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Given that $F_1$ and $F_2$ are the left and right foci of the ellipse $\frac{x^{2}}{16}{+}\frac{y^{2}}{b^{2}}{=}1$, and the line $l$ passing through $F_1$ intersects the ellipse at points $A$ and $B$. If the maximum value of $|AF_2|+|BF_2|$ is $10$, find the eccentricity of the ellipse.
|
\frac{1}{2}
| |
A three-digit number has different digits in each position. By writing a 2 to the left of this three-digit number, we get a four-digit number; and by writing a 2 to the right of this three-digit number, we get another four-digit number. The difference between these two four-digit numbers is 945. What is this three-digit number?
|
327
| |
There is a point inside an equilateral triangle with side length \( d \) whose distances from the vertices are 3, 4, and 5 units. Find the side length \( d \).
|
\sqrt{25 + 12 \sqrt{3}}
| |
A cone is inscribed in a regular quadrilateral pyramid. Find the ratio of the total surface area of the cone to the lateral surface area of the cone, given that the side length of the pyramid's base is 4 and the angle between the pyramid's height and the plane of its lateral face is $30^{\circ}$.
|
\frac{3 + \sqrt{3}}{3}
| |
Cara is sitting at a circular table with six friends. Assume there are three males and three females among her friends. How many different possible pairs of people could Cara sit between if each pair must include at least one female friend?
|
12
| |
There are several positive numbers written on the board. The sum of the five largest numbers is 0.29 of the sum of all the numbers, and the sum of the five smallest numbers is 0.26 of the sum of all the numbers. How many numbers in total are written on the board?
|
18
| |
Parallelogram $ABCD$ has vertices $A(3,4)$, $B(-2,1)$, $C(-5,-2)$, and $D(0,1)$. If a point is selected at random from the region determined by the parallelogram, what is the probability that the point is left of the $y$-axis? Express your answer as a common fraction.
|
\frac{1}{2}
| |
How many unordered pairs of edges in a regular tetrahedron determine a plane?
|
12
| |
The function $f(x)$ satisfies
\[f(x - y) = f(x) f(y)\]for all real numbers $x$ and $y,$ and $f(x) \neq 0$ for all real numbers $x.$ Find $f(3).$
|
1
| |
A bug crawls along a number line, starting at $-2$. It crawls to $-6$, then turns around and crawls to $5$. How many units does the bug crawl altogether?
|
15
|
1. **Calculate the distance from $-2$ to $-6$:**
The distance on a number line is the absolute difference between the two points. Thus, the distance from $-2$ to $-6$ is:
\[
|-6 - (-2)| = |-6 + 2| = |-4| = 4 \text{ units}
\]
2. **Calculate the distance from $-6$ to $5$:**
Similarly, the distance from $-6$ to $5$ is:
\[
|5 - (-6)| = |5 + 6| = |11| = 11 \text{ units}
\]
3. **Add the distances to find the total distance crawled:**
The total distance the bug crawls is the sum of the distances calculated in steps 1 and 2:
\[
4 \text{ units} + 11 \text{ units} = 15 \text{ units}
\]
Thus, the total distance the bug crawls is $\boxed{\textbf{(E)}\ 15}$.
|
Elisenda has a piece of paper in the shape of a triangle with vertices $A, B$, and $C$ such that $A B=42$. She chooses a point $D$ on segment $A C$, and she folds the paper along line $B D$ so that $A$ lands at a point $E$ on segment $B C$. Then, she folds the paper along line $D E$. When she does this, $B$ lands at the midpoint of segment $D C$. Compute the perimeter of the original unfolded triangle.
|
168+48 \sqrt{7}
|
Let $F$ be the midpoint of segment $D C$. Evidently $\angle A D B=60^{\circ}=\angle B D E=\angle E D C$. Moreover, we have $B D=D F=F C, A D=D E$, and $A B=B E$. Hence angle bisector on $B D C$ gives us that $B E=42, E C=84$, and hence angle bisector on $A B C$ gives us that if $A D=x$ then $C D=3 x$. Now this gives $B D=3 x / 2$, so thus the Law of Cosines on $A D B$ gives $x=12 \sqrt{7}$. Hence, $B C=42+84=126$ and $A C=4 x=48 \sqrt{7}$. The answer is $42+126+48 \sqrt{7}=168+48 \sqrt{7}$.
|
Given a function $f(x)=\log _{a}\left(\sqrt {x^{2}+1}+x\right)+\dfrac{1}{a^{x}-1}+\dfrac{3}{2}$, where $a > 0$ and $a \neq 1$. If $f\left(\log _{3}b\right)=5$ for $b > 0$ and $b \neq 1$, find the value of $f\left(\log _{\frac{1}{3}}b\right)$.
|
-3
| |
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?
|
3120
|
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 $\boxed{3120 \ \textbf{(C)}}$.
|
Each of 8 balls is randomly and independently painted either black or white with equal probability. Calculate the probability that every ball is different in color from at least half of the other 7 balls.
|
\frac{35}{128}
| |
We divide the height of a cone into three equal parts, and through the division points, we lay planes parallel to the base. How do the volumes of the resulting solids compare to each other?
|
1:7:19
| |
A card is chosen at random from a standard deck of 52 cards, and then it is replaced and another card is chosen. What is the probability that at least one of the cards is a heart, a spade, or a king?
|
\frac{133}{169}
| |
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
| |
Given a finite sequence $S=(a_1,a_2,\ldots ,a_n)$ of $n$ real numbers, let $A(S)$ be the sequence
$\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)$
of $n-1$ real numbers. Define $A^1(S)=A(S)$ and, for each integer $m$, $2\le m\le n-1$, define $A^m(S)=A(A^{m-1}(S))$. Suppose $x>0$, and let $S=(1,x,x^2,\ldots ,x^{100})$. If $A^{100}(S)=\left(\frac{1}{2^{50}}\right)$, then what is $x$?
|
\sqrt{2}-1
|
1. **Understanding the Sequence Transformation**:
Given a sequence $S = (a_1, a_2, \ldots, a_n)$, the sequence $A(S)$ is defined as:
\[
A(S) = \left(\frac{a_1 + a_2}{2}, \frac{a_2 + a_3}{2}, \ldots, \frac{a_{n-1} + a_n}{2}\right)
\]
This transformation reduces the length of the sequence by 1 each time it is applied.
2. **Recursive Application of $A$**:
The sequence $A^m(S)$ is defined recursively as $A(A^{m-1}(S))$. We need to understand the form of $A^m(S)$ after multiple applications.
3. **Pattern in Coefficients**:
From the problem statement, we know:
\[
A^2(S) = \left(\frac{a_1 + 2a_2 + a_3}{4}, \frac{a_2 + 2a_3 + a_4}{4}, \ldots, \frac{a_{n-2} + 2a_{n-1} + a_n}{4}\right)
\]
The coefficients of $a_i$ in $A^2(S)$ resemble binomial coefficients scaled by $2^2$. This pattern continues such that the coefficients in $A^m(S)$ are binomial coefficients $\binom{m}{k}$ scaled by $2^m$.
4. **Applying to the Given Sequence $S$**:
For $S = (1, x, x^2, \ldots, x^{100})$, the sequence $A^{100}(S)$ simplifies to:
\[
A^{100}(S) = \left(\frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m} x^m\right)
\]
This is because each term $x^k$ in $S$ contributes to the sum in $A^{100}(S)$ according to the binomial theorem.
5. **Equating to Given Value**:
We know $A^{100}(S) = \left(\frac{1}{2^{50}}\right)$. Therefore:
\[
\frac{1}{2^{100}} (1 + x)^{100} = \frac{1}{2^{50}}
\]
Simplifying, we get:
\[
(1 + x)^{100} = 2^{50}
\]
Taking the 100th root of both sides:
\[
1 + x = 2^{1/2} = \sqrt{2}
\]
Solving for $x$:
\[
x = \sqrt{2} - 1
\]
6. **Conclusion**:
The value of $x$ that satisfies the given conditions is $\boxed{\sqrt{2} - 1}$, which corresponds to choice $\mathrm{(B)}$.
|
Given a fixed circle \\(F_{1}:(x+2)^{2}+y^{2}=24\\) and a moving circle \\(N\\) passing through point \\(F_{2}(2,0)\\) and tangent to circle \\(F_{1}\\), denote the locus of the center of circle \\(N\\) as \\(E\\).
\\((I)\\) Find the equation of the locus \\(E\\);
\\((II)\\) If a line \\(l\\) not coincident with the x-axis passes through point \\(F_{2}(2,0)\\) and intersects the locus \\(E\\) at points \\(A\\) and \\(B\\), is there a fixed point \\(M\\) on the x-axis such that \\(\overrightarrow{MA}^{2}+ \overrightarrow{MA}· \overrightarrow{AB}\\) is a constant? If it exists, find the coordinates of point \\(M\\) and the constant value; if not, explain why.
|
-\frac{5}{9}
| |
Right $\triangle PQR$ has sides $PQ = 5$, $QR = 12$, and $PR = 13$. Rectangle $ABCD$ is inscribed in $\triangle PQR$ such that $A$ and $B$ are on $\overline{PR}$, $D$ on $\overline{PQ}$, and $C$ on $\overline{QR}$. If the height of the rectangle (parallel to side $\overline{PQ}$) is half its length (parallel to side $\overline{PR}$), find the length of the rectangle.
|
7.5
| |
If \( m \) and \( n \) are the roots of the quadratic equation \( x^2 + 1994x + 7 = 0 \), then the value of \((m^2 + 1993m + 6)(n^2 + 1995n + 8)\) is
|
1986
| |
Convert $3206_7$ to a base 10 integer.
|
1133
| |
A square piece of paper has sides of length $100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upped edges, can be written in the form $\sqrt[n]{m}$, where $m$ and $n$ are positive integers, $m<1000$, and $m$ is not divisible by the $n$th power of any prime. Find $m+n$.
[asy]import cse5; size(200); pathpen=black; real s=sqrt(17); real r=(sqrt(51)+s)/sqrt(2); D((0,2*s)--(0,0)--(2*s,0)); D((0,s)--r*dir(45)--(s,0)); D((0,0)--r*dir(45)); D((r*dir(45).x,2*s)--r*dir(45)--(2*s,r*dir(45).y)); MP("30^\circ",r*dir(45)-(0.25,1),SW); MP("30^\circ",r*dir(45)-(1,0.5),SW); MP("\sqrt{17}",(0,s/2),W); MP("\sqrt{17}",(s/2,0),S); MP("\mathrm{cut}",((0,s)+r*dir(45))/2,N); MP("\mathrm{cut}",((s,0)+r*dir(45))/2,E); MP("\mathrm{fold}",(r*dir(45).x,s+r/2*dir(45).y),E); MP("\mathrm{fold}",(s+r/2*dir(45).x,r*dir(45).y));[/asy]
|
871
|
In the final pyramid, let $ABCD$ be the smaller square and let $A'B'C'D'$ be the larger square such that $AA'$, etc. are edges.
It is obvious from the diagram that $\angle A'AB = \angle A'AD = 105^\circ$.
Let $AB$ and $AD$ be the positive $x$ and $y$ axes in a 3-d coordinate system such that $A'$ has a positive $z$ coordinate. Let $\alpha$ be the angle made with the positive $x$ axis. Define $\beta$ and $\gamma$ analogously.
It is easy to see that if $P: = (x,y,z)$, then $x = AA'\cdot \cos\alpha$. Furthermore, this means that $\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.
We have that $\alpha = \beta = 105^\circ$, so $\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt [4]{\frac {3}{4}}$.
It is easy to see from the Law of Sines that $\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}$.
Now, $z = AA'\cdot \cos\gamma = \sqrt [4]{34^2\cdot \frac {3}{4}} = \sqrt [4]{867}$.
It follows that the answer is $867 + 4 = \boxed{871}$.
|
Let the set \( A = \{1, 2, 3, 4, 5, 6\} \) and the mapping \( f: A \rightarrow A \). If the triple composition \( f \cdot f \cdot f \) is an identity mapping, how many such functions \( f \) are there?
|
81
| |
$ S$ is a non-empty subset of the set $ \{ 1, 2, \cdots, 108 \}$, satisfying:
(1) For any two numbers $ a,b \in S$ ( may not distinct), there exists $ c \in S$, such that $ \gcd(a,c)\equal{}\gcd(b,c)\equal{}1$.
(2) For any two numbers $ a,b \in S$ ( may not distinct), there exists $ c' \in S$, $ c' \neq a$, $ c' \neq b$, such that $ \gcd(a, c') > 1$, $ \gcd(b,c') >1$.
Find the largest possible value of $ |S|$.
|
79
|
Let \( S \) be a non-empty subset of the set \( \{ 1, 2, \ldots, 108 \} \) satisfying the following conditions:
1. For any two numbers \( a, b \in S \) (not necessarily distinct), there exists \( c \in S \) such that \( \gcd(a, c) = \gcd(b, c) = 1 \).
2. For any two numbers \( a, b \in S \) (not necessarily distinct), there exists \( c' \in S \), \( c' \neq a \), \( c' \neq b \), such that \( \gcd(a, c') > 1 \) and \( \gcd(b, c') > 1 \).
We claim that the largest possible value of \( |S| \) is \( \boxed{79} \).
To construct such a set \( S \), consider the set \( T \) of positive integers less than or equal to 108 which have either 1 or 2 prime divisors among the set \( \{2, 3, 5, 7, 11\} \). By removing the elements 55 and 77 from \( T \) and adding the elements 30, 60, 90, 42, and 84, we obtain a set of 79 elements that satisfies the given conditions.
To show that this is optimal, we proceed with the following lemmas and cases:
**Lemma 1:** There are at most 2 primes in \( S \) which are greater than 7.
- **Proof:** Suppose primes \( p_1, p_2 > 7 \) were both in \( S \). Applying the second condition on them leads to a contradiction.
**Lemma 2:** \( 1 \notin S \).
- **Proof:** Applying the second condition on \( a = b = 1 \) leads to a contradiction.
Using Lemmas 1 and 2, we can bound \( |S| \) by 84. We now consider two main cases:
**Case 1:** There is no prime \( p > 7 \) in \( S \).
- Among the pairs \( (6, 35), (10, 21), (14, 15), (2, 105), (3, 70), (5, 42), (7, 30) \), at least one number in each pair must not be in \( S \). This reduces the upper bound from 84 to 77, which is less than 79.
**Case 2:** There is a prime \( p > 7 \) in \( S \).
- We examine subcases where one of \( 2, 3 \) is not in \( S \). If \( 2 \notin S \), then either one of \( 4, 8, 16, 32, 64 \) is in \( S \) or \( |S| \leq 79 \). If \( 3 \notin S \), then either one of \( 9, 27, 81 \) is in \( S \) or \( |S| \leq 79 \). By similar logic, we can assume \( 2, 3 \in S \).
- We further consider subcases where \( 2, 3 \) are in \( S \) but \( 5, 7 \) may or may not be in \( S \). Each subcase analysis shows that the upper bound is reduced to 79.
Thus, by exhausting all cases and subcases, we have shown that the maximum size of \( S \) is 79.
The answer is \(\boxed{79}\).
|
A frustum of a right circular cone is formed by cutting a small cone off of the top of a larger cone. If a particular frustum has a lower base radius of 6 inches, an upper base radius of 3 inches, and a height of 4 inches, what is its lateral surface area? (The lateral surface area of a cone or frustum is the curved surface excluding the base(s).)
[asy]size(200);
import three; defaultpen(linewidth(.8)); currentprojection = orthographic(0,-3,0.5); pen dots = linetype("0 3") + linewidth(1);
real h = 2.3, ratio = (91-24)/(171-24);
picture p1, p2; /* p1 is left-hand picture */
triple A = (0,0,0), B = (0,0,h); draw(p1,(-1,0,0)..(0,-1,0)..(1,0,0)); draw(p1,(-1,0,0)..(0,1,0)..(1,0,0),dots); draw(p1,(-1,0,0)--B--(1,0,0));
add(p1);
triple vlift = (0,0,0.5);
path3 toparc1 = shift((0,0,h*(1-ratio)))*scale3(ratio)*((-1,0,0)..(0,1,0)..(1,0,0)), toparc2 = shift((0,0,h*(1-ratio)))*scale3(ratio)*((1,0,0)..(0,-1,0)..(-1,0,0));
draw(p2,(-1,0,0)..(0,-1,0)..(1,0,0)); draw(p2,(-1,0,0)..(0,1,0)..(1,0,0),dots);
draw(p2,(-1,0,0)--ratio*(-1,0,0)+(1-ratio)*B^^ratio*(1,0,0)+(1-ratio)*B--(1,0,0));
draw(p2,shift(vlift)*(ratio*(-1,0,0)+(1-ratio)*B--B--ratio*(1,0,0)+(1-ratio)*B));
draw(p2,toparc1--toparc2); draw(p2,shift(vlift)*toparc1,dots); draw(p2,shift(vlift)*toparc2);
draw(p2,shift(vlift)*((1-ratio)*B--B),linewidth(0.7)); dot(p2,shift(vlift)*((1-ratio)*B),linewidth(1.5));
label(p2,"frustum",(0,0,h/4));
add(shift((3.4,0,0))*p2);
[/asy]
|
45\pi
| |
Given the function $f(x)=x^{2}\cos \frac {πx}{2}$, the sequence {a<sub>n</sub>} is defined as a<sub>n</sub> = f(n) + f(n+1) (n ∈ N*), find the sum of the first 40 terms of the sequence {a<sub>n</sub>}, denoted as S<sub>40</sub>.
|
1680
| |
Let $\alpha, \beta$, and $\gamma$ be three real numbers. Suppose that $\cos \alpha+\cos \beta+\cos \gamma =1$ and $\sin \alpha+\sin \beta+\sin \gamma =1$. Find the smallest possible value of $\cos \alpha$.
|
\frac{-1-\sqrt{7}}{4}
|
Let $a=\cos \alpha+i \sin \alpha, b=\cos \beta+i \sin \beta$, and $c=\cos \gamma+i \sin \gamma$. We then have $a+b+c=1+i$ where $a, b, c$ are complex numbers on the unit circle. Now, to minimize $\cos \alpha=\operatorname{Re}[a]$, consider a triangle with vertices $a, 1+i$, and the origin. We want $a$ as far away from $1+i$ as possible while maintaining a nonnegative imaginary part. This is achieved when $b$ and $c$ have the same argument, so $|b+c|=|1+i-a|=2$. Now $a, 0$, and $1+i$ form a $1-2-\sqrt{2}$ triangle. The value of $\cos \alpha$ is now the cosine of the angle between the 1 and $\sqrt{2}$ sides plus the $\frac{\pi}{4}$ angle from $1+i$. Call the first angle $\delta$. Then $\cos \delta =\frac{1^{2}+(\sqrt{2})^{2}-2^{2}}{2 \cdot 1 \cdot \sqrt{2}} =\frac{-1}{2 \sqrt{2}}$ and $\cos \alpha =\cos \left(\frac{\pi}{4}+\delta\right) =\cos \frac{\pi}{4} \cos \delta-\sin \frac{\pi}{4} \sin \delta =\frac{\sqrt{2}}{2} \cdot \frac{-1}{2 \sqrt{2}}-\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{7}}{2 \sqrt{2}} =\frac{-1-\sqrt{7}}{4}$.
|
What is the smallest positive integer that can be written in the form $2002m + 44444n$, where $m$ and $n$ are integers?
|
2
| |
Maria is 54 inches tall, and Samuel is 72 inches tall. Using the conversion 1 inch = 2.54 cm, how tall is each person in centimeters? Additionally, what is the difference in their heights in centimeters?
|
45.72
| |
Find all the real solutions to
\[\frac{(x - 1)(x - 2)(x - 3)(x - 4)(x - 3)(x - 2)(x - 1)}{(x - 2)(x - 4)(x - 2)} = 1.\]Enter all the solutions, separated by commas.
|
2 + \sqrt{2}, 2 - \sqrt{2}
| |
Find the root of the following equation to three significant digits:
$$
(\sqrt{5}-\sqrt{2})(1+x)=(\sqrt{6}-\sqrt{3})(1-x)
$$
|
-0.068
| |
Given that the scores X of 10000 people approximately follow a normal distribution N(100,13^2), it is given that P(61 < X < 139)=0.997, find the number of people scoring no less than 139 points in this exam.
|
15
| |
In the Cartesian coordinate system $xOy$, the parametric equation of curve $C$ is $$\begin{cases} x=3\cos \theta \\ y=2\sin \theta \end{cases} (\theta \text{ is the parameter}),$$ and the parametric equation of the line $l$ is $$\begin{cases} x=t-1 \\ y=2t-a-1 \end{cases} (t \text{ is the parameter}).$$
(Ⅰ) If $a=1$, find the length of the line segment cut off by line $l$ from curve $C$.
(Ⅱ) If $a=11$, find a point $M$ on curve $C$ such that the distance from $M$ to line $l$ is minimal, and calculate the minimum distance.
|
2\sqrt{5}-2\sqrt{2}
| |
Determine the slope \(m\) of the asymptotes for the hyperbola given by the equation
\[
\frac{y^2}{16} - \frac{x^2}{9} = 1.
\]
|
\frac{4}{3}
| |
Compute the value of $\left(81\right)^{0.25} \cdot \left(81\right)^{0.2}$.
|
3 \cdot \sqrt[5]{3^4}
| |
Let $m$ and $n$ be positive integers with $m\le 2000$ and $k=3-\frac{m}{n}$. Find the smallest positive value of $k$.
|
$\boxed{ \frac{1}{667}} .$
|
Given the problem with positive integers \( m \) and \( n \) such that \( m \leq 2000 \), and \( k = 3 - \frac{m}{n} \). We are tasked to find the smallest positive value of \( k \).
Firstly, to ensure \( k \) is positive, we need:
\[
k = 3 - \frac{m}{n} > 0,
\]
which implies:
\[
3 > \frac{m}{n}.
\]
Rearranging gives:
\[
n > \frac{m}{3}.
\]
Since \( m \) and \( n \) are integers, \( n \) must be greater than \(\frac{m}{3}\), so we set:
\[
n \geq \left\lceil \frac{m}{3} \right\rceil.
\]
The expression for \( k \) becomes:
\[
k = \frac{3n - m}{n}.
\]
To minimize the positive value of \( k \), we need the smallest possible \( n \) that satisfies the condition. Setting the equality \( n = \left\lceil \frac{m}{3} \right\rceil \) leads to:
\[
n = \left\lceil \frac{m}{3} \right\rceil.
\]
Substitute \( n = \left\lceil \frac{m}{3} \right\rceil \) into the expression for \( k \):
\[
k = \frac{3\left\lceil \frac{m}{3} \right\rceil - m}{\left\lceil \frac{m}{3} \right\rceil}.
\]
To find the smallest positive \( k \), consider the smallest value for which this fraction can exist. For \(\left\lceil \frac{m}{3} \right\rceil\) to be as close as possible to \(\frac{m}{3}\), let \( m = 3t + 1 \) or \( m = 3t + 2 \) for the smallest integer change.
By setting \( m = 2000 \), we have:
\[
n = \left\lceil \frac{2000}{3} \right\rceil = 667.
\]
This gives:
\[
k = \frac{3 \times 667 - 2000}{667} = \frac{2001 - 2000}{667} = \frac{1}{667}.
\]
Thus, the smallest positive value of \( k \) is:
\[
\boxed{\frac{1}{667}}.
\]
|
How many ways can you mark 8 squares of an $8 \times 8$ chessboard so that no two marked squares are in the same row or column, and none of the four corner squares is marked? (Rotations and reflections are considered different.)
|
21600
|
In the top row, you can mark any of the 6 squares that is not a corner. In the bottom row, you can then mark any of the 5 squares that is not a corner and not in the same column as the square just marked. Then, in the second row, you have 6 choices for a square not in the same column as either of the two squares already marked; then there are 5 choices remaining for the third row, and so on down to 1 for the seventh row, in which you make the last mark. Thus, altogether, there are $6 \cdot 5 \cdot(6 \cdot 5 \cdots 1)=30 \cdot 6!=30 \cdot 720=21600$ possible sets of squares.
|
Let \(x\) and \(y\) be real numbers such that \(2(x^3 + y^3) = x + y\). Find the maximum value of \(x - y\).
|
\frac{\sqrt{2}}{2}
| |
Given $0 \leq x_0 < 1$, for all integers $n > 0$, let
$$
x_n = \begin{cases}
2x_{n-1}, & \text{if } 2x_{n-1} < 1,\\
2x_{n-1} - 1, & \text{if } 2x_{n-1} \geq 1.
\end{cases}
$$
Find the number of initial values of $x_0$ such that $x_0 = x_6$.
|
64
| |
Given the function $f(x)=2\sin \omega x\cos \omega x-2\sqrt{3}\cos^{2}\omega x+\sqrt{3}$ ($\omega > 0$), and the distance between two adjacent axes of symmetry of the graph of $y=f(x)$ is $\frac{\pi}{2}$.
(Ⅰ) Find the interval of monotonic increase for the function $f(x)$;
(Ⅱ) Given that in $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, angle $C$ is acute, and $f(C)=\sqrt{3}$, $c=3\sqrt{2}$, $\sin B=2\sin A$, find the area of $\triangle ABC$.
|
3\sqrt{3}
| |
Given a parabola with vertex \( V \) and a focus \( F \), and points \( B \) and \( C \) on the parabola such that \( BF=25 \), \( BV=24 \), and \( CV=20 \), determine the sum of all possible values of the length \( FV \).
|
\frac{50}{3}
| |
Solve for $x$ if $\frac{2}{x+3} + \frac{3x}{x+3} - \frac{4}{x+3} = 4$.
|
-14
| |
Solve for $x>0$ in the following arithmetic sequence: $1^2, x^2, 3^2, \ldots$.
|
\sqrt{5}
| |
Evaluate \(\left(a^a - a(a-2)^a\right)^a\) when \( a = 4 \).
|
1358954496
| |
Let $A,$ $B,$ $C$ be the angles of a non-right triangle. Compute
\[\begin{vmatrix} \tan A & 1 & 1 \\ 1 & \tan B & 1 \\ 1 & 1 & \tan C \end{vmatrix}.\]
|
2
| |
Two distinct numbers $a$ and $b$ are chosen randomly from the set $\{2, 2^2, 2^3, ..., 2^{25}\}$. What is the probability that $\log_a b$ is an integer?
|
\frac{31}{300}
|
1. **Define the problem in terms of logarithms and powers of 2:**
Let $a = 2^x$ and $b = 2^y$ where $x$ and $y$ are integers such that $1 \leq x, y \leq 25$. We are given that $\log_a b$ is an integer, say $z$. Therefore, we have:
\[
\log_a b = z \implies a^z = b \implies (2^x)^z = 2^y \implies 2^{xz} = 2^y \implies xz = y.
\]
This implies that $x$ divides $y$ (i.e., $x|y$).
2. **Count the valid pairs $(x, y)$:**
For each $x$, we need to count the number of $y$ values such that $x$ divides $y$ and $y \neq x$ (since $a$ and $b$ must be distinct). The possible values of $y$ are $x, 2x, 3x, \ldots, kx$ where $kx \leq 25$. The largest $k$ for a given $x$ is $\left\lfloor \frac{25}{x} \right\rfloor$.
However, since $y \neq x$, we exclude $x$ itself, so the count of valid $y$ values for each $x$ is $\left\lfloor \frac{25}{x} \right\rfloor - 1$.
3. **Sum the counts for each $x$:**
\[
\sum_{x=1}^{25} \left(\left\lfloor \frac{25}{x} \right\rfloor - 1\right)
\]
We only need to consider $x$ up to 12 because for $x > 12$, $\left\lfloor \frac{25}{x} \right\rfloor$ would be 1 or 0, contributing nothing to the sum after subtracting 1. Calculating:
\[
\sum_{x=1}^{12} \left(\left\lfloor \frac{25}{x} \right\rfloor - 1\right) = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62.
\]
4. **Calculate the total number of ways to pick two distinct numbers from the set:**
The total number of ways to choose two distinct numbers from a set of 25 numbers is given by $\binom{25}{2} = \frac{25 \times 24}{2} = 300$.
5. **Compute the probability:**
The probability that $\log_a b$ is an integer is the ratio of the number of valid pairs to the total number of pairs:
\[
\frac{62}{300} = \frac{31}{150} = \boxed{\textbf{(B)}\ \frac{31}{300}}.
\]
|
The year 2009 has the property that rearranging its digits never results in a smaller four-digit number (numbers do not start with zero). In which year will this property first repeat?
|
2022
| |
Ten points are given in the plane, and no three points are collinear. Four distinct segments connecting pairs of these points are chosen at random, all with the same probability. What is the probability that three of the chosen segments will form a triangle?
|
16/473
| |
Rewrite $\sqrt[3]{2^6\cdot3^3\cdot11^3}$ as an integer.
|
132
| |
Find $\log _{n}\left(\frac{1}{2}\right) \log _{n-1}\left(\frac{1}{3}\right) \cdots \log _{2}\left(\frac{1}{n}\right)$ in terms of $n$.
|
(-1)^{n-1}
|
Using $\log \frac{1}{x}=-\log x$ and $\log _{b} a=\frac{\log a}{\log b}$, we get that the product equals $\frac{(-\log 2)(-\log 3) \cdots(-\log n)}{\log n \cdots \log 3 \log 2}=(-1)^{n-1}$.
|
A hexagon is obtained by joining, in order, the points $(0,1)$, $(1,2)$, $(2,2)$, $(2,1)$, $(3,1)$, $(2,0)$, and $(0,1)$. The perimeter of the hexagon can be written in the form $a+b\sqrt{2}+c\sqrt{5}$, where $a$, $b$ and $c$ are whole numbers. Find $a+b+c$.
|
6
| |
Let the set \( T = \{0, 1, \dots, 6\} \),
$$
M = \left\{\left.\frac{a_1}{7}+\frac{a_2}{7^2}+\frac{a_3}{7^3}+\frac{a_4}{7^4} \right\rvert\, a_i \in T, i=1,2,3,4\right\}.
$$
If the elements of the set \( M \) are arranged in decreasing order, what is the 2015th number?
|
\frac{386}{2401}
| |
Mrs. Thompson gives extra points on tests to her students with test grades that exceed the class median. Given that 101 students take the same test, what is the largest number of students who can be awarded extra points?
|
50
| |
Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids:
$\circ$ Roger's cookies are rectangles:
$\circ$ Paul's cookies are parallelograms:
$\circ$ Trisha's cookies are triangles:
How many cookies will be in one batch of Trisha's cookies?
|
24
|
To solve this problem, we need to determine the number of cookies Trisha can make from the same amount of dough as Art, given that all cookies have the same thickness and thus the same volume per unit area.
1. **Calculate the area of one of Art's cookies**:
Art's cookies are trapezoids. The formula for the area of a trapezoid is:
\[
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
Given that the bases are 3 inches and 5 inches (assuming the smaller base is 3 inches and the larger base is 5 inches), and the height is 3 inches, the area of one of Art's cookies is:
\[
\text{Area} = \frac{1}{2} \times (3 + 5) \times 3 = \frac{1}{2} \times 8 \times 3 = 12 \text{ in}^2
\]
Since there are 12 cookies in one of Art's batches, the total area of dough used by Art is:
\[
12 \times 12 = 144 \text{ in}^2
\]
2. **Calculate the area of one of Trisha's cookies**:
Trisha's cookies are triangles. The formula for the area of a triangle is:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
Given that the base is 3 inches and the height is 4 inches, the area of one of Trisha's cookies is:
\[
\text{Area} = \frac{1}{2} \times 3 \times 4 = 6 \text{ in}^2
\]
3. **Calculate the number of Trisha's cookies per batch**:
Using the same total area of dough as Art, the number of cookies Trisha can make is:
\[
\frac{144 \text{ in}^2}{6 \text{ in}^2} = 24
\]
Therefore, Trisha can make 24 cookies per batch.
Thus, the number of cookies in one batch of Trisha's cookies is $\boxed{\textbf{(E)}\ 24}$.
|
5 people are standing in a row for a photo, among them one person must stand in the middle. There are ways to arrange them.
|
24
| |
In the diagram below, how many distinct paths are there from January 1 to December 31, moving from one adjacent dot to the next either to the right, down, or diagonally down to the right?
|
372
|
For each dot in the diagram, we can count the number of paths from January 1 to it by adding the number of ways to get to the dots to the left of it, above it, and above and to the left of it, starting from the topmost leftmost dot. This yields the following numbers of paths: 372.
|
Find all the pairs of prime numbers $ (p,q)$ such that $ pq|5^p\plus{}5^q.$
|
(2, 3), (2, 5), (3, 2), (5, 2), (5, 5), (5, 313), (313, 5)
|
To find all pairs of prime numbers \((p, q)\) such that \(pq \mid 5^p + 5^q\), we consider several cases:
1. **Case 1: Either \(p\) or \(q\) is 2.**
- Assume \(p = 2\). Then \(2q \mid 5^2 + 5^q\).
- This simplifies to \(2q \mid 25 + 5^q\).
- Since \(25 \equiv 1 \pmod{2}\), we have \(5^q + 1 \equiv 0 \pmod{q}\), implying \(q \mid 30\).
- The prime divisors of 30 are 2, 3, and 5. Since \(q \neq 2\), the possible values for \(q\) are 3 and 5.
- Thus, the pairs \((p, q)\) are \((2, 3)\) and \((2, 5)\).
2. **Case 2: \(p = q = 5\).**
- Clearly, \(5^2 \mid 5^5 + 5^5\), so \((5, 5)\) is a solution.
3. **Case 3: \(p = 5\) and \(q \neq 5\).**
- Then \(5q \mid 5^5 + 5^q\).
- This simplifies to \(5q \mid 3125 + 5^q\).
- Since \(3125 \equiv 0 \pmod{5}\), we have \(5^q + 3125 \equiv 0 \pmod{q}\), implying \(q \mid 3126\).
- The prime divisors of 3126 are 2 and 313. Thus, the pairs \((p, q)\) are \((5, 2)\) and \((5, 313)\).
4. **Case 4: \(p \neq 5\) and \(q \neq 5\).**
- Assume \(p > q\). Then \(pq \mid 5^p + 5^q\).
- This implies \(5^q(5^{p-q} + 1) \equiv 0 \pmod{pq}\).
- Therefore, \(5^{p-q} \equiv -1 \pmod{q}\) and \(\text{ord}_q(5) \mid 2(p-q)\).
- Since \(\text{ord}_q(5) \nmid p-q\), we get a contradiction in the valuation of 2, leading to no further solutions.
Combining all cases, the pairs of prime numbers \((p, q)\) that satisfy the condition are:
\[
\boxed{(2, 3), (2, 5), (3, 2), (5, 2), (5, 5), (5, 313), (313, 5)}
\]
|
$f : \mathbb{Z} \rightarrow \mathbb{Z}$ satisfies $m+f(m+f(n+f(m))) = n + f(m)$ for every integers $m,n$. Given that $f(6) = 6$, determine $f(2012)$.
|
-2000
| |
Consider a sequence $F_0=2$ , $F_1=3$ that has the property $F_{n+1}F_{n-1}-F_n^2=(-1)^n\cdot2$ . If each term of the sequence can be written in the form $a\cdot r_1^n+b\cdot r_2^n$ , what is the positive difference between $r_1$ and $r_2$ ?
|
\frac{\sqrt{17}}{2}
| |
Given complex numbers $z_1 = \cos\theta - i$ and $z_2 = \sin\theta + i$, the maximum value of the real part of $z_1 \cdot z_2$ is \_\_\_\_\_\_, and the maximum value of the imaginary part is \_\_\_\_\_\_.
|
\sqrt{2}
| |
Suppose that $x$ and $y$ are real numbers that satisfy the two equations: $x^{2} + 3xy + y^{2} = 909$ and $3x^{2} + xy + 3y^{2} = 1287$. What is a possible value for $x+y$?
|
27
|
Since $x^{2} + 3xy + y^{2} = 909$ and $3x^{2} + xy + 3y^{2} = 1287$, then adding these gives $4x^{2} + 4xy + 4y^{2} = 2196$. Dividing by 4 gives $x^{2} + xy + y^{2} = 549$. Subtracting this from the first equation gives $2xy = 360$, so $xy = 180$. Substituting $xy = 180$ into the first equation gives $x^{2} + 2xy + y^{2} = 729$, which is $(x+y)^{2} = 27^{2}$. Therefore, $x+y = 27$ or $x+y = -27$. This also shows that $x+y$ cannot equal any of 39, 29, 92, and 41.
|
The reciprocal of the opposite number of \(-(-3)\) is \(\frac{1}{3}\).
|
-\frac{1}{3}
| |
Find the probability that the chord $\overline{AB}$ does not intersect with chord $\overline{CD}$ when four distinct points, $A$, $B$, $C$, and $D$, are selected from 2000 points evenly spaced around a circle.
|
\frac{2}{3}
| |
There are lily pads in a row numbered $0$ to $11$, in that order. There are predators on lily pads $3$ and $6$, and a morsel of food on lily pad $10$. Fiona the frog starts on pad $0$, and from any given lily pad, has a $\frac{1}{2}$ chance to hop to the next pad, and an equal chance to jump $2$ pads. What is the probability that Fiona reaches pad $10$ without landing on either pad $3$ or pad $6$?
|
\frac{15}{256}
|
1. **Define the problem**: Fiona the frog starts at lily pad $0$ and can hop to the next pad or jump two pads forward with equal probability of $\frac{1}{2}$. She must reach pad $10$ without landing on pads $3$ or $6$.
2. **Calculate probability to reach pad $2$**:
- Fiona can reach pad $2$ by either:
- Hopping to pad $1$ and then to pad $2$: Probability = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
- Jumping directly to pad $2$: Probability = $\frac{1}{2}$
- Total probability to reach pad $2$ = $\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$
3. **Calculate probability to reach pad $4$ from pad $2$**:
- Fiona must jump from pad $2$ to pad $4$ (since hopping to pad $3$ is not allowed): Probability = $\frac{1}{2}$
- Total probability to reach pad $4$ = $\frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$
4. **Calculate probability to reach pad $5$ from pad $4$**:
- Fiona must hop from pad $4$ to pad $5$: Probability = $\frac{1}{2}$
- Total probability to reach pad $5$ = $\frac{3}{8} \times \frac{1}{2} = \frac{3}{16}$
5. **Calculate probability to reach pad $7$ from pad $5$**:
- Fiona must jump from pad $5$ to pad $7$ (since hopping to pad $6$ is not allowed): Probability = $\frac{1}{2}$
- Total probability to reach pad $7$ = $\frac{3}{16} \times \frac{1}{2} = \frac{3}{32}$
6. **Calculate probability to reach pad $10$ from pad $7$**:
- Fiona has several options to reach pad $10$:
- Three consecutive $1$-jumps: Probability = $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
- One $1$-jump to pad $8$, then a $2$-jump to pad $10$: Probability = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
- One $2$-jump to pad $9$, then a $1$-jump to pad $10$: Probability = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
- Total probability to reach pad $10$ = $\frac{1}{8} + \frac{1}{4} + \frac{1}{4} = \frac{5}{8}$
7. **Final probability calculation**:
- Multiply the probability of reaching pad $7$ by the probability of reaching pad $10$ from pad $7$:
- $\frac{3}{32} \times \frac{5}{8} = \frac{15}{256}$
$\boxed{\textbf{(A) } \frac{15}{256}}$
|
Let $f(x)=x^{2}+a x+b$ and $g(x)=x^{2}+c x+d$ be two distinct real polynomials such that the $x$-coordinate of the vertex of $f$ is a root of $g$, the $x$-coordinate of the vertex of $g$ is a root of $f$ and both $f$ and $g$ have the same minimum value. If the graphs of the two polynomials intersect at the point (2012, - 2012), what is the value of $a+c$ ?
|
-8048
|
It is clear, by symmetry, that 2012 is the equidistant from the vertices of the two quadratics. Then it is clear that reflecting $f$ about the line $x=2012$ yields $g$ and vice versa. Thus the average of each pair of roots is 2012 . Thus the sum of the four roots of $f$ and $g$ is 8048 , so $a+c=-8048$.
|
Consider sequences of positive real numbers of the form $x, 2000, y, \dots$ in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of $x$ does the term 2001 appear somewhere in the sequence?
|
4
| |
For any positive real numbers \(a\) and \(b\), define \(a \circ b=a+b+2 \sqrt{a b}\). Find all positive real numbers \(x\) such that \(x^{2} \circ 9x=121\).
|
\frac{31-3\sqrt{53}}{2}
|
Since \(a \circ b=(\sqrt{a}+\sqrt{b})^{2}\), we have \(x^{2} \circ 9x=(x+3\sqrt{x})^{2}\). Moreover, since \(x\) is positive, we have \(x+3\sqrt{x}=11\), and the only possible solution is that \(\sqrt{x}=\frac{-3+\sqrt{53}}{2}\), so \(x=\frac{31-3\sqrt{53}}{2}\).
|
Given that a rectangular room is 15 feet long and 108 inches wide, calculate the area of the new extended room after adding a 3 feet wide walkway along the entire length of one side, in square yards, where 1 yard equals 3 feet and 1 foot equals 12 inches.
|
20
| |
How many ways are there to arrange the numbers $1,2,3,4,5,6$ on the vertices of a regular hexagon such that exactly 3 of the numbers are larger than both of their neighbors? Rotations and reflections are considered the same.
|
8
|
Label the vertices of the hexagon $a b c d e f$. The numbers that are larger than both of their neighbors can't be adjacent, so assume (by rotation) that these numbers take up slots ace. We also have that 6 and 5 cannot be smaller than both of their neighbors, so assume (by rotation and reflection) that $a=6$ and $c=5$. Now, we need to insert $1,2,3,4$ into $b, d, e, f$ such that $e$ is the largest among $d, e, f$. There are 4 ways to choose $b$, which uniquely determines $e$, and 2 ways to choose the ordering of $d, f$, giving $4 \cdot 2=8$ total ways.
|
Given an ellipse $C:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1(a>b>0)$ with foci $F_{1}$ and $F_{2}$, point $A$ lies on $C$, point $B$ lies on the $y$-axis, and satisfies $\overrightarrow{A{F}_{1}}⊥\overrightarrow{B{F}_{1}}$, $\overrightarrow{A{F}_{2}}=\frac{2}{3}\overrightarrow{{F}_{2}B}$. What is the eccentricity of $C$?
|
\frac{\sqrt{5}}{5}
| |
How many four-digit numbers starting with the digit $2$ and having exactly three identical digits are there?
|
27
| |
In rectangle $ABCD$, $DC = 2 \cdot CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$?
|
\frac{3\sqrt{3}}{16}
|
1. **Assigning Lengths**: Let the length of $AD$ be $x$. Since $DC = 2 \cdot CB$ and $ABCD$ is a rectangle, $AB = 2x$ and $BC = x$. Therefore, the area of rectangle $ABCD$ is $AB \times BC = 2x \times x = 2x^2$.
2. **Angle Analysis**: In rectangle $ABCD$, $\angle ADC = 90^\circ$. Since $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$, each smaller angle is $\frac{90^\circ}{3} = 30^\circ$. Thus, $\angle ADE = \angle EDF = \angle FDC = 30^\circ$.
3. **Triangle Properties**:
- $\triangle ADE$ is a $30-60-90$ triangle with $\angle ADE = 30^\circ$ and $\angle DAE = 60^\circ$.
- $\angle DEF = 180^\circ - \angle EDF - \angle FDC = 180^\circ - 30^\circ - 30^\circ = 120^\circ$.
- $\angle EFD = 180^\circ - \angle DEF - \angle FDC = 180^\circ - 120^\circ - 30^\circ = 30^\circ$.
4. **Calculating Side Lengths**:
- In $\triangle ADE$, using the properties of $30-60-90$ triangles, the length of $AE$ (opposite the $30^\circ$ angle) is $\frac{x}{2}$, and the length of $DE$ (opposite the $60^\circ$ angle) is $\frac{x\sqrt{3}}{2}$.
5. **Area of $\triangle DEF$**:
- Drop an altitude from $E$ to $DF$ at point $G$. Since $\angle EFD = 30^\circ$, $\triangle EFG$ is also a $30-60-90$ triangle.
- The length of $EG$ (opposite the $30^\circ$ angle) is $\frac{1}{2} \times \text{base} = \frac{1}{2} \times \frac{x\sqrt{3}}{2} = \frac{x\sqrt{3}}{4}$.
- The length of $DF$ (hypotenuse) is $x$.
- The area of $\triangle DEF$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times \frac{x\sqrt{3}}{4} = \frac{x^2\sqrt{3}}{8}$.
6. **Ratio of Areas**:
- The ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$ is $\frac{\frac{x^2\sqrt{3}}{8}}{2x^2} = \frac{x^2\sqrt{3}}{8} \div 2x^2 = \frac{\sqrt{3}}{16}$.
7. **Conclusion**:
- The correct answer is $\boxed{\textbf{(C) }\frac{3\sqrt{3}}{16}}$.
|
Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$.
|
295
|
This solution refers to the Remarks section.
By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ \gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{12mm} &&\text{by }\textbf{Claim 1} \\ \gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ 2^{\gcd(2m,n)}-1&>2^{\gcd(m,n)}-1 &&\text{by }\textbf{Claim 2} \\ \gcd(2m,n)&>\gcd(m,n), \end{align*} which implies that $n$ must have more factors of $2$ than $m$ does.
We construct the following table for the first $30$ positive integers: \[\begin{array}{c|c|c} && \\ [-2.5ex] \boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\ \hline && \\ [-2.25ex] 0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\ && \\ [-2.25ex] 1 & 2,6,10,14,18,22,26,30 & 8 \\ && \\ [-2.25ex] 2 & 4,12,20,28 & 4 \\ && \\ [-2.25ex] 3 & 8,24 & 2 \\ && \\ [-2.25ex] 4 & 16 & 1 \\ \end{array}\] To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has:
If $m$ has $0$ factors of $2,$ then $m$ has $15$ options and $n$ has $8+4+2+1=15$ options. So, this case has $15\cdot15=225$ ordered pairs.
If $m$ has $1$ factor of $2,$ then $m$ has $8$ options and $n$ has $4+2+1=7$ options. So, this case has $8\cdot7=56$ ordered pairs.
If $m$ has $2$ factors of $2,$ then $m$ has $4$ options and $n$ has $2+1=3$ options. So, this case has $4\cdot3=12$ ordered pairs.
If $m$ has $3$ factors of $2,$ then $m$ has $2$ options and $n$ has $1$ option. So, this case has $2\cdot1=2$ ordered pairs.
Together, the answer is $225+56+12+2=\boxed{295}.$
~Lcz ~MRENTHUSIASM
Remarks
Claim 1 (GCD Property)
If $\boldsymbol{r,s,}$ and $\boldsymbol{t}$ are positive integers such that $\boldsymbol{\gcd(r,s)=1,}$ then $\boldsymbol{\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).}$
As $r$ and $s$ are relatively prime (have no prime divisors in common), this property is intuitive.
~MRENTHUSIASM
Claim 2 (Olympiad Number Theory Lemma)
If $\boldsymbol{u,a,}$ and $\boldsymbol{b}$ are positive integers such that $\boldsymbol{u\geq2,}$ then $\boldsymbol{\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.}$
There are two proofs to this claim, as shown below.
~MRENTHUSIASM
Claim 2 Proof 1 (Euclidean Algorithm)
If $a=b,$ then $\gcd(a,b)=a=b,$ from which the claim is clearly true.
Otherwise, let $a>b$ without the loss of generality. For all integers $p$ and $q$ such that $p>q>0,$ the Euclidean Algorithm states that \[\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(p\operatorname{mod}q,q).\] We apply this result repeatedly to reduce the larger number: \[\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^a-1-u^{a-b}\left(u^b-1\right),u^b-1\right)=\gcd\left(u^{a-b}-1,u^b-1\right).\] Continuing, we have \begin{align*} \gcd\left(u^a-1,u^b-1\right)&=\gcd\left(u^{a-b}-1,u^b-1\right) \\ & \ \vdots \\ &=\gcd\left(u^{\gcd(a,b)}-1,u^{\gcd(a,b)}-1\right) \\ &=u^{\gcd(a,b)}-1, \end{align*} from which the proof is complete.
~MRENTHUSIASM
Claim 2 Proof 2 (Bézout's Identity)
Let $d=\gcd\left(u^a-1,u^b-1\right).$ It follows that $u^a\equiv1\pmod{d}$ and $u^b\equiv1\pmod{d}.$
By Bézout's Identity, there exist integers $x$ and $y$ such that $ax+by=\gcd(a,b),$ so \[u^{\gcd(a,b)}=u^{ax+by}=(u^a)^x\cdot(u^b)^y\equiv1\pmod{d},\] from which $u^{\gcd(a,b)}-1\equiv0\pmod{d}.$ We know that $u^{\gcd(a,b)}-1\geq d.$
Next, we notice that \begin{align*} u^a-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{a-\gcd{(a,b)}}+u^{a-2\gcd{(a,b)}}+u^{a-3\gcd{(a,b)}}+\cdots+1\right), \\ u^b-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{b-\gcd{(a,b)}}+u^{b-2\gcd{(a,b)}}+u^{b-3\gcd{(a,b)}}+\cdots+1\right). \end{align*} Since $u^{\gcd(a,b)}-1$ is a common divisor of $u^a-1$ and $u^b-1,$ we conclude that $u^{\gcd(a,b)}-1=d,$ from which the proof is complete.
~MRENTHUSIASM
~MathProblemSolvingSkills.com
~Interstigation
|
Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$
|
578
|
We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$
Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$, $MN=AM-AN$, and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$
From the third equation, we get $CN=\frac{168} {25}.$
By the Pythagorean Theorem in $\Delta ACN,$ we have
$AN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.$
Thus, $MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.$
In $\Delta ADM$, we use the Pythagorean Theorem to get $DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.$
Thus, $[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.$
Hence, the answer is $527+11+40=\boxed{578}.$
|
Real numbers \(a, b, c\) and a positive number \(\lambda\) such that \(f(x)=x^3 + ax^2 + bx + c\) has three real roots \(x_1, x_2, x_3\), satisfying
1. \(x_2 - x_1 = \lambda\);
2. \(x_3 > \frac{1}{2}(x_1 + x_2)\).
Find the maximum value of \(\frac{2a^3 + 27c - 9ab}{\lambda^3}\).
|
\frac{3 \sqrt{3}}{2}
| |
Let \( a, \) \( b, \) \( c \) be positive real numbers such that
\[
\left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right) + \left( \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right) = 9.
\]
Find the minimum value of
\[
\left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right) \left( \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right).
\]
|
57
| |
Given that \( f(x) \) is an odd function defined on \( \mathbf{R} \), and for any \( x \in \mathbf{R} \), the following holds:
$$
f(2+x) + f(2-x) = 0.
$$
When \( x \in [-1, 0) \), it is given that
$$
f(x) = \log_{2}(1-x).
$$
Find \( f(1) + f(2) + \cdots + f(2021) \).
|
-1
| |
The circle centered at $(2,-1)$ and with radius $4$ intersects the circle centered at $(2,5)$ and with radius $\sqrt{10}$ at two points $A$ and $B$. Find $(AB)^2$.
|
15
| |
In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$, where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$
|
450
|
We will solve for $\cos B$ using $\triangle CBD$, which gives us $\cos B = \frac{29^3}{BC}$. By the Pythagorean Theorem on $\triangle CBD$, we have $BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6$. Trying out factors of $29^6$, we can either guess and check or just guess to find that $BC + DC = 29^4$ and $BC - DC = 29^2$ (The other pairs give answers over 999). Adding these, we have $2BC = 29^4 + 29^2$ and $\frac{29^3}{BC} = \frac{2*29^3}{29^2 (29^2 +1)} = \frac{58}{842} = \frac{29}{421}$, and our answer is $\boxed{450}$.
|
For how many integer values of $a$ does the equation $$x^2 + ax + 12a = 0$$ have integer solutions for $x$?
|
16
| |
Let $f : N \to N$ be a strictly increasing function such that $f(f(n))= 3n$ , for all $n \in N$ . Find $f(2010)$ .
Note: $N = \{0,1,2,...\}$
|
3015
| |
The spinner shown is divided into 6 sections of equal size. Determine the probability of landing on a section that contains the letter Q using this spinner.
|
\frac{2}{6}
| |
Compute
$3(1+3(1+3(1+3(1+3(1+3(1+3(1+3(1+3(1+3)))))))))$
|
88572
| |
Find $2^{\frac{1}{2}} \cdot 4^{\frac{1}{4}} \cdot 8^{\frac{1}{8}} \cdot 16^{\frac{1}{16}} \dotsm.$
|
4
| |
If the equation $\frac{m}{x-3}-\frac{1}{3-x}=2$ has a positive root with respect to $x$, then the value of $m$ is ______.
|
-1
| |
The numbers $1, 2, 3, 4, 5$ are to be arranged in a circle. An arrangement is $\textit{bad}$ if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n$. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
|
2
|
To solve this problem, we need to determine how many arrangements of the numbers $1, 2, 3, 4, 5$ in a circle are "bad," meaning that it is not possible to find a subset of numbers that appear consecutively on the circle that sum to every integer from $1$ to $15$.
#### Step 1: Count the total number of distinct arrangements
Since the numbers are arranged in a circle, arrangements that differ only by rotation or reflection are considered the same. We can fix one number (say $1$) in a specific position to eliminate equivalent rotations. This leaves $4! = 24$ ways to arrange the remaining four numbers. Considering reflections, each arrangement has exactly one reflection, so we divide by $2$ to get:
$$ \frac{4!}{2} = \frac{24}{2} = 12 \text{ distinct arrangements under rotation and reflection.} $$
#### Step 2: Check each arrangement for "badness"
To be a "bad" arrangement, there must be at least one integer $n$ between $1$ and $15$ for which no subset of consecutive numbers sums to $n$. We note:
- Single numbers give sums $1, 2, 3, 4, 5$.
- The entire circle sums to $15$.
- Omitting one number from the circle gives sums $14, 13, 12, 11, 10$ (since $15 - n$ for $n = 1, 2, 3, 4, 5$).
We need to check for sums $6, 7, 8, 9$. If a subset sums to $6$, its complement in the circle sums to $9$, and similarly, a subset summing to $7$ complements a subset summing to $8$. Thus, we only need to check for the existence of subsets summing to $6$ and $7$.
#### Step 3: Identify "bad" arrangements
We systematically check each of the $12$ cases to see if they can form subsets that sum to $6$ and $7$. We find:
- **For $6$:** Possible subsets are $\{4, 2\}$, $\{3, 2, 1\}$, $\{5, 1\}$. We need to check if any arrangement prevents these subsets.
- **For $7$:** Possible subsets are $\{3, 4\}$, $\{5, 2\}$, $\{4, 2, 1\}$. We need to check if any arrangement prevents these subsets.
Through detailed examination (as outlined in the initial solution), we find that there are exactly two arrangements that prevent forming subsets that sum to $6$ or $7$. These arrangements are distinct under rotation and reflection.
#### Conclusion:
The number of different "bad" arrangements is $\boxed{\textbf{(B)}\ 2}$.
|
Compute
\[
\log_2 \left( \prod_{a=1}^{2015} \prod_{b=1}^{2015} (1+e^{2\pi i a b/2015}) \right)
\]
Here $i$ is the imaginary unit (that is, $i^2=-1$).
|
13725
|
The answer is $13725$.
We first claim that if $n$ is odd, then $\prod_{b=1}^{n} (1+e^{2\pi i ab/n}) = 2^{\gcd(a,n)}$. To see this, write $d = \gcd(a,n)$ and $a = da_1$, $n=dn_1$ with $\gcd(a_1,n_1) = 1$. Then
$a_1, 2a_1,\dots,n_1 a_1$ modulo $n_1$ is a permutation of $1,2,\dots,n_1$ modulo $n_1$, and so $\omega^{a_1},\omega^{2a_1},\dots,\omega^{n_1 a_1}$ is a permutation of $\omega,\omega^2,\ldots,\omega^{n_1}$; it follows that for $\omega = e^{2\pi i/n_1}$,
\[
\prod_{b=1}^{n_1} (1+e^{2\pi i a b/n}) =
\prod_{b=1}^{n_1} (1+e^{2\pi i a_1 b/n_1}) = \prod_{b=1}^{n_1} (1+\omega^b).
\]
Now since the roots of $z^{n_1}-1$ are $\omega,\omega^2,\ldots,\omega^{n_1}$, it follows that
$z^{n_1}-1 = \prod_{b=1}^{n_1} (z-\omega^b)$. Setting $z=-1$ and using the fact that $n_1$ is odd gives $\prod_{b=1}^{n_1} (1+\omega^b) = 2$.
Finally,
$\prod_{b=1}^{n} (1+e^{2\pi i ab/n}) = (\prod_{b=1}^{n_1} (1+e^{2\pi i ab/n}))^d = 2^d$, and we have proven the claim.
From the claim, we find that
\begin{align*}
&\log_2 \left( \prod_{a=1}^{2015} \prod_{b=1}^{2015} (1+e^{2\pi i a b/2015}) \right) \\
&= \sum_{a=1}^{2015} \log_2 \left(\prod_{b=1}^{2015} (1+e^{2\pi i a b/2015}) \right) \\
&= \sum_{a=1}^{2015} \gcd(a,2015).
\end{align*}
Now for each divisor $d$ of $2015$, there are $\phi(2015/d)$ integers between $1$ and $2015$ inclusive whose $\gcd$ with $2015$ is $d$. Thus
\[
\sum_{a=1}^{2015} \gcd(a,2015) = \sum_{d|2015} d\cdot \phi(2015/d).
\]
We factor $2015 = pqr$ with $p=5$, $q=13$, and $r=31$, and calculate
\begin{align*}
&\sum_{d|pqr} d\cdot \phi(pqr/d) \\
&= 1 \cdot (p-1)(q-1)(r-1) + p \cdot (q-1)(r-1) \\
&\quad + q\cdot (p-1)(r-1) + r\cdot (p-1)(q-1) + pq \cdot (r-1) \\
& \quad + pr\cdot (q-1) + qr\cdot (p-1) + pqr \cdot 1 \\
&\quad = (2p-1)(2q-1)(2r-1).
\end{align*}
When $(p,q,r) = (5,13,31)$, this is equal to $13725$.
|
In triangle $ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given vectors $\overrightarrow{m} = (\cos B, \cos C)$ and $\overrightarrow{n} = (2a + c, b)$, and $\overrightarrow{m} \perp \overrightarrow{n}$.
(I) Find the measure of angle $B$ and the range of $y = \sin 2A + \sin 2C$;
(II) If $b = \sqrt{13}$ and $a + c = 4$, find the area of triangle $ABC$.
|
\frac{3\sqrt{3}}{4}
| |
$(1)$ Calculate: $2^{-1}+|\sqrt{6}-3|+2\sqrt{3}\sin 45^{\circ}-\left(-2\right)^{2023}\cdot (\frac{1}{2})^{2023}$.
$(2)$ Simplify and then evaluate: $\left(\frac{3}{a+1}-a+1\right) \div \frac{{{a}^{2}}-4}{{{a}^{2}}+2a+1}$, where $a$ takes a suitable value from $-1$, $2$, $3$ for evaluation.
|
-4
| |
What is the smallest positive integer $n$ such that $2n$ is a perfect square and $3n$ is a perfect cube?
|
72
| |
The value of \( \frac{1}{2} + \frac{2}{4} + \frac{4}{8} + \frac{8}{16} \) is what?
|
2
|
In the given sum, each of the four fractions is equivalent to \( \frac{1}{2} \). Therefore, the given sum is equal to \( \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2 \).
|
A number $x$ is randomly chosen from the interval $[-1, 1]$. What is the probability that the value of $\cos \frac{\pi x}{2}$ lies between $0$ and $\frac{1}{2}$?
|
$\frac{1}{3}$
| |
If the value of the expression $(\square + 121 \times 3.125) \div 121$ is approximately 3.38, what natural number should be placed in $\square$?
|
31
| |
In triangle $ABC$, $AB = 10$, $BC = 14$, and $CA = 16$. Let $D$ be a point in the interior of $\overline{BC}$. Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$, respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$. The maximum possible area of $\triangle BPC$ can be expressed in the form $a - b\sqrt {c}$, where $a$, $b$, and $c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a + b + c$.
Diagram
[asy] defaultpen(fontsize(11)+0.8); size(300); pair A,B,C,D,Ic,Ib,P; A=MP("A",origin,down+left); B=MP("B",8*right,down+right); C=MP("C",IP(CR(A,5), CR(B,7)),2*up); real t=0.505; D=MP("",B+t*(C-B),SW); draw(A--B--C--A--D); path c1=incircle(A,D,C); path c2=incircle(A,D,B); draw(c1, gray+0.25); draw(c2, gray+0.25); Ic=MP("I_C",incenter(A,D,C),down+left); Ib=MP("I_B",incenter(A,D,B),left); path c3=circumcircle(Ic,D,C); path c4=circumcircle(Ib,D,B); draw(c3, fuchsia+0.2); draw(c4, fuchsia+0.2); P=MP("P",OP(c3,c4),up); draw(arc(circumcenter(B,C,P),B,C), royalblue+0.5+dashed); draw(C--Ic--D--P--C^^P--Ic, black+0.3); draw(B--Ib--D--P--B^^P--Ib, black+0.3); label("10",A--B,down); label("16",A--C,left); [/asy]
|
150
|
Proceed as in Solution 2 until you find $\angle CPB = 150$. The locus of points $P$ that give $\angle CPB = 150$ is a fixed arc from $B$ to $C$ ($P$ will move along this arc as $D$ moves along $BC$) and we want to maximise the area of [$\triangle BPC$]. This means we want $P$ to be farthest distance away from $BC$ as possible, so we put $P$ in the middle of the arc (making $\triangle BPC$ isosceles). We know that $BC=14$ and $\angle CPB = 150$, so $\angle PBC = \angle PCB = 15$. Let $O$ be the foot of the perpendicular from $P$ to line $BC$. Then the area of [$\triangle BPC$] is the same as $7OP$ because base $BC$ has length $14$. We can split $\triangle BPC$ into two $15-75-90$ triangles $BOP$ and $COP$, with $BO=CO=7$ and $OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3$. Then, the area of [$\triangle BPC$] is equal to $7 \cdot OP=98-49\sqrt{3}$, and so the answer is $98+49+3=\boxed{150}$.
|
In the diagram, $JKLM$ and $NOPM$ are squares each of area 25. If $Q$ is the midpoint of both $KL$ and $NO$, find the total area of polygon $JMQPON$.
[asy]
unitsize(3 cm);
pair J, K, L, M, N, O, P, Q;
O = (0,0);
P = (1,0);
M = (1,1);
N = (0,1);
Q = (N + O)/2;
J = reflect(M,Q)*(P);
K = reflect(M,Q)*(O);
L = reflect(M,Q)*(N);
draw(J--K--L--M--cycle);
draw(M--N--O--P--cycle);
label("$J$", J, N);
label("$K$", K, W);
label("$L$", L, S);
label("$M$", M, NE);
label("$N$", N, NW);
label("$O$", O, SW);
label("$P$", P, SE);
label("$Q$", Q, SW);
[/asy]
|
25
| |
Consider the cards $A, 2, \cdots, J, Q, K$ as the numbers $1, 2, \cdots, 11, 12, 13$. If we take the 13 cards of spades and 13 cards of hearts together and randomly draw 2 cards, what is the probability that the two cards are of the same suit and the product of the two numbers is a perfect square?
|
2/65
| |
Arrange the numbers from 1 to 25 in a random order. Then subtract them sequentially from 1, 2, 3, ..., 25, always subtracting the smaller number from the larger number. What is the maximum number of even differences that can appear among these 25 differences?
|
25
| |
Jessica's six assignment scores are 87, 94, 85, 92, 90, and 88. What is the arithmetic mean of these six scores?
|
89.3
| |
A manufacturer built a machine which will address $500$ envelopes in $8$ minutes. He wishes to build another machine so that when both are operating together they will address $500$ envelopes in $2$ minutes. The equation used to find how many minutes $x$ it would require the second machine to address $500$ envelopes alone is:
|
$\frac{1}{8}+\frac{1}{x}=\frac{1}{2}$
|
1. **Understanding the problem**: The first machine addresses 500 envelopes in 8 minutes. We need to find the time $x$ it would take for a second machine to address 500 envelopes alone such that both machines working together can address 500 envelopes in 2 minutes.
2. **Representing the rate of the first machine**: The rate of the first machine is $\frac{500 \text{ envelopes}}{8 \text{ minutes}} = 62.5 \text{ envelopes per minute}$.
3. **Combined rate needed**: Since both machines together need to address 500 envelopes in 2 minutes, their combined rate must be $\frac{500 \text{ envelopes}}{2 \text{ minutes}} = 250 \text{ envelopes per minute}$.
4. **Rate of the second machine**: Let the rate of the second machine be $r$ envelopes per minute. Then, the combined rate of both machines is $62.5 + r = 250$. Solving for $r$, we get $r = 250 - 62.5 = 187.5 \text{ envelopes per minute}$.
5. **Finding $x$**: The time $x$ for the second machine to address 500 envelopes alone at a rate of $187.5 \text{ envelopes per minute}$ is $x = \frac{500}{187.5} = \frac{500}{187.5} = \frac{40}{15} = \frac{8}{3}$ minutes.
6. **Checking the options**: We need to find an equation that correctly represents the relationship between the rates and times of the two machines:
- Option (B) $\frac{1}{8} + \frac{1}{x} = \frac{1}{2}$ translates to $\frac{1}{8} + \frac{3}{8} = \frac{1}{2}$, which simplifies to $\frac{4}{8} = \frac{1}{2}$, a true statement.
7. **Conclusion**: The correct equation that models the situation where both machines working together can address 500 envelopes in 2 minutes, with the second machine alone taking $x$ minutes to do the same, is given by option (B).
$\boxed{\textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2}}$
|
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