}$ in the denominator. Simplifying this gives:
\[
\frac{2^{2001}}{2^{2002}} = 2^{2001-2002} = 2^{-1} = \frac{1}{2}
\]
- For the powers of $3$, we have $3^{2003}$ in the numerator and $3^{2002}$ in the denominator. Simplifying this gives:
\[
\frac{3^{2003}}{3^{2002}} = 3^{2003-2002} = 3^1 = 3
\]
5. **Combine the simplified terms:**
\[
\frac{1}{2} \cdot 3 = \frac{3}{2}
\]
6. **Conclude with the final answer:**
\[
\boxed{\mathrm{(E) \ } \frac{3}{2}}
\]
|
\frac{3}{2}
|
deepscale
| 1,299
| |
In a tournament, there are 16 chess players. Determine the number of different possible schedules for the first round (schedules are considered different if they differ by the participants of at least one match; the color of the pieces and the board number are not considered).
|
2027025
|
deepscale
| 7,419
| ||
In right triangle $A B C$, a point $D$ is on hypotenuse $A C$ such that $B D \perp A C$. Let $\omega$ be a circle with center $O$, passing through $C$ and $D$ and tangent to line $A B$ at a point other than $B$. Point $X$ is chosen on $B C$ such that $A X \perp B O$. If $A B=2$ and $B C=5$, then $B X$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100 a+b$.
|
Note that since $A D \cdot A C=A B^{2}$, we have the tangency point of $\omega$ and $A B$ is $B^{\prime}$, the reflection of $B$ across $A$. Let $Y$ be the second intersection of $\omega$ and $B C$. Note that by power of point, we have $B Y \cdot B C=B B^{\prime 2}=4 A B^{2} \Longrightarrow B Y=\frac{4 A B^{2}}{B C}$. Note that $A X$ is the radical axis of $\omega$ and the degenerate circle at $B$, so we have $X B^{2}=X Y \cdot X C$, so $$B X^{2}=(B C-B X)(B Y-B X)=B X^{2}-B X(B C+B Y)+B C \cdot B Y$$ This gives us $$B X=\frac{B C \cdot B Y}{B C+B Y}=\frac{4 A B^{2} \cdot B C}{4 A B^{2}+B C^{2}}=\frac{80}{41}$$
|
8041
|
deepscale
| 3,996
| |
Given the price of Product A was set at 70 yuan per piece in the first year, with an annual sales volume of 118,000 pieces, starting from the second year, the price per piece increased by $$\frac {70\cdot x\%}{1-x\%}$$ yuan due to a management fee, and the annual sales volume decreased by $10^4x$ pieces, calculate the maximum value of x such that the management fee collected in the second year is not less than 1.4 million yuan.
|
10
|
deepscale
| 8,612
| ||
Group the set of positive odd numbers $\{1, 3, 5, \cdots\}$ in increasing order such that the $n$-th group has $(2n-1)$ odd numbers:
\[
\{1\}, \quad \{3, 5, 7\}, \quad \{9, 11, 13, 15, 17\}, \cdots
\]
(first group)(second group)(third group)
Determine which group 1991 belongs to.
|
32
|
deepscale
| 11,435
| ||
Divide 6 volunteers into 4 groups for service at four different venues of the 2012 London Olympics. Among these groups, 2 groups will have 2 people each, and the other 2 groups will have 1 person each. How many different allocation schemes are there? (Answer with a number)
|
540
|
deepscale
| 32,666
| ||
Anca and Bruce left Mathville at the same time. They drove along a straight highway towards Staton. Bruce drove at $50 \mathrm{~km} / \mathrm{h}$. Anca drove at $60 \mathrm{~km} / \mathrm{h}$, but stopped along the way to rest. They both arrived at Staton at the same time. For how long did Anca stop to rest?
|
Since Bruce drove 200 km at a speed of $50 \mathrm{~km} / \mathrm{h}$, this took him $\frac{200}{50}=4$ hours. Anca drove the same 200 km at a speed of $60 \mathrm{~km} / \mathrm{h}$ with a stop somewhere along the way. Since Anca drove 200 km at a speed of $60 \mathrm{~km} / \mathrm{h}$, the time that the driving portion of her trip took was $\frac{200}{60}=3 \frac{1}{3}$ hours. The length of Anca's stop is the difference in driving times, or $4-3 \frac{1}{3}=\frac{2}{3}$ hours. Since $\frac{2}{3}$ hours equals 40 minutes, then Anca stops for 40 minutes.
|
40 \text{ minutes}
|
deepscale
| 5,862
| |
Given the function $f(x)=\sin \omega x\cos \omega x- \sqrt{3}\cos^2\omega x+ \frac{\sqrt{3}}{2} (\omega > 0)$, the two adjacent axes of symmetry of its graph are $\frac{\pi}{2}$.
$(1)$ Find the equation of the axis of symmetry for the function $y=f(x)$.
$(2)$ If the zeros of the function $y=f(x)- \frac{1}{3}$ in the interval $(0,\pi)$ are $x_{1}$ and $x_{2}$, find the value of $\cos (x_{1}-x_{2})$.
|
\frac{1}{3}
|
deepscale
| 18,576
| ||
Given that $\tan x = -\frac{12}{5}$ and $x \in (\frac{\pi}{2}, \pi)$, find the value of $\cos(-x + \frac{3\pi}{2})$.
|
-\frac{12}{13}
|
deepscale
| 24,042
| ||
Currently, 7 students are to be assigned to participate in 5 sports events, with the conditions that students A and B cannot participate in the same event, each event must have at least one participant, and each student can only participate in one event. How many different ways can these conditions be satisfied? (Answer with a number)
|
15000
|
deepscale
| 7,542
| ||
In the diagram, $D$ and $E$ are the midpoints of $\overline{AB}$ and $\overline{BC}$ respectively. Find the sum of the slope and $y$-intercept of the line passing through the points $C$ and $D.$ [asy]
size(180); defaultpen(linewidth(.7pt)+fontsize(10pt));
pair A, B, C, D, E, F;
A=(0,6);
B=(0,0);
C=(8,0);
D=(0,3);
E=(4,0);
F=(8/3,2);
draw(E--A--C--D);
draw((-1,0)--(10,0), EndArrow);
draw((0,-1)--(0,8), EndArrow);
label("$A(0,6)$", A, W);
label("$B(0,0)$", B, SW);
label("$C(8,0)$", C, S);
label("$D$", D, W);
label("$E$", E, S);
label("$F$", F, SW);
label("$x$", (10,0), dir(0));
label("$y$", (0,8), dir(90));
[/asy]
|
\frac{21}{8}
|
deepscale
| 34,296
| ||
Find $\sin \frac{11 \pi}{3}.$
|
-\frac{\sqrt{3}}{2}
|
deepscale
| 39,952
| ||
The symphony orchestra has more than 200 members but fewer than 300 members. When they line up in rows of 6, there are two extra members; when they line up in rows of 8, there are three extra members; and when they line up in rows of 9, there are four extra members. How many members are in the symphony orchestra?
|
260
|
deepscale
| 28,313
| ||
Let $x$ and $y$ be positive real numbers. Find the minimum value of
\[\left( x + \frac{1}{y} \right) \left( x + \frac{1}{y} - 2018 \right) + \left( y + \frac{1}{x} \right) \left( y + \frac{1}{x} - 2018 \right).\]
|
-2036162
|
deepscale
| 36,763
| ||
Given in $\bigtriangleup ABC$, $AB = 75$, and $AC = 120$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover, $\overline{BX}$ and $\overline{CX}$ have integer lengths. Find the length of $BC$.
|
117
|
deepscale
| 14,664
| ||
A circle has a radius of 3 units. There are many line segments of length 4 units that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.
A) $3\pi$
B) $5\pi$
C) $4\pi$
D) $7\pi$
E) $6\pi$
|
4\pi
|
deepscale
| 20,153
| ||
Solve the equations:
(1) $(x-2)^2=25$;
(2) $x^2+4x+3=0$;
(3) $2x^2+4x-1=0$.
|
\frac{-2-\sqrt{6}}{2}
|
deepscale
| 32,885
| ||
\[ y = x + \cos(2x) \] in the interval \((0, \pi / 4)\).
|
\frac{\pi}{12} + \frac{\sqrt{3}}{2}
|
deepscale
| 8,295
| ||
Find the smallest number $n \geq 5$ for which there can exist a set of $n$ people, such that any two people who are acquainted have no common acquaintances, and any two people who are not acquainted have exactly two common acquaintances.
*Bulgaria*
|
11
|
deepscale
| 27,467
| ||
Given the triangular pyramid P-ABC, PA is perpendicular to the base ABC, AB=2, AC=AP, BC is perpendicular to CA. If the surface area of the circumscribed sphere of the triangular pyramid P-ABC is $5\pi$, find the value of BC.
|
\sqrt{3}
|
deepscale
| 22,280
| ||
The side lengths \(a, b, c\) of triangle \(\triangle ABC\) satisfy the conditions:
1. \(a, b, c\) are all integers;
2. \(a, b, c\) form a geometric sequence;
3. At least one of \(a\) or \(c\) is equal to 100.
Find all possible sets of the triplet \((a, b, c)\).
|
10
|
deepscale
| 25,414
| ||
What are the last three digits of \(2003^N\), where \(N = 2002^{2001}\)?
|
241
|
deepscale
| 13,165
| ||
If the integer solutions to the system of inequalities
\[
\begin{cases}
9x - a \geq 0, \\
8x - b < 0
\end{cases}
\]
are only 1, 2, and 3, find the number of ordered pairs \((a, b)\) of integers that satisfy this system.
|
72
|
deepscale
| 7,786
| ||
What is the least positive integer greater than 1 that leaves a remainder of 2 when divided by each of 3, 4, 5, 6, 7, 8, 9, and 11?
|
27722
|
deepscale
| 19,239
| ||
Find $AX$ in the diagram if $CX$ bisects $\angle ACB$.
[asy]
import markers;
real t=.56;
pair A=(0,0);
pair B=(3,2);
pair C=(.5,1.5);
pair X=t*A+(1-t)*B;
draw(C--A--B--C--X);
label("$A$",A,SW);
label("$B$",B,E);
label("$C$",C,N);
label("$X$",X,SE);
//markangle(n=1,radius=15,A,C,X,marker(markinterval(stickframe(n=1),true)));
//markangle(n=1,radius=15,X,C,B,marker(markinterval(stickframe(n=1),true)));
label("$28$",.5*(B+X),SE);
label("$30$",.5*(B+C),N);
label("$21$",.5*(A+C),NW);
[/asy]
|
\frac{98}5
|
deepscale
| 35,805
| ||
The constant term in the expansion of the binomial \\((x \sqrt {x}- \dfrac {1}{x})^{5}\\) is \_\_\_\_\_\_ . (Answer with a number)
|
-10
|
deepscale
| 20,887
| ||
Two parallel chords of a circle have lengths 24 and 32 respectively, and the distance between them is 14. What is the length of another parallel chord midway between the two chords?
|
2\sqrt{249}
|
deepscale
| 11,195
| ||
John drove continuously from 8:30 a.m. until 2:15 p.m. of the same day and covered a distance of 246 miles. What was his average speed in miles per hour?
|
42.78
|
deepscale
| 14,486
| ||
The first four terms of an arithmetic sequence are $a, x, b, 2x$. The ratio of $a$ to $b$ is
|
1. **Identify the common difference**: Given an arithmetic sequence, the common difference $d$ can be calculated using any two consecutive terms. From the terms $x$ and $2x$, we have:
\[
d = 2x - x = x
\]
However, since $2x$ is the fourth term and $x$ is the second term, the difference $x$ accounts for two steps in the sequence. Therefore, the common difference for one step is:
\[
d = \frac{x}{2}
\]
2. **Write the terms using the common difference**: The first term is $a$, and the second term is $x$. Using the common difference $d = \frac{x}{2}$, the first term $a$ can be expressed as:
\[
a = x - d = x - \frac{x}{2} = \frac{x}{2}
\]
Similarly, the third term $b$ is:
\[
b = x + d = x + \frac{x}{2} = \frac{3x}{2}
\]
3. **Calculate the ratio of $a$ to $b$**: The ratio $\frac{a}{b}$ is:
\[
\frac{a}{b} = \frac{\frac{x}{2}}{\frac{3x}{2}} = \frac{x/2}{3x/2} = \frac{x}{2} \cdot \frac{2}{3x} = \frac{1}{3}
\]
4. **Conclusion**: The ratio of $a$ to $b$ is $\frac{1}{3}$, which corresponds to choice $\textbf{(B)}$.
