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Richard and Shreyas are arm wrestling against each other. They will play $10$ rounds, and in each round, there is exactly one winner. If the same person wins in consecutive rounds, these rounds are considered part of the same “streak”. How many possible outcomes are there in which there are strictly more than $3$ strea... | To solve this problem, we need to count the number of possible outcomes in which there are strictly more than 3 streaks in 10 rounds of arm wrestling. We will use the complement principle to simplify our calculations.
1. **Total Possible Outcomes**:
Each round can be won by either Richard (R) or Shreyas (S). There... | 932 | Combinatorics | math-word-problem | aops_forum | cf608fdb-b115-5801-bd68-837ff4103f83 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given $y=f(x)$, $x \in (-a, a)$, $F(x) = f(x) + f(-x)$, then $F(x)$ is ( )
A: Odd function
B: Even function
C: Both odd and even function
D: Neither odd nor even function | Since $x \in (-a, a)$, and $F(x) = f(x) + f(-x)$, we have $F(-x) = f(-x) + f(x) = F(x)$,
thus, $F(x)$ is an even function,
therefore, the correct choice is $\boxed{\text{B}}$. | \text{B} | Algebra | MCQ | cn_k12 | 9eb0b5fa-0ebc-5627-b456-f93617275e71 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
In acute triangle $ABC$, the lengths of the sides opposite to angles $A$, $B$, $C$ are $a$, $b$, $c$ respectively, and $b= \sqrt{2}a\sin \;B$.
(1) Find the measure of angle $A$;
(2) If $b= \sqrt{6}$ and $c= \sqrt{3}+1$, find $a.21$ (this might be a typo, I assume it means $a \times 21$). | (1) From $b= \sqrt{2}a\sin \;B$, using the sine rule we get $\sin B= \sqrt{2}\sin A\sin B$.
Since $\sin B \neq 0$ in triangle $ABC$, we have $\sin A= \dfrac{\sqrt{2}}{2}$.
As $ABC$ is an acute triangle, $A= \dfrac{\pi}{4}$.
(2) Given $b= \sqrt{6}$ and $c= \sqrt{3}+1$, and using $\cos A = \dfrac{\sqrt{2}}{2}$, we app... | 42 | Geometry | math-word-problem | cn_k12 | a2b3e35f-14c9-5400-864e-ff59d6321da1 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
In a round robin chess tournament each player plays every other player exactly once. The winner of each game gets $ 1$ point and the loser gets $ 0$ points. If the game is tied, each player gets $ 0.5$ points. Given a positive integer $ m$, a tournament is said to have property $ P(m)$ if the following holds: among eve... | To determine the minimum value of \( n \) such that in every \( n \)-player round robin chess tournament with property \( P(m) \), the final scores of the \( n \) players are all distinct, we need to analyze the properties and constraints given in the problem.
1. **Definition of Property \( P(m) \)**:
- Among every... | 2m-3 | Combinatorics | math-word-problem | aops_forum | 829b4b20-2415-5edc-a5c5-3e831aea831a | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
4、The weight ratio of two bags of candy, A and B, is $4: 1$. If 10 grams are taken from bag A and put into bag B, the weight ratio of A and B becomes $7: 8$. Therefore, the total weight of the two bags of candy is $\qquad$ grams. | 4. Solution: Let the weight of the first bag of candy be $4 x$ grams, and the weight of the second bag of candy be $x$ grams, then $(4 x-10):(x+10)=7: 8$ Solving for $x$ gives $x=6$, the total weight is $5 x=30$ (grams). | 30 | Algebra | math-word-problem | olympiads | 46b1d58c-e11f-588a-bb1c-972acc257758 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given the function $f(x) = -|x-a|$, then "f(x)$ satisfies $f(1+x) = f(1-x)$" is a condition for "a=1" to be ( )
A: A sufficient but not necessary condition
B: A necessary but not sufficient condition
C: A necessary and sufficient condition
D: Neither a sufficient nor a necessary condition | For the function $f(x) = -|x-a|$, it satisfies $f(1+x) = f(1-x)$,
$\therefore -|1+x-a| = -|1-x-a|$,
Since $x \in \mathbb{R}$, $\therefore 1-a = 0$
$\therefore a = 1$
When $a=1$, $f(x) = -|x|$, in this case, it satisfies $f(1+x) = f(1-x)$;
Therefore, "f(x) satisfies $f(1+x) = f(1-x)$" is a necessary and suffic... | \text{C} | Algebra | MCQ | cn_k12 | f8147552-4779-5b81-924e-69c4c1d57f7a | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given that the first three terms of a geometric sequence $\{a_n\}$ are $a-1$, $a+1$, $a+2$, then the general formula for this sequence is \_\_\_\_\_\_. | Since the first three terms of the geometric sequence $\{a_n\}$ are $a-1$, $a+1$, $a+2$,
we have $(a+1)^2 = (a-1)(a+2)$. Solving this equation, we get $a = -3$.
Therefore, the first three terms of the geometric sequence $\{a_n\}$ are $-4$, $-2$, $-1$, and thus the common ratio $q = \frac{1}{2}$.
Hence, $a_n = (-4) \... | - \frac{1}{2^{n-3}} | Algebra | math-word-problem | cn_k12 | 8155b3a4-4149-56e8-bab3-ac47dc582615 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Example 1. Find $\int\left(x^{2}-x+1\right) \cos 2 x d x$. | Solution. Here $P(x)=x^{2}-x+1$ is a polynomial of the second degree, which we take as $u$ and integrate by parts twice. The continuity of the solution will be ensured by the following presentation:
$$
\begin{gathered}
\int\left(x^{2}-x+1\right) \cos 2 x d x=\left\{\begin{array}{cc}
x^{2}-x+1=u, d u=(2 x-1) d x \\
\co... | \frac{2x^{2}-2x+1}{4}\sin2x+\frac{2x-1}{4}\cos2x+C | Calculus | math-word-problem | olympiads | 3c1c8ee1-37f5-5beb-9961-791a49e6f474 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given that the median to the leg of an isosceles triangle divides its perimeter into two parts, 15cm and 12cm, find the length of the base of this isosceles triangle. | Since the perimeter of the isosceles triangle is $15\text{cm} + 12\text{cm} = 27\text{cm}$,
let the length of the leg and the base of the isosceles triangle be $x\text{cm}$ and $y\text{cm}$, respectively. According to the problem, we have
$$
\begin{cases}
\frac{1}{2}x + x = 15 \\
\frac{1}{2}x + y = 12
\end{cases}
$$... | 7\text{cm} \text{ or } 11\text{cm} | Geometry | math-word-problem | cn_k12 | 9fba4d49-837b-5320-a9bb-fa59b5646939 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Consider the function $$f(x)=\sum_{k=1}^{m}(x-k)^{4}~, \qquad~ x \in \mathbb{R}$$ where $m>1$ is an integer. Show that $f$ has a unique minimum and find the point where the minimum is attained.
| 1. **Existence of Minima:**
To determine if the function \( f(x) = \sum_{k=1}^{m}(x-k)^{4} \) has a minimum, we first compute the second derivative \( f''(x) \).
\[
f(x) = \sum_{k=1}^{m}(x-k)^{4}
\]
The first derivative \( f'(x) \) is:
\[
f'(x) = \sum_{k=1}^{m} 4(x-k)^{3}
\]
The second d... | \frac{m+1}{2} | Calculus | math-word-problem | aops_forum | 2f5626f1-5d09-56c2-af32-34f79d9228d3 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
11. (14 points) In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=1, a_{n+1}=2 a_{n}-n+2\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the sum of the first $n$ terms $S_{n}$ of the sequence $\left\{a_{n}\right\}$. | II. 11. From $a_{n+1}=2 a_{n}-n+2$, we get
$$
a_{n+1}-n=2\left[a_{n}-(n-1)\right] \text {. }
$$
Therefore, the sequence $\left\{a_{n}-(n-1)\right\}$ is a geometric sequence with the first term 1 and common ratio 2.
Thus, $a_{n}=2^{n-1}+n-1$.
Hence, the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ ... | S_{n}=2^{n}-1+\frac{n(n-1)}{2} | Algebra | math-word-problem | olympiads | aca39921-4027-5e17-84d6-e9255daf9677 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
16.21 In $\triangle A B C$, $\angle A>\angle B>\angle C, \angle A \neq 90^{\circ}$, draw a line to divide $\triangle A B C$ into two parts, such that one part is similar to $\triangle A B C$. The number of such non-parallel lines is
(A) 3.
(B) 4.
(C) 5.
(D) 6.
(China Guangzhou, Wuhan, Fuzhou, and other five cities juni... | ［Solution］Such straight lines can be divided into three categories:
(1) Those parallel to one side of the triangle, totaling three;
(2) Two lines starting from the largest angle $\angle A$ meet the requirement, see the following figure (1) (note $\angle A$ $\left.\neq 90^{\circ}\right):$
(3) One line starting from the ... | 6 | Geometry | MCQ | olympiads | c7eebef4-f329-5700-ae98-3a77240431db | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Eddy draws $6$ cards from a standard $52$-card deck. What is the probability that four of the cards that he draws have the same value? | 1. **Calculate the total number of ways to draw 6 cards from a 52-card deck:**
\[
\binom{52}{6} = \frac{52!}{6!(52-6)!} = \frac{52!}{6! \cdot 46!}
\]
2. **Determine the number of ways to get four cards of the same value:**
- There are 13 different values (ranks) in a deck (2 through 10, Jack, Queen, King, ... | \frac{3}{4165} | Combinatorics | math-word-problem | aops_forum | 165e2152-9f9e-566d-bc66-a0fd39943006 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given the sequence $\{a_n\}$, where $a_1=2$ and $a_{n+1}-2a_n=0$, and $b_n=\log_2a_n$, then the sum of the first $10$ terms of the sequence $\{b_n\}$ equals ( ).
A: $130$
B: $120$
C: $55$
D: $50$ | In the sequence $\{a_n\}$, since $a_1=2$ and $a_{n+1}-2a_n=0$, it follows that $\dfrac{a_{n+1}}{a_n}=2$,
$\therefore$ the sequence $\{a_n\}$ is a geometric sequence with the first term $2$ and common ratio $2$,
$\therefore a_n=2\times2^{n-1}=2^n$.
$\therefore b_n=\log_2 2^n=n$.
$\therefore$ the sum of the first $10... | C | Algebra | MCQ | cn_k12 | 9977dda0-0b0a-5230-bffe-85c546b9028c | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Let the sequence $\{a_n\}$ satisfy $a_1+3a_2+\ldots+(2n-1)a_n=2n$.
(1) Find the general formula for $\{a_n\}$.
(2) Calculate the sum of the first $n$ terms of the sequence $\left\{ \frac{a_n}{2n+1} \right\}$. | (1) The sequence $\{a_n\}$ satisfies $a_1+3a_2+\ldots+(2n-1)a_n=2n$.
For $n\geqslant 2$, $a_1+3a_2+\ldots+(2n-3)a_{n-1}=2(n-1)$.
Therefore, $(2n-1)a_n=2$,
Thus, $a_n= \frac{2}{2n-1}$.
When $n=1$, $a_1=2$, the above formula also holds.
Therefore, $a_n= \frac{2}{2n-1}$;
(2) $\frac{a_n}{2n+1} = \frac{2}{(2n-1... | \frac{2n}{2n+1} | Algebra | math-word-problem | cn_k12 | bf70c08a-142a-5ccb-8b43-948f03614ae3 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Example 2 Find the maximum value of the positive integer $r$ that satisfies the following condition: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. ${ }^{\text {[2] }}$
(2013, Romanian National Team Selection Exam) | 【Analysis】Similarly, map the five subsets of 500 elements each to five vectors in a 1000-dimensional linear space. Since the requirement is the number of elements rather than their parity, we can consider the Euclidean space.
Let $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ be the five vectors after transformation.
Notice,
$$
\... | 200 | Combinatorics | math-word-problem | cn_contest | d59d83d7-649b-5bf7-b91a-cf18c3653ac8 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Let $a$, $b$, $c$ be positive integers with $a \le 10$. Suppose the parabola $y = ax^2 + bx + c$ meets the $x$-axis at two distinct points $A$ and $B$. Given that the length of $\overline{AB}$ is irrational, determine, with proof, the smallest possible value of this length, across all such choices of $(a, b, c)$. | 1. **Understanding the Problem:**
We are given a quadratic equation \( y = ax^2 + bx + c \) where \( a, b, c \) are positive integers, \( a \leq 10 \), and the parabola intersects the x-axis at two distinct points \( A \) and \( B \). The length of \( \overline{AB} \) is irrational, and we need to find the smallest ... | \frac{\sqrt{13}}{9} | Geometry | math-word-problem | aops_forum | fc693eca-7e02-5eab-9e11-188d735ad4a8 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Starting from an equilateral triangle like the one in figure a, trisect each side of the triangle, and then construct new equilateral triangles outward using the middle segments as sides, as shown in figure b, to get a "snowflake hexagon". Next, trisect each of the 12 sides of the "snowflake hexagon" and construct new ... | 17. Answer: $40: 27$
Analysis: Let the area of the equilateral triangle in figure $\mathrm{a}$ be 1. In figure $\mathrm{b}$, the area of each additional equilateral triangle added to each side is $\frac{1}{9}$. A total of 3 equilateral triangles are added, so the ratio of the area of figure $\mathrm{b}$ to the area of... | 40:27 | Geometry | math-word-problem | olympiads | 84127acd-d1e5-5c40-befe-0622904120c6 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
4. Given three distinct points $Z_{1}, Z_{2}, Z_{3}$ in the complex plane corresponding to the complex numbers $z_{1}, z_{2}, z_{3}$. If $\left(z_{2}-z_{1}\right)^{2}+\left(z_{3}-z_{1}\right)^{2}=0$, then $\triangle Z_{1} Z_{2} Z_{3}$ is ( ).
(A) an equilateral triangle
(B) an isosceles right triangle
(C) a right trian... | 4.B.
From the given equation, we get $z_{2}-z_{1}= \pm \mathrm{i}\left(z_{3}-z_{1}\right)$. Therefore, $Z_{1} Z_{2}=Z_{1} Z_{3}$, and $Z_{1} Z_{2} \perp Z_{1} Z_{3}$. | B | Geometry | MCQ | cn_contest | a42a61cb-5bce-5f04-896f-be85a0dd50be | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
4. Consider some intermediate step in Kuzia's movement. If she is at point $A$ at this step, the probability of being in $A$ on the next step is zero. If, however, she is at any of the remaining points, $B, C$ or $D$, the probability of being in $A$ on the next step is $1 / 3$, since from each such point there are thre... | Answer: $\quad P(A)=\frac{3^{2019}+1}{4 \cdot 3^{2019}}$. | \frac{3^{2019}+1}{4\cdot3^{2019}} | Algebra | proof | olympiads | 1a7a2d5f-bbcb-5da1-8849-fcd592af9ea4 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given sets $A=\{x|2 \lt x\leqslant 6\}$ and $B=\{x|x^{2}-4x \lt 0\}$.
$(1)$ Find $A\cap B$ and $\complement _{R}(A\cup B)$;
$(2)$ Given set $C=\{x|m+1 \lt x \lt 2m-1\}$, if satisfying ______, find the range of real number $m$.
Please choose one from $C\subseteq \left(C\cap B\right)$, $C_{R}C\supseteq C_{R}B$, $B\... | ### Solution:
#### Part (1)
Given $A=\{x|2 \lt x\leqslant 6\}$ and $B=\{x|x^{2}-4x \lt 0\}$, we can rewrite set $B$ by factoring the quadratic inequality:
- For $B$, we have $x(x-4) 0$ to ensure the lower bound is positive.
- $2m-1 \leqslant 4$ to ensure the upper bound of $C$ does not exceed that of $B$.
- $... | m\leqslant \frac{5}{2} | Algebra | math-word-problem | cn_k12 | 5a0215aa-ef60-567f-aa95-b54c0be3d460 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
3. The minimum value of the area of a right-angled triangle circumscribed around a circle with radius 1 is ( ).
(A) $3-2 \sqrt{2}$
(B) $3+2 \sqrt{2}$
(C) $6-4 \sqrt{2}$
(D) $6+4 \sqrt{2}$ | 3. (B).
Let the lengths of the two legs of a right triangle be $a$ and $b$, and the area be $\mathrm{S}$. Then we have
$$
\frac{1}{2}\left(a+b+\sqrt{a^{2}+b^{2}}\right) \times 1=\frac{1}{2} a b .
$$
(The area of the triangle is equal to the product of the semi-perimeter and the inradius.)
Simplifying, we get $a b-2 a-... | B | Geometry | MCQ | cn_contest | 877dadbc-c8a0-5d74-b11a-f0790d5c1cd2 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
## Task A-2.2.
Let $a, b \in \mathbb{R}$. The solutions to the quadratic equation $a x^{2} + b x + 1 = 0$ are real. If each of these solutions is decreased by 1, we get the solutions to the quadratic equation $b x^{2} + x + a = 0$.
Determine all such real numbers $a, b$. | ## First Solution.
Let $x_{1}$ and $x_{2}$ be the solutions of the quadratic equation $b x^{2}+x+a=0$.
From Viète's formulas, we have
$$
x_{1}+x_{2}=-\frac{1}{b} \quad \text { and } \quad x_{1} x_{2}=\frac{a}{b}
$$
From the condition of the problem, we know that $x_{1}+1$ and $x_{2}+1$ are the solutions of the quad... | (-1,1)(1,-1-\sqrt{2}) | Algebra | math-word-problem | olympiads | f6727adb-6cf2-50e0-8bd2-a7ba46d0082d | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Let the sequence \\(\{a_n\}\) have a sum of the first \\(n\\) terms \\(S_n = 2^{n+1} - 2\\), and the sequence \\(\{b_n\}\) satisfies \\(b_n = \frac{1}{(n+1)\log_2{a_n}}\).
\\((1)\\) Find the general formula for the sequence \\(\{a_n\}\).
\\((2)\\) Find the sum of the first \\(n\\) terms \\(T_n\\) of the sequence \\(\... | Solution:
\\((1)\\) When \\(n=1\\), \\(a_1 = S_1 = 2\\),
Given \\(S_n = 2^{n+1} - 2\\), \\(\therefore S_{n-1} = 2^n - 2\\) (for \\(n \geqslant 2\\))
\\(\therefore a_n = S_n - S_{n-1} = 2^n\\) (for \\(n \geqslant 2\\)),
Therefore, the general formula for the sequence \\(\{a_n\}\) is: \\(\boxed{a_n = 2^n}.\\)
\\((2)... | T_n = \frac{n}{n+1} | Algebra | math-word-problem | cn_k12 | 98b19cc6-ae08-5b4d-b46a-5e189a292d3c | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Example 3. Compute the integral $\int_{C} e^{\bar{z}} d z$, where $C$ is the line segment $y=-x$, connecting the points $z_{1}=0$ and $z_{2}=\pi-i \pi$. | Solution. The parametric equations of the curve $C$ are
$$
x=t, \quad y=-t
$$
or in complex form
$$
z=t-i t
$$
where the real variable $t$ varies from 0 to $\pi$.
Applying formula (2), we get
$$
\int_{C} \mathrm{e}^{\bar{z}} d z=\int_{0}^{\pi} e^{t+i t}(1-i) d t=(1-i) \int_{0}^{\pi} e^{(1+i) t} d t=\left.\frac{1-... | (e^{\pi}+1)i | Calculus | math-word-problem | olympiads | 95d7bbae-2351-58ca-843c-27107e9f3fd5 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given that the sum of all odd-numbered terms in the binomial expansion of ${\left(\sqrt{x}-\sqrt[3]{x}\right)}^{n}$ is $512$,
(1) Find all rational terms (terms with integral exponents) in the expansion.
(2) Find the coefficient of the ${x}^{2}$ term in the expansion of ${\left(1-x\right)}^{3}+{\left(1-x\right)}^{4}+... | (1) From the given condition, we have ${2}^{n-1}=512$, so $n=10$.
Thus, ${T}_{r+1}=C_{10}^{r}{\left(\sqrt{x}\right)}^{10-r}{\left(-\sqrt[3]{x}\right)}^{r}={(-1)}^{r}C_{10}^{r}{x}^{5-\frac{r}{6}}\quad (r=0,1,\ldots ,10)$
Since $5-\frac{r}{6}\in Z$, we have $r=0,6$.
The rational terms are ${T}_{1}=C_{10}^{0}{x}^{5}={x... | 164 | Algebra | math-word-problem | cn_k12 | dfc6423b-35bc-5a37-bc89-57ee1b16891a | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
8. Solve the system $\left\{\begin{array}{l}x \log _{2} 3+y=\log _{2} 18 ; \\ 5^{x}=25^{y} .\end{array}\right.$
 | # Solution:
$\left\{\begin{array}{l}\log _{2} 3^{x}+y=\log _{2} 18, \\ 5^{x}=5^{2 y},\end{array}\left\{\begin{array}{l}2 y \log _{2} 3+y=18 \\ x=2 y,\end{array},\left\{\begin{array}{l}y\left(\log _{2} 3^{2}+1\right)=18, \\ x=2 y,\end{array}\left\{\begin{array}{l}y\left(\log _{2}\left(3^{2} \cdot 2\right)\right)=18, \\... | {2;1} | Algebra | math-word-problem | olympiads | 94065d16-7148-5617-a963-af88184d1d11 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
[Trigonometric Identities]
Let $\alpha$ and $\beta$ be acute and positive angles satisfying the equations
$$
\begin{aligned}
& 3 \sin ^{2} \alpha+2 \sin ^{2} \beta=1 \\
& 3 \sin 2 \alpha-2 \sin 2 \beta=0
\end{aligned}
$$
Prove that $\alpha+2 \beta=\frac{\pi}{2}$. | From the given relations, we find:
$$
\sin 2 \beta=\frac{3}{2} \sin 2 \alpha, \quad 3 \sin ^{2} \alpha=1-2 \sin ^{2} \beta=\cos 2 \beta
$$
From this,
$$
\cos (\alpha+2 \beta)=\cos \alpha \cdot 3 \sin ^{2} \alpha-\sin \alpha \frac{3}{2} \sin 2 \alpha=0
$$ | \alpha+2\beta=\frac{\pi}{2} | Algebra | proof | olympiads | 5b94b236-0133-5e1c-a0db-50da68ef35eb | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
(1) Given $f(x)=\frac{x+1}{x-1}(x \in \mathbf{R}$ and $x \neq \pm 1)$, which of the following equations always holds?.
(A) $f(x) \cdot f(-x)=1$
(B) $f(x)+f(-x)=0$
(C) $f(x) f(-x)=-1$
(D) $f(x)+f(-x)=1$ | (1) A | A | Algebra | MCQ | olympiads | 92718c9f-9b6a-56e4-a652-bead5c9ea0d3 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Let $ p$ be an odd prime. Determine positive integers $ x$ and $ y$ for which $ x \leq y$ and $ \sqrt{2p} \minus{} \sqrt{x} \minus{} \sqrt{y}$ is non-negative and as small as possible. | 1. We start by considering the expression \(\sqrt{2p} - \sqrt{x} - \sqrt{y}\) and aim to minimize it while ensuring it is non-negative. This means we need to find \(x\) and \(y\) such that \(x \leq y\) and \(\sqrt{2p} \geq \sqrt{x} + \sqrt{y}\).
2. We claim that the minimum value occurs when \(x = \frac{p-1}{2}\) and ... | (x, y) = \left(\frac{p-1}{2}, \frac{p+1}{2}\right) | Inequalities | math-word-problem | aops_forum | 5dc5c07a-18cf-5973-b8fa-d0ae62133745 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
1. The smallest positive odd number that cannot be expressed as $7^{x}-3 \times 2^{y}\left(x 、 y \in \mathbf{Z}_{+}\right)$ is $\qquad$ | -1.3 .
Since $x, y \in \mathbf{Z}_{+}$, therefore, $7^{x}-3 \times 2^{y}$ is always an odd number, and $7^{1}-3 \times 2^{1}=1$.
If $7^{x}-3 \times 2^{y}=3$, then $317^{x}$.
And $7^{x}=(1+6)^{x}=1(\bmod 3)$, thus, there do not exist positive integers $x, y$ such that
$$
7^{x}-3 \times 2^{y}=3 \text {. }
$$
Therefore, ... | 3 | Number Theory | math-word-problem | cn_contest | 44f587ce-347f-55a3-8048-294f9af9f6fa | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
After the graph of the quadratic function $y=\frac{1}{2}x^2+3x+\frac{5}{2}$ is shifted $2$ units to the right and then $3$ units up, the vertex of the resulting function graph is ( ).
A: $(-5,1)$
B: $(1,-5)$
C: $(-1,1)$
D: $(-1,3)$ | To solve for the vertex of the transformed quadratic function, we start with the given function and apply the transformations step by step:
1. **Given Function**:
$$y=\frac{1}{2}x^2+3x+\frac{5}{2}$$
2. **Completing the Square**:
To make it easier to identify the vertex, we complete the square:
\begin{align*... | \text{C} | Algebra | MCQ | cn_k12 | d969dc8c-7e0c-5a07-87c5-c17b8d1c1b30 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
7. If real numbers $x, y$ satisfy $2 x^{2}+3 x y+2 y^{2}=1$, then the minimum value of $x+y+x y$ is $\qquad$ . | 7. $-\frac{9}{8}$.
Let $k=x+y+xy$.
From $2x^{2}+3xy+2y^{2}=1 \Rightarrow 2(x+y)^{2}=1+xy$
$\Rightarrow 2(x+y)^{2}=k-(x+y)+1$
$\Rightarrow 2(x+y)^{2}+(x+y)-(1+k)=0$.
Consider the above equation as a quadratic equation in terms of $x+y$.
Since $x+y$ is a real number, we have
$$
\Delta=1+8(1+k) \geqslant 0 \Rightarrow k ... | -\frac{9}{8} | Algebra | math-word-problem | olympiads | 96c221bb-dfa3-5c89-98e0-b48285a5692c | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
We write 1s at the two endpoints of a segment. Then, step by step, we bisect the resulting segments and write the sum of the numbers at the two ends of the segments to the midpoint. Repeating the bisection of the segments one million times, how many times will the number $1978$ appear among the written numbers? | Consider the numbers $a$ and $b$ (in this order) as neighbors if, during the process, a segment is formed where $a$ is at the left endpoint and $b$ is at the right endpoint. For example, according to Figure 1, 1 and 2, 3 and 2, and 3 and 5 are all neighbors. First, we will show that neighboring numbers are relatively p... | 924 | Number Theory | math-word-problem | olympiads | 76ffd87c-959a-5e81-8aa3-6e2a1ea030c3 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
1. Among the 95 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 95^{2}$, the numbers with an odd digit in the tens place total $\qquad$.
untranslated part: $\qquad$ | In $1^{2}, 2^{2}, \cdots, 10^{2}$, it is found through calculation that the tens digit is odd only for $4^{2}=16, 6^{2}=36$. A two-digit square number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2},
$$
Therefore, $b$ can only be 4 or 6. That is, in every 10 consecutive numbers, there are two numbers whose ... | 19 | Number Theory | math-word-problem | cn_contest | 586ef2b7-73e0-56c3-a2fe-877d5461234e | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given the sets $A=\{x|-2<x<2\}$ and $B=\{x|x^2-2x\leq0\}$, then $A\cap B$ equals to ( )
A: $(0,2)$
B: $(0,2]$
C: $[0,2)$
D: $[0,2]$ | According to the problem, we have: $B=\{x|x^2-2x\leq0\}=\{x|0\leq x\leq2\}$
Therefore, $A\cap B=[0,2)$
Hence, the correct choice is $\boxed{C}$ | C | Algebra | MCQ | cn_k12 | bbbfbcf8-cad6-586d-b560-cc02463333da | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
In a psychology experiment, an image of a cat or an image of a dog is flashed briefly onto a screen and then Anna is asked to guess whether the image showed a cat or a dog. This process is repeated a large number of times with an equal number of images of cats and images of dogs shown. If Anna is correct $95 \%$ of the... | Suppose that Anna guesses "cat" $c$ times and guesses "dog" $d$ times.
When she guesses "dog", she is correct $95 \%$ of the time.
When she guesses "cat", she is correct $90 \%$ of the time.
Thus, when she guesses "cat", she is shown $0.9 c$ images of cats and so $c-0.9 c=0.1 c$ images of dogs.
Thus, when she guess... | 8:9 | Combinatorics | math-word-problem | olympiads | e5660444-70bc-5da1-8429-e76bd324c72e | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
Solve the equation $\frac{2003 x}{2004}=2003^{\log _{x} 2004}$. | Solution. Both sides are positive, so we can write the logarithm of both sides of the equation with base $x$ ($x>0$, $x \neq 1$):
$$
\log _{x} 2003+\log _{x} x-\log _{x} 2004=\log _{x} 2004 \cdot \log _{x} 2003
$$
We know that $\log _{x} x=1$, and let $\log _{x} 2003$ be $a$, and $\log _{x} 2004$ be $b$. Then we get ... | \frac{1}{2003}or2004 | Algebra | math-word-problem | olympiads | 9cd317fb-c75a-5427-99e4-916451b180b2 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
12. Find the unit digit of $17^{17} \times 19^{19} \times 23^{23}$. | 12. Answer: 1
$$
\begin{array}{l}
17^{17}=7^{17}=7(\bmod 10) \\
19^{19} \equiv 9^{19} \equiv 9(\bmod 10) \\
23^{23} \equiv 3^{23} \equiv 7(\bmod 10)
\end{array}
$$
Since $7 \times 9 \times 7=441=1(\bmod 10)$, the unit digit is 1 . | 1 | Number Theory | math-word-problem | olympiads | 8321ce94-0dcb-55d2-b561-66bfef4a57f2 | open-r1/OpenR1-Math-220k | Dummy value | [
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Let $A_{1}, \ldots, A_{2022}$ be the vertices of a regular 2022-gon in the plane. Alice and Bob play a game. Alice secretly chooses a line and colors all points in the plane on one side of the line blue, and all points on the other side of the line red. Points on the line are colored blue, so every point in the plane i... | The answer is 22 . To prove the lower bound, note that there are $2022 \cdot 2021+2>2^{21}$ possible colorings. If Bob makes less than 22 queries, then he can only output $2^{21}$ possible colorings, which means he is wrong on some coloring.
Now we show Bob can always win in 22 queries. A key observation is that the s... | 22 | Combinatorics | math-word-problem | olympiads | 8968e30e-8f67-573b-a2ca-7b9ed900caea | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
Using the numbers 1, 2, and 3 to form a number (without repeating any digit), the largest number obtained is ( )
A: 321
B: $21^3$
C: $3^{21}$
D: $2^{31}$ | **Solution**: Given that $21^3 = (3 \times 7)^3$,
$= 27 \times 289$
Therefore, $27 \times 289 > 321$.
Thus, $321 512$, and $3^{21} = 3^6 \times 3^6 \times 3^6 \times 3^3$
$= 729 \times 729 \times 729 \times 27$;
$2^{31} = 2^9 \times 2^9 \times 2^9 \times 2^4$
$= 512 \times 512 \times 512 \times 16$,
Ther... | C | Combinatorics | MCQ | cn_k12 | 52d12581-f492-5220-906c-b71802a45b05 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Consider a circle of radius 1. Describe a regular $n$-sided polygon around it and inscribe a regular $n$-sided polygon in it. Denote their perimeters by $P_{n}$ (for the circumscribed) and $p_{n}$ (for the inscribed).
a) Find $P_{4}, p_{4}, P_{6}$, and $p_{6}$.
b) Prove that the following recurrence relations hold: $... | b) Let $a_{n}$ and $b_{n}$ denote the sides of the inscribed and circumscribed regular $n$-gons. The midpoints $K, L, M$ of the sides $A B, B C, C D$ of the circumscribed regular $2 n$-gon are consecutive vertices of the inscribed regular $2 n$-gon. Moreover, $K M$ is a side of the inscribed regular $n$-gon, the point ... | P_{96}\approx6.285429,p_{96}\approx6.282064 | Geometry | math-word-problem | olympiads | 4d946a92-1f9f-5be8-8453-d05c7bd86a22 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
Given that $z$ is a complex number, both $z+2i$ and $\frac{z}{2-i}$ are real numbers, where $i$ is the imaginary unit.
(1) Find the complex number $z$ and its modulus $|z|$;
(2) If $z_1 = \overline{z}+3m+(m^2−6)i$ lies in the fourth quadrant, determine the range of values for the real number $m$. | (1) Let $z=a+bi$ where $a,b\in\mathbb{R}$. Since $z+2i=a+(b+2)i$ is real, it follows that $b+2=0$, and hence $b=-2$.
Given that $\frac {z}{2-i}$ is real, we have:
$$\frac {z}{2-i} = \frac {a-2i}{2-i} = \frac {(a-2i)(2+i)}{(2-i)(2+i)} = \frac {2a+2}{5}+\frac {a-4}{5}i.$$
For this to be real, the imaginary part must be... | -\frac{4}{3} < m < 2 | Algebra | math-word-problem | cn_k12 | d9ef744a-5b97-5b0d-9bc4-cc355274281f | open-r1/OpenR1-Math-220k | Dummy value | [
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Find all ordered pairs $ (m,n)$ where $ m$ and $ n$ are positive integers such that $ \frac {n^3 \plus{} 1}{mn \minus{} 1}$ is an integer. | 1. We start with the given condition that $\frac{n^3 + 1}{mn - 1}$ is an integer. This implies that $mn - 1$ divides $n^3 + 1$.
2. Note that $\gcd(mn-1, n) = 1$ because $mn-1$ and $n$ are coprime. This is because $mn-1$ is not divisible by $n$.
3. We can rewrite the expression as:
\[
\frac{n^3 + 1}{mn - 1} =... | (1, 2), (1, 3), (2, 1), (2, 2), (2, 5), (3, 1), (3, 5), (5, 2), (5, 3) | Number Theory | math-word-problem | aops_forum | b3b008ed-0542-524c-9f41-c42b4c2ae00b | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-13.5,13.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 756
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-28 \leqslant y-1 \leqslant 26$ and $-25 \leqslant 2-x \leqslant 29$. Therefore, $(y-1)(2-x)+2 \leqslant 26 \cdot 29+2=756$. ... | 756 | Algebra | math-word-problem | olympiads | fc66fff4-e14c-5f29-9b8b-9c8647839e04 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
8.2 Find all three-digit numbers $\mathrm{N}$ such that the sum of the digits of the number $\mathrm{N}$ is 11 times smaller than the number $\mathrm{N}$ itself. | Solution. Let a, b, c be the digits of the number N: $\mathrm{N}=100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}$, where $1 \leq \mathrm{a} \leq 9$, $0 \leq \mathrm{b}, \mathrm{c} \leq 9$. According to the condition, $100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}=11(\mathrm{a}+\mathrm{b}+\mathrm{c})$, i.e., $89 \mathrm{a}=\mathrm{... | 198 | Number Theory | math-word-problem | olympiads | b804d4ee-8ff9-5fbd-9530-adaac7625b72 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
Example 1 Solve the system of congruences $\left\{\begin{array}{l}x \equiv 1(\bmod 7), \\ x \equiv 1(\bmod 8), \\ x \equiv 3(\bmod 9) .\end{array}\right.$ | Since $7, 8, 9$ are pairwise coprime, the system of congruences has a solution.
$$
\begin{array}{l}
M=7 \cdot 8 \cdot 9=504, M_{1}=\frac{504}{7}=72, M_{2}=\frac{504}{8}=63, M_{3}=\frac{504}{9}=56 . \text { Thus } \\
M_{1}^{\prime} \equiv \frac{1}{72} \equiv \frac{1}{7 \cdot 10+2} \equiv \frac{1}{2} \equiv \frac{4}{8} \... | 57 | Number Theory | math-word-problem | olympiads | fda1ba93-aae0-57f3-83e6-4ad71303a270 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
3. If the quadratic equation with real coefficients $a x^{2}+b x+c=0$ has 2 imaginary roots $x_{1}, x_{2}$, and $x_{1}^{3} \in R$, then $\frac{a c}{b^{2}}=$ | Answer: 1
Explanation: Note that $x_{2}=\overline{x_{1}}$, from $x_{1}^{3} \in R \Rightarrow x_{1}^{3}=\overline{x_{1}^{3}}=\left(\overline{x_{1}}\right)^{3}=x_{2}^{3} \Rightarrow\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)=0$
$$
\Rightarrow x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=0 \Rightarrow\left(x_{... | 1 | Algebra | math-word-problem | olympiads | e7aca063-7e44-53f1-944e-e836e0c3220b | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
Given that $0 < α < \dfrac {π}{2}$, $- \dfrac {π}{2} < β < 0$, $\cos ( \dfrac {π}{4}+α)= \dfrac {1}{3}$, and $\cos ( \dfrac {π}{4}-β)= \dfrac { \sqrt {3}}{3}$, find $\cos (α+β)$. | Since $0 < α < \dfrac {π}{2}$ and $- \dfrac {π}{2} < β < 0$, we have $\dfrac {π}{4} < \dfrac {π}{4}+α < \dfrac {3π}{4}$ and $\dfrac {π}{4} < \dfrac {π}{4}-β < \dfrac {3π}{4}$.
From $\cos ( \dfrac {π}{4}+α)= \dfrac {1}{3}$ and $\cos ( \dfrac {π}{4}-β)= \dfrac { \sqrt {3}}{3}$, we can derive $\sin ( \dfrac {π}{4}+α)= \d... | \dfrac {5 \sqrt {3}}{9} | Algebra | math-word-problem | cn_k12 | f5971cc8-10df-5901-b84d-e75ec8a504ec | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
Three. (25 points) Given the quadratic trinomial $a x^{2}+b x+c$ $(a>0)$.
(1) When $c<0$, find the maximum value of the function
$$
y=-2\left|a x^{2}+b x+c\right|-1
$$
(2) For any real number $k$, the line $y=k(x-$
1) $-\frac{k^{2}}{4}$ intersects the parabola $y=a x^{2}+b x+c$ at exactly one point, find the value of ... | (1) Given $a>0, c<0$, we know that $y^{\prime}=a x^{2}+b x+c$ intersects the $x$-axis, and $y_{\text {nuin }}^{\prime}<0$. Therefore,
$$
\left|y^{\prime}\right|=\left|a x^{2}+b x+c\right| \geqslant 0 \text {. }
$$
Thus, the minimum value of $\left|y^{\prime}\right|$ is 0.
At this point, $y_{\text {man }}=-2 \times 0-1... | 0 | Algebra | math-word-problem | cn_contest | 5e1110c4-02c4-58b5-87d6-d7191d73f862 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
Given the function $f(x) = \begin{cases} x & \text{if } x\geq 0 \\ x^2 & \text{if } x<0 \end{cases}$, then the value of $f(f(-2))$ is ( )
A: 2
B: -2
C: 4
D: -4 | Since the given function is $f(x) = \begin{cases} x & \text{if } x\geq 0 \\ x^2 & \text{if } x<0 \end{cases}$,
then $f(-2) = (-2)^2$,
thus $f(f(-2)) = f(4) = 4$,
therefore, the correct choice is $\boxed{C}$.
Given $f(x)$ is a piecewise function, substitute $x = -2$ into the expression $y = x^2$ to find $f(-2)$.... | C | Algebra | MCQ | cn_k12 | b52a7227-379c-52b4-91a4-8842e8c4ea3c | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
II. (50 points) On an infinite grid paper, some cells are colored red, and the rest are colored blue. In every $2 \times$ 3 rectangle of six cells, there are exactly two red cells. How many red cells are there in a $9 \times 11$ rectangle of 99 cells? | II. Solution: 33 red squares. As shown in the figure, take any red square $K_{0}$ as the center of a $3 \times 3$ square; it is not possible to color $K$ red. If $K$ is colored red, then in the $2 \times 3$ rectangles $A F H D$, $A B S T$, and $M N C D$, there will be two red squares each. To ensure that the rectangle ... | 33 | Combinatorics | math-word-problem | olympiads | d305bd13-50cc-563c-8f1d-5a4cbbed5a47 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
If a series of functions have the same analytical expression and the same range, but different domains, then these functions are called "homogeneous functions". How many "homogeneous functions" are there whose analytical expression is $y = x^2$ and whose range is $\{1, 2\}$? | To find the number of "homogeneous functions" with the given conditions, we need to consider the conditions for the range and the analytical expression $y = x^2$.
The range $\{1, 2\}$ corresponds to the $y$ values of the function. For $y = x^2$, the $x$ values that give $y = 1$ are $x = \pm 1$, and the $x$ values that... | 9 | Algebra | math-word-problem | cn_k12 | ba748bd3-3eb0-5b3d-ad79-12fd81b68d25 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
3. (15 points) Determine the mass $m$ of helium needed to fill an empty balloon of mass $m_{1}=10$ g so that the balloon will rise. Assume the temperature and pressure of the gas in the balloon are equal to the temperature and pressure of the air. The molar mass of helium $M_{\mathrm{r}}=4$ g/mol, the molar mass of air... | 3. For the balloon to take off, the condition is $F_{\text {Arch }}=\left(m+m_{1}\right) g$
$F_{\text {Arch }}=\rho_{0} g V$
From the Clapeyron-Mendeleev equation, the density of air is
$\rho_{0}=\frac{P_{0} M_{\mathrm{B}}}{R T_{0}}, \quad$ and the volume of the balloon $V=\frac{m}{M_{\mathrm{r}}} \frac{R T_{0}}{P_{... | 1.6 | Algebra | math-word-problem | olympiads | ba4a4bf2-aa5c-5afb-8437-e6ca8d7258ba | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
1. Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+$ $\frac{1}{1+b^{n}}$ is $\qquad$. | $$
\begin{array}{l}
\text { 1. } \because a b>0 \\
\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, \quad a^{n} b^{n} \leqslant 1 . \\
\text { So } \quad \frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1 .
\end{array}
$$
When $a=b=1$, the above expression $=1$. ... | 1 | Inequalities | math-word-problem | olympiads | 6c63eb2c-98e5-5f4f-89c7-fd689cbece7d | open-r1/OpenR1-Math-220k | Dummy value | [
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## Task B-3.1.
Calculate: $\frac{1}{\left(337^{\frac{1}{2}}+6^{\frac{1}{2}}\right)^{-2}}-\left(\frac{\sqrt{337}-\sqrt{6}}{337^{\frac{3}{2}}-6^{\frac{3}{2}}}\right)^{-1}$. | ## Solution.
Let's write the given expression without negative exponents, and express powers with rational exponents using roots (or vice versa). Then we will apply the formulas for the square of a binomial and the difference of cubes. We then get step by step:
$$
\begin{array}{cc}
\frac{1}{\left(337^{\frac{1}{2}}+6^... | \sqrt{2022} | Algebra | math-word-problem | olympiads | 571c7a6e-5915-5247-9681-4fd4332370cf | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
Adam and Mada are playing a game of one-on-one basketball, in which participants may take $2$-point shots (worth $2$ points) or $3$-point shots (worth $3$ points). Adam makes $10$ shots of either value while Mada makes $11$ shots of either value. Furthermore, Adam made the same number of $2$-point shots as Mada made $3... | 1. Let \( x \) be the number of 2-point shots Adam made. Since Adam made a total of 10 shots, the number of 3-point shots Adam made is \( 10 - x \).
2. Let \( y \) be the number of 3-point shots Mada made. Since Mada made a total of 11 shots, the number of 2-point shots Mada made is \( 11 - y \).
3. According to the ... | 52 | Algebra | math-word-problem | aops_forum | e6f3792d-86f6-51c7-bb78-1c126d7e9b4b | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
According to the following known conditions, the unique $\triangle ABC$ that can be drawn is ( )
A: $\angle A=60^{\circ}$, $\angle B=45^{\circ}$, $AB=4$
B: $\angle C=90^{\circ}$, $\angle B=30^{\circ}$, $\angle A=60^{\circ}$
C: $AB=3$, $BC=4$, $CA=8$
D: $AB=4$, $BC=3$, $\angle A=60^{\circ}$ | To determine the unique $\triangle ABC$ that can be drawn based on the given conditions, let's analyze each option step by step:
**Option A:** $\angle A=60^{\circ}$, $\angle B=45^{\circ}$, $AB=4$
Given two angles and a side, specifically the side between the two angles, we can apply the Law of Sines or use the fact t... | A | Geometry | MCQ | cn_k12 | 8070fbc9-97e0-5c87-b87d-320a25fe2b12 | open-r1/OpenR1-Math-220k | Dummy value | [
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] |
2. A three-digit number $X$ was written with three different digits $A B C$. Four schoolchildren made the following statements. Petya: “The largest digit in the number $X$ is $B$”. Vasya: “$C=8$”. Tolya: “The largest digit is $C$”. Dima: “$C$ is the arithmetic mean of $A$ and $B$”. Find the number $X$, given that exact... | Answer: 798. Solution. If Tolya told the truth and the largest number is $-C$, then immediately two people (Petya and Dima) would be wrong, which cannot be the case, so Tolya must be wrong, and the others told the truth. Thus, the number $C=8 ; B$ is the largest number, i.e., $B=9 ; C$ is the average of numbers $A$ and... | 798 | Logic and Puzzles | math-word-problem | olympiads | 51917d8e-fa33-5244-9ff9-5bb8cac16821 | open-r1/OpenR1-Math-220k | Dummy value | [
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2・108 $2 \cdot 108 p^{2} \geqslant 4 q$ is the condition for the real-coefficient equation $x^{4}+p x^{2}+q=0$ to have real roots
(A) A necessary condition but not a sufficient condition.
(B) A sufficient condition but not a necessary condition.
(C) A necessary and sufficient condition.
(D) Neither a sufficient conditi... | [Solution] If we let $x^{2}=t$, the given equation can be transformed into $t^{2}+p t+q=0$, which has real roots if and only if $\Delta=p^{2}-4 q \geqslant 0$, i.e., $p^{2} \geqslant 4 q$.
However, the real roots of this quadratic equation do not necessarily guarantee that the original equation has real roots.
Converse... | A | Algebra | MCQ | olympiads | 5cf414be-3792-59cb-a790-170e0765c2a2 | open-r1/OpenR1-Math-220k | Dummy value | [
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Factorization: $x^{2}-16=\_\_\_\_\_\_.$ | To factorize the given expression $x^{2}-16$, we recognize it as a difference of squares. The difference of squares formula is $a^2 - b^2 = (a+b)(a-b)$. Applying this formula to our expression:
1. Identify $a^2$ and $b^2$ in the expression $x^{2}-16$. Here, $a^2 = x^2$ and $b^2 = 16$.
2. Find $a$ and $b$. Since $a^2 =... | (x+4)(x-4) | Algebra | math-word-problem | cn_k12 | c4bc049b-c4fb-5900-9a4c-c4502f0d63e7 | open-r1/OpenR1-Math-220k | Dummy value | [
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61. Given positive real numbers $x, y, z$ satisfying: $x y + y z + z x \neq 1$, and $\frac{(x^2 - 1)(y^2 - 1)}{xy} + \frac{(y^2 - 1)(z^2 - 1)}{yz} + \frac{(z^2 - 1)(x^2 - 1)}{zx} = 4$, find the value of $\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}$. | Answer: 1 | 1 | Algebra | math-word-problem | olympiads | 4d5c883f-aff9-589e-987e-1c05b9595dae | open-r1/OpenR1-Math-220k | Dummy value | [
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Given that the sum of the first $n$ terms of an arithmetic sequence ${a_n}$ is $S_n$, if $\overrightarrow{OB}=a_3 \overrightarrow{OA}+a_{2014} \overrightarrow{OC}$, and points $A$, $B$, $C$ are collinear ($O$ is a point outside this line), then $S_{2016} =$ \_\_\_\_\_\_. | Since $\overrightarrow{OB}=a_3 \overrightarrow{OA}+a_{2014} \overrightarrow{OC}$, and points $A$, $B$, $C$ are collinear ($O$ is a point outside this line),
We have $a_3+a_{2014}=1$,
This implies that $a_1+a_{2016}=a_3+a_{2014}=1$,
Thus, $S_{2016}= \frac{2016(a_1+a_{2016})}{2}=1008$.
Therefore, the answer is $\boxe... | 1008 | Algebra | math-word-problem | cn_k12 | e8ffbcf0-e196-52b1-8a4d-055d542361b6 | open-r1/OpenR1-Math-220k | Dummy value | [
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In order to estimate the value of $x-y$ where $x$ and $y$ are real numbers with $x>y>0$, Xiaoli rounded $x$ up by a small amount, rounded $y$ down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct?
$\textbf{(A) } \text{Her estimate is larger than } x - ... | The original expression $x-y$ now becomes $(x+k) - (y-k)=(x-y)+2k>x-y$, where $k$ is a positive constant, hence the answer is $\boxed{\textbf{(A)}}$. | \textbf{(A)} | Algebra | MCQ | amc_aime | 2aafab31-0305-5a63-9d16-8c2d12ec4643 | open-r1/OpenR1-Math-220k | Dummy value | [
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8. On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5(x-(\sqrt{3} / 2))^{2}$ pass, the angle between which is $60^{\circ} . \quad$ (12 points) | # Solution:
On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5 \cdot(x-\sqrt{3} / 2)^{2}$ pass, with the angle between them being $60^{\circ}$.
Solution (without using derivatives).
$y=0.5 \cdot(x-\sqrt{3} / 2)^{2}, M\left(0 ; y_{0}\right) \cdot U$ The equation $0.5 ... | M(0;0)orM(0;-\frac{5}{3}) | Algebra | math-word-problem | olympiads | 4ae6be19-63f5-5405-acac-cd58541db19f | open-r1/OpenR1-Math-220k | Dummy value | [
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Calculate: $(\frac{1}{2})^{-1}+\sqrt{12}-4\sin 60^{\circ}$. | To calculate the given expression $(\frac{1}{2})^{-1}+\sqrt{12}-4\sin 60^{\circ}$ step by step, we follow the operations as outlined:
1. **Exponentiation**: $(\frac{1}{2})^{-1}$ simplifies to $2$ because raising a fraction to the power of $-1$ inverts the fraction. Thus, $(\frac{1}{2})^{-1} = 2$.
2. **Square Root**: ... | 2 | Algebra | math-word-problem | cn_k12 | ee9353f9-89fe-561b-b788-21e561015894 | open-r1/OpenR1-Math-220k | Dummy value | [
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Problem 9.6. In triangle $ABC$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $AC$, point $K$ is marked, and on side $BC$, points $L$ and $M$ are marked such that $KL=KM$ (point $L$ lies on segment $BM$).
Find the length of segment $LM$, if it is known that $AK=4$, $BL=31$, and $MC=3$.
... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | olympiads | 5c28b636-44a2-558f-9d71-0b8ca3a5db6f | open-r1/OpenR1-Math-220k | Dummy value | [
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The negation of the proposition "For any $x \in \mathbb{R}$, $x^2 \geq 0$" is ( )
A: For any $x \in \mathbb{R}$, $x^2 < 0$
B: There does not exist $x \in \mathbb{R}$ such that $x^2 < 0$
C: There exists $x_0 \in \mathbb{R}$ such that $x_0^2 \geq 0$
D: There exists $x_0 \in \mathbb{R}$ such that $x_0^2 < 0$ | Since the negation of "For all $x \in M$, $p(x)$" is "There exists $x \in M$ such that not $p(x)$", the negation of "For any $x \in \mathbb{R}$, $x^2 \geq 0$" is "There exists $x_0 \in \mathbb{R}$ such that $x_0^2 < 0$".
Therefore, the correct answer is $\boxed{\text{D}}$. | \text{D} | Inequalities | MCQ | cn_k12 | 4efc63f2-3639-591d-8c27-44a8c37e6c99 | open-r1/OpenR1-Math-220k | Dummy value | [
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A plane has a special point $O$ called the origin. Let $P$ be a set of 2021 points in the plane, such that
(i) no three points in $P$ lie on a line and
(ii) no two points in $P$ lie on a line through the origin.
A triangle with vertices in $P$ is fat, if $O$ is strictly inside the triangle. Find the maximum number of ... | We will count the minimal number of triangles that are not fat. Let $F$ be the set of fat triangles, and $S$ the set of triangles that are not fat. If triangle $X Y Z \in S$, we call $X$ and $Z$ good vertices if $O Y$ is located between $O X$ and $O Z$. For $A \in P$, let $S_{A} \subseteq S$ be the set of triangles in ... | 2021 \cdot 505 \cdot 337 | Combinatorics | math-word-problem | olympiads_ref | NaN | open-r1/OpenR1-Math-220k | Dummy value | [
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## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\sqrt{1-x^{2}}+\arcsin x, 0 \leq x \leq \frac{7}{9}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\sqrt{1-x^{2}}+\arcsin x\right)^{\prime}=-\frac{2 x}{... | \frac{2\sqrt{2}}{3} | Calculus | math-word-problem | olympiads | b47f4841-2b05-52eb-8e39-029b20be6113 | open-r1/OpenR1-Math-220k | Dummy value | [
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Among the following functions, the one that is an even function with the smallest positive period of $\pi$, and is increasing in the interval $(0, \frac{\pi}{2})$ is ( )
A: $y=\sin |x|$
B: $y=|\sin x|$
C: $y=|\cos x|$
D: $y=\cos |x|$ | Solution:
A. $y=\sin |x|$ is an even function and is increasing in the interval $(0, \frac{\pi}{2})$, but it is not a periodic function.
B. $y=|\sin x|$ is an even function and is increasing in the interval $(0, \frac{\pi}{2})$, with a period of $\pi$, and is a periodic function. It meets the conditions.
C. $y=|\... | B | Algebra | MCQ | cn_k12 | 6de0572d-7af2-5e3d-aefa-e064690d6666 | open-r1/OpenR1-Math-220k | Dummy value | [
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Given the universal set $U=\{1,2,3,4,5,6,7,8\}$, $M=\{1,3,5,7\}$, $N=\{2,5,8\}$, then $(C_{\cup}M) \cap N=$ （ ）
A: $\{5\}$
B: $\{2,8\}$
C: $\{1,3,7\}$
D: $\{4,6\}$ | Since the universal set $U=\{1,2,3,4,5,6,7,8\}$, and $M=\{1,3,5,7\}$,
then $C_{\cup}M=\{2,4,6,8\}$. Since $N=\{2,5,8\}$,
then $(C_{\cup}M) \cap N=\{2,8\}$.
Therefore, the correct answer is $\boxed{B}$. | B | Logic and Puzzles | MCQ | cn_k12 | c0bf0c1c-69bb-5117-891c-f2fef94f8bb1 | open-r1/OpenR1-Math-220k | Dummy value | [
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Reasoning: "① A rectangle is a parallelogram, ② A square is a rectangle, ③ Therefore, a square is a parallelogram." The minor premise is ( )
A: ①
B: ②
C: ③
D: ①② | In the reasoning: "① A rectangle is a parallelogram, ② A square is a rectangle, ③ Therefore, a square is a parallelogram."
- Major premise: A rectangle is a parallelogram;
- Minor premise: A square is a rectangle;
- Conclusion: Therefore, a square is a parallelogram.
The minor premise is: ② A square is a rectangle.
T... | \text{B} | Logic and Puzzles | MCQ | cn_k12 | 172a216f-0a5e-528d-b581-fa329fb25317 | open-r1/OpenR1-Math-220k | Dummy value | [
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If $0 < x < 1$, then the minimum value of $\frac{1}{x}+\frac{9}{1-x}$ is ( )
A: $20$
B: $18$
C: $16$
D: $14$ | **Analysis**
This problem involves the application of derivatives in solving extremum problems. The key to solving the problem is to use derivatives to study the monotonicity of the function, and then determine the location where the extremum is attained based on the monotonicity of the function.
**Solution**
Given:... | C | Inequalities | MCQ | cn_k12 | 20026c48-a345-5568-9bdb-262bf3592ab1 | open-r1/OpenR1-Math-220k | Dummy value | [
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6. The range of the function $f(x)=\sqrt{x-3}+\sqrt{12-3 x}$ is ( ).
(A) $[1, \sqrt{2}]$
(B) $[1, \sqrt{3}]$
(C) $\left[1, \frac{3}{2}\right]$
(D) $[1,2]$ | 6.D.
The domain of $f(x)$ is $3 \leqslant x \leqslant 4$, so $0 \leqslant x-3 \leqslant 1$.
Let $x-3=\sin ^{2} \theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$, then
$$
\begin{array}{l}
f(x)=\sqrt{x-3}+\sqrt{3(4-x)} \\
=\sin \theta+\sqrt{3\left(1-\sin ^{2} \theta\right)}=\sin \theta+\sqrt{3} \cos \theta \... | D | Algebra | MCQ | cn_contest | f53b6aed-9c35-52e0-a7e5-aacf99d42e98 | open-r1/OpenR1-Math-220k | Dummy value | [
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2. If the decimal parts of $7+\sqrt{7}$ and $7-\sqrt{7}$ are $a$ and $b$ respectively, then $a b-3 a+2 b+1=$ $\qquad$ . | 2.0.
Since $2<\sqrt{7}<3$, therefore, the decimal part of $7+\sqrt{7}$ is $a=\sqrt{7}-2$. Also, because $-3<-\sqrt{7}<-2$, then $0<3-\sqrt{7}<1$. Therefore, the decimal part of $7-\sqrt{7}$ is $b=3-\sqrt{7}$. Hence
$$
\begin{array}{l}
a b-3 a+2 b+1 \\
=(a+2)(b-3)+7=\sqrt{7} \times(-\sqrt{7})+7=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | cn_contest | 0f3c156f-c9f6-545a-9b18-fb507b144b1c | open-r1/OpenR1-Math-220k | Dummy value | [
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12.130. The development of the lateral surface of a cylinder is a rectangle, in which the diagonal is equal to $a$ and forms an angle $\alpha$ with the base. Find the volume of the cylinder. | ## Solution.
The volume of the cylinder $V=\pi R^{2} H$ (Fig. 12.130). The height of the cylinder from $\triangle A B C$ is $H=C B=a \sin \alpha$. The length of the circumference of the base $c=A B=a \cos \alpha$. On the other hand, $c=2 \pi R$. Then $R=\frac{a \cos \alpha}{2 \pi}$. Substituting, we get: $V=\pi \frac{... | \frac{^{3}\cos^{2}\alpha\sin\alpha}{4\pi} | Geometry | math-word-problem | olympiads | 3705936a-f45a-531d-a111-55d7a6f0bc64 | open-r1/OpenR1-Math-220k | Dummy value | [
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9・15 If the relation is
$$
\log _{3}\left[\log _{\frac{1}{2}}\left(\log _{2} x\right)\right]=\log _{3}\left[\log _{\frac{1}{3}}\left(\log _{3} y\right)\right]=\log _{3}\left[\log _{\frac{1}{5}}\left(\log _{5} z\right)\right]=0 .
$$
then the size relationship of $x, y, z$ is
(A) $x<y<z$.
(B) $y<z<x$.
(C) $z<x<y$.
(D) $... | [Solution 1] From $\log _{3}\left[\log _{\frac{1}{2}}\left(\log _{2} x\right)\right]=0$, we get
$$
\log _{\frac{1}{2}}\left(\log _{2} x\right)=1, \log _{2} x=\frac{1}{2},
$$
so
$$
x=2^{\frac{1}{2}}.
$$
Similarly, we get $y=3^{\frac{1}{3}}, \quad z=5^{\frac{1}{5}}$. Converting to the same root, we have
$$
\begin{array... | z<x<y | Algebra | MCQ | olympiads | b1f26240-e3fa-52e5-8af4-7eacbdef7cfd | open-r1/OpenR1-Math-220k | Dummy value | [
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In the arithmetic sequence $\left\{a_{n}\right\}$ and the geometric sequence $\left\{b_{n}\right\}$, $a_{1}=b_{1}=1$, $a_{2}=b_{2}$, and $2+a_{4}=b_{3}$.
1. Find the general formulas for $\left\{a_{n}\right\}$ and $\left\{b_{n}\right}$.
2. Find the sum $S_{n}$ of the first $n$ terms of the sequence $\left\{a_{n}+b_{n}... | ### Step-by-Step Solution
#### Part 1: Finding the General Formulas
1. **Given Information**:
- The first term of both sequences, $a_{1} = b_{1} = 1$.
- The second terms are equal, $a_{2} = b_{2}$.
- The relation between the fourth term of the arithmetic sequence and the third term of the geometric sequence... | S_{n} = n^{2} + \frac{3^{n} - 1}{2} | Algebra | math-word-problem | cn_k12 | 31946b05-e0e4-58a3-8611-efde85cb23de | open-r1/OpenR1-Math-220k | Dummy value | [
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3. [4 points] Solve the equation $\frac{1}{2}(x+5) \sqrt{x^{3}-16 x+25}=x^{2}+3 x-10$. | Answer: $3 ; \frac{\sqrt{13}+1}{2}$.
Solution. Factoring the right side, we get $(x+5) \sqrt{x^{3}-16 x+25}=(x+$ 5) $(2 x-4)$. From here, there are two possibilities: either $x+5=0$ (then $x=-5$, which does not fit the domain of definition, as the expression under the root is negative), or $\sqrt{x^{3}-16 x+25}=2 x-4$... | 3;\frac{\sqrt{13}+1}{2} | Algebra | math-word-problem | olympiads | f5b3c4a6-6aeb-596d-8521-2ab3ed9d81f6 | open-r1/OpenR1-Math-220k | Dummy value | [
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# Problem 5. (2 points)
The inequality $x^{2}<n$ has exactly $n$ integer solutions. Find $n$. If there are multiple answers, list them in ascending order separated by commas or semicolons.
# | # Problem 5. (2 points)
The inequality $x^{2}<n$ has exactly $n$ integer solutions. Find $n$. If there are multiple answers, list them in ascending order separated by commas or semicolons.
Answer: $0,1,3,5$ | 0,1,3,5 | Inequalities | math-word-problem | olympiads | ce551228-5875-54d9-b63f-9a80b7894824 | open-r1/OpenR1-Math-220k | Dummy value | [
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4. A national football association stipulates: In the league, a team gets $a$ points for a win, $b$ points for a draw, and 0 points for a loss, where real numbers $a>b>0$. If a team has exactly 2015 possible total scores after $n$ matches, find the minimum value of $n$. | 4. Suppose a team wins $x$ games, draws $y$ games, and loses $z$ games in $n$ matches, where $x, y, z \in \mathbf{N}$, and $x+y+z=n$. Then, the total score of the team after $n$ matches is $a x + b y (0 \leqslant x + y \leqslant n)$.
Therefore, to consider the number of possible total scores the team can achieve after... | 62 | Combinatorics | math-word-problem | olympiads | 49fe0b45-fe3c-5130-9ba9-2ff9f723a2aa | open-r1/OpenR1-Math-220k | Dummy value | [
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Example 1 (1994 National High School Mathematics League Question) Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$,
and $\left\{\begin{array}{l}x^{3}+\sin x-2 a=0, \\ 4 y^{3}+\sin y \cos y+a=0 .\end{array}\right.$ Find the value of $\cos (x+2 y)$. | Analysis: The feature of this problem is a very small population. The required solution is not closely related to the given conditions. Pay attention to comparing the structure of the two expressions, think boldly, and by constructing a function, a successful path can be found.
Solution: The conditions can be transfor... | 1 | Algebra | math-word-problem | olympiads | b4365cfa-134b-5d0d-8800-f1c85eaeb35d | open-r1/OpenR1-Math-220k | Dummy value | [
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8. Given that the sum of the distance from any point on curve $C$ to point $A(0,0)$ and the line $x=4$ is equal to 5, for a given point $B(b, 0)$, there are exactly three pairs of distinct points on the curve that are symmetric with respect to point $B$, then the range of $b$ is $\qquad$ | $8.2<b<4$
Analysis: Let a point on curve $C$ be $(p, q)$, then $\sqrt{p^{2}+q^{2}}+|p-4|=5$, simplifying to: when $p \geq 4$, $p=\frac{81-q^{2}}{18}$; when $p<4$, $p=\frac{q^{2}-1}{2}$; since both segments are parabolas and are symmetric about the $x$-axis, the remaining two pairs of symmetric points must be distribut... | 8.2<b<4 | Geometry | math-word-problem | olympiads | f69c018b-9793-59ba-99d5-cd68979f7d14 | open-r1/OpenR1-Math-220k | Dummy value | [
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4. Let $p(x)$ be a polynomial of degree $2n$, $p(0)=p(2)=\cdots=p(2n)=0, p(1)=p(3)=\cdots=$ $p(2n-1)=2, p(2n+1)=-30$. Find $n$. | 4. From $\Delta^{2 n+1} p(0)=\sum_{b=0}^{2 n+1}(-1)^{k} \cdot C_{2 n+1}^{k} \cdot p(2 n+1-k)=0$, we have $(-30)+\sum_{j=1}^{n} C_{2 n+1}^{2 j} \cdot 2=0$, which simplifies to $(-30)+2 \cdot\left(\frac{1}{2} \cdot 2^{2 n+1}-1\right)=0$, solving for $n$ yields $n=2$. | 2 | Algebra | math-word-problem | olympiads | ab14470c-154c-5fa1-9ee3-9f62fc980c67 | open-r1/OpenR1-Math-220k | Dummy value | [
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21. If the length of the upper base of a trapezoid is $l$, and the length of the line segment connecting the midpoints of the two non-parallel sides is $m$, then the length of the line segment connecting the midpoints of the two diagonals is ( ).
(A) $m-2 l$
(B) $\frac{m}{2}-l$
(C) $2 m-l$
(D) $m-l$ | 21. D.
Let the lower base of the trapezoid be $a$, and the length of the required line segment be $b$. Then $\frac{a+l}{2}=m, \frac{a-l}{2}=b$.
Solving for $b$ gives $b=m-l$. | D | Geometry | MCQ | cn_contest | eeec849d-a88c-5991-9cb9-5c953f104638 | open-r1/OpenR1-Math-220k | Dummy value | [
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8. The circle $\rho=D \cos \theta+E \sin \theta$ is tangent to the line of the polar axis if and only if ( ).
(A) $D \cdot E=0$
(B) $D \cdot E \neq 0$
(C) $D=0, E \neq 0$
(D) $D \neq 0, E=0$ | 8.(C).
From $\rho=D \cos \theta+E \sin \theta$, the rectangular coordinate equation is $\left(x-\frac{D}{2}\right)^{2}+\left(y-\frac{E}{2}\right)^{2}=\frac{1}{4}\left(D^{2}+E^{2}\right)$, or from $\rho=E \sin \theta(E \neq 0)$, it represents a circle tangent to the line of the polar axis. | C | Geometry | MCQ | cn_contest | 62198a89-7580-58c9-b119-953cb9e301c8 | open-r1/OpenR1-Math-220k | Dummy value | [
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Find the standard equation of the circle whose center lies on the line $y = -4x$ and is tangent to the line $x + y - 1 = 0$ at point $P(3, -2)$. | Let the center of the circle be $O(x, -4x)$. The slope of the line $OP$ is calculated by using the coordinates of point $P$ and the center $O$ as:
$$k_{OP} = \frac{(-2) - (-4x)}{3 - x} = \frac{2 - 4x}{3 - x}.$$
Since the given line $x + y - 1 = 0$ has a slope $k_L = -1$ and the circle is tangent to this line at point... | (x - 1)^2 + (y + 4)^2 = 8 | Geometry | math-word-problem | cn_k12 | b292c413-bfb5-5f00-aa8b-2b1c1c88eab6 | open-r1/OpenR1-Math-220k | Dummy value | [
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In the geometric sequence $\{a\_n\}$ where all terms are positive, if $a\_5a\_6=9$, then $\log \_{3}a\_{1}+\log \_{3}a\_{2}+…+\log \_{3}a\_{10}=($ $)$
A: $12$
B: $2+\log \_{3}5$
C: $8$
D: $10$ | Given that all terms in the geometric sequence are positive and $a\_5a\_6=9$, we can use the property of geometric sequences that states $a\_1a\_{10}=a\_2a\_9=…=a\_5a\_6$.
Applying this property, we get $a\_5a\_6=9 \Rightarrow a\_1a\_{10}=9$.
Now, let's consider the given expression $\log \_{3}a\_{1}+\log \_{3}a\_{2}... | D | Algebra | MCQ | cn_k12 | e1f89d88-f54c-555e-8cb9-4c20c40a50bc | open-r1/OpenR1-Math-220k | Dummy value | [
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2. Let $X_{0}$ be the interior of a triangle with side lengths 3,4 , and 5 . For all positive integers $n$, define $X_{n}$ to be the set of points within 1 unit of some point in $X_{n-1}$. The area of the region outside $X_{20}$ but inside $X_{21}$ can be written as $a \pi+b$, for integers $a$ and $b$. Compute $100 a+b... | Answer: 4112
Solution:
$X_{n}$ is the set of points within $n$ units of some point in $X_{0}$. The diagram above shows $X_{0}, X_{1}, X_{2}$, and $X_{3}$. As seen above it can be verified that $X_{n}$ is the union of
- $X_{0}$,
- three rectangles of height $n$ with the sides of $X_{0}$ as bases, and
- three sectors of ... | 4112 | Geometry | math-word-problem | olympiads | 0f225d82-9ac5-57ad-bae1-96c9ef3a430e | open-r1/OpenR1-Math-220k | Dummy value | [
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In the figure, $ABCD$ is a $2 \times 2$ square, $E$ is the midpoint of $\overline{AD}$, and $F$ is on $\overline{BE}$. If $\overline{CF}$ is perpendicular to $\overline{BE}$, then the area of quadrilateral $CDEF$ is
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3-\frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \frac{11}{5}\qquad\textbf{... | Since $\angle EBA = \angle FCB$ and $\angle FBC = \angle AEB$, we have $\triangle ABE \sim \triangle FCB$.
$\frac{AB}{FC} = \frac{BE}{CB} = \frac{EA}{BF}$
$\frac{2}{FC} = \frac{\sqrt{5}}{2} = \frac{1}{BF}$
From those two equations, we find that $CF = \frac{4}{\sqrt{5}}$ and $BF = \frac{2}{\sqrt{5}}$
Now that we have $B... | \frac{11}{5} | Geometry | MCQ | amc_aime | 41c830ea-8dd3-5e72-8782-f99c9e62bcbe | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
2. Let $n=9+99+\cdots+\underbrace{99 \cdots 9}_{99 \uparrow}$. Then in the decimal representation of $n$, the digit 1 appears ( ) times.
(A) 50
(B) 90
(C) 99
(D) 100 | 2.C.
Since $n=(10-1)+(100-1)+\cdots+(1 \underbrace{00 \cdots 0}_{99 \uparrow}-1)$ $=\underbrace{11 \cdots 10}_{99}-99=\underbrace{11 \cdots 1}_{91 \uparrow} 011$,
therefore, in the decimal representation of $n$, the digit 1 appears $97+2=99$ (times). | C | Number Theory | MCQ | cn_contest | 4ad46810-f97b-58ab-a08e-937feb0a32a5 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
An arc of a circle measures $300^{\circ}$ and its length is $2 \mathrm{~km}$. What is the nearest integer to the measure of the radius of the circle, in meters?
(a) 157
(b) 284
(c) 382
(d) 628
(e) 764 | The correct option is (c).
Solution 1: If the radius is $r$, then the length of an arc of $\theta$ degrees is $\frac{2 \pi}{360} \theta r$. Thus, in the given problem, we have that
$$
2000 \mathrm{~m}=\frac{2 \pi}{360} 300 r=\frac{5 \pi}{3} r
$$
therefore $r=2000 \times(3 / 5 \pi) \approx 382.17 \mathrm{~m}$.
Solut... | 382 | Geometry | MCQ | olympiads | 7b70d17d-e8f2-52da-a797-de9b611988d1 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
The sum of the first $n$ terms of an arithmetic sequence $\{a_{n}\}$ is $S_{n}$, and $a_{4}+a_{5}=10$. What is $S_{8}$?
A: $10$
B: $20$
C: $30$
D: $40$ | Given that $\{a_{n}\}$ is an arithmetic sequence and $a_{4}+a_{5}=10$, we aim to find $S_{8}$, the sum of the first $8$ terms.
1. In an arithmetic sequence, the sum of the first $n$ terms, $S_{n}$, can be expressed as $S_{n} = \frac{n(a_{1}+a_{n})}{2}$.
2. Given $a_{4}+a_{5}=10$, we can use this to find $S_{8}$.
3. Th... | D | Algebra | MCQ | cn_k12 | a373ea0d-c8d0-5b5b-94fd-2b238fc63fe0 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
[ Product of lengths of chord segments and lengths of secant segments ] [ Auxiliary similar triangles Area of a triangle (using height and base). ]
Diameter $A B$ and chord $C D$ of a circle intersect at point $E$, and $C E=D E$. Tangents to the circle at points $B$ and $C$ intersect at point $K$. Segments $A K$ and $... | The height of triangle $C K M$ is equal to the segment $B E$. Calculate the tangent of angle $C A E$ and consider similar triangles $А М Е$ and $А К В$.
## Solution
By the theorem of the equality of the products of the segments of intersecting chords, $C E^{2}=9$. Additionally, $C E \perp A B$.
Let $O$ be the center... | \frac{27}{4} | Geometry | math-word-problem | olympiads | e06e6966-32f8-57de-b074-45a482426e1e | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
(7) Let $x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}} \geqslant 1$, then $x^{2}+y^{2}=$ | (7) 1 Hint: Use trigonometric substitution.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The second paragraph seems to be an instruction rather than part of the text to be translated. I will provide the translatio... | 1 | Inequalities | math-word-problem | olympiads | 41c84dd2-d9e1-5ac2-a279-2de0eaf97f4c | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
A square area of size $100 \times 100$ is paved with $1 \times 1$ square tiles of four colors: white, red, black, and gray - such that no two tiles of the same color touch each other (that is, they do not share a side or a vertex). How many red tiles can there be? | We will divide the area into 2,500 squares of $2 \times 2$, each of which consists of four tiles sharing a common vertex. Therefore, each of these squares can contain no more than one red tile, and thus the number of red tiles cannot exceