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Simplify first, then evaluate: $-2(-x^2y+xy^2)-[-3x^2y^2+3x^2y+(3x^2y^2-3xy^2)]$, where $x=-1$, $y=2$.
|
-6
|
deepscale
| 19,331
| ||
Let $D$ be the determinant of the matrix whose column vectors are $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}.$ Find the determinant of the matrix whose column vectors are $\mathbf{a} + \mathbf{b},$ $\mathbf{b} + \mathbf{c},$ and $\mathbf{c} + \mathbf{a},$ in terms of $D.$
|
2D
|
deepscale
| 40,073
| ||
When a certain unfair die is rolled, an even number is $5$ times as likely to appear as an odd number. The die is rolled twice. Calculate the probability that the sum of the numbers rolled is odd.
|
\frac{5}{18}
|
deepscale
| 27,313
| ||
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
|
1. **Formulate the congruences:** We are given that a number $n$ leaves a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11. We can express these conditions using congruences:
\[
n \equiv 2 \pmod{6}, \quad n \equiv 5 \pmod{9}, \quad n \equiv 7 \pmod{11}
\]
2. **Express in terms of variables:** Let's express $n$ in terms of variables corresponding to each modulus:
\[
n = 6x + 2, \quad n = 9y + 5, \quad n = 11z + 7
\]
where $x, y, z$ are integers.
3. **Analyze the range of $n$:** Since $n$ is a three-digit number, we have:
\[
100 \leq n < 1000
\]
Substituting $n = 11z + 7$ into the inequality:
\[
100 \leq 11z + 7 < 1000
\]
Simplifying, we get:
\[
93 \leq 11z < 993 \implies \frac{93}{11} \leq z < \frac{993}{11} \implies 8.45 \leq z < 90.27
\]
Since $z$ must be an integer, we have:
\[
9 \leq z \leq 90
\]
4. **Find common values satisfying all conditions:** We need to find $n$ such that $6x + 2 = 9y + 5 = 11z + 7$. Simplifying the first and third equations:
\[
6x + 2 = 11z + 7 \implies 6x = 11z + 5
\]
This implies:
\[
6x \equiv 5 \pmod{11}
\]
Similarly, from the second and third equations:
\[
9y + 5 = 11z + 7 \implies 9y = 11z + 2
\]
This implies:
\[
9y \equiv 2 \pmod{11}
\]
5. **Solve using the Chinese Remainder Theorem (CRT):** We need to find a common solution to these congruences. We know $z + 1$ must be a multiple of both 6 and 9 (the least common multiple of 6 and 9 is 18). Thus, let $z + 1 = 18p$:
\[
z = 18p - 1
\]
Substituting back into the range for $z$:
\[
9 \leq 18p - 1 \leq 90 \implies 10 \leq 18p \leq 91 \implies \frac{10}{18} \leq p \leq \frac{91}{18} \implies 1 \leq p \leq 5
\]
Since $p$ must be an integer, $p$ can take 5 values.
6. **Conclusion:** There are 5 possible values for $p$, and hence 5 three-digit integers that satisfy all the given conditions.
\[
\boxed{\textbf{(E) }5}
\]
|
5
|
deepscale
| 554
| |
Let $ABCD$ be a regular tetrahedron, and let $O$ be the centroid of triangle $BCD$. Consider the point $P$ on $AO$ such that $P$ minimizes $PA+2(PB+PC+PD)$. Find $\sin \angle PBO$.
|
We translate the problem into one about 2-D geometry. Consider the right triangle $ABO$, and $P$ is some point on $AO$. Then, the choice of $P$ minimizes $PA+6PB$. Construct the line $\ell$ through $A$ but outside the triangle $ABO$ so that $\sin \angle(AO, \ell)=\frac{1}{6}$. For whichever $P$ chosen, let $Q$ be the projection of $P$ onto $\ell$, then $PQ=\frac{1}{6}AP$. Then, since $PA+6PB=6(PQ+PB)$, it is equivalent to minimize $PQ+PB$. Observe that this sum is minimized when $B, P, Q$ are collinear and the line through them is perpendicular to $\ell$ (so that $PQ+PB$ is simply the distance from $B$ to $\ell$). Then, $\angle AQB=90^{\circ}$, and since $\angle AOB=90^{\circ}$ as well, we see that $A, Q, P, B$ are concyclic. Therefore, $\angle PBO=\angle OPA=\angle(AO, \ell)$, and the sine of this angle is therefore $\frac{1}{6}$.
|
\frac{1}{6}
|
deepscale
| 3,282
| |
A certain point has rectangular coordinates $(10,3)$ and polar coordinates $(r, \theta).$ What are the rectangular coordinates of the point with polar coordinates $(r^2, 2 \theta)$?
|
(91,60)
|
deepscale
| 40,240
| ||
In $\triangle ABC$, the sides opposite to angles $A$, $B$, $C$ are denoted as $a$, $b$, $c$ respectively. Given that $b=3a$ and $c=2$, find the area of $\triangle ABC$ when angle $A$ is at its maximum value.
|
\frac { \sqrt {2}}{2}
|
deepscale
| 28,827
| ||
Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$
|
Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follows that the equation of $R(x)$ is \[R(x)=-\frac12x+c\] for some constant $c,$ and we wish to find $R(0)=c.$
We substitute $x=20$ into this equation to get $106=-\frac12\cdot20+c,$ from which $c=\boxed{116}.$
~MRENTHUSIASM
|
116
|
deepscale
| 7,331
| |
How many integers are there in $\{0,1, 2,..., 2014\}$ such that $C^x_{2014} \ge C^{999}{2014}$ ?
Note: $C^{m}_{n}$ stands for $\binom {m}{n}$
|
17
|
deepscale
| 31,703
| ||
Let $n, k \geq 3$ be integers, and let $S$ be a circle. Let $n$ blue points and $k$ red points be chosen uniformly and independently at random on the circle $S$. Denote by $F$ the intersection of the convex hull of the red points and the convex hull of the blue points. Let $m$ be the number of vertices of the convex polygon $F$ (in particular, $m=0$ when $F$ is empty). Find the expected value of $m$.
|
We prove that $$E(m)=\frac{2 k n}{n+k-1}-2 \frac{k!n!}{(k+n-1)!}$$ Let $A_{1}, \ldots, A_{n}$ be blue points. Fix $i \in\{1, \ldots, n\}$. Enumerate our $n+k$ points starting from a blue point $A_{i}$ counterclockwise as $A_{i}, X_{1, i}, X_{2, i}, \ldots, X_{(n+k-1), i}$. Denote the minimal index $j$ for which the point $X_{j, i}$ is blue as $m(i)$. So, $A_{i} X_{m(i), i}$ is a side of the convex hull of blue points. Denote by $b_{i}$ the following random variable: $$b_{i}= \begin{cases}1, & \text { if the chord } A_{i} X_{m(i), i} \text { contains a side of } F \\ 0, & \text { otherwise. }\end{cases}$$ Define analogously $k$ random variables $r_{1}, \ldots, r_{k}$ for the red points. Clearly, $$m=b_{1}+\ldots+b_{n}+r_{1}+\ldots+r_{k}$$ We proceed with computing the expectation of each $b_{i}$ and $r_{j}$. Note that $b_{i}=0$ if and only if all red points lie on the side of the line $A_{i} X_{m(i), i}$. This happens either if $m(i)=1$, i.e., the point $X_{i, 1}$ is blue (which happens with probability $\frac{n-1}{k+n-1}$ ), or if $i=k+1$, points $X_{1, i}, \ldots, X_{k, i}$ are red, and points $X_{k+1, i}, \ldots, X_{k+n-1, i}$ are blue (which happens with probability $1 /\binom{k+n-1}{k}$ ), since all subsets of size $k$ of $\{1,2, \ldots, n+k-1\}$ have equal probabilities to correspond to the indices of red points between $\left.X_{1, i}, \ldots, X_{n+k-1, i}\right)$. Thus the expectation of $b_{i}$ equals $1-\frac{n-1}{k+n-1}-1 /\binom{k+n-1}{k}=\frac{k}{n+k-1}-\frac{k!(n-1)!}{(k+n-1)!}$. Analogously, the expectation of $r_{j}$ equals $\frac{n}{n+k-1}-\frac{n!(k-1)!}{(k+n-1)!}$. It remains to use ( $\mathcal{C}$ ) and linearity of expectation.
|
\frac{2 k n}{n+k-1}-2 \frac{k!n!}{(k+n-1)!
|
deepscale
| 4,949
| |
The segment connecting the centers of two intersecting circles is divided by their common chord into segments equal to 5 and 2. Find the common chord, given that the radius of one circle is twice the radius of the other.
|
2 \sqrt{3}
|
deepscale
| 13,068
| ||
Calculate the sum of 5739204.742 and -176817.835, and round the result to the nearest integer.
|
5562387
|
deepscale
| 19,059
| ||
Given vectors $\overrightarrow{a} = (\cos \frac{3x}{2}, \sin \frac{3x}{2})$ and $\overrightarrow{b} = (\cos \frac{x}{2}, -\sin \frac{x}{2})$, with $x \in \left[-\frac{\pi}{3}, \frac{\pi}{4}\right]$,
(Ⅰ) Find $\overrightarrow{a} \cdot \overrightarrow{b}$ and $|\overrightarrow{a} + \overrightarrow{b}|$.
(Ⅱ) Let $f(x) = \overrightarrow{a} \cdot \overrightarrow{b} - |\overrightarrow{a} + \overrightarrow{b}|$, find the maximum and minimum values of $f(x)$.
|
-\frac{3}{2}
|
deepscale
| 26,894
| ||
In the Chinese length measurement units, 1 meter = 3 chi, 1 zhang = 10 chi, and 1 kilometer = 2 li. How many zhang are in 1 li?
|
150
|
deepscale
| 9,095
| ||
Among 6 courses, if person A and person B each choose 3 courses, the number of ways in which exactly 1 course is chosen by both A and B is \_\_\_\_\_\_.
|
180
|
deepscale
| 10,945
| ||
Find the number of pairs of integers \((a, b)\) with \(1 \leq a<b \leq 57\) such that \(a^{2}\) has a smaller remainder than \(b^{2}\) when divided by 57.
|
There are no such pairs when \(b=57\), so we may only consider pairs with \(1 \leq a<b \leq 56\). The key idea is that unless \(a^{2} \bmod 57=b^{2} \bmod 57,(a, b)\) can be paired with \((57-b, 57-a)\) and exactly one of them satisfies \(a^{2} \bmod 57<b^{2} \bmod 57\). Hence if \(X\) is the number of pairs \((a, b)\) with \(1 \leq a<b \leq 56\) and \(a^{2} \equiv b^{2}(\bmod 57)\), then the answer is \(\frac{1}{2}\left(\binom{56}{2}-X\right)\). To count \(X\), let's first count the number of pairs \((a, b)\) with \(1 \leq a, b \leq 57\) and \(a^{2} \equiv b^{2}(\bmod 57)\). By the Chinese Remainder Theorem, the condition is equivalent to \((a-b)(a+b) \equiv 0(\bmod 3)\) and \((a-b)(a+b) \equiv 0(\bmod 19)\). There are \(2 \cdot 3-1=5\) pairs of residues modulo 3 where \((a-b)(a+b) \equiv 0\) \((\bmod 3)\), namely \((0,0),(1,1),(2,2),(1,2),(2,1)\). Similarly, there are \(2 \cdot 19-1=37\) pairs of residues modulo 19 where \((a-b)(a+b) \equiv 0(\bmod 19)\). By the Chinese Remainder Theorem, each choice of residues modulo 3 for \(a\) and \(b\) and residues modulo 19 for \(a\) and \(b\) corresponds to unique residues modulo 57 for \(a\) and \(b\). It follows that there are \(5 \cdot 37=185\) such pairs. To get the value of \(X\), we need to subtract the 57 pairs where \(a=b\) and divide by 2 for the pairs with \(a>b\), for a value of \(X=\frac{1}{2}(185-57)=64\). Therefore the final answer is \(\frac{1}{2}\left(\binom{56}{2}-64\right)=738\).
|
738
|
deepscale
| 5,038
| |
Let $a,$ $b,$ $c,$ $d$ be distinct integers, and let $\omega$ be a complex number such that $\omega^4 = 1$ and $\omega \neq 1.$ Find the smallest possible value of
\[|a + b \omega + c \omega^2 + d \omega^3|.\]
|
\sqrt{4.5}
|
deepscale
| 31,596
| ||
Given a cone with a base radius of $5$ and a lateral area of $65π$, let the angle between the slant height and the height of the cone be $θ$. Find the value of $\sinθ$.
|
\frac{5}{13}
|
deepscale
| 16,951
| ||
A designer has 3 fabric colors he may use for a dress: red, green, and blue. Four different patterns are available for the dress. If each dress design requires exactly one color and one pattern, how many different dress designs are possible?
|
12
|
deepscale
| 39,074
| ||
Given points $A(-3, -4)$ and $B(6, 3)$ in the xy-plane; point $C(1, m)$ is taken so that $AC + CB$ is a minimum. Find the value of $m$.
|
-\frac{8}{9}
|
deepscale
| 30,343
| ||
Find the largest real number $\lambda$ with the following property: for any positive real numbers $p,q,r,s$ there exists a complex number $z=a+bi$($a,b\in \mathbb{R})$ such that $$ |b|\ge \lambda |a| \quad \text{and} \quad (pz^3+2qz^2+2rz+s) \cdot (qz^3+2pz^2+2sz+r) =0.$$
|
To find the largest real number \(\lambda\) such that for any positive real numbers \(p, q, r, s\), there exists a complex number \(z = a + bi\) (\(a, b \in \mathbb{R}\)) satisfying
\[
|b| \ge \lambda |a|
\]
and
\[
(pz^3 + 2qz^2 + 2rz + s) \cdot (qz^3 + 2pz^2 + 2sz + r) = 0,
\]
we proceed as follows:
The answer is \(\lambda = \sqrt{3}\). This value is obtained when \(p = q = r = s = 1\).
To verify that \(\lambda = \sqrt{3}\) works, consider the polynomial equations:
\[
(pz^3 + 2qz^2 + 2rz + s) = 0 \quad \text{or} \quad (qz^3 + 2pz^2 + 2sz + r) = 0.
\]
For \(z = a + bi\), we need to show that \(|b| \ge \sqrt{3} |a|\).
Suppose \(z\) is a root of one of the polynomials. Without loss of generality, assume \(z\) is a root of \(pz^3 + 2qz^2 + 2rz + s = 0\). Then we have:
\[
p(a + bi)^3 + 2q(a + bi)^2 + 2r(a + bi) + s = 0.
\]
Separating real and imaginary parts and considering the magnitudes, we derive the inequality:
\[
|b| \ge \sqrt{3} |a|.
\]
Thus, the largest real number \(\lambda\) satisfying the given conditions is:
\[
\boxed{\sqrt{3}}.
|
\sqrt{3}
|
deepscale
| 2,908
| |
How many total days were there in the years 2001 through 2004?
|
1461
|
deepscale
| 34,767
| ||
The vertical coordinate of the intersection point of the new graph obtained by shifting the graph of the quadratic function $y=x^{2}+2x+1$ $2$ units to the left and then $3$ units up is ______.
|
12
|
deepscale
| 26,251
| ||
Given that there are $m$ distinct positive even numbers and $n$ distinct positive odd numbers such that their sum is 2015. Find the maximum value of $20m + 15n$.
|
1105
|
deepscale
| 28,850
| ||
Let $F_k(a,b)=(a+b)^k-a^k-b^k$ and let $S={1,2,3,4,5,6,7,8,9,10}$ . For how many ordered pairs $(a,b)$ with $a,b\in S$ and $a\leq b$ is $\frac{F_5(a,b)}{F_3(a,b)}$ an integer?
|
22
|
deepscale
| 9,742
| ||
Given the power function $f(x) = (m^2 - m - 1)x^{m^2 + m - 3}$ on the interval $(0, +\infty)$, determine the value of $m$ that makes it a decreasing function.
|
-1
|
deepscale
| 27,172
| ||
In $\triangle ABC$, $BC= \sqrt {5}$, $AC=3$, $\sin C=2\sin A$. Find:
1. The value of $AB$.
2. The value of $\sin(A- \frac {\pi}{4})$.
|
- \frac { \sqrt {10}}{10}
|
deepscale
| 22,460
| ||
In the XY-plane, mark all the lattice points $(x, y)$ where $0 \leq y \leq 10$. For an integer polynomial of degree 20, what is the maximum number of these marked lattice points that can lie on the polynomial?
|
20
|
deepscale
| 27,298
| ||
On an old-fashioned bicycle the front wheel has a radius of $2.5$ feet and the back wheel has a radius of $4$ inches. If there is no slippage, how many revolutions will the back wheel make while the front wheel makes $100$ revolutions?
|
750
|
deepscale
| 35,994
| ||
A bag contains red and yellow balls. When 60 balls are taken out, exactly 56 of them are red. Thereafter, every time 18 balls are taken out, 14 of them are always red, until the last batch of 18 balls is taken out. If the total number of red balls in the bag is exactly four-fifths of the total number of balls, how many red balls are in the bag?
|
336
|
deepscale
| 8,180
| ||
Given that $a>0$, $b>1$, and $a+b=2$, find the minimum value of $$\frac{1}{2a}+\frac{2}{b-1}$$.
|
\frac{9}{2}
|
deepscale
| 23,196
| ||
Given a right triangular prism $ABC-A_{1}B_{1}C_{1}$ whose side edge length is equal to the base edge length, find the sine value of the angle formed by $AB_{1}$ and the side face $ACC_{1}A_{1}$.
|
\frac{\sqrt{6}}{4}
|
deepscale
| 13,617
| ||
For positive integers $N$ and $k$, define $N$ to be $k$-nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $500$ that are neither $3$-nice nor $5$-nice.
|
266
|
deepscale
| 22,507
| ||
What is the smallest integer $n$ , greater than one, for which the root-mean-square of the first $n$ positive integers is an integer?
$\mathbf{Note.}$ The root-mean-square of $n$ numbers $a_1, a_2, \cdots, a_n$ is defined to be \[\left[\frac{a_1^2 + a_2^2 + \cdots + a_n^2}n\right]^{1/2}\]
|
Let's first obtain an algebraic expression for the root mean square of the first $n$ integers, which we denote $I_n$ . By repeatedly using the identity $(x+1)^3 = x^3 + 3x^2 + 3x + 1$ , we can write \[1^3 + 3\cdot 1^2 + 3 \cdot 1 + 1 = 2^3,\] \[1^3 + 3 \cdot(1^2 + 2^2) + 3 \cdot (1 + 2) + 1 + 1 = 3^3,\] and \[1^3 + 3\cdot(1^2 + 2^2 + 3^2) + 3 \cdot (1 + 2 + 3) + 1 + 1 + 1 = 4^3.\] We can continue this pattern indefinitely, and thus for any
positive integer $n$ , \[1 + 3\sum_{j=1}^n j^2 + 3 \sum_{j=1}^n j^1 + \sum_{j=1}^n j^0 = (n+1)^3.\] Since $\sum_{j=1}^n j = n(n+1)/2$ , we obtain \[\sum_{j=1}^n j^2 = \frac{2n^3 + 3n^2 + n}{6}.\] Therefore, \[I_n = \left(\frac{1}{n} \sum_{j=1}^n j^2\right)^{1/2} = \left(\frac{2n^2 + 3n + 1}{6}\right)^{1/2}.\] Requiring that $I_n$ be an integer, we find that \[(2n+1 ) (n+1) = 6k^2,\] where $k$ is an integer. Using the Euclidean algorithm, we see that $\gcd(2n+1, n+1) = \gcd(n+1,n) = 1$ , and so $2n+1$ and $n+1$ share no
factors greater than 1. The equation above thus implies that $2n+1$ and $n+1$ is each proportional to a perfect square. Since $2n+1$ is
odd, there are only two possible cases:
Case 1: $2n+1 = 3 a^2$ and $n+1 = 2b^2$ , where $a$ and $b$ are integers.
Case 2: $2n+1 = a^2$ and $n+1 = 6b^2$ .
In Case 1, $2n+1 = 4b^2 -1 = 3a^2$ . This means that $(4b^2 -1)/3 = a^2$ for some integers $a$ and $b$ . We proceed by checking whether $(4b^2-1)/3$ is a perfect square for $b=2, 3, 4, \dots$ . (The solution $b=1$ leads to $n=1$ , and we are asked to find a value of $n$ greater than 1.) The smallest positive integer $b$ greater than 1 for
which $(4b^2-1)/3$ is a perfect square is $b=13$ , which results in $n=337$ .
In Case 2, $2n+1 = 12b^2 - 1 = a^2$ . Note that $a^2 = 2n+1$ is an odd square, and hence is congruent to $1 \pmod 4$ . But $12b^2 -1 \equiv 3 \pmod 4$ for any $b$ , so Case 2 has no solutions.
Alternatively, one can proceed by checking whether $12b^2 -1$ is a perfect square for $b=1, 2 ,3 ,\dots$ . We find that $12b^2 -1$ is not a perfect square for $b = 1,2, 3, ..., 7, 8$ , and $n= 383$ when $b=8$ . Thus the smallest positive integers $a$ and $b$ for which $12b^2- 1 = a^2$ result in a value of $n$ exceeding the value found in Case 1, which was 337.
In summary, the smallest value of $n$ greater than 1 for which $I_n$ is an integer is $\boxed{337}$ .
|
\(\boxed{337}\)
|
deepscale
| 2,968
| |
Given a rectangular grid constructed with toothpicks of equal length, with a height of 15 toothpicks and a width of 12 toothpicks, calculate the total number of toothpicks required to build the grid.
|
387
|
deepscale
| 24,048
| ||
It is known that the 3 sides of a triangle are consecutive positive integers and the largest angle is twice the smallest angle. Find the perimeter of this triangle.
|
15
|
deepscale
| 7,698
| ||
What is the smallest four-digit positive integer that is divisible by 47?
|
1034
|
deepscale
| 37,837
| ||
The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory?
|
1. **Identify Satisfactory Grades**: According to the problem, grades A, B, C, and D are considered satisfactory.
2. **Count the Number of Satisfactory Grades**:
- Number of students with grade A = 5
- Number of students with grade B = 4
- Number of students with grade C = 3
- Number of students with grade D = 3
- Total number of satisfactory grades = $5 + 4 + 3 + 3 = 15$.
3. **Determine the Total Number of Students**:
- Since 5 students received grades that are not satisfactory (grade F), the total number of students is the sum of students with satisfactory grades and those with unsatisfactory grades.
- Total number of students = Number of satisfactory grades + Number of unsatisfactory grades = $15 + 5 = 20$.
4. **Calculate the Fraction of Satisfactory Grades**:
- The fraction of students with satisfactory grades is given by the ratio of the number of satisfactory grades to the total number of grades.
- Fraction of satisfactory grades = $\frac{\text{Number of satisfactory grades}}{\text{Total number of students}} = \frac{15}{20}$.
5. **Simplify the Fraction**:
- Simplify $\frac{15}{20}$ by dividing the numerator and the denominator by their greatest common divisor, which is 5.
- $\frac{15}{20} = \frac{15 \div 5}{20 \div 5} = \frac{3}{4}$.
6. **Conclusion**:
- The fraction of grades that are satisfactory is $\frac{3}{4}$.
- From the given options, $\boxed{\text{C}}$ $\frac{3}{4}$ is the correct answer.
|
\frac{3}{4}
|
deepscale
| 619
| |
There are 5 blue chips, 4 red chips and 3 yellow chips in a bag. One chip is drawn from the bag. That chip is placed back into the bag, and a second chip is drawn. What is the probability that the two selected chips are of different colors? Express your answer as a common fraction.
|
\frac{47}{72}
|
deepscale
| 35,287
| ||
A unit square $A B C D$ and a circle $\Gamma$ have the following property: if $P$ is a point in the plane not contained in the interior of $\Gamma$, then $\min (\angle A P B, \angle B P C, \angle C P D, \angle D P A) \leq 60^{\circ}$. The minimum possible area of $\Gamma$ can be expressed as $\frac{a \pi}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100 a+b$.
|
Note that the condition for $\Gamma$ in the problem is equivalent to the following condition: if $\min (\angle A P B, \angle B P C, \angle C P D, \angle D P A)>60^{\circ}$, then $P$ is contained in the interior of $\Gamma$. Let $X_{1}, X_{2}, X_{3}$, and $X_{4}$ be the four points in $A B C D$ such that $A B X_{1}, B C X_{2}, C D X_{3}$, and $D A X_{4}$ are all equilateral triangles. Now, let $\Omega_{1}, \Omega_{2}, \Omega_{3}$, and $\Omega_{4}$ be the respective circumcircles of these triangles, and let the centers of these circles be $O_{1}, O_{2}, O_{3}$, and $O_{4}$. Note that the set of points $P$ such that $\angle A P B, \angle B P C, \angle C P D, \angle D P A>60^{\circ}$ is the intersection of $\Omega_{1}, \Omega_{2}, \Omega_{3}$, and $\Omega_{4}$. We want to find the area of the minimum circle containing this intersection. Let $\Gamma_{1}$ and $\Gamma_{2}$ intersect at $B$ and $B^{\prime}$. Define $C^{\prime}, D^{\prime}$ and $A^{\prime}$ similarly. It is not hard to see that the circumcircle of square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the desired circle. Now observe that $\angle A B^{\prime} D^{\prime}=\angle A B^{\prime} D=60^{\circ}$. Similarly, $\angle A D^{\prime} B^{\prime}=60^{\circ}$, so $A B^{\prime} D^{\prime}$ is equilateral. Its height is the distance from $A$ to $B^{\prime} D^{\prime}$, which is $\frac{1}{\sqrt{2}}$, so its side length is $\frac{\sqrt{6}}{3}$. This is also the diameter of the desired circle, so its area is $\frac{\pi}{4} \cdot \frac{6}{9}=\frac{\pi}{6}$.
|
106
|
deepscale
| 3,930
| |
During a break, a fly entered the math classroom and began crawling on a poster depicting the graph of a quadratic function \( y = f(x) \) with a leading coefficient of -1. Initially, the fly moved exactly along the parabola to the point with an abscissa of 2, but then started moving along a straight line until it again reached the parabola at the point with an abscissa of 4. Find \( f(3) \), given that the line \( y = 2023x \) intersects the fly's path along the straight segment at its midpoint.
|
6070
|
deepscale
| 12,088
| ||
What is the sum of the ten terms in the arithmetic sequence $-3, 4, \dots, 40$?
|
285
|
deepscale
| 22,812
| ||
Find the largest integer $n$ such that $2007^{1024}-1$ is divisible by $2^n$.
|
14
|
deepscale
| 11,174
| ||
Let $x$ be a real number such that $x^3+4x=8$. Determine the value of $x^7+64x^2$.
|
128
|
deepscale
| 36,961
| ||
Find the measure of the angle
$$
\delta=\arccos \left(\left(\sin 2903^{\circ}+\sin 2904^{\circ}+\cdots+\sin 6503^{\circ}\right)^{\cos 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6480^{\circ}}\right)
$$
|
67
|
deepscale
| 15,370
| ||
The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
|
1. **Calculate the number of games within the conference:**
- There are 8 teams in the conference.
- Each team plays every other team twice (once at home and once away).
- The number of ways to choose 2 teams from 8 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items to choose from, and $k$ is the number of items to choose. Here, $n=8$ and $k=2$:
\[
\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28
\]
- Since each pair of teams plays twice, the total number of games within the conference is:
\[
28 \times 2 = 56
\]
2. **Calculate the number of games outside the conference:**
- Each team plays 4 games against non-conference opponents.
- There are 8 teams in the conference, so the total number of non-conference games played by all teams is:
\[
4 \times 8 = 32
\]
3. **Calculate the total number of games in the season:**
- Add the number of games within the conference to the number of games outside the conference:
\[
56 + 32 = 88
\]
4. **Conclusion:**
- The total number of games in a season involving the "Middle School Eight" teams is $\boxed{88}$, corresponding to choice $\boxed{\text{(B)}}$.
|
88
|
deepscale
| 386
| |
Mary's top book shelf holds five books with the following widths, in centimeters: $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$. What is the average book width, in centimeters?
|
1. **List the widths of the books**: The widths of the books are given as $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$ centimeters.
2. **Calculate the total sum of the widths**:
\[
6 + \dfrac{1}{2} + 1 + 2.5 + 10 = 6 + 0.5 + 1 + 2.5 + 10
\]
\[
= 6 + 0.5 + 1 + 2.5 + 10 = 20
\]
3. **Count the number of books**: There are $5$ books on the shelf.
4. **Compute the average width**:
\[
\text{Average width} = \frac{\text{Total sum of widths}}{\text{Number of books}} = \frac{20}{5} = 4
\]
5. **Conclusion**: The average book width is $4$ centimeters.
\[
\boxed{D}
\]
|
4
|
deepscale
| 44
| |
Find all odd natural numbers greater than 500 but less than 1000, each of which has the property that the sum of the last digits of all its divisors (including 1 and the number itself) is equal to 33.
|
729
|
deepscale
| 13,766
| ||
What is the largest three-digit multiple of 8 whose digits' sum is 24?
|
888
|
deepscale
| 19,278
| ||
A spherical scoop of vanilla ice cream with radius of 2 inches is dropped onto the surface of a dish of hot chocolate sauce. As it melts, the ice cream spreads out uniformly forming a cylindrical region 8 inches in radius. Assuming the density of the ice cream remains constant, how many inches deep is the melted ice cream? Express your answer as a common fraction.
|
\frac{1}{6}
|
deepscale
| 35,713
| ||
In triangle $\triangle ABC$, $sinB=\sqrt{2}sinA$, $∠C=105°$, and $c=\sqrt{3}+1$. Calculate the area of the triangle.
|
\frac{\sqrt{3} + 1}{2}
|
deepscale
| 9,023
| ||
How many four-digit numbers are there in which at least one digit occurs more than once?
|
4464. There are 9000 four-digit numbers altogether. If we consider how many four-digit numbers have all their digits distinct, there are 9 choices for the first digit (since we exclude leading zeroes), and then 9 remaining choices for the second digit, then 8 for the third, and 7 for the fourth, for a total of $9 \cdot 9 \cdot 8 \cdot 7=4536$. Thus the remaining $9000-4536=4464$ numbers have a repeated digit.
|
4464
|
deepscale
| 3,573
| |
Add $518_{12} + 276_{12}$. Express your answer in base 12, using $A$ for $10$ and $B$ for $11$ if necessary.
|
792_{12}
|
deepscale
| 17,324
| ||
Find all real parameters $a$ for which the equation $x^8 +ax^4 +1 = 0$ has four real roots forming an arithmetic progression.
|
-\frac{82}{9}
|
deepscale
| 14,127
| ||
A number is reduced by 5 times and then increased by 20 times to get 40. What is this number?
|
10
|
deepscale
| 15,149
| ||
A point $(x, y)$ is randomly selected such that $0 \leq x \leq 4$ and $0 \leq y \leq 5$. What is the probability that $x + y \leq 5$? Express your answer as a common fraction.
|
\frac{3}{5}
|
deepscale
| 30,045
| ||
Let
\[f(x) = x^3 + 6x^2 + 16x + 28.\]The graphs of $y = f(x)$ and $y = f^{-1}(x)$ intersect at exactly one point $(a,b).$ Enter the ordered pair $(a,b).$
|
(-4,-4)
|
deepscale
| 37,360
| ||
Solve for $x$:
$$x^2 + 4x + 3 = -(x + 3)(x + 5).$$
|
-3
|
deepscale
| 33,858
| ||
Determine the height of a tower from a 20-meter distant building, given that the angle of elevation to the top of the tower is 30° and the angle of depression to the base of the tower is 45°.
|
20 \left(1 + \frac {\sqrt {3}}{3}\right)
|
deepscale
| 20,612
| ||
It is now 12:00:00 midnight, as read on a 12-hour digital clock. In 122 hours, 39 minutes and 44 seconds the time will be $A:B:C$. What is the value of $A + B + C$?
|
85
|
deepscale
| 37,856
| ||
Miyuki texted a six-digit integer to Greer. Two of the digits of the six-digit integer were 3s. Unfortunately, the two 3s that Miyuki texted did not appear and Greer instead received the four-digit integer 2022. How many possible six-digit integers could Miyuki have texted?
|
The six-digit integer that Miyuki sent included the digits 2022 in that order along with two 3s. If the two 3s were consecutive digits, there are 5 possible integers: 332022, 233022, 203322, 202332, 202233. If the two 3s are not consecutive digits, there are 10 possible pairs of locations for the 3s: 1st/3rd, 1st/4th, 1st/5th, 1st/6th, 2nd/4th, 2nd/5th, 2nd/6th, 3rd/5th, 3rd/6th, 4th/6th. These give the following integers: 323022, 320322, 320232, 320223, 230322, 230232, 230223, 203232, 203223, 202323. In total, there are thus \(5 + 10 = 15\) possible six-digit integers that Miyuki could have texted.
|
15
|
deepscale
| 5,521
| |
Calculate:<br/>$(1)(\sqrt{50}-\sqrt{8})÷\sqrt{2}$;<br/>$(2)\sqrt{\frac{3}{4}}×\sqrt{2\frac{2}{3}}$.
|
\sqrt{2}
|
deepscale
| 19,912
| ||
Alan, Beth, Carla, and Dave weigh themselves in pairs. Together, Alan and Beth weigh 280 pounds, Beth and Carla weigh 230 pounds, Carla and Dave weigh 250 pounds, and Alan and Dave weigh 300 pounds. How many pounds do Alan and Carla weigh together?
|
250
|
deepscale
| 30,976
| ||
Calculate \( \frac{2}{1} \times \frac{2}{3} \times \frac{4}{3} \times \frac{4}{5} \times \frac{6}{5} \times \frac{6}{7} \times \frac{8}{7} \). Express the answer in decimal form, accurate to two decimal places.
|
1.67
|
deepscale
| 15,872
| ||
How many zeros are in the expansion of $999,\!999,\!999,\!998^2$?
|
11
|
deepscale
| 33,650
| ||
Consider a cube $A B C D E F G H$, where $A B C D$ and $E F G H$ are faces, and segments $A E, B F, C G, D H$ are edges of the cube. Let $P$ be the center of face $E F G H$, and let $O$ be the center of the cube. Given that $A G=1$, determine the area of triangle $A O P$.
|
From $A G=1$, we get that $A E=\frac{1}{\sqrt{3}}$ and $A C=\frac{\sqrt{2}}{\sqrt{3}}$. We note that triangle $A O P$ is located in the plane of rectangle $A C G E$. Since $O P \| C G$ and $O$ is halfway between $A C$ and $E G$, we get that $[A O P]=\frac{1}{8}[A C G E]$. Hence, $[A O P]=\frac{1}{8}\left(\frac{1}{\sqrt{3}}\right)\left(\frac{\sqrt{2}}{\sqrt{3}}\right)=\frac{\sqrt{2}}{24}$.
|
$\frac{\sqrt{2}}{24}$
|
deepscale
| 4,532
| |
Given the function \( f(x) = \frac{2+x}{1+x} \), let \( f(1) + f(2) + \cdots + f(1000) = m \) and \( f\left(\frac{1}{2}\right) + f\left(\frac{1}{3}\right) + \cdots + f\left(\frac{1}{1000}\right) = n \). What is the value of \( m + n \)?
|
2998.5
|
deepscale
| 12,227
| ||
What is the value of $x$ if $x=\frac{2023^2 - 2023 + 1}{2023}$?
|
2022 + \frac{1}{2023}
|
deepscale
| 17,104
| ||
Given that θ is an acute angle and $\sqrt {2}$sinθsin($θ+ \frac {π}{4}$)=5cos2θ, find the value of tanθ.
|
\frac {5}{6}
|
deepscale
| 23,088
| ||
In the Cartesian coordinate plane $xOy$, a circle with center $C(1,1)$ is tangent to the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Points $M$ and $N$ lie on the line segments $OA$ and $OB$, respectively. If $MN$ is tangent to circle $C$, find the minimum value of $|MN|$.
|
2\sqrt{2} - 2
|
deepscale
| 22,198
| ||
Given a sequence $\{a_{n}\}$ such that $a_{2}=2a_{1}=4$, and $a_{n+1}-b_{n}=2a_{n}$, where $\{b_{n}\}$ is an arithmetic sequence with a common difference of $-1$.
$(1)$ Investigation: Determine whether the sequence $\{a_{n}-n\}$ is an arithmetic sequence or a geometric sequence, and provide a justification.
$(2)$ Find the smallest positive integer value of $n$ that satisfies $a_{1}+a_{2}+\ldots +a_{n} \gt 2200$.
|
12
|
deepscale
| 16,990
| ||
Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$, $23578$, and $987620$ are monotonous, but $88$, $7434$, and $23557$ are not. How many monotonous positive integers are there?
|
1. **Understanding Monotonous Numbers**: A monotonous number is defined as a number whose digits are either strictly increasing or strictly decreasing when read from left to right. This includes all one-digit numbers.
2. **Counting One-Digit Monotonous Numbers**: There are 9 one-digit numbers (1 through 9). Each of these is trivially monotonous.
3. **Counting Multi-Digit Monotonous Numbers**:
- **Increasing Sequence**: For any set of $n$ digits chosen from $\{1, 2, \ldots, 9\}$, there is exactly one way to arrange them in increasing order. The number of ways to choose $n$ digits from 9 is given by $\binom{9}{n}$.
- **Decreasing Sequence**: Similarly, for any set of $n$ digits chosen, there is exactly one way to arrange them in decreasing order. This also counts as $\binom{9}{n}$ ways.
- **Special Case for Decreasing Sequence with Zero**: Adding a zero to the end of a decreasing sequence of digits from $\{1, 2, \ldots, 9\}$ still forms a valid monotonous number (e.g., $3210$). This is valid for any $n$-digit decreasing sequence where $n \geq 1$.
4. **Total Count for Multi-Digit Monotonous Numbers**:
- For $n \geq 2$, each choice of $n$ digits can form one increasing and one decreasing number, so $2 \cdot \binom{9}{n}$.
- For $n = 1$, we have 9 increasing sequences (the digits themselves) and 9 decreasing sequences with zero added (10, 20, ..., 90), giving $9 + 9 = 18$.
5. **Summing Up All Cases**:
- Summing for $n \geq 2$: $\sum_{n=2}^{9} 2 \cdot \binom{9}{n}$
- Adding the $n = 1$ case: $18$
- Adding the one-digit numbers: $9$
6. **Calculating the Total**:
\[
\text{Total} = \sum_{n=2}^{9} 2 \cdot \binom{9}{n} + 18 + 9
\]
\[
= 2 \cdot \left(\sum_{n=1}^{9} \binom{9}{n} - \binom{9}{1}\right) + 18 + 9
\]
\[
= 2 \cdot (2^9 - 1 - 9) + 18 + 9
\]
\[
= 2 \cdot (511 - 9) + 27
\]
\[
= 2 \cdot 502 + 27
\]
\[
= 1004 + 27 = 1031
\]
7. **Revising the Calculation**:
- The calculation above seems incorrect as it does not match any of the given options. Revisiting the solution, we realize that the special case for decreasing sequences with zero was not correctly accounted for. Each decreasing sequence (for $n \geq 1$) can also have a zero appended, effectively doubling the count for decreasing sequences.
- Correcting this:
\[
\text{Total} = \sum_{n=1}^{9} 3 \cdot \binom{9}{n} + 9
\]
\[
= 3 \cdot (2^9 - 1) + 9
\]
\[
= 3 \cdot 511 + 9 = 1533 + 9 = 1542
\]
8. **Final Answer**:
- The correct calculation should match one of the options. Rechecking the original solution, it seems there was a mistake in my revised calculation. The original solution correctly calculates the total as:
\[
3 \cdot (2^9 - 1) - 9 = 3 \cdot 511 - 9 = 1533 - 9 = 1524
\]
\[
\boxed{\textbf{(B)}\ 1524}
\]
|
1524
|
deepscale
| 1,093
| |
Consider a polynomial $P(x) \in \mathbb{R}[x]$ , with degree $2023$ , such that $P(\sin^2(x))+P(\cos^2(x)) =1$ for all $x \in \mathbb{R}$ . If the sum of all roots of $P$ is equal to $\dfrac{p}{q}$ with $p, q$ coprime, then what is the product $pq$ ?
|
4046
|
deepscale
| 20,384
| ||
How many odd numbers between $100$ and $999$ have distinct digits?
|
320
|
deepscale
| 35,339
| ||
Two concentric circles have radii of 24 and 36 units, respectively. A shaded region is formed between these two circles. A new circle is to be drawn such that its diameter is equal to the area of the shaded region. What must the diameter of this new circle be? Express your answer in simplest radical form.
|
720 \pi
|
deepscale
| 9,055
| ||
Given vectors satisfying $\overrightarrow{a}\cdot (\overrightarrow{a}-2\overrightarrow{b})=3$ and $|\overrightarrow{a}|=1$, with $\overrightarrow{b}=(1,1)$, calculate the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.
|
\dfrac {3\pi}{4}
|
deepscale
| 29,783
| ||
Given the function $f(x)=\sqrt{3}\sin \omega x+\cos \omega x (\omega > 0)$, where the x-coordinates of the points where the graph of $f(x)$ intersects the x-axis form an arithmetic sequence with a common difference of $\frac{\pi}{2}$, determine the probability of the event "$g(x) \geqslant \sqrt{3}$" occurring, where $g(x)$ is the graph of $f(x)$ shifted to the left along the x-axis by $\frac{\pi}{6}$ units, when a number $x$ is randomly selected from the interval $[0,\pi]$.
|
\frac{1}{6}
|
deepscale
| 23,384
| ||
Given that α is in (0, π) and cosα = -$$\frac{15}{17}$$, find the value of sin($$\frac{π}{2}$$ + α) • tan(π + α).
|
\frac{8}{17}
|
deepscale
| 9,161
| ||
In a store, there are 50 light bulbs in stock, 60% of which are produced by Factory A and 40% by Factory B. The first-class rate of the light bulbs produced by Factory A is 90%, and the first-class rate of the light bulbs produced by Factory B is 80%.
(1) If one light bulb is randomly selected from these 50 light bulbs (each light bulb has an equal chance of being selected), what is the probability that it is a first-class product produced by Factory A?
(2) If two light bulbs are randomly selected from these 50 light bulbs (each light bulb has an equal chance of being selected), and the number of first-class products produced by Factory A among these two light bulbs is denoted as $\xi$, find the value of $E(\xi)$.
|
1.08
|
deepscale
| 19,542
| ||
Calculate the difference between 12.358 and 7.2943, keeping the answer in decimal form.
|
5.0637
|
deepscale
| 16,889
| ||
A box contains 4 labels marked with the numbers $1$, $2$, $3$, and $4$. Two labels are randomly selected according to the following conditions. Find the probability that the numbers on the two labels are consecutive integers:
1. The selection is made without replacement;
2. The selection is made with replacement.
|
\frac{3}{16}
|
deepscale
| 25,802
| ||
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy \[f(x^2-y)+2yf(x)=f(f(x))+f(y)\] for all $x,y\in\mathbb{R}$ .
|
Plugging in $y$ as $0:$ \begin{equation}
f(x^2)=f(f(x))+f(0) \text{ } (1)
\end{equation}
Plugging in $x, y$ as $0:$ \[f(0)=f(f(0))+f(0)\] or \[f(f(0))=0\] Plugging in $x$ as $0:$ \[f(-y)+2yf(0)=f(f(0))+f(y),\] but since $f(f(0))=0,$ \begin{equation}
f(-y)+2yf(0)=f(y) \text{ } (2)
\end{equation}
Plugging in $y^2$ instead of $y$ in the given equation: \[f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)\] Replacing $y$ and $x$ : \[f(y^2-x^2)+2x^2f(y)=f(f(y))+f(x^2)\] The difference would be:
\begin{equation}
f(x^2-y^2)-f(y^2-x^2)+2y^2f(x)-2x^2f(y)=f(f(x))-f(x^2)-f(f(y))-f(y^2) \text{ } (3)
\end{equation}
The right-hand side would be $f(0)-f(0)=0$ by $(1).$ Also, \[f(x^2-y^2)-f(y^2-x^2)=2(x^2-y^2)f(0)\] by $(2)$ So, $(3)$ is reduced to: \[2(x^2-y^2)f(0)+2y^2f(x)-2x^2f(y)=0\] Regrouping and dividing by 2: \[y^2(f(x)-f(0))=x^2(f(y)-f(0))\] \[\frac{f(x)-f(0)}{x^2}=\frac{f(y)-f(0)}{y^2}\] Because this holds for all x and y, $\frac{f(x)-f(0)}{x^2}$ is a constant. So, $f(x)=cx^2+f(0)$ .
This function must be even, so $f(y)-f(-y)=0$ .
So, along with $(2)$ , $2yf(0)=0$ for all $y$ , so $f(0)=0$ , and $f(x)=cx^2$ .
Plugging in $cx^2$ for $f(x)$ in the original equation, we get: \[c(x^4-2x^2y+y^2)+2cx^2y=c^3x^4+cy^2\] \[c(x^4+y^2)=c(c^2x^4+y^2)\] So, $c=0$ or $c^2=1.$ All of these solutions work, so the solutions are $f(x)=-x^2, 0, x^2$ .
-codemaster11
|
\[ f(x) = -x^2, \quad f(x) = 0, \quad f(x) = x^2 \]
|
deepscale
| 3,013
| |
Compute the number of positive integers less than 10! which can be expressed as the sum of at most 4 (not necessarily distinct) factorials.
|
Since $0!=1!=1$, we ignore any possible 0!'s in our sums. Call a sum of factorials reduced if for all positive integers $k$, the term $k$! appears at most $k$ times. It is straightforward to show that every positive integer can be written uniquely as a reduced sum of factorials. Moreover, by repeatedly replacing $k+1$ occurrences of $k$! with $(k+1)$!, every non-reduced sum of factorials is equal to a reduced sum with strictly fewer terms, implying that the aforementioned reduced sum associated to a positive integer $n$ in fact uses the minimum number of factorials necessary. It suffices to compute the number of nonempty reduced sums involving $\{1!, 2!, \ldots, 9!\}$ with at most 4 terms. By stars and bars, the total number of such sums, ignoring the reduced condition, is $\binom{13}{9}=714$. The sums that are not reduced must either contain two copies of 1!, three copies of 2!, or four copies of 3!. Note that at most one of these conditions is true, so we can count them separately. If $k$ terms are fixed, there are $\binom{13-k}{9}$ ways to choose the rest of the terms, meaning that we must subtract $\binom{11}{9}+\binom{10}{9}+\binom{9}{9}=66$. Our final answer is $714-66=648$.
|
648
|
deepscale
| 5,184
| |
A traffic light runs repeatedly through the following cycle: green for 45 seconds, then yellow for 5 seconds, and then red for 50 seconds. Mark picks a random five-second time interval to watch the light. What is the probability that the color changes while he is watching?
|
\frac{3}{20}
|
deepscale
| 10,295
| ||
What is the least possible value of
\[(x+2)(x+3)(x+4)(x+5) + 2024\] where \( x \) is a real number?
A) 2022
B) 2023
C) 2024
D) 2025
E) 2026
|
2023
|
deepscale
| 19,126
| ||
In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
|
1. **Identify the fixed elements and set up the problem**: Given that points $A$ and $B$ are $10$ units apart, we can place them at coordinates $(0,0)$ and $(10,0)$ respectively without loss of generality. This simplifies the problem to finding a point $C$ such that the perimeter of $\triangle ABC$ is $50$ units and the area is $100$ square units.
2. **Express the area condition**: The area of $\triangle ABC$ can be expressed using the determinant formula for the area of a triangle formed by points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|
\]
Substituting $A = (0,0)$, $B = (10,0)$, and $C = (x, y)$, the area becomes:
\[
\text{Area} = \frac{1}{2} \left| 0(0-y) + 10(y-0) + x(0-0) \right| = \frac{1}{2} \left| 10y \right| = 5|y|
\]
Setting this equal to $100$ gives:
\[
5|y| = 100 \implies |y| = 20
\]
Therefore, $y = 20$ or $y = -20$.
3. **Check the perimeter condition**: The perimeter of $\triangle ABC$ is the sum of the lengths of sides $AB$, $AC$, and $BC$. We know $AB = 10$. The lengths of $AC$ and $BC$ are:
\[
AC = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}
\]
\[
BC = \sqrt{(x-10)^2 + (y-0)^2} = \sqrt{(x-10)^2 + y^2}
\]
Substituting $y = 20$ or $y = -20$, we get:
\[
AC = BC = \sqrt{x^2 + 400}
\]
The perimeter is then:
\[
P = 10 + 2\sqrt{x^2 + 400}
\]
Setting this equal to $50$:
\[
10 + 2\sqrt{x^2 + 400} = 50 \implies 2\sqrt{x^2 + 400} = 40 \implies \sqrt{x^2 + 400} = 20 \implies x^2 + 400 = 400
\]
This equation simplifies to $x^2 = 0$, so $x = 0$.
4. **Conclusion**: The only possible coordinates for $C$ are $(0, 20)$ and $(0, -20)$. However, substituting these into the perimeter formula:
\[
P = 10 + 2\sqrt{0^2 + 400} = 10 + 2\cdot 20 = 50
\]
This calculation shows that the perimeter is indeed $50$ units, but we need to check if this is the minimal perimeter. Since the perimeter is exactly $50$ when $C$ is directly above or below the midpoint of $AB$, and moving $C$ any other way increases the perimeter, these are the only configurations possible. Thus, there are exactly two points $C$ that satisfy the conditions.
Therefore, the answer is $\boxed{\textbf{(B) }2}$.
|
2
|
deepscale
| 2,487
| |
Alexio now has 100 cards numbered from 1 to 100. He again randomly selects one card from the box. What is the probability that the number on the chosen card is a multiple of 3, 5, or 7? Express your answer as a common fraction.
|
\frac{11}{20}
|
deepscale
| 12,178
| ||
The positive integers \(a\), \(b\) are such that \(15a + 16b\) and \(16a - 15b\) are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?
|
481
|
deepscale
| 30,318
| ||
Suppose two distinct integers are chosen from between 1 and 29, inclusive. What is the probability that their product is neither a multiple of 2 nor 3?
|
\dfrac{45}{406}
|
deepscale
| 21,518
| ||
If $3n=9+9+9$, what is the value of $n$?
|
Since $3n=9+9+9=3 imes 9$, then $n=9$. Alternatively, we could note that $9+9+9=27$ and so $3n=27$ which gives $n= rac{27}{3}=9$.
|
9
|
deepscale
| 5,259
| |
A fair coin is tossed 4 times. What is the probability of at least two consecutive heads?
|
\frac{5}{8}
|
deepscale
| 25,455
| ||
Juan, Carlos and Manu take turns flipping a coin in their respective order. The first one to flip heads wins. What is the probability that Manu will win? Express your answer as a common fraction.
|
\frac{1}{7}
|
deepscale
| 33,392
| ||
If $64$ is divided into three parts proportional to $2$, $4$, and $6$, the smallest part is:
|
1. **Identify the Proportions**: The problem states that $64$ is divided into three parts proportional to $2$, $4$, and $6$. We can simplify these ratios by dividing each by the smallest number, which is $2$. This gives us the simplified ratio $1:2:3$.
2. **Set Up the Equations**: Let the three parts be $x$, $2x$, and $3x$, respectively, where $x$ corresponds to the part proportional to $1$, $2x$ to the part proportional to $2$, and $3x$ to the part proportional to $3$.
3. **Formulate the Total Sum Equation**: The sum of these parts must equal $64$. Therefore, we write the equation:
\[
x + 2x + 3x = 6x
\]
\[
6x = 64
\]
4. **Solve for $x$**: Divide both sides of the equation by $6$ to isolate $x$:
\[
x = \frac{64}{6} = \frac{32}{3}
\]
Converting $\frac{32}{3}$ to a mixed number:
\[
x = 10 \frac{2}{3}
\]
5. **Identify the Smallest Part**: Since $x$ is the smallest part (being proportional to the smallest number in the ratio $1:2:3$), the smallest part is $10 \frac{2}{3}$.
Thus, the smallest part is $\boxed{\textbf{(C)}\ 10\frac{2}{3}}$.
|
$10\frac{2}{3}$
|
deepscale
| 1,280
| |
It takes Mary 30 minutes to walk uphill 1 km from her home to school, but it takes her only 10 minutes to walk from school to home along the same route. What is her average speed, in km/hr, for the round trip?
|
3
|
deepscale
| 33,267
| ||
What is the value of $\sqrt{36 \times \sqrt{16}}$?
|
12
|
deepscale
| 39,280
| ||
Rectangle \(ABCD\) is divided into four parts by \(CE\) and \(DF\). It is known that the areas of three of these parts are \(5\), \(16\), and \(20\) square centimeters, respectively. What is the area of quadrilateral \(ADOE\) in square centimeters?
|
19
|
deepscale
| 29,510
| ||
A sports lottery stipulates that 7 numbers are drawn from a total of 36 numbers, ranging from 01 to 36, for a single entry, which costs 2 yuan. A person wants to select the lucky number 18 first, then choose 3 consecutive numbers from 01 to 17, 2 consecutive numbers from 19 to 29, and 1 number from 30 to 36 to form an entry. If this person wants to purchase all possible entries that meet these requirements, how much money must they spend at least?
|
2100
|
deepscale
| 20,580
| ||
Calculate $6!-5\cdot5!-5!$.
|
0
|
deepscale
| 34,995
| ||
When $\{a,0,-1\} = \{4,b,0\}$, find the values of $a$ and $b$.
|
-1
|
deepscale
| 24,395
| ||
For each even positive integer $x$, let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square.
|
First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even. so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.$ Further noting that $S_0=1$ we can see that $S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1).$ which is the same as above. To simplify the process of finding the largest square $S_n$ we can note that if $n-1$ is odd then $n+1$ must be exactly divisible by an odd power of $2$. However, this means $n+1$ is even but it cannot be. Thus $n-1$ is even and $n+1$ is a large even square. The largest even square $< 1000$ is $900$ so $n+1= 900 => n= \boxed{899}$
|
899
|
deepscale
| 6,863
|
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