Thus, the correct answer is $\boxed{\text{B}}$.
|
\frac{1}{3}
|
deepscale
| 1,695
| |
The area of a triangle \(ABC\) is \(\displaystyle 40 \text{ cm}^2\). Points \(D, E\) and \(F\) are on sides \(AB, BC\) and \(CA\) respectively. If \(AD = 3 \text{ cm}, DB = 5 \text{ cm}\), and the area of triangle \(ABE\) is equal to the area of quadrilateral \(DBEF\), find the area of triangle \(AEC\) in \(\text{cm}^2\).
|
15
|
deepscale
| 32,022
| ||
The ratio of the area of a square inscribed in a semicircle to the area of a square inscribed in a full circle is:
|
2: 5
|
deepscale
| 17,370
| ||
Let $a_n = n(2n+1)$ . Evaluate
\[
\biggl | \sum_{1 \le j < k \le 36} \sin\bigl( \frac{\pi}{6}(a_k-a_j) \bigr) \biggr |.
\]
|
18
|
deepscale
| 27,274
| ||
A square flag has a green cross of uniform width with a yellow square in the center on a white background. The cross is symmetric with respect to each of the diagonals of the square. If the entire cross (both the green arms and the yellow center) occupies 49% of the area of the flag, what percent of the area of the flag is yellow?
|
25.14\%
|
deepscale
| 12,735
| ||
In the prime factorization of $24!$, what is the exponent of $3$? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\cdot 4\cdot3\cdot2\cdot 1= 120$.)
|
10
|
deepscale
| 39,150
| ||
Given the equation \((7+4 \sqrt{3}) x^{2}+(2+\sqrt{3}) x-2=0\), calculate the difference between the larger root and the smaller root.
|
6 - 3 \sqrt{3}
|
deepscale
| 22,898
| ||
The function $f$ takes nonnegative integers to real numbers, such that $f(1) = 1,$ and
\[f(m + n) + f(m - n) = \frac{f(2m) + f(2n)}{2}\]for all nonnnegative integers $m \ge n.$ Find the sum of all possible values of $f(10).$
|
100
|
deepscale
| 36,573
| ||
A circle has radius $52$ and center $O$ . Points $A$ is on the circle, and point $P$ on $\overline{OA}$ satisfies $OP = 28$ . Point $Q$ is constructed such that $QA = QP = 15$ , and point $B$ is constructed on the circle so that $Q$ is on $\overline{OB}$ . Find $QB$ .
*Proposed by Justin Hsieh*
|
11
|
deepscale
| 24,273
| ||
In the diagram, \(ABCD\) is a square with a side length of \(8 \, \text{cm}\). Point \(E\) is on \(AB\) and point \(F\) is on \(DC\) so that \(\triangle AEF\) is right-angled at \(E\). If the area of \(\triangle AEF\) is \(30\%\) of the area of \(ABCD\), what is the length of \(AE\)?
|
4.8
|
deepscale
| 11,888
| ||
What is the arithmetic mean of the integers from -4 through 5, inclusive? Express your answer as a decimal to the nearest tenth.
|
0.5
|
deepscale
| 38,591
| ||
The absolute value of -9 is ; the reciprocal of -3 is .
|
-\frac{1}{3}
|
deepscale
| 17,344
| ||
We are given $2n$ natural numbers
\[1, 1, 2, 2, 3, 3, \ldots, n - 1, n - 1, n, n.\]
Find all $n$ for which these numbers can be arranged in a row such that for each $k \leq n$, there are exactly $k$ numbers between the two numbers $k$.
|
We are given \(2n\) natural numbers:
\[
1, 1, 2, 2, 3, 3, \ldots, n-1, n-1, n, n.
\]
and we need to find all values of \(n\) for which these numbers can be arranged such that there are exactly \(k\) numbers between the two occurrences of the number \(k\).
First, consider the positions of the number \( k \) in a valid arrangement. If the first occurrence of \( k \) is at position \( p \), then the second occurrence must be at \( p + k + 1 \).
### Example Verification:
To validate which \( n \) allow such an arrangement, start from small values:
- **\( n = 3 \):**
- Attempt to arrange: \(1, 1, 2, 2, 3, 3\).
- Possible arrangement: \( 3, 1, 2, 1, 3, 2 \) where
- \( 1 \) with 1 number between its occurrences,
- \( 2 \) with 2 numbers between its occurrences, and
- \( 3 \) with 3 numbers between its occurrences.
- **\( n = 4 \):**
- Arrangement required: \( 1, 1, 2, 2, 3, 3, 4, 4 \).
- Possible arrangement: \( 4, 1, 3, 1, 2, 4, 3, 2 \) where
- \( 1 \) has 1 number between its occurrences,
- \( 2 \) with 2,
- \( 3 \) with 3,
- \( 4 \) with 4.
Continue pattern checking up to the constraint given in the problem:
- **\( n = 5, 6 \):**
- Attempt to construct arrangements for \( n = 5 \) and \( n = 6 \) encounter problems fitting the pairs within the constraint, as not enough space or overlap issues.
- **\( n = 7 \):**
- Arrangement: \( 1, 1, 2, 3, 4, 5, 6, 7, 2, 3, 4, 5, 6, 7 \), satisfying all conditions for numbers counting correctly between occurrences.
- **\( n = 8 \):**
- Arrangement similarly should also validate with careful arrangement like above, though requires careful calculations to place each number pair the correctly given spacing.
### Conclusion
By examining possible arrangements and constraints, it can be concluded that the valid values of \( n \) are:
\[
\boxed{3, 4, 7, 8}
\]
These values of \( n \) allow for the construction of an arrangement fitting all \( k \) such that exactly \( k \) numbers are placed between the two occurrences of each \( k \).
|
$n=3,4,7,8$
|
deepscale
| 6,352
| |
Antônio needs to find a code with 3 different digits \( A, B, C \). He knows that \( B \) is greater than \( A \), \( A \) is less than \( C \), and also:
\[
\begin{array}{cccc}
& B & B \\
+ & A & A \\
\hline
& C & C \\
\end{array} = 242
\]
What is the code that Antônio is looking for?
|
232
|
deepscale
| 25,007
| ||
Given $f(x)= \sqrt {3}\sin x\cos (x+ \dfrac {π}{6})+\cos x\sin (x+ \dfrac {π}{3})+ \sqrt {3}\cos ^{2}x- \dfrac { \sqrt {3}}{2}$.
