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A line $l$ passes through the focus of the parabola $y^{2}=4x$ and intersects the parabola at points $A$ and $B$. If the x-coordinate of the midpoint of $AB$ is $3$, then the length of segment $AB$ is ( ).
A: $5$
B: $6$
C: $7$
D: $8$ | Let the focus of the parabola $y^{2}=4x$ be $F$, and its directrix be $l_{0}$. Let $C$ be the midpoint of $AB$.
Draw perpendiculars from points $A$ and $B$ to the directrix $l_{0}$, with the feet of the perpendiculars being $M$ and $N$, respectively.
According to the definition of a parabola,
we have $|AB|=|AF|+|BF|... | cn_k12 | 33f5bce8-5dda-5905-ba26-0db95404b248 | open-r1/OpenR1-Math-220k | [
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Given the circle \\(x^{2}+y^{2}+2x-2y+a=0\\) and the chord formed by the intersection with the line \\(x+y+2=0\\) has a length of \\(4\\), then the value of the real number \\(a\\) is \\((\\) \\()\\)
A: \\(-10\\)
B: \\(-8\\)
C: \\(-4\\)
D: \\(-2\\) | The standard equation of the circle is \\((x+1)^{2}+(y-1)^{2}=2-a\\),
so the coordinates of the center of the circle are \\((-1,1)\\), and the radius \\(r= \sqrt{2-a}\\),
Since the chord formed by the intersection of the circle \\(x^{2}+y^{2}+2x-2y+a=0\\) with the line \\(x+y+2=0\\) has a length of \\(4\\),
the d... | cn_k12 | 86ccc819-12bb-525b-b187-7098091d4c8f | open-r1/OpenR1-Math-220k | [
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1. Simplify:
$$
\sum_{k=1}^{2016}(k \sqrt{k+1}+(k+1) \sqrt{k})^{-1}=
$$
$\qquad$ | $$
-1.1-\frac{1}{\sqrt{2017}} .
$$
Notice,
$$
\begin{array}{l}
(k \sqrt{k+1}+(k+1) \sqrt{k})^{-1} \\
=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} .
\end{array}
$$
Therefore, the required result is $1-\frac{1}{\sqrt{2017}}$. | cn_contest | 7aea0e52-331f-56d5-a6bf-15e41facc20d | open-r1/OpenR1-Math-220k | [
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] |
24. Three people, A, B, and C, play a game: each person chooses a real number from the interval $[0,1]$, and the one who selects a number between the numbers chosen by the other two wins. A randomly selects a number from the interval $[0,1]$, B randomly selects a number from the interval $\left[\frac{1}{2}, \frac{2}{3}... | 24. B.
Let the number chosen by C be $x$.
According to the problem, the probability that the number chosen by C is greater than the number chosen by A and less than the number chosen by B is $x\left(\frac{2}{3}-x\right)$.
Similarly, the probability that the number chosen by C is greater than the number chosen by B an... | cn_contest | f580149b-d411-5b3c-8256-e615e0891179 | open-r1/OpenR1-Math-220k | [
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] |
Given that the direction vector of line $l$ is $\left(2,m,1\right)$, the normal vector of plane $\alpha$ is $(1,\frac{1}{2},2)$, and $l$ is parallel to $\alpha$, find the value of $m$. | Given that the direction vector of line $l$ is $\overrightarrow{m}=\left(2,m,1\right)$, and the normal vector of plane $\alpha$ is $\overrightarrow{n}=(1,\frac{1}{2},2)$, and knowing that $l$ is parallel to $\alpha$, we can deduce that $\overrightarrow{m}$ is perpendicular to $\overrightarrow{n}$. This implies that the... | cn_k12 | 4e590694-2b22-5320-8359-0d114e9e2978 | open-r1/OpenR1-Math-220k | [
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## Task 4
Add the numbers 1 to 10. | $1+2+3+4+5+6+7+8+9+10=55$. The sum is 55. | olympiads | e6d0f049-80f4-544e-8df5-5bea731426bb | open-r1/OpenR1-Math-220k | [
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] |
The solution set of the inequality $(x+2)(1-x)>0$ is ( )
A: $\{x|x1\}$
B: $\{x|x2\}$
C: $\{x|-2<x<1\}$
D: $\{x|-1<x<2\}$ | **Analysis**: To solve the inequality $(x+2)(1-x)>0$, which is equivalent to $(x+2)(x-1)<0$, the solution set is $\{x|-1<x<2\}$. Therefore, the correct option is $\boxed{C}$. | cn_k12 | 8d84f6f2-eb13-5ba7-adc4-f7685c27f640 | open-r1/OpenR1-Math-220k | [
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] |
Jose, Thuy, and Kareem each start with the number 10. Jose subtracts 1 from the number 10, doubles his answer, and then adds 2. Thuy doubles the number 10, subtracts 1 from her answer, and then adds 2. Kareem subtracts 1 from the number 10, adds 2 to his number, and then doubles the result. Who gets the largest fin... | Jose gets $10 - 1 = 9$, then $9 \cdot 2 = 18$, then $18 + 2 = 20$.
Thuy gets $10 \cdot 2 = 20$, then $20 - 1 = 19$, and then $19 + 2 = 21$.
Kareem gets $10 - 1 = 9$, then $9 + 2 = 11$, and then $11\cdot 2 = 22$.
Thus, Kareem gets the highest number, and the answer is $\boxed{C}$. | amc_aime | 75fb5a90-a9e8-5800-ab2b-836572689718 | open-r1/OpenR1-Math-220k | [
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] |
For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$. Determine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences
$$ t(n+a)-t(n), \quad t(n+a+1)-t(n+1), \quad \ldots, \quad t(n+2 a-1)-t(n+a-1) $$
are divisible by 4. Answer. $a=1,3$, or 5. | A pair $(a, n)$ satisfying the condition of the problem will be called a winning pair. It is straightforward to check that the pairs $(1,1),(3,1)$, and $(5,4)$ are winning pairs. Now suppose that $a$ is a positive integer not equal to 1, 3, and 5. We will show that there are no winning pairs $(a, n)$ by distinguishing ... | olympiads_ref | NaN | open-r1/OpenR1-Math-220k | [
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] |
In rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. If $AC=6$ and $BD=8$, then the area of rhombus $ABCD$ is
A: $6$
B: $12$
C: $24$
D: $48$ | To find the area of rhombus $ABCD$, we use the formula for the area of a rhombus, which is given by the product of its diagonals divided by 2. Given that the diagonals $AC$ and $BD$ intersect at point $O$ and their lengths are $AC = 6$ and $BD = 8$, we can calculate the area as follows:
\[
\text{Area of rhombus } ABCD... | cn_k12 | f2f5d044-55e6-5a85-80ed-235984eab6c9 | open-r1/OpenR1-Math-220k | [
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## Zadatak 1.
Odredi sve funkcije $f: \mathbb{R} \rightarrow \mathbb{R}$ za koje vrijedi
$$
f(x f(x)+f(x y))=f\left(x^{2}\right)+y f(x), \quad \text { za sve } x, y \in \mathbb{R}
$$
| ## Rješenje.
Uvrstimo li $x=0$ u jednadžbu, dobivamo $f(f(0))=f(0)+y f(0)$ za sve $y \in \mathbb{R}$. To povlači $f(0)=0$. Neka je $f(1)=c$ i uvrstimo $x=1$. Slijedi $f(c+f(y))=c(1+y)$ za sve $y \in \mathbb{R}$.
Ako je $c=0$, onda je $f(f(y))=0$ za sve $y \in \mathbb{R}$. Primijenimo li $f$ na danu jednadžbu, slijedi... | olympiads | 6b8f874e-af7e-5b86-b980-3a6f597285e7 | open-r1/OpenR1-Math-220k | [
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Define a function on the positive integers recursively by $f(1) = 2$, $f(n) = f(n-1) + 1$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than $1$. What is $f(2017)$?
$\textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036$ | This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, $f(1)=2$. We also know that when $n$ is odd, $f(n)=f(n-2)+2$. Thus we know that $f(2017)=f(2015)+2$. Thus we know that n will always be odd in the recursion of $f(20... | amc_aime | 249b2920-0129-5511-84a1-b555e263e124 | open-r1/OpenR1-Math-220k | [
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] |
# Task No. 7.2
Condition:
In the city of Abracodabra, funtics, tubrics, and santics are in circulation. One funtic can be exchanged for 1 tubric or 1 santic, 1 tubric for 5 funtics, and 1 santic for 2 funtics. No other exchanges are allowed. Lunatic, initially having 1 funtic, made 24 exchanges and now has 40 funtics... | Answer: 9
Exact match of the answer -1 point
Solution by analogy with task No. 7.1.
## Condition:
On the planet Mon Calamari, dataries, flans, and pegats are in circulation. One datary can be exchanged for 1 flan or 1 pegat, 1 flan - for 2 dataries, and 1 pegat - for 4 dataries. No other exchanges are allowed. Mera... | olympiads | 06010ec8-df5d-5af3-9183-0ce34e0a1910 | open-r1/OpenR1-Math-220k | [
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] |
Calculate:$(1)(-3a^{2})^{3}-4a^{2}\cdot a^{4}+5a^{9}\div a^{3}$.$(2)\left[\left(x+1\right)\left(x+2\right)+2\left(x-1\right)\right]\div x$. | ### Problem 1: Calculate $(1)(-3a^{2})^{3}-4a^{2}\cdot a^{4}+5a^{9}\div a^{3}$.
#### Step-by-Step Solution:
1. Evaluate each term individually:
- The first term: $(-3a^{2})^{3} = (-3)^{3} \cdot (a^{2})^{3} = -27a^{6}$.
- The second term: $-4a^{2} \cdot a^{4} = -4a^{6}$.
- The third term: $5a^{9} \div a^{3} =... | cn_k12 | 5f1ff009-8102-5a10-9773-f16a43469a44 | open-r1/OpenR1-Math-220k | [
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] |
Among the following four numbers, the largest one is ( ):
A: $$ln \sqrt[3]{3}$$
B: $$\frac {1}{e}$$
C: $$\frac {lnπ}{\pi }$$
D: $$\frac { \sqrt {15}ln15}{30}$$ | Let $f(x) = \frac{lnx}{x}$, then $f'(x) = \frac{1 - lnx}{x^2}$.
Thus, when $x \geq e$, $f'(x) \leq 0$. Therefore, $f(x)$ is decreasing on $[e, +\infty)$.
Given that $e f(3) > f(\pi) > f(15)$,
Which implies $e^{\frac{1}{e}} > \sqrt[3]{3} > \pi^{\frac{1}{\pi}} > 15^{\frac{1}{15}} > 15^{\frac{\sqrt{15}}{30}}$,
Or equ... | cn_k12 | 1ef2978f-6bc9-5b36-8bfe-0f0b6715f0a0 | open-r1/OpenR1-Math-220k | [
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] |
Given vectors $\overrightarrow{a} = (\sqrt{3}\sin 3x, -y)$ and $\overrightarrow{b} = (m, \cos 3x - m)$ (where $m \in \mathbb{R}$), and it is known that $\overrightarrow{a} + \overrightarrow{b} = \overrightarrow{0}$. Let $y = f(x)$.
(1) Find the expression of $f(x)$ and the coordinate of the lowest point M on the graph ... | (1) Since $\overrightarrow{a} + \overrightarrow{b} = \overrightarrow{0}$, we have the following system of equations:
$$
\begin{cases}
\sqrt{3}\sin 3x + m = 0 \\
-y + \cos 3x - m = 0
\end{cases}
$$
Eliminating $m$, we obtain
$$
y = \sqrt{3}\sin 3x + \cos 3x
$$
Therefore,
$$
f(x) = \sqrt{3}\sin 3x + \cos 3x = 2\sin\left... | cn_k12 | d907aaff-76ca-5c81-9639-e4abe6d84868 | open-r1/OpenR1-Math-220k | [
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6. Let $g(x)$ be a strictly increasing function defined for all $x \geq 0$. It is known that the range of $t$ satisfying
$$
\mathrm{g}\left(2 t^{2}+t+5\right)<\mathrm{g}\left(t^{2}-3 t+2\right)
$$
is $b<t<a$. Find $a-b$. | 6. Answer: 2
Solution. Note that $2 t^{2}+t+5=2\left(t+\frac{1}{4}\right)^{2}+\frac{39}{8}>0$. Hence $\mathrm{g}\left(2 t^{2}+t+5\right)<\mathrm{g}\left(t^{2}-3 x+2\right)$ is true if and only if
$$
2 t^{2}+t+5<t^{2}-3 t+2 \text {, }
$$
which is equivalent to $(t+3)(t+1)<0$. Hence the range of $t$ satisfying the given ... | olympiads | b57034c0-e338-547d-af0a-e9f90cc67134 | open-r1/OpenR1-Math-220k | [
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] |
Let the function $f(x) = \overrightarrow{m} \cdot \overrightarrow{n}$, where vector $\overrightarrow{m} = (2\cos x, 1)$, $\overrightarrow{n} = (\cos x, \sqrt{3}\sin 2x)$, and $x \in \mathbb{R}$.
1. Find the interval of monotonic increase for $f(x)$.
2. In $\triangle ABC$, $a$, $b$, and $c$ are the sides opposite angles... | Solution:
1. $f(x) = 2\cos^2 x + \sqrt{3}\sin 2x = \cos 2x + \sqrt{3}\sin 2x + 1 = 2\sin(2x + \frac{\pi}{6}) + 1$,
Let $- \frac{\pi}{2} + 2k\pi \leq 2x + \frac{\pi}{6} \leq \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z}$,
Solving this, we get: $- \frac{\pi}{3} + k\pi \leq x \leq \frac{\pi}{6} + k\pi, k \in \mathbb{Z}$.
... | cn_k12 | 687ad899-2295-5705-a204-a28b67c2dd0f | open-r1/OpenR1-Math-220k | [
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] |
## Task 6 - 080736
The great German mathematician Carl Friedrich Gauss was born on April 30, 1777, in Braunschweig.
On which day of the week did his birthday fall?
(April 30, 1967, was a Sunday; the years 1800 and 1900 were not leap years). | From one day of the year 1777 to the same day of the year 1967, it is 190 years, and specifically 45 years with 366 days and 145 years with 365 days.
In the 45 years, the weekday moved forward by 90 weekdays, and in the 145 years by 145 weekdays. Together, this is 235 weekdays, i.e., 33 times a week and 4 weekdays. Th... | olympiads | 3881354c-7324-5f74-813a-55723eabd56a | open-r1/OpenR1-Math-220k | [
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It takes 18 doodads to make 5 widgets. It takes 11 widgets to make 4 thingamabobs. How many doodads does it take to make 80 thingamabobs? | Since $80=20 \cdot 4$, then to make $80=20 \cdot 4$ thingamabobs, it takes $20 \cdot 11=220$ widgets.
Since $220=44 \cdot 5$, then to make $220=44 \cdot 5$ widgets, it takes $44 \cdot 18=792$ doodads.
Therefore, to make 80 thingamabobs, it takes 792 doodads.
ANSWER: 792 | olympiads | c60f1b8b-ff24-5135-89f8-310ed3a4231e | open-r1/OpenR1-Math-220k | [
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How many natural divisors does 121 have? How many natural divisors does 1000 have? How many natural divisors does 1000000000 have? | $121=11^{2}$. Therefore, the natural divisors of 121 are exactly 1, which has no prime factors, and the naturals for which 11 is the only prime factor and this prime factor appears 1 or 2 times in the prime factorization, which are 11 and 121. Thus, 121 has exactly 3 natural divisors.
$1000=2^{3} \cdot 5^{3}$. The nat... | olympiads | 388bf918-fac4-5ba6-bbd6-0cdb4987c5d3 | open-r1/OpenR1-Math-220k | [
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] |
Given the function $f(x) = x^2 - 6x - 9$, determine the range of $f(x)$ when $x \in (1, 4)$. | Since $f(x)$ can be rewritten as
$$
f(x) = x^2 - 6x - 9 = (x - 3)^2 - 18,
$$
the axis of symmetry of the parabola is at $x = 3$, which lies within the interval $(1, 4)$.
Therefore, the minimum value of $f(x)$ within $x \in (1, 4)$ is at the vertex, where $x = 3$, so we get
$$
f_{\min}(x) = f(3) = (3 - 3)^2 - 18 = -18.... | cn_k12 | 5891eab3-f4d9-548b-bdfc-08803ff9fa5c | open-r1/OpenR1-Math-220k | [
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18 Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\left(a, b \in \mathbf{R}^{+}\right)$ with semi-focal distance $c$, and $b^{2}=a c . P$ and $Q$ are any two points on the hyperbola, $M$ is the midpoint of $P Q$, when the slopes $k_{P Q}$ and $k_{O M}$ of $P Q$ and $O M$ both exist, find the value of $k_... | Given that $M$ is the midpoint of $P Q$, we can set the coordinates of point $M$ as $\left(x_{0}, y_{0}\right)$, the coordinates of point $P$ as $\left(x_{0}+\alpha, y_{0}+\beta\right)$, and the coordinates of point $Q$ as $\left(x_{0}-\alpha, y_{0}-\beta\right)$. Thus, $k_{O M}=\frac{y_{0}}{x_{0}}, k_{P Q}=\frac{\beta... | olympiads | 13a728e6-2838-5659-921b-352c96cd04e6 | open-r1/OpenR1-Math-220k | [
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] |
4. (20 points) In math class, each dwarf needs to find a three-digit number such that when 198 is added to it, the result is a number with the same digits but in reverse order. For what maximum number of dwarfs could all the numbers they find be different? | Answer: 70.
Solution: By the Pigeonhole Principle, the maximum number of gnomes is equal to the number of numbers that satisfy the condition of the problem.
Let the three-digit number be denoted as $\overline{x y z}$, where $x$ is the hundreds digit, $y$ is the tens digit, and $z$ is the units digit. Since the number... | olympiads | 3444f9d0-c65f-5936-8c58-3354aad8095d | open-r1/OpenR1-Math-220k | [
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Given the function $f(x) = \log_{a}(x+1) - \log_{a}(1-x)$, where $a > 0$ and $a \neq 1$.
(1) Find the domain of $f(x)$;
(2) Determine the parity of $f(x)$ and provide a proof;
(3) When $a > 1$, find the range of $x$ that satisfies $f(x) > 0$. | (1) The domain of $f(x)$:
Since $f(x) = \log_{a}(x+1) - \log_{a}(1-x)$, both $\log_{a}(x+1)$ and $\log_{a}(1-x)$ must be defined. This implies:
$$
\begin{cases}
x + 1 > 0 \\
1 - x > 0
\end{cases}
$$
Solving the inequalities, we get $-1 0$ when $a > 1$:
Since $a > 1$, $\log_{a}$ is an increasing function. Consequentl... | cn_k12 | 6be68d9b-fc53-5d22-b540-fc0b59615d72 | open-r1/OpenR1-Math-220k | [
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If non-empty sets A, B, C satisfy $A \cup B = C$, and B is not a subset of A, then ( )
A: "x ∈ C" is a sufficient but not necessary condition for "x ∈ A"
B: "x ∈ C" is a necessary but not sufficient condition for "x ∈ A"
C: "x ∈ C" is a necessary and sufficient condition for "x ∈ A"
D: "x ∈ C" is neither a sufficient ... | Given that $A \cup B = C$ and B is not a subset of A, we can deduce the following:
- If $x ∈ A$, then $x$ must belong to the union of A and B, which is C. Thus, $x ∈ A$ implies $x ∈ C$.
However, since B is not a subset of A, there are elements in B that do not belong to A. So, if $x ∈ C$, $x$ could be in either A or ... | cn_k12 | 0a26db01-feae-583b-b982-4a4ba3e656e6 | open-r1/OpenR1-Math-220k | [
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] |
2. Let $a, b, c, d$ be non-negative real numbers satisfying $a b+b c+c d+d a=1$. Prove that: $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}$. | 2. Let the left side of the inequality to be proved be $N$, consider the function $f(t)=2(a b+a c+a b+b c+b d+c d) t^{2}-2\left(a^{2}+b^{2}+\right.$ $\left.c^{2}+d^{2}\right) t+N=[a(b+c+d)+b(c+d+a)+c(a+b+d)+d(a+b+c)] t^{2}-2\left(a^{2}+b^{2}+\right.$ $\left.c^{2}+d^{2}\right) t+N=\left[\sqrt{a(b+c+d)} t-\sqrt{\frac{a^{... | olympiads | 2e1df224-5baf-50fd-a7c5-39ed1e48f41b | open-r1/OpenR1-Math-220k | [
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] |
If $\sin(\pi - A) = \frac{1}{2}$, then $\cos\left(\frac{\pi}{2} - A\right) =$ （ ）
A: $-\frac{1}{2}$
B: $\frac{1}{2}$
C: $-\frac{\sqrt{3}}{2}$
D: $\frac{\sqrt{3}}{2}$ | Given $\sin(\pi - A) = \frac{1}{2}$, we can deduce that $\sin A = \frac{1}{2}$.
Therefore, $\cos\left(\frac{\pi}{2} - A\right) = \sin A = \frac{1}{2}$.
Hence, the correct option is: $\boxed{\text{B}}$.
This problem can be directly solved by simplifying the function value using the trigonometric identities. It tests ... | cn_k12 | 207de32c-923e-51d4-b3a0-db9a408e49df | open-r1/OpenR1-Math-220k | [
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] |
In the interior of triangle $ABC$, point $P$ is positioned such that $\angle PAC = 10^{\circ}, \angle PCA = 20^{\circ}, \angle PAB = 30^{\circ}$, and $\angle ABC = 40^{\circ}$. What is the measure of $\angle BPC$ in degrees? | The triangle ABC is clearly isosceles. Let's draw its axis of symmetry $t$, and reflect point $P$ over $t$, with its reflection being $P^{\prime}$. Due to the reflection, $\angle ACP = \angle BCP^{\prime} = 20^{\circ}$. From this, using the fact that $\angle ACB = 100^{\circ}$, we get $\angle PCP^{\prime} = 60^{\circ}$... | olympiads | 6a3440b8-8c9d-51eb-af03-829432f3d0f8 | open-r1/OpenR1-Math-220k | [
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] |
Given the set $A=\{1,3,5,7\}$ and $B=\{x|x^{2}-2x-5\leqslant 0\}$, then $A\cap B=\left(\ \ \right)$
A: $\{1,3\}$
B: $\{1,5\}$
C: $(\{5,7\}$
D: $\{1,7\}$ | To find the intersection $A\cap B$ given $A=\{1,3,5,7\}$ and $B=\{x|x^{2}-2x-5\leqslant 0\}$, we need to check which elements of $A$ satisfy the inequality defining set $B$.
1. **For $x=1$:**
- Plug $x=1$ into the inequality:
$$
1^{2}-2(1)-5 = 1-2-5 = -6 \leqslant 0
$$
This shows that $1$ satisfies th... | cn_k12 | bd248f08-9833-5deb-b83b-a45922ceb98a | open-r1/OpenR1-Math-220k | [
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] |
The graph of the function $y=\log_{2}(x-1)$ is denoted as C. To obtain the graph of the function $y=\log_{2}(x+1)$, all points on C need to be ( )
A: moved 1 unit to the right
B: moved 1 unit to the left
C: moved 2 units to the right
D: moved 2 units to the left | Since $y=\log_{2}(x+1)=\log_{2}(x+2-1)$,
we can obtain the graph of $y=\log_{2}(x+1)$ by moving the graph of $y=\log_{2}(x-1)$ 2 units to the left.
Therefore, the correct answer is $\boxed{\text{D}}$. | cn_k12 | 389871bd-6f27-5858-be1c-7846681c57ca | open-r1/OpenR1-Math-220k | [
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] |
Nicki spent the last year running a lot. For the first half of the year, she ran a total of 20 miles per week. For the second half of the year, she increased the mileage to 30 miles per week. How many miles total did she run for the year? | Each half of the year consists of 52 / 2 = <<52/2=26>>26 weeks.
For the first half of the year, Nicki ran a total of 20 * 26 = <<20*26=520>>520 miles.
For the second half of the year, Nicki ran a total of 30 * 26 = <<30*26=780>>780 miles.
For the entire year, Nicki ran a total of 520 + 780 = <<520+780=1300>>1,300 miles... | null | null | openai/gsm8k | [
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] |
Given $a$ and $b$ with $a > b$, which of the following statements is correct?
A: $\frac{a}{b} > 1$
B: $\frac{1}{a} |b|$
D: $a^3 > b^3$ | For option A, if we take $a=1$ and $b=-1$, then $\frac{a}{b} = -1$, which means option A is not always true.
For option D, considering the power function $y=x^3$ is an increasing function, if $a > b$, then $a^3 > b^3$. Therefore, option D is correct.
For option B, if we take $a=1$ and $b=-1$, then $\frac{1}{a} = 1 > ... | cn_k12 | 47000889-7b91-580a-b638-6529cdb4ecd3 | open-r1/OpenR1-Math-220k | [
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] |
15. (12 points) For the function $f(x)$, if $f(x)=x$, then $x$ is called a "fixed point" of $f(x)$; if $f(f(x))=x$, then $x$ is called a "stable point" of $f(x)$. The sets of "fixed points" and "stable points" of the function $f(x)$ are denoted as $A$ and $B$, respectively, i.e., $A=\{x \mid f(x)=x\}, B=\{x \mid f(f(x)... | (1) If $A=\varnothing$, then $A \subseteq B$ is obviously true.
If $A \neq \varnothing$, let $t \in A$, then
$$
f(t)=t, f(f(t))=f(t)=t \text {. }
$$
Thus, $t \in B$, hence $A \subseteq B$.
(2) The elements of $A$ are the real roots of the equation $f(x)=x$, i.e., $a x^{2}-1=x$. Since $A \neq \varnothing$, we have
$$
a... | cn_contest | 068775e7-fc57-5b35-a2e9-38cff6c44050 | open-r1/OpenR1-Math-220k | [
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] |
Example 1 (Question from the 11th "Hope Cup" Invitational Competition) Let $a > b > c, n \in \mathbf{N}$, and $\frac{1}{a-b}+\frac{1}{b-c} \geqslant \frac{n}{a-c}$ always holds, then the maximum value of $n$ is ( ).
A. 2
B. 3
C. 4
D. 5 | Solution: Choose C. Reason: $\frac{1}{a-b}+\frac{1}{b-c} \geqslant \frac{n}{a-c} \Leftrightarrow \frac{a-c}{a-b}+\frac{a-c}{b-c} \geqslant n$, thus $n \leqslant\left[\frac{a-c}{a-b}+\frac{a-c}{b-c}\right]_{\text {min }}$. And $\frac{a-c}{a-b}+\frac{a-c}{b-c}=2+\frac{b-c}{a-b}+\frac{a-b}{b-c} \geqslant 4$, and when $2 b... | olympiads | b841d64a-fd5d-5cb4-9b84-20d8bf9df44c | open-r1/OpenR1-Math-220k | [
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] |
Hagrid has 100 animals. Among these animals,
- each is either striped or spotted but not both,
- each has either wings or horns but not both,
- there are 28 striped animals with wings,
- there are 62 spotted animals, and
- there are 36 animals with horns.
How many of Hagrid's spotted animals have horns?
(A) 8
(B) 10
... | Each of the animals is either striped or spotted, but not both.
Since there are 100 animals and 62 are spotted, then there are $100-62=38$ striped animals. Each striped animal must have wings or a horn, but not both.
Since there are 28 striped animals with wings, then there are $38-28=10$ striped animals with horns.
... | olympiads | 79020226-3232-5348-8558-024a0664666a | open-r1/OpenR1-Math-220k | [
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] |
In Rt $\triangle A B C$, it is known that $\angle A=$ $20^{\circ}, \angle B=90^{\circ}, A D$ is the bisector of $\angle B A C$, point $E$ is on side $A B$, and lines $C E$ and $D E$ are connected. If $\angle D C E=30^{\circ}$, find the degree measure of $\angle A D E$. | Solve As shown in Figure 2, construct $\angle A E F=20^{\circ}, E F$ intersects $A D$ at point $F$. Take point $G$ on side $A C$ such that $A G=A E$, and connect $F G$, $E G$, and $C F$.
Since $A D$ bisects $\angle E A G$, by its symmetry we know
$$
F E=F G,
$$
and
$$
\begin{array}{l}
\angle G F D=\angle E F D \\
=\an... | cn_contest | f8b0dfa2-155b-5be1-ab3d-bcf3edbe6747 | open-r1/OpenR1-Math-220k | [
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] |
Let $s_k$ denote the sum of the $\textit{k}$th powers of the roots of the polynomial $x^3-5x^2+8x-13$. In particular, $s_0=3$, $s_1=5$, and $s_2=9$. Let $a$, $b$, and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$, $3$, $....$ What is $a+b+c$?
$\textbf{(A)} \; -6 \qquad \te... | Applying [Newton's Sums](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/Newton's_Sums), we have\[s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,\]so\[s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},\]we get the answer as $5+(-8)+13=10$. | amc_aime | 4fdc7f09-f7b0-50f0-89d8-0c1bc4fc2fc1 | open-r1/OpenR1-Math-220k | [
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] |
6.15 Let $k$ be a positive integer, and the quadratic equation
$$
(k-1) x^{2}-p x+k=0
$$
has two positive integer roots. Find the value of $k^{k p}\left(p^{p}+k^{k}\right)$.
(China Beijing Junior High School Grade 2 Mathematics Competition, 1984) | ［Solution］Since $k$ is a positive integer, and the equation
$$
(k-1) x^{2}-p x+k=0
$$
is a quadratic equation, then $k \geqslant 2$.
Let the two positive integer roots of equation (1) be $x_{1}$ and $x_{2}$. By Vieta's formulas, we have
$$
x_{1} x_{2}=\frac{k}{k-1} \text {. }
$$
If $k-1 \neq 1$, then $(k-1, k)=1$, in... | olympiads | 576d9805-f2a2-5211-b243-a0d1933221e7 | open-r1/OpenR1-Math-220k | [
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1. Given $P A 、 P B 、 P C$ are three non-coplanar rays emanating from point $P$, and the angle between any two rays is $60^{\circ}$. There is a sphere with a radius of 1 that is tangent to all three rays. Find the distance from the center of the sphere $O$ to point $P$. | Consider the three face diagonals originating from a vertex of the cube, thus, the center of the sphere in the problem is the center of the inscribed sphere of the cube. $O P=\sqrt{3}$. | olympiads | 24c9bf00-5f55-5632-8986-47678a929059 | open-r1/OpenR1-Math-220k | [
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] |
Given $\overrightarrow{a}=(1,1)$ and $\overrightarrow{b}=(2,n)$, if $|\overrightarrow{a}+ \overrightarrow{b}|= \overrightarrow{a}\cdot \overrightarrow{b}$, then $n=$ ______. | Since we know $\overrightarrow{a}=(1,1)$ and $\overrightarrow{b}=(2,n)$, we have $\overrightarrow{a}+ \overrightarrow{b}=(3,1+n)$ and $\overrightarrow{a}\cdot \overrightarrow{b}=2+n$.
From the condition $|\overrightarrow{a}+ \overrightarrow{b}|= \overrightarrow{a}\cdot \overrightarrow{b}$, we get $\sqrt{9+(1+n)^{2}}=2... | cn_k12 | 93634a68-333b-5347-8624-4290aff279f1 | open-r1/OpenR1-Math-220k | [
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] |
Given that the sum of the first $n$ terms of a geometric series $\{a_n\}$ is $S_n = 3^{n+1} + a$, find the value of $a$. | We know that the sum of the first $n$ terms of a geometric series $\{a_n\}$ is given by $S_n = 3^{n+1} + a$.
Let's find the first term $a_1$ and second term $a_2$ using the given sum $S_n$:
$$ a_1 = S_1 = 3^{1+1} + a = 9 + a $$
The second term $a_2$ can be found using the sum up to the second term and subtracting the... | cn_k12 | 62ca34e1-d6e5-55dd-8136-adf7d5b04c3a | open-r1/OpenR1-Math-220k | [
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] |
For a positive integer $n$, there is a school with $2n$ people. For a set $X$ of students in this school, if any two students in $X$ know each other, we call $X$ [i]well-formed[/i]. If the maximum number of students in a well-formed set is no more than $n$, find the maximum number of well-formed set.
Here, an empty s... | 1. **Graph Representation and Definitions**:
- Let $\mathcal{G}$ be the graph representing the school where vertices represent students and edges represent the acquaintance between students.
- A set $X$ of students is called *well-formed* if any two students in $X$ know each other, i.e., $X$ forms a clique in $\m... | aops_forum | b8dccd36-32d2-5de5-bf58-bdb161a59e91 | open-r1/OpenR1-Math-220k | [
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] |
The equation of the tangent line to the curve $y=\frac{2x-1}{x+2}$ at the point $\left(-1,-3\right)$ is ______. | To find the equation of the tangent line to the curve $y=\frac{2x-1}{x+2}$ at the point $\left(-1,-3\right)$, we first need to calculate the derivative of $y$ with respect to $x$, which gives us the slope of the tangent line at any point on the curve.
Given $y=\frac{2x-1}{x+2}$, we apply the quotient rule for differen... | cn_k12 | ac353fd2-e401-59af-9a57-af014f631444 | open-r1/OpenR1-Math-220k | [
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] |
8.1. In a circle, 58 balls of two colors - red and blue - are arranged. It is known that the number of triples of consecutive balls, among which there are more red ones, is the same as the number of triples with a majority of blue ones. What is the smallest number of red balls that could be present?
![](https://cdn.ma... | Answer: 20
Evaluation
There are a total of 58 threes, so the number of threes with dominance of each of the two colors is 29. Therefore, the number of red balls should be no less than $\frac{2}{3} * 29$, which is no less than 20.
Example
![](https://cdn.mathpix.com/cropped/2024_05_06_6c94238ddc8917ebef5dg-8.jpg?hei... | olympiads | 4ab34745-b1cd-51fc-af10-10750ccbc246 | open-r1/OpenR1-Math-220k | [
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] |
If $\sqrt{a+1}+\sqrt{b-1}=0$, then the value of $a^{1011}+b^{1011}$ is:
A: $2$
B: $0$
C: $1$
D: $-2$ | Given that $\sqrt{a+1}+\sqrt{b-1}=0$, we aim to find the value of $a^{1011}+b^{1011}$.
First, we observe that for the sum of two square roots to equal zero, each square root must itself be zero or the square roots must be negatives of each other. However, since square roots are non-negative for real numbers, we conclu... | cn_k12 | f3202167-5f80-5aa3-b44c-816cae620744 | open-r1/OpenR1-Math-220k | [
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] |
Example 5 If real numbers $a, b, c$ satisfy $a+b+c=1, a^{2}+b^{2}+c^{2}=2, a^{3}+b^{3}+c^{3}=3$, find the values of $a b c$ and $a^{4}+b^{4}+c^{4}$. | Analysis Let $a, b, c$ be the roots of the polynomial $f(t)=t^{3}-\sigma_{1} t^{2}+\sigma_{2} t-\sigma_{3}$. Denote $s_{k}=a^{k}+b^{k}+c^{k} (k=1,2,3,4)$. By Newton's formulas, we can find $\sigma_{1}, \sigma_{2}, \sigma_{3}$, and then solve the problem.
Solution Let $a, b, c$ be the roots of the polynomial $f(t)=t^{3... | olympiads | 3e9f25c6-22f3-5c0c-964b-29a2d3e5be5d | open-r1/OpenR1-Math-220k | [
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The following figure shows a [i]walk[/i] of length 6:
[asy]
unitsize(20);
for (int x = -5; x <= 5; ++x)
for (int y = 0; y <= 5; ++y)
dot((x, y));
label("$O$", (0, 0), S);
draw((0, 0) -- (1, 0) -- (1, 1) -- (0, 1) -- (-1, 1) -- (-1, 2) -- (-1, 3));
[/asy]
This walk has three interesting properties:
[li... | To solve the problem of finding the number of northern walks of length 6, we need to consider the properties of such walks and use combinatorial methods to count them.
1. **Understanding the Walk Structure**:
- Each northern walk starts at the origin.
- Each step is 1 unit north, east, or west, with no south ste... | aops_forum | 7dd61727-8883-5232-85be-468d63d4d426 | open-r1/OpenR1-Math-220k | [
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] |
In the Cartesian coordinate plane $(xOy)$, it is given that $\overrightarrow{OA}=(-1,t)$ and $\overrightarrow{OB}=(2,2)$. If $\angle ABO=90^{\circ}$, find the value of the real number $t$. | Given $\overrightarrow{OA}=(-1,t)$ and $\overrightarrow{OB}=(2,2)$, we can find $\overrightarrow{AB}$ by subtracting the components of $\overrightarrow{OA}$ from $\overrightarrow{OB}$:
$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2, 2) - (-1, t) = (3, 2-t)$$
Since $\angle ABO = 90^{\circ}$, th... | cn_k12 | 9f846c87-2da2-557c-8a63-51c5004362c6 | open-r1/OpenR1-Math-220k | [
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] |
The asymptotes of the hyperbola $$\frac {x^{2}}{9}- \frac {y^{2}}{4}=1$$ are tangent to the circle $(x-3)^{2}+y^{2}=r^{2}$ ($r>0$). Then, $r=$ （ ）
A: $$\frac {6 \sqrt {13}}{13}$$
B: $$\frac {6 \sqrt {7}}{7}$$
C: $$\frac {6 \sqrt {11}}{11}$$
D: $$\sqrt {3}$$ | The equations of the asymptotes of the hyperbola are $y=\pm \frac {2}{3}x$, which can also be written as $x\pm \frac {3}{2}y=0$.
The distance $d$ from the center of the circle $(3,0)$ to the line is $d= \frac {|3|}{ \sqrt {1+( \frac {3}{2})^{2}}} = \frac {6 \sqrt {13}}{13}$.
Since the asymptotes of the hyperbola $\fr... | cn_k12 | f7f1f460-b4a4-5a8e-a1d3-86606581c486 | open-r1/OpenR1-Math-220k | [
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] |
1. Let $i_{1}, i_{2}, \cdots, i_{10}$ be a permutation of $1,2, \cdots, 10$. Define $S=\left|i_{1}-i_{2}\right|+\left|i_{3}-i_{4}\right|+\cdots+\left|i_{9}-i_{10}\right|$. Find all possible values of $S$.
[2] | Since $S \geqslant 1+1+1+1+1=5$, $S \leqslant 6+7+\cdots+10-(1+2+\cdots+5)=25$,
and $S \equiv \sum_{k=1}^{10} k(\bmod 2) \equiv 1(\bmod 2)$,
therefore, it only needs to be proven that $S$ can take all odd numbers from 5 to 25. | cn_contest | c3dad042-0f2b-5eae-9ef1-e65c378a3f6d | open-r1/OpenR1-Math-220k | [
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] |
The distance from a fixed point $P$ on the plane to two vertices $A, B$ of an equilateral triangle $A B C$ are $A P=2 ; B P=3$. Determine the maximum value that the segment $P C$ can have.
# | Let $A, B, C$ and $P$ be points on a plane such that $AB = BC = CA$, $AP = 2$, and $BP = 3$. Draw a ray $BM$ from point $B$ such that $\angle CBM = \angle ABP$, and mark a segment $BP' = PB$ on this ray. From the equality of angles: $\angle CBM = \angle ABP$, it follows that $\angle PBP' = \angle ABC = 60^\circ$, and t... | olympiads | 7dd26e20-ed80-5ba6-8387-2bf2ff7dcb2a | open-r1/OpenR1-Math-220k | [
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Given the equation about $x$: $$( \frac {1}{2})^{x}-x^{ \frac {1}{3}}=0$$, which of the following intervals contains the root of the equation?
A: $$(0， \frac {1}{3})$$
B: $$( \frac {1}{3}， \frac {1}{2})$$
C: $$( \frac {1}{2}， \frac {2}{3})$$
D: $$( \frac {2}{3}，1)$$ | Let $f(x) = ( \frac {1}{2})^{x} - x^{ \frac {1}{3}}$. It is obvious that $f(x)$ is decreasing in the interval $(0, +\infty)$.
Since $f(\frac {1}{3}) \cdot f(\frac {1}{2}) < 0$,
it implies that $f(x)$ has a zero point in the interval $(\frac {1}{3}, \frac {1}{2})$.
Therefore, the equation $$( \frac {1}{2})^{x}-x^{ \f... | cn_k12 | 37882bf5-4f1e-5163-b6cd-11438eede99d | open-r1/OpenR1-Math-220k | [
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] |
Find, with proof, all real numbers $ x \in \lbrack 0, \frac {\pi}{2} \rbrack$, such that $ (2 \minus{} \sin 2x)\sin (x \plus{} \frac {\pi}{4}) \equal{} 1$. | 1. We start with the given equation:
\[
(2 - \sin 2x) \sin \left( x + \frac{\pi}{4} \right) = 1
\]
2. Using the double-angle identity for sine, we have:
\[
\sin 2x = 2 \sin x \cos x
\]
Substituting this into the equation, we get:
\[
(2 - 2 \sin x \cos x) \sin \left( x + \frac{\pi}{4} \right)... | aops_forum | 090fc074-3417-5cf2-8d40-d501a3eb0b6f | open-r1/OpenR1-Math-220k | [
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] |
Example 23. Solve the system
$$
\left\{\begin{array}{l}
4 \log _{2}^{2} x+1=2 \log _{2} y \\
\log _{2} x^{2} \geqslant \log _{2} y
\end{array}\right.
$$ | Solution. The domain of admissible values of the system is defined by the system of inequalities $x>0, y>0$. The second inequality of the system on the domain of admissible values is equivalent to the inequality $2 \log _{2} x \geqslant \log _{2} y$, replacing $2 \log _{2} y$ in which with $4 \log _{2}^{2} x+1$, we obt... | olympiads | 384be002-a140-5c90-8e4c-be41f0ef705f | open-r1/OpenR1-Math-220k | [
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] |
124 Given that the inverse function of $y=f(x)$ is $g(x)=\log _{\sin ^{2} \theta}\left(\frac{1}{x}-\cos ^{2} \theta\right)$, where the constant $\theta \in$ $\left(0, \frac{\pi}{2}\right)$, then the solution to the equation $f(x)=1$ is $\qquad$ . | 124 1. The solution to $f(x)=1$ is
$$
g(1)=\log _{\sin ^{2} \theta}\left(1-\cos ^{2} \theta\right)=1 .
$$ | olympiads | 0a04ee78-dd2d-5f9e-adfe-80adf6b34db2 | open-r1/OpenR1-Math-220k | [
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Let $S$ be the set of [ordered pairs](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) $(x, y)$ such that $0 < x \le 1, 0<y\le 1,$ and $\left[\log_2{\left(\frac 1x\right)}\right]$ and $\left[\log_5{\left(\frac 1y\right)}\right]$ are both even. Given that the area of the graph of $S$ is $m/n,$ where $m$ and ... | $\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor$ is even when
\[x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots\]
Likewise:
$\left\lfloor\log_5\left(\frac{1}{y}\right)\right\rfloor$ is even when
\[y \in \left(\frac{1}{5},1\right... | amc_aime | e16f1695-d30d-534c-b784-feec0520adc8 | open-r1/OpenR1-Math-220k | [
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] |
52. If the solution to the linear equation in one variable $a x+b-5=0$ is $x=2$, then $4 a^{2}+b^{2}+4 a b-2 a-b$ $+3=$ $\qquad$ . | Reference answer: 23 | olympiads | 2534ed25-249c-54f8-a58b-d37883c0ae1d | open-r1/OpenR1-Math-220k | [
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] |
1. Find all four-digit numbers $\overline{a b c d}$, for which $\overline{a b c d}=20 \cdot \overline{a b}+16 \cdot \overline{c d}$. | 1. In the equation from the assignment
$$
1000 a+100 b+10 c+d=20(10 a+b)+16(10 c+d)
$$
the unknown digits $a$ and $b$ have larger coefficients on the left side, while the digits $c$ and $d$ have larger coefficients on the right side. Therefore, we rearrange the equation to the form $800 a+80 b=150 c+15 d$, which afte... | olympiads | a2108f18-da9c-5fc4-b931-9867d0f40a5a | open-r1/OpenR1-Math-220k | [
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] |
1. A domino has a left end and a right end, each of a certain color. Alice has four dominos, colored red-red, red-blue, blue-red, and blue-blue. Find the number of ways to arrange the dominos in a row end-to-end such that adjacent ends have the same color. The dominos cannot be rotated. | Answer:
4
Solution: Without loss of generality assume that the the left end of the first domino is red. Then, we have two cases:
If the first domino is red-red, this forces the second domino to be red-blue. The third domino cannot be blue-red, since the fourth domino would then be forced to be blue-blue, which is impo... | olympiads | 5f39f7ca-b5a1-5b81-9cdc-c298c6ef7d02 | open-r1/OpenR1-Math-220k | [
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Given that the function $f(x)$ is an even function on $(-\infty,+\infty)$, and for $x \geq 0$, $f(x+2)=f(x)$ holds. Also, when $x \in [0, 2)$, $f(x)=\log_{2}(x+1)$. Find the value of $f(-2010)+f(2011)$. | Since $f(x)$ is an even function, we have $f(-x) = f(x)$ for all $x$. Therefore, $f(-2010) = f(2010)$.
Given $f(x+2) = f(x)$ for $x \geq 0$, we can apply this property repeatedly to find the value of $f(2010)$ and $f(2011)$ by reducing these arguments until they fall within the interval $[0, 2)$.
For $f(2010)$, sin... | cn_k12 | abaf567c-3e9c-5e6c-978d-c9de0c6a1d21 | open-r1/OpenR1-Math-220k | [
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Two planes that are perpendicular to the same plane are parallel. ( )
A: True
B: False
C:
D: | **Analysis**
This question examines the positional relationship between planes in space, and it can be directly solved based on the relationship between the positions of the planes.
**Solution**
Solution: Two planes that are perpendicular to the same plane can either be parallel or perpendicular.
Therefore, the sta... | cn_k12 | d4e35bc6-2786-576d-a6e0-3a25402d273d | open-r1/OpenR1-Math-220k | [
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] |
If all terms of the geometric sequence $\{a_n\}$ are positive, and $a_{10}a_{11}+a_{9}a_{12}=2e^{5}$, then $\ln a_{1}+\ln a_{2}+\ldots+\ln a_{20}$ equals \_\_\_\_\_\_\_\_. | **Analysis**
This problem examines the operational properties of geometric sequences, the operational properties of logarithms, and computational ability. It is a basic question.
**Solution**
Given that $\{a_n\}$ is a geometric sequence, and $a_{10}a_{11}+a_{9}a_{12}=2e^{5}$,
Therefore, $a_{10}a_{11}+a_{9}a_{12}=2a... | cn_k12 | 68a4b8e4-e6ee-5c0a-8aae-06ac16c3a2d6 | open-r1/OpenR1-Math-220k | [
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] |
Given matrices A = $$\begin{bmatrix} 1 & 0 \\ 0 & 2\end{bmatrix}$$, B = $$\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$$, if line $l$ undergoes transformations $T\_A$, $T\_B$ in sequence to obtain line $l'$: 2x + y - 2 = 0, find the equation of line $l$. | Let point $P(x, y)$ be any point on line $l$, which undergoes transformations $T\_A$, $T\_B$ in sequence to obtain point $P'(x', y')$.
Then we have: $$\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 \\ 0 & 2\end{bmatrix}$$ $$\begin{bmatrix} x \\ y\end{bmatrix}$$ = $$\begin{bmatrix} x' \\ y'\end{b... | cn_k12 | 830b674c-ac40-5e03-9e81-8dc6d170716a | open-r1/OpenR1-Math-220k | [
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] |
A triangle $ABC$ is given, in which the segment $BC$ touches the incircle and the corresponding excircle in points $M$ and $N$. If $\angle BAC = 2 \angle MAN$, show that $BC = 2MN$.
(N.Beluhov) | 1. **Assume \(AB < AC\)**: This assumption helps us to orient the triangle and the points correctly for the argument that follows.
2. **Define points and properties**:
- Let \(D\) be the antipode of \(M\) with respect to the incircle \(\omega\) of \(\triangle ABC\). This means \(D\) is the point on \(\omega\) such ... | aops_forum | a279a94b-ae23-5660-8d40-e0beb6545a7c | open-r1/OpenR1-Math-220k | [
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] |
Given sets $A=\{x|x^{2}-5x-6 \lt 0\}$ and $B=\{x|202{2}^{x}>\sqrt{2022}\}$, then $A\cap B=\left(\ \ \right)$
A: $(\frac{1}{2}, 1)$
B: $(\frac{1}{2}, 6)$
C: $(-1, \frac{1}{2})$
D: $(\frac{1}{2}, 3)$ | To solve the problem, we need to find the sets $A$ and $B$ first, and then find their intersection $A \cap B$.
**Step 1: Find the set $A$**
Given $A=\{x|x^{2}-5x-6 \lt 0\}$, we factor the quadratic inequality:
\begin{align*}
x^{2}-5x-6 &\sqrt{2022}\}$, we can rewrite the inequality to find the range of $x$:
\begin{al... | cn_k12 | a4a532cd-f64c-508e-a8e6-f0c6bd6112db | open-r1/OpenR1-Math-220k | [
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13.030. Three brigades of workers built an embankment. The entire work is valued at 325500 rubles. What salary will each brigade receive if the first one consisted of 15 people and worked for 21 days, the second one - of 14 people and worked for 25 days, and the number of workers in the third brigade, which worked for ... | ## Solution.
Let $x$ rubles be received by one person for one day of work. Then the first team will receive $x \cdot 15 \cdot 21$ rubles; the second - $x \cdot 14 \cdot 25$ rubles; the third $x \cdot 1.4 \cdot 15 \cdot 20$ rubles. According to the condition $15 \cdot 21 x + 14 \cdot 25 x + 1.4 \cdot 15 \cdot 20 x = 32... | olympiads | 7a51662b-51a8-52df-bddb-704c6c2bad5b | open-r1/OpenR1-Math-220k | [
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] |
The correspondence f given below can constitute a function from set A=(-1,1) to set B=(-1,1) ( )
A: f: x→2x
B: f: x→|x|
C: f: x→$x^{ \frac {1}{2}}$
D: f: x→tanx | For option A, when x=-1 and x=1, there is no unique value in set B that corresponds to it through the correspondence f: x→2x, hence it cannot constitute a function from set A=(-1,1) to set B=(-1,1).
For option B, when x=-1 and x=1, there is a unique value 1 in set B that corresponds to it through the correspondence f:... | cn_k12 | 25d979f7-e1ea-5e02-942b-34520fff53cd | open-r1/OpenR1-Math-220k | [
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19.4.3 ** For any positive integer $q_{0}$, consider the sequence $q_{1}, q_{2}, \cdots, q_{n}$ defined by $q_{i}=\left(q_{i-1}-1\right)^{3}+3, i=1,2, \cdots, n$. If each $q_{i}(i=1,2, \cdots, n)$ is a power of a prime, find the largest possible value of $n$. | Since $m^{3}-m=(m-1) m(m+1)$ is divisible by 3, we have $m^{3} \equiv m(\bmod 3)$. Therefore, $q_{i}-\left(q_{i-1}-1\right)^{3}+3 \equiv q_{i-1}-1(\bmod 3)$. Thus, one of $q_{1}, q_{2}, q_{3}$ must be divisible by 3, making it a power of 3. However, $\left(q_{i}-1\right)^{3}+3$ is a power of 3 only when $q_{i}=1$, so $... | olympiads | 89f7286c-3932-5cad-9726-dfebbee4f2d8 | open-r1/OpenR1-Math-220k | [
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Example 2.3.5 (Bulgarian MO 2002) Find the smallest number $k$, such that $\frac{t_{a}+t_{b}}{a+b}<k$, where $a, b$ are two sides of a triangle, and $t_{a}, t_{b}$ are the lengths of the angle bisectors corresponding to these two sides. | Prove by first considering the extreme case: $\triangle A B C$ satisfies $a=b$, and the base angle $\angle A \rightarrow 0$ when $c \rightarrow 2 b$, and
\[ c2\left(\frac{2}{3}-\frac{b}{a+2 b}-\frac{a}{2 a+b}\right)=\frac{2(a-b)^{2}}{3(a+2 b)(2 a+b)} \geqslant 0 . \]
Therefore, $\frac{t_{a}+t_{b}}{a+b}<\frac{4}{3}$, ... | olympiads | 9994db55-f9db-5ebe-b7f5-5fcee0e1542b | open-r1/OpenR1-Math-220k | [
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Solve the following equations:
(1) $(x+2)^2=2x+4$;
(2) $x^2-2x-5=0$;
(3) $x^2-5x-6=0$;
(4) $(x+3)^2=(1-2x)^2$. | Let's solve each equation step by step:
### Equation (1): $(x+2)^2=2x+4$
First, we expand and simplify the equation:
\[
(x+2)^2 - 2x - 4 = 0 \implies x^2 + 4x + 4 - 2x - 4 = 0 \implies x^2 + 2x = 0
\]
Factor out $x$:
\[
x(x + 2) = 0
\]
Setting each factor equal to zero gives us the solutions:
\[
x = 0 \quad \text{or}... | cn_k12 | 3b86e1ea-d82a-5f8b-8a62-a5a6284544cb | open-r1/OpenR1-Math-220k | [
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16. Let $\left\{a_{n}\right\}$ be a sequence of positive integers such that $a_{1}=1, a_{2}=2009$ and for $n \geq 1$, $a_{n+2} a_{n}-a_{n+1}^{2}-a_{n+1} a_{n}=0$. Determine the value of $\frac{a_{993}}{100 a_{991}}$. | 16. Answer: 89970
$$
a_{n+2} a_{n}-a_{n+1}^{2}-a_{n+1} a_{n}=0 \quad \Rightarrow \quad \frac{a_{n+2} a_{n}-a_{n+1}^{2}-a_{n+1} a_{n}}{a_{n+1} a_{n}}=0
$$
$\frac{a_{n+2}}{a_{n+1}}-\frac{a_{n+1}}{a_{n}}=1$. From here, we see that $\left\{\frac{a_{n+1}}{a_{n}}\right\}$ is an arithmetic sequence with first term 2009 and co... | olympiads | 8955b3cf-d48c-5bde-9c42-f19840523fe9 | open-r1/OpenR1-Math-220k | [
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Find the imaginary part of the complex number $\dfrac{1 + 2i}{1 + i}$ ($i$ is the imaginary unit). | We first simplify the given complex number by multiplying both the numerator and denominator by the conjugate of the denominator:
$$\dfrac{1 + 2i}{1 + i} = \dfrac{(1 + 2i)(1 - i)}{(1 + i)(1 - i)} = \dfrac{3 + i}{2} = \dfrac{3}{2} + \dfrac{1}{2}i$$
The imaginary part of a complex number $a + bi$ is given by $b$, where... | cn_k12 | a4ee1b9a-abd1-5a00-93c4-9e32f1b254d2 | open-r1/OpenR1-Math-220k | [
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13. From the sequence of positive integers $1,2,3,4, \cdots$, remove the multiples of 3 and 4, but retain all multiples of 5 (for example, $15,20,60,120$ are not removed). After the removal, the remaining numbers form a sequence in their original order: $a_{1}=1, a_{2}=$ $2, a_{3}=5, a_{4}=7, \cdots$. Find $a_{2003}$. | 13. Since the least common multiple of $3,4,5$ is 60, let $S=\{1,2, \cdots, 60\}, A_{i}=\{k \mid k \in S$ and $k$ is divisible by $i\}(i=3,4,5)$, then the set of numbers in $S$ that are not divided is $\left(\bar{A}_{3} \cap \bar{A}_{4} \cap \bar{A}_{5}\right) \cup A_{5}$. By the principle of inclusion-exclusion, we ca... | olympiads | 7147ae0c-2b8d-51b8-b44d-b3f8489bb4bb | open-r1/OpenR1-Math-220k | [
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If there exists a positive number $x$ such that $e^{x}(x+a) \lt 1$ holds, then the range of values for $a$ is:
A: $(-\infty ,+\infty )$
B: $(-\infty ,1)$
C: $({-∞，\frac{1}{e}-1})$
D: $(-\infty ,-1)$ | Given that there exists a positive number $x$ such that $e^{x}(x+a) 0$, we can evaluate $f(x)$ at $x = 0$ to find its maximum value in the interval $(0, +\infty)$:
\[f(0) = \frac{1}{e^{0}} - 0 = 1 - 0 = 1\]
Since $f(x)$ is monotonically decreasing for $x > 0$, for all $x > 0$, $f(x) < f(0)$, which means:
\[f(x) < 1\]... | cn_k12 | 99eeed74-ae7f-510e-9494-a7545952d754 | open-r1/OpenR1-Math-220k | [
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A3. In a class, $55 \%$ of students scored at least $55 \%$ on a test. $65 \%$ of students scored at most $65 \%$ on the same test. What percentage of students scored between $55 \%$ and $65 \%$ (inclusive) on the test? | SolUtion
$20 \%$
As $55 \%$ of students scored at least $55 \%$ on the test, $45 \%$ of the students scored less than $55 \%$. Also, $65 \%$ of students scored $65 \%$ or less on the test, so the percentage of students who scored between $55 \%$ and $65 \%$ (inclusive) is $65 \%-45 \%=20 \%$. | olympiads | ccfe953e-6a49-5782-9475-26a3dbb512bc | open-r1/OpenR1-Math-220k | [
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7. Given the parabola $C: y^{2}=8 x$ with focus $F$, the directrix intersects the $x$-axis at point $K$. Point $A$ is on the parabola $C$, and $|A K|=\sqrt{2}|A F|$. Then the area of $\triangle A F K$ is ( ).
(A) 4
(B) 8
(C) 16
(D) 32 | 7. B.
Since the focus of the parabola $C: y^{2}=8 x$ is $F(2,0)$, and the directrix is $x=-2$, we have $K(-2,0)$.
Let $A\left(x_{0}, y_{0}\right)$. Draw a perpendicular line $A B$ from point $A$ to the directrix, then $B\left(-2, y_{0}\right)$.
Given that $|A K|=\sqrt{2}|A F|$, and
$$
A F=A B=x_{0}-(-2)=x_{0}+2 \text... | cn_contest | 01716340-14fa-5769-bc1d-75d55a340296 | open-r1/OpenR1-Math-220k | [
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11.4. Inside a right-angled triangle with sides 3, 4, and 5 cm, there are two circles, the ratio of whose radii is 9 to 4. The circles touch each other externally, both touch the hypotenuse, one touches one leg, and the other touches the other leg. Find the radii of the circles. | Answer: $\frac{20}{47}$ cm and $\frac{45}{47}$ cm.
Solution. Let the radii of the circles be $4x$ and $9x$ respectively, and they touch the legs of the triangle as shown in Figure 2. Express through $x$ the distance $y$ between the points of tangency of the circles with the hypotenuse of the triangle (see Figure 1).
... | olympiads | a21d4f09-ba8e-55b0-80a0-867c52d7d47a | open-r1/OpenR1-Math-220k | [
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In the book "Nine Chapters on the Mathematical Art," there is a problem recorded as follows: A group of people went to buy an item together. If each person contributes $8$ coins, there will be an excess of $3$ coins; if each person contributes $7$ coins, there will be a shortage of $4$ coins. How many people are there?... | To solve the problem, we start by translating the given conditions into mathematical equations.
1. If each person contributes $8$ coins, there will be an excess of $3$ coins. This can be represented as the total amount collected being $3$ coins more than what is needed. If $x$ is the number of people, then the total a... | cn_k12 | ca5058ba-8eff-5271-b11d-9aa106cb6190 | open-r1/OpenR1-Math-220k | [
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A particle is launched from the surface of a uniform, stationary spherical planet at an angle to the vertical. The particle travels in the absence of air resistance and eventually falls back onto the planet. Spaceman Fred describes the path of the particle as a parabola using the laws of projectile motion. Spacewoman K... | 1. **Understanding the problem**: We need to determine why there is a discrepancy between Fred's description of the particle's path as a parabola and Kate's description of the path as an ellipse, based on Kepler's laws.
2. **Fred's perspective**: Fred uses the laws of projectile motion, which are typically applied in ... | aops_forum | ef1cfafb-64da-58e7-95b1-54a8435cae3d | open-r1/OpenR1-Math-220k | [
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"The condition 'a=1' is" sufficient but not necessary for the function $f(x) = \log(ax)$ to be monotonically increasing on $(0, +\infty)$." ( )
A: A sufficient but not necessary condition
B: A sufficient and necessary condition
C: A necessary but not sufficient condition
D: Neither sufficient nor necessary | When $a=1$, the function $f(x) = \log(ax) = \log x$ is monotonically increasing on $(0, +\infty)$, which proves sufficiency.
When the function $f(x) = \log(ax)$ is monotonically increasing on $(0, +\infty)$, it implies that $a > 0$. However, this does not lead to the conclusion that $a=1$, indicating that the necessi... | cn_k12 | 449ff3c5-6197-5408-ad71-33a9982675c1 | open-r1/OpenR1-Math-220k | [
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$19 \cdot 24$ If the sum of all but one of the angles of a convex polygon is $2190^{\circ}$, then the number of sides of the polygon should be
(A) 13.
(B) 15.
(C) 17.
(D) 19.
(E) 21.
(24th American High School Mathematics Examination, 1973) | [Solution] Let $n$ be the number of sides (angles) of the known convex polygon, and $x$ be the degree of the removed angle.
Then $180(n-2)=2190+x$,
so $n-2=\frac{2190+x}{180}=12+\frac{30}{180}+\frac{x}{180}=12+1$, noting that $0^{\circ}<x<180^{\circ}$.
Thus, $x=150^{\circ}, \therefore n=15$.
Therefore, the answer is $(... | olympiads | 58d96c80-9d2a-5233-876a-ac2429541dd0 | open-r1/OpenR1-Math-220k | [
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Given a point (1, 2), there always exist two lines that are tangent to the circle $x^2+y^2+kx+2y+k^2-15=0$. The range of values for $k$ is ( )
A: $k2$
B: $k2$ or $-\frac{8}{3}\sqrt{3}<k<-3$
D: $-\frac{8}{3}\sqrt{3}<k<-3$ or $2<k<\frac{8}{3}\sqrt{3}$ | First, convert the equation of the circle to its standard form: $(x+\frac{1}{2}k)^2+(y+1)^2=16-\frac{3}{4}k^2$.
Therefore, $16-\frac{3}{4}k^2>0$, which gives $-\frac{8}{3}\sqrt{3}0$, which simplifies to $(k-2)(k+3)>0$.
Solving this inequality, we find $k>2$ or $k<-3$.
Thus, the range of values for $k$ is $(-\frac{8}... | cn_k12 | f8a4bb8f-11e2-5b33-aba7-88d6d252a425 | open-r1/OpenR1-Math-220k | [
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Prove that for all real $x > 0$ holds the inequality $$\sqrt{\frac{1}{3x+1}}+\sqrt{\frac{x}{x+3}}\ge 1.$$
For what values of $x$ does the equality hold? | To prove the inequality
\[ \sqrt{\frac{1}{3x+1}} + \sqrt{\frac{x}{x+3}} \ge 1 \]
for all \( x > 0 \), we will use the method of substitution and algebraic manipulation.
1. **Substitution and Simplification:**
Let \( y = \sqrt{\frac{1}{3x+1}} \). Then, we have:
\[ y^2 = \frac{1}{3x+1} \]
which implies:
\[ ... | aops_forum | 92903904-da53-5f46-b069-69ffdeabf15a | open-r1/OpenR1-Math-220k | [
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Mom gave Vasya money for 30 pencils. It turned out that the pencil factory was running a promotional campaign in the store: in exchange for a receipt for a set of 20 pencils, they return $25\%$ of the set's cost, and for a receipt for a set of 5 pencils, they return $10\%$. What is the maximum number of pencils Vasya c... | Note that $25 \%$ of the cost of 20 pencils is the cost of 5 pencils, and $10 \%$ of the cost of 5 pencils is half the cost of a pencil. It is clear that to get the maximum discount, Vasya should act as follows: as long as he has enough money, he should buy a set of 20 pencils and immediately exchange the receipt; if h... | olympiads | 6d4f4884-616b-5117-b270-4dd8af4361eb | open-r1/OpenR1-Math-220k | [
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On the sides $AB$ and $AC$ of triangle $ABC$, points $K$ and $L$ are located such that $AK: KB = 4: 7$ and $AL: LC = 3: 2$. The line $KL$ intersects the extension of side $BC$ at point $M$. F... | Through point $A$, draw a line parallel to $B C$. Let $T$ be the point of its intersection with line $K L$. From the similarity of triangles $A L T$ and $C L M$, we find that $A T = \frac{3}{2} C M$, and from the similarity of triangles $A K T$ and $B K M - B M = \frac{7}{4} A T = \frac{7}{4} \cdot \frac{3}{2} C M = \f... | olympiads | ff81a7dd-d781-500c-9565-39d59dc8cdd6 | open-r1/OpenR1-Math-220k | [
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8. On the line $y=-13 / 6$ find the point $M$, through which two tangents to the graph of the function $y=x^{2} / 2$ pass, the angle between which is $60^{\circ}$. | Solution (without using derivatives).
$$
y=x^{2} / 2, M\left(x_{0} ;-13 / 6\right)
$$
The equation $\frac{1}{2} x^{2}=-\frac{13}{6}+k\left(x-x_{0}\right)$, or $x^{2}-2 k x+2 k x_{0}+\frac{13}{3}=0$, has a unique solution if $\frac{D}{4}=k^{2}-2 k x_{0}-\frac{13}{3}=0$. The two values of $k$ found from this equation m... | olympiads | 098b0e9a-447b-50ed-8fb4-29b779faaa87 | open-r1/OpenR1-Math-220k | [
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A college has three majors, A, B, and C, with a total of 1500 students. To investigate the situation of students working part-time to support their studies, it is planned to use stratified sampling to draw a sample of 150 students. It is known that there are 420 students in major A and 580 students in major B. Therefor... | Solution: According to the standard of stratified sampling, the number of students to be drawn from major C is calculated as $\frac{150}{1500} \times (1500 - 420 - 580) = 50$.
Therefore, the answer is $\boxed{50}$.
By establishing a proportional relationship based on the definition of stratified sampling, the conclus... | cn_k12 | 761653e9-08ae-5f8d-b741-655ddb69208c | open-r1/OpenR1-Math-220k | [
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2. Each of the equations $a x^{2}-b x+c=0$ and $c x^{2}-a x+b=0$ has two distinct real roots. The sum of the roots of the first equation is non-negative, and the product of the roots of the first equation is 9 times the sum of the roots of the second equation. Find the ratio of the sum of the roots of the first equatio... | # Solution.
From the condition, it follows that the coefficients $a, c \neq 0$.
By Vieta's theorem, from the condition it follows that $\frac{c}{a}=9 \frac{a}{c}$. Hence, $c^{2}=9 a^{2}$, which means $\left[\begin{array}{l}c=3 a, \\ c=-3 a .\end{array}\right.$
1 case.
$c=3 a$.
We get the equations $a x^{2}-b x+3 a... | olympiads | 62c6643f-7cbe-5c3c-a15b-b2048fe130fd | open-r1/OpenR1-Math-220k | [
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Four, (50 points) Given several rectangular boxes, the lengths of their edges are positive integers not greater than an odd positive integer $n$ (allowing the three edge lengths to be the same), and the thickness of the box walls is negligible. Each box has its three pairs of opposite faces painted red, blue, and yello... | Let the maximum number of harmonious boxes be $f(n)$.
In a three-dimensional Cartesian coordinate system, the coordinate planes
$$
\begin{array}{l}
\{(x, y, z) \mid x=0, y \geqslant 0, z \geqslant 0\}, \\
\{(x, y, z) \mid x \geqslant 0, y=0, z \geqslant 0\}, \\
\{(x, y, z) \mid x \geqslant 0, y \geqslant 0, z=0\}
\end... | cn_contest | b2373227-8034-5cfd-a9b9-b9cd12780962 | open-r1/OpenR1-Math-220k | [
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Given set A={x|x^2^-x-2≤0}, B={x|1≤2^x≤8, x∈Z}, find A∩B=（ ）
A: [-1，3]
B: {0，1}
C: [0，2]
D: {0，1，2} | Solution: Since set A={x|x^2^-x-2≤0}={x|-1≤x≤2},
B={x|1≤2^x≤8, x∈Z}={x|0≤x≤3, x∈Z}={0，1，2，3},
Hence, A∩B={0，1，2}.
So, the answer is: $\boxed{\text{D}}$.
We solve the quadratic inequality to find A, solve the exponential inequality to find B, and then find A∩B according to the definition of the intersection of two set... | cn_k12 | 7a0576a0-3113-54fc-bd46-e655f9b9990d | open-r1/OpenR1-Math-220k | [
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[ Sphere inscribed in a pyramid ]
[Volume helps solve the problem.]
In a triangular pyramid, two opposite edges are equal to 12 and 4, and the other edges are equal to 7. A sphere is inscribed in the pyramid. Find the distance from the center of the sphere to the edge equal to 12. | Let $A B C D$ be a triangular pyramid, where $A B=12, C D=4, A C=B C=A D=B D=7$. Since $D A = D B$, the orthogonal projection $P$ of vertex $D$ onto the plane $A B C$ is equidistant from points $A$ and $B$, so point $P$ lies on the perpendicular bisector of side $A B$ of the isosceles triangle $A B C$. Let $K$ be the m... | olympiads | b08f0f6e-911c-5264-9ea2-c4ddf0e506d1 | open-r1/OpenR1-Math-220k | [
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In a scalene triangle, where $a$ is the longest side, to conclude that the angle $A$ opposite to side $a$ is obtuse, the sides $a$, $b$, $c$ must satisfy _________. | By the cosine rule, $\cos A = \frac{b^2 + c^2 - a^2}{2bc} b^2 + c^2$. Thus, the sides $a$, $b$, $c$ must satisfy $\boxed{a^2 > b^2 + c^2}$. | cn_k12 | 840130ca-10ce-57cb-9072-261048c28ba8 | open-r1/OpenR1-Math-220k | [
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4. Given $a+b=5, a b=3$, then the value of the algebraic expression $a^{3} b-2 a^{2} b^{2}+a b^{3}$ is
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Answer: 39 | olympiads | 7508342c-9fd2-58ff-af54-cdc209d54577 | open-r1/OpenR1-Math-220k | [
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The sum of all positive integers less than 100 that can be divided by 7 is __________. | The sum of all positive integers less than 100 that can be divided by 7 is $\boxed{735}$. | cn_k12 | bda23eb0-9e4e-5088-8faf-dbeb375e7aca | open-r1/OpenR1-Math-220k | [
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Given a sequence $\{a_n\}$ whose sum of the first $n$ terms $S_n = 2n^2 - 2n$, then the sequence $\{a_n\}$ is ( )
A: An arithmetic sequence with a common difference of 4
B: An arithmetic sequence with a common difference of 2
C: A geometric sequence with a common ratio of 4
D: A geometric sequence with a common ratio o... | Since $S_n = 2n^2 - 2n$,
then $S_n - S_{n-1} = a_n = 2n^2 - 2n - [2(n-1)^2 - 2(n-1)] = 4n - 4$
Therefore, the sequence $\{a_n\}$ is an arithmetic sequence with a common difference of 4.
Hence, the correct answer is $\boxed{\text{A}}$. | cn_k12 | 18269efd-c628-5306-8edf-e2c6bc489ee1 | open-r1/OpenR1-Math-220k | [
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Example 10 Find $A^{2}$, where $A$ is the sum of the absolute values of all roots of the following equation:
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}
$$
(9th American Invitational Mathematics Examination) | Solution: According to the structural characteristics of the original equation, set
$$
\begin{array}{ll}
x=\sqrt{19}+\frac{91}{y}, & y=\sqrt{19}+\frac{91}{z}, \\
z=\sqrt{19}+\frac{91}{u}, & u=\sqrt{19}+\frac{91}{v}, \\
v=\sqrt{19}+\frac{91}{x} . &
\end{array}
$$
Assume $x>y$ and $x<y$ respectively, then
$$
\begin{arra... | cn_contest | cbe159fd-0bd7-5b27-b734-8b596e3db99e | open-r1/OpenR1-Math-220k | [
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The binomial coefficients of the third and fourth terms in the expansion of \\((x- \frac {2}{x})^{n}\\) are equal. The area enclosed by the line $y=nx$ and the curve $y=x^{2}$ is \_\_\_\_\_\_. | The binomial coefficients of the third and fourth terms in the expansion of \\((x- \frac {2}{x})^{n}\\) are equal,
i.e., \\( C\_{ n }^{ 2 }= C\_{ n }^{ 3 }\\). Solving for $n$, we get $n=5$.
Thus, the line is $y=5x$ and the curve is $y=x^{2}$.
The coordinates of the points of intersection of the line and the curve are ... | cn_k12 | 93a41106-6f1f-5611-af8c-627f86aef2c6 | open-r1/OpenR1-Math-220k | [
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Let $ABCD$ be a quadrilateral with side lengths $AB = 2$, $BC = 3$, $CD = 5$, and $DA = 4$. What is the maximum possible radius of a circle inscribed in quadrilateral $ABCD$? | 1. **Identify the problem**: We need to find the maximum possible radius of a circle inscribed in a quadrilateral \(ABCD\) with given side lengths \(AB = 2\), \(BC = 3\), \(CD = 5\), and \(DA = 4\).
2. **Determine the type of quadrilateral**: The area of a quadrilateral is maximized when it is cyclic. A cyclic quadril... | aops_forum | 201f1cfd-9c20-596c-be6b-66313188eab7 | open-r1/OpenR1-Math-220k | [
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# 3.2. Solve the system of equations:
$$
\left\{\begin{array}{l}
2^{x+y}-2^{x-y}=4 \\
2^{x+y}-8 \cdot 2^{y-x}=6
\end{array}\right.
$$ | Answer: $x=\frac{5}{2}, y=\frac{1}{2}$.
Solution.
$$
\begin{aligned}
& 2^{x+y}=u(>0), \quad 2^{y-x}=v(>0) \\
& \left\{\begin{array}{l}
u-v=4 \\
u-\frac{8}{v}=6
\end{array}\right. \\
& \left\{\begin{array}{l}
u=v+4 \\
v+4-\frac{8}{v}=6
\end{array}\right. \\
& v^{2}+4 v-6 v-8=0 \\
& v^{2}-2 v-8=0 \\
& {\left[\begin{arr... | olympiads | 51dcc02d-c809-51dc-8fe0-3a190483e1cc | open-r1/OpenR1-Math-220k | [
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] |
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