(I) Find the range of $f(x)$ when $x\in(0, \dfrac {π}{2})$;
(II) Given $\dfrac {π}{12} < α < \dfrac {π}{3}$, $f(α)= \dfrac {6}{5}$, $- \dfrac {π}{6} < β < \dfrac {π}{12}$, $f(β)= \dfrac {10}{13}$, find $\cos (2α-2β)$.
|
-\dfrac{33}{65}
|
deepscale
| 30,674
| ||
Connecting the centers of adjacent faces of a cube forms a regular octahedron. What is the volume ratio of this octahedron to the cube?
|
$\frac{1}{6}$
|
deepscale
| 18,501
| ||
On a $10000 order a merchant has a choice between three successive discounts of 20%, 20%, and 10% and three successive discounts of 40%, 5%, and 5%. By choosing the better offer, he can save:
|
To determine which discount option provides the greater savings, we need to calculate the final price for each option and then compare them.
#### Option 1: Three successive discounts of 20%, 20%, and 10%
1. **First discount of 20%**:
\[
\$10,000 \times (1 - 0.20) = \$10,000 \times 0.80 = \$8,000
\]
2. **Second discount of 20%**:
\[
\$8,000 \times (1 - 0.20) = \$8,000 \times 0.80 = \$6,400
\]
3. **Third discount of 10%**:
\[
\$6,400 \times (1 - 0.10) = \$6,400 \times 0.90 = \$5,760
\]
#### Option 2: Three successive discounts of 40%, 5%, and 5%
1. **First discount of 40%**:
\[
\$10,000 \times (1 - 0.40) = \$10,000 \times 0.60 = \$6,000
\]
2. **Second discount of 5%**:
\[
\$6,000 \times (1 - 0.05) = \$6,000 \times 0.95 = \$5,700
\]
3. **Third discount of 5%**:
\[
\$5,700 \times (1 - 0.05) = \$5,700 \times 0.95 = \$5,415
\]
#### Comparison of Savings:
To find out which option saves more, we compare the final prices:
\[
\$5,760 \text{ (Option 1)} - \$5,415 \text{ (Option 2)} = \$345
\]
Thus, by choosing the second option, the merchant saves an additional $\boxed{\textbf{(D)}\ \$345}$.
|
$345
|
deepscale
| 1,923
| |
What is the largest $n$ for which the numbers $1,2, \ldots, 14$ can be colored in red and blue so that for any number $k=1,2, \ldots, n$, there are a pair of blue numbers and a pair of red numbers, each pair having a difference equal to $k$?
|
11
|
deepscale
| 26,545
| ||
Find the value(s) of $x$ such that $8xy-12y+2x-3=0$ is true for all values of $y$.
|
\frac{3}{2}
|
deepscale
| 36,819
| ||
What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1, k_2, \dots, k_n$ for which
\[k_1^2 + k_2^2 + \dots + k_n^2 = 2002?\]
|
The problem asks for the maximum number of distinct positive integers $k_1, k_2, \ldots, k_n$ such that the sum of their squares equals 2002. We need to find the largest possible $n$ for which this is possible.
1. **Understanding the constraint**: The sum of squares $k_1^2 + k_2^2 + \ldots + k_n^2 = 2002$ must be achieved with each $k_i$ being a distinct positive integer.
2. **Estimating an upper bound for $n$**: We know that the squares of integers grow quite rapidly. To maximize $n$, we should start with the smallest squares. The smallest sum of squares for $n$ distinct integers starts with $1^2, 2^2, \ldots, n^2$.
3. **Calculating the sum of squares for increasing $n$**:
- The sum of the squares of the first $n$ integers is given by the formula $\frac{n(n+1)(2n+1)}{6}$.
- We need to find the largest $n$ such that $\frac{n(n+1)(2n+1)}{6} \leq 2002$.
4. **Checking values of $n$**:
- For $n = 14$, calculate $\frac{14 \times 15 \times 29}{6} = 1015$.
- For $n = 15$, calculate $\frac{15 \times 16 \times 31}{6} = 1240$.
- For $n = 16$, calculate $\frac{16 \times 17 \times 33}{6} = 1496$.
- For $n = 17$, calculate $\frac{17 \times 18 \times 35}{6} = 1785$.
- For $n = 18$, calculate $\frac{18 \times 19 \times 37}{6} = 2109$.
Here, $2109 > 2002$, so $n = 18$ is too large.
5. **Checking feasibility for $n = 17$**:
- We have already seen that the sum of squares up to $17^2$ is 1785, which is less than 2002. We need to check if we can find distinct integers such that their squares sum up to exactly 2002.
- The difference $2002 - 1785 = 217$.
- We check if there is a perfect square close to 217, and indeed $15^2 = 225$ is close but slightly over. However, $14^2 = 196$.
- Replacing $17^2$ with $14^2$ in our sum, we get $1785 - 289 + 196 = 1692$.
- We need to find a square that, when added to 1692, equals 2002. The difference is $2002 - 1692 = 310$.
- The square closest to 310 is $18^2 = 324$, which is slightly over. However, $17^2 = 289$ fits perfectly.
- Thus, replacing $14^2$ with $17^2$, we get $1692 + 289 = 1981$.
- The difference now is $2002 - 1981 = 21$, and $21$ is not a square of any integer.
Therefore, we cannot achieve 2002 exactly with 17 squares.
6. **Conclusion**:
- The maximum value of $n$ for which the sum of squares of distinct positive integers equals 2002 is $n = 16$.
$\boxed{\text{(C) }16}$
|
16
|
deepscale
| 2,270
| |
A circle has 2017 distinct points $A_{1}, \ldots, A_{2017}$ marked on it, and all possible chords connecting pairs of these points are drawn. A line is drawn through the point $A_{1}$, which does not pass through any of the points $A_{2}, \ldots A_{2017}$. Find the maximum possible number of chords that can intersect this line in at least one point.
|
1018080
|
deepscale
| 13,486
| ||
A particle starts from the origin on the number line, and at each step, it can move either 1 unit in the positive direction or 1 unit in the negative direction. After 10 steps, if the distance between the particle and the origin is 4, then the total number of distinct ways the particle can move is (answer in digits).
|
240
|
deepscale
| 19,124
| ||
Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?
|
1. **Convert Etienne's euros to dollars**: Given that 1 euro is equivalent to 1.3 dollars, we can calculate the dollar value of Etienne's 400 euros as follows:
\[
400 \text{ euros} \times 1.3 \frac{\text{dollars}}{\text{euro}} = 520 \text{ dollars}
\]
2. **Compare Etienne's dollar amount to Diana's dollar amount**: Diana has 500 dollars. Now, we compare Etienne's amount in dollars to Diana's:
\[
\text{Etienne's amount} = 520 \text{ dollars}, \quad \text{Diana's amount} = 500 \text{ dollars}
\]
3. **Calculate the percentage increase**: To find by what percent Etienne's money is greater than Diana's, we use the formula for percentage increase:
\[
\text{Percentage Increase} = \left(\frac{\text{New Value} - \text{Old Value}}{\text{Old Value}}\right) \times 100\%
\]
Plugging in the values:
\[
\text{Percentage Increase} = \left(\frac{520 - 500}{500}\right) \times 100\% = \left(\frac{20}{500}\right) \times 100\% = 4\%
\]
4. **Conclusion**: The value of Etienne's money is greater than the value of Diana's money by 4 percent.
Thus, the correct answer is $\boxed{\textbf{(B)}\ 4}$.
|
4
|
deepscale
| 1,518
| |
If $A$ is an angle such that $\tan A + \sec A = 2,$ enter all possible values of $\cos A,$ separated by commas.
|
\frac{4}{5}
|
deepscale
| 39,660
| ||
Simplify $\dfrac{3+4i}{1+2i}$. Your answer should be of the form $a+bi$, where $a$ and $b$ are both real numbers and written as improper fractions (if necessary).
|
\dfrac{11}{5} - \dfrac{2}{5}i
|
deepscale
| 34,614
| ||
Let
\[\mathbf{A} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}.\]Compute $\mathbf{A}^{95}.$
|
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}
|
deepscale
| 39,610
| ||
$(1)$ Solve the fractional equation: $\frac{x}{x+1}=\frac{2x}{3x+3}+1$;<br/>$(2)$ Simplify first ($\frac{x+2}{{x}^{2}-2x}-\frac{x-1}{{x}^{2}-4x+4})\div \frac{x+2}{{x}^{3}-4x}$, then choose a suitable number from $2$, $0$, $-1$ to substitute into the simplified result for evaluation.
|
\frac{5}{3}
|
deepscale
| 8,823
| ||
\(1.25 \times 67.875 + 125 \times 6.7875 + 1250 \times 0.053375\).
|
1000
|
deepscale
| 11,140
| ||
Done with her new problems, Wendy takes a break from math. Still without any fresh reading material, she feels a bit antsy. She starts to feel annoyed that Michael's loose papers clutter the family van. Several of them are ripped, and bits of paper litter the floor. Tired of trying to get Michael to clean up after himself, Wendy spends a couple of minutes putting Michael's loose papers in the trash. "That seems fair to me," confirms Hannah encouragingly.
While collecting Michael's scraps, Wendy comes across a corner of a piece of paper with part of a math problem written on it. There is a monic polynomial of degree $n$, with real coefficients. The first two terms after $x^n$ are $a_{n-1}x^{n-1}$ and $a_{n-2}x^{n-2}$, but the rest of the polynomial is cut off where Michael's page is ripped. Wendy barely makes out a little of Michael's scribbling, showing that $a_{n-1}=-a_{n-2}$. Wendy deciphers the goal of the problem, which is to find the sum of the squares of the roots of the polynomial. Wendy knows neither the value of $n$, nor the value of $a_{n-1}$, but still she finds a [greatest] lower bound for the answer to the problem. Find the absolute value of that lower bound.
|
1
|
deepscale
| 37,495
| ||
Find $\tan G$ in the right triangle shown below.
[asy]
pair H,F,G;
H = (0,0);
G = (15,0);
F = (0,8);
draw(F--G--H--F);
draw(rightanglemark(F,H,G,20));
label("$H$",H,SW);
label("$G$",G,SE);
label("$F$",F,N);
label("$17$",(F+G)/2,NE);
label("$15$",G/2,S);
[/asy]
|
\frac{8}{15}
|
deepscale
| 35,764
| ||
A sequence $(a_n)$ of real numbers is defined by $a_0=1$, $a_1=2015$ and for all $n\geq1$, we have
$$a_{n+1}=\frac{n-1}{n+1}a_n-\frac{n-2}{n^2+n}a_{n-1}.$$
Calculate the value of $\frac{a_1}{a_2}-\frac{a_2}{a_3}+\frac{a_3}{a_4}-\frac{a_4}{a_5}+\ldots+\frac{a_{2013}}{a_{2014}}-\frac{a_{2014}}{a_{2015}}$.
|
We start by examining the sequence \((a_n)\) given by the recurrence relations \(a_0 = 1\) and \(a_1 = 2015\), with the following recursive formula for \(n \geq 1\):
\[
a_{n+1} = \frac{n-1}{n+1}a_n - \frac{n-2}{n^2+n}a_{n-1}.
\]
The goal is to evaluate the expression:
\[
S = \frac{a_1}{a_2} - \frac{a_2}{a_3} + \frac{a_3}{a_4} - \frac{a_4}{a_5} + \cdots + \frac{a_{2013}}{a_{2014}} - \frac{a_{2014}}{a_{2015}}.
\]
To understand the behavior of this sequence and simplify \(S\), notice that the form of the sequence allows possible telescoping. Define:
\[
b_n = \frac{a_n}{a_{n+1}}.
\]
Thus, \(S = b_1 - b_2 + b_3 - b_4 + \cdots + b_{2013} - b_{2014}.\)
To evaluate \(b_n = \frac{a_n}{a_{n+1}}\), consider substituting using the recurrence relation:
\[
b_n = \frac{a_n}{\frac{n-1}{n+1}a_n - \frac{n-2}{n^2+n}a_{n-1}}
\]
This calculation is complex, so let's consider the pattern generated by each fraction \(b_n\). We seek to reveal any possible simplification or telescopic nature in the expression of \(S\).
Next, evaluate specific terms or attempt to find a recognizable pattern. Rewrite \(b_n\) using the sequence properties:
\[
b_n = a_n \times \left(\frac{n+1}{n-1}\right) \frac{1}{a_n} \times \left(1 + \frac{a_{n-1}(n-2)}{a_n(n+1)} \right) = \frac{n+1}{n-1} \times \left(1 + \frac{a_{n-1}(n-2)}{a_n(n+1)} \right).
\]
The complexity in determining the explicit values of \(b_n\) once simplified suggests focusing on establishing any identity or reduction of pattern to simplify \(S\).
The given recursive structure favors that \(b_n\) forms a simple identity or cancellation across sequences:
\[
S = b_1 - b_2 + b_3 - b_4 + \cdots + b_{2013} - b_{2014} = (b_1 - b_{2014}).
\]
Given initial assumptions or calculations for smaller terms, compute these values directly or examine whether they simplify or cancel within the context designed in smaller segments.
However, the given reference answer \(\boxed{3021}\) is derived recognizing intricacies resolving many prior steps noticing sequences' structured collapses yielding reductions in exact terms, matching this value.
Thus, the sum simplifies to:
\[
\boxed{3021}.
\]
|
3021
|
deepscale
| 6,078
| |
Consider the function $f(x) = 2x^2 - 4x + 9$. Evaluate $2f(3) + 3f(-3)$.
|
147
|
deepscale
| 33,873
| ||
When $\left( 1 - \frac{1}{a} \right)^6$ is expanded the sum of the last three coefficients is:
|
1. **Rewrite the expression**: The given expression $\left(1 - \frac{1}{a}\right)^6$ can be rewritten using the property of exponents as $\frac{(a-1)^6}{a^6}$.
2. **Expand using the Binomial Theorem**: The expansion of $(a-1)^6$ using the Binomial Theorem is:
\[
(a-1)^6 = \sum_{k=0}^6 \binom{6}{k} a^{6-k} (-1)^k
\]
This expansion results in terms with coefficients $\binom{6}{k}(-1)^k$ for $k = 0, 1, 2, \ldots, 6$.
3. **Identify the last three terms**: The last three terms in the expansion of $(a-1)^6$ correspond to $k = 0, 1, 2$. These terms are:
- For $k=0$: $\binom{6}{0} a^6 (-1)^0 = 1$
- For $k=1$: $\binom{6}{1} a^5 (-1)^1 = -6a^5$
- For $k=2$: $\binom{6}{2} a^4 (-1)^2 = 15a^4$
4. **Calculate the sum of the coefficients of the last three terms**: The coefficients are $1$, $-6$, and $15$. Their sum is:
\[
1 - 6 + 15 = 10
\]
5. **Conclusion**: The sum of the last three coefficients in the expansion of $\left(1 - \frac{1}{a}\right)^6$ is $\boxed{10 \textbf{ (C)}}$.
|
10
|
deepscale
| 2,303
| |
In $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are denoted as $a$, $b$, and $c$ respectively. The radius of the circumcircle is $1$, and it is given that $\frac{\tan A}{\tan B} = \frac{2c-b}{b}$. Find the maximum value of the area of $\triangle ABC$.
|
\frac{\sqrt{3}}{2}
|
deepscale
| 26,574
| ||
In $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are denoted as $a$, $b$, and $c$ respectively, and it is given that $2c-2a\cos B=b$.
$(1)$ Find the size of angle $A$;
$(2)$ If the area of $\triangle ABC$ is $\frac{\sqrt{3}}{4}$, and $c^{2}+ab\cos C+a^{2}=4$, find $a$.
|
\frac{\sqrt{7}}{2}
|
deepscale
| 16,622
| ||
Given that $\sum_{k=1}^{40}\sin 4k=\tan \frac{p}{q},$ where angles are measured in degrees, and $p$ and $q$ are relatively prime positive integers that satisfy $\frac{p}{q}<90,$ find $p+q.$
|
85
|
deepscale
| 26,544
| ||
Given that every high school in the town of Pythagoras sent a team of 3 students to a math contest, and Andrea's score was the median among all students, and hers was the highest score on her team, and Andrea's teammates Beth and Carla placed 40th and 75th, respectively, calculate the number of schools in the town.
|
25
|
deepscale
| 24,472
| ||
In $\triangle ABC$, $\overrightarrow {AD}=3 \overrightarrow {DC}$, $\overrightarrow {BP}=2 \overrightarrow {PD}$, if $\overrightarrow {AP}=λ \overrightarrow {BA}+μ \overrightarrow {BC}$, then $λ+μ=\_\_\_\_\_\_$.
|
- \frac {1}{3}
|
deepscale
| 19,092
| ||
A furniture store received a batch of office chairs that were identical except for their colors: 15 chairs were black and 18 were brown. The chairs were in demand and were being bought in a random order. At some point, a customer on the store's website discovered that only two chairs were left for sale. What is the probability that these two remaining chairs are of the same color?
|
0.489
|
deepscale
| 24,893
| ||
If $\{1, a, \frac{b}{a}\} = \{0, a^{2}, a+b\}$, find the value of $a^{2009} + b^{2009}$.
|
-1
|
deepscale
| 20,254
| ||
For how many integers $n$ is $\frac n{20-n}$ the square of an integer?
|
We are given the expression $\frac{n}{20-n}$ and need to determine for how many integers $n$ this expression is a perfect square.
1. **Examine the domain of $n$:**
- If $n < 0$ or $n > 20$, the fraction $\frac{n}{20-n}$ is negative, and thus cannot be a perfect square since squares of real numbers are non-negative.
- If $n = 20$, the fraction is undefined (division by zero).
- If $n \in \{1, 2, \dots, 9\}$, the fraction $\frac{n}{20-n}$ results in a positive number less than 1, which cannot be a perfect square of any integer other than 0. However, $\frac{n}{20-n} = 0$ is impossible for $n \neq 0$.
2. **Check $n = 0$:**
- $\frac{0}{20-0} = 0$, which is $0^2$ (a perfect square).
3. **Check $n \in \{10, 11, \dots, 19\}$:**
- For $n = 10$, $\frac{10}{20-10} = \frac{10}{10} = 1$, which is $1^2$ (a perfect square).
- For $n = 11$ to $n = 19$, we need to check if $\frac{n}{20-n}$ simplifies to a perfect square.
4. **Examine remaining values of $n$:**
- For prime $n$, $\frac{n}{20-n}$ will not be an integer unless $20-n$ is a multiple of $n$, which is not the case here.
- For composite $n$, we check if $\frac{n}{20-n}$ simplifies to a perfect square:
- $n = 12$: $\frac{12}{20-12} = \frac{12}{8} = \frac{3}{2}$ (not a perfect square).
- $n = 14$: $\frac{14}{20-14} = \frac{14}{6} = \frac{7}{3}$ (not a perfect square).
- $n = 15$: $\frac{15}{20-15} = \frac{15}{5} = 3$ (not a perfect square).
- $n = 16$: $\frac{16}{20-16} = \frac{16}{4} = 4$, which is $2^2$ (a perfect square).
- $n = 18$: $\frac{18}{20-18} = \frac{18}{2} = 9$, which is $3^2$ (a perfect square).
5. **Conclusion:**
- The values of $n$ for which $\frac{n}{20-n}$ is a perfect square are $n = 0, 10, 16, 18$.
Thus, there are $\boxed{\mathrm{(D)}\ 4}$ integers $n$ for which $\frac{n}{20-n}$ is the square of an integer.
|
4
|
deepscale
| 1,792
| |
On a party with 99 guests, hosts Ann and Bob play a game (the hosts are not regarded as guests). There are 99 chairs arranged in a circle; initially, all guests hang around those chairs. The hosts take turns alternately. By a turn, a host orders any standing guest to sit on an unoccupied chair $c$. If some chair adjacent to $c$ is already occupied, the same host orders one guest on such chair to stand up (if both chairs adjacent to $c$ are occupied, the host chooses exactly one of them). All orders are carried out immediately. Ann makes the first move; her goal is to fulfill, after some move of hers, that at least $k$ chairs are occupied. Determine the largest $k$ for which Ann can reach the goal, regardless of Bob's play.
|
Answer. $k=34$. Solution. Preliminary notes. Let $F$ denote the number of occupied chairs at the current position in the game. Notice that, on any turn, $F$ does not decrease. Thus, we need to determine the maximal value of $F$ Ann can guarantee after an arbitrary move (either hers or her opponent's). Say that the situation in the game is stable if every unoccupied chair is adjacent to an occupied one. In a stable situation, we have $F \geq 33$, since at most $3 F$ chairs are either occupied or adjacent to such. Moreover, the same argument shows that there is a unique (up to rotation) stable situation with $F=33$, in which exactly every third chair is occupied; call such stable situation bad. If the situation after Bob's move is stable, then Bob can act so as to preserve the current value of $F$ indefinitely. Namely, if $A$ puts some guest on chair $a$, she must free some chair $b$ adjacent to $a$. Then Bob merely puts a guest on $b$ and frees $a$, returning to the same stable position. On the other hand, if the situation after Bob's move is unstable, then Ann may increase $F$ in her turn by putting a guest on a chair having no adjacent occupied chairs. Strategy for Ann, if $k \leq 34$. In short, Ann's strategy is to increase $F$ avoiding appearance of a bad situation after Bob's move (conversely, Ann creates a bad situation in her turn, if she can). So, on each her turn, Ann takes an arbitrary turn increasing $F$ if there is no danger that Bob reaches a bad situation in the next turn (thus, Ann always avoids forcing any guest to stand up). The exceptional cases are listed below. Case 1. After possible Ann's move (consisting in putting a guest on chair a), we have $F=32$, and Bob can reach a bad situation by putting a guest on some chair. This means that, after Ann's move, every third chair would be occupied, with one exception. But this means that, by her move, Ann could put a guest on a chair adjacent to $a$, avoiding the danger. Case 2. After possible Ann's move (by putting a guest on chair a), we have $F=33$, and Bob can reach a stable situation by putting a guest on some chair $b$ and freeing an adjacent chair $c$. If $a=c$, then Ann could put her guest on $b$ to create a stable situation after her turn; that enforces Bob to break stability in his turn. Otherwise, as in the previous case, Ann could put a guest on some chair adjacent to $a$, still increasing the value of $F$, but with no danger of bad situation arising. So, acting as described, Ann increases the value of $F$ on each turn of hers whenever $F \leq 33$. Thus, she reaches $F=34$ after some her turn. Strategy for Bob, if $k \geq 35$. Split all chairs into 33 groups each consisting of three consecutive chairs, and number the groups by $1,2, \ldots, 33$ so that Ann's first turn uses a chair from group 1. In short, Bob's strategy is to ensure, after each his turn, that $(*)$ In group 1, at most two chairs are occupied; in every other group, only the central chair may be occupied. If $(*)$ is satisfied after Bob's turn, then $F \leq 34<k$; thus, property $(*)$ ensures that Bob will not lose. It remains to show that Bob can always preserve $(*)$. after any his turn. Clearly, he can do that oat the first turn. Suppose first that Ann, in her turn, puts a guest on chair $a$ and frees an adjacent chair $b$, then Bob may revert her turn by putting a guest on chair $b$ and freeing chair $a$. Suppose now that Ann just puts a guest on some chair $a$, and the chairs adjacent to $a$ are unoccupied. In particular, group 1 still contains at most two occupied chairs. If the obtained situation satisfies $(*)$, then Bob just makes a turn by putting a guest into group 1 (preferably, on its central chair), and, possibly, removing another guest from that group. Otherwise, $a$ is a non-central chair in some group $i \geq 2$; in this case Bob puts a guest to the central chair in group $i$ and frees chair $a$. So Bob indeed can always preserve $(*)$.
|
34
|
deepscale
| 4,406
| |
Express $1.\overline{27}$ as a common fraction in lowest terms.
|
\dfrac{14}{11}
|
deepscale
| 39,309
| ||
Given the function $$f(x)=4\sin(x- \frac {π}{6})\cos x+1$$.
(Ⅰ) Find the smallest positive period of f(x);
(Ⅱ) Find the maximum and minimum values of f(x) in the interval $$\[-\frac {π}{4}, \frac {π}{4}\]$$ .
|
-2
|
deepscale
| 32,220
| ||
For each positive integer $n$ , let $g(n)$ be the sum of the digits when $n$ is written in binary. For how many positive integers $n$ , where $1\leq n\leq 2007$ , is $g(n)\geq 3$ ?
|
1941
|
deepscale
| 11,107
| ||
The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:
|
1. **Start by squaring both sides of the equation**:
\[
\left(\sqrt{5x-1} + \sqrt{x-1}\right)^2 = 2^2
\]
Expanding the left side using the identity $(a+b)^2 = a^2 + 2ab + b^2$, we get:
\[
\sqrt{5x-1}^2 + 2\sqrt{5x-1}\sqrt{x-1} + \sqrt{x-1}^2 = 4
\]
Simplifying the squares of the square roots, we have:
\[
5x - 1 + 2\sqrt{(5x-1)(x-1)} + x - 1 = 4
\]
Combining like terms:
\[
6x - 2 + 2\sqrt{5x^2 - 6x + 1} = 4
\]
2. **Isolate the square root term**:
\[
2\sqrt{5x^2 - 6x + 1} = 4 - 6x + 2
\]
Simplifying the right side:
\[
2\sqrt{5x^2 - 6x + 1} = -6x + 6
\]
Dividing both sides by 2:
\[
\sqrt{5x^2 - 6x + 1} = -3x + 3
\]
3. **Square both sides again to eliminate the square root**:
\[
(5x^2 - 6x + 1) = (-3x + 3)^2
\]
Expanding the right side:
\[
5x^2 - 6x + 1 = 9x^2 - 18x + 9
\]
Rearranging all terms to one side:
\[
5x^2 - 6x + 1 - 9x^2 + 18x - 9 = 0
\]
Simplifying:
\[
-4x^2 + 12x - 8 = 0
\]
Factoring out the common factor:
\[
-4(x^2 - 3x + 2) = 0
\]
Factoring the quadratic:
\[
-4(x-1)(x-2) = 0
\]
Thus, $x = 1$ or $x = 2$.
4. **Verify the solutions in the original equation**:
- For $x = 1$:
\[
\sqrt{5(1)-1} + \sqrt{1-1} = \sqrt{4} + \sqrt{0} = 2 + 0 = 2
\]
This holds true.
- For $x = 2$:
\[
\sqrt{5(2)-1} + \sqrt{2-1} = \sqrt{9} + \sqrt{1} = 3 + 1 = 4
\]
This does not hold true.
5. **Conclude that the only solution is**:
\[
\boxed{\textbf{(D)}\ x=1}
\]
|
$x=1$
|
deepscale
| 763
| |
Compute
\[ e^{2 \pi i/17} + e^{4 \pi i/17} + e^{6 \pi i/17} + \dots + e^{32 \pi i/17}. \]
|
-1
|
deepscale
| 26,044
| ||
Observation: Given $\sqrt{5}≈2.236$, $\sqrt{50}≈7.071$, $\sqrt[3]{6.137}≈1.8308$, $\sqrt[3]{6137}≈18.308$; fill in the blanks:<br/>① If $\sqrt{0.5}\approx \_\_\_\_\_\_.$<br/>② If $\sqrt[3]{x}≈-0.18308$, then $x\approx \_\_\_\_\_\_$.
|
-0.006137
|
deepscale
| 19,928
| ||
The clock shows 00:00, and the hour and minute hands coincide. Considering this coincidence to be number 0, determine after what time interval (in minutes) they will coincide the 19th time. If necessary, round the answer to two decimal places following the rounding rules.
|
1243.64
|
deepscale
| 32,218
| ||
Let $\omega$ be the unit circle centered at the origin of $R^2$ . Determine the largest possible value for the radius of the circle inscribed to the triangle $OAP$ where $ P$ lies the circle and $A$ is the projection of $P$ on the axis $OX$ .
|
\frac{\sqrt{2} - 1}{2}
|
deepscale
| 8,384
| ||
The sides $PQ$ and $PR$ of triangle $PQR$ are respectively of lengths $4$ inches, and $7$ inches. The median $PM$ is $3\frac{1}{2}$ inches. Then $QR$, in inches, is:
|
1. **Using the Median Formula**: The formula for the length of a median in a triangle, which connects a vertex to the midpoint of the opposite side, is given by:
\[
PM = \frac{1}{2}\sqrt{2PQ^2 + 2PR^2 - QR^2}
\]
where $PQ$, $PR$, and $QR$ are the lengths of the sides of the triangle, and $PM$ is the median from vertex $P$ to the midpoint $M$ of side $QR$.
2. **Substitute the given values**: We know $PQ = 4$ inches, $PR = 7$ inches, and $PM = 3\frac{1}{2} = 3.5$ inches. Plugging these values into the formula, we get:
\[
3.5 = \frac{1}{2}\sqrt{2(4^2) + 2(7^2) - QR^2}
\]
3. **Simplify the equation**:
\[
3.5 = \frac{1}{2}\sqrt{2(16) + 2(49) - QR^2}
\]
\[
3.5 = \frac{1}{2}\sqrt{32 + 98 - QR^2}
\]
\[
3.5 = \frac{1}{2}\sqrt{130 - QR^2}
\]
4. **Eliminate the fraction and square both sides**:
\[
7 = \sqrt{130 - QR^2}
\]
Squaring both sides:
\[
49 = 130 - QR^2
\]
5. **Solve for $QR^2$**:
\[
QR^2 = 130 - 49
\]
\[
QR^2 = 81
\]
6. **Find $QR$**:
\[
QR = \sqrt{81}
\]
\[
QR = 9
\]
Thus, the length of side $QR$ is $\boxed{9}$ inches, corresponding to choice $\textbf{(D)}$.
|
9
|
deepscale
| 805
| |
Thirty clever students from 6th, 7th, 8th, 9th, and 10th grades were tasked with creating forty problems for an olympiad. Any two students from the same grade came up with the same number of problems, while any two students from different grades came up with a different number of problems. How many students came up with one problem each?
|
26
|
deepscale
| 23,569
|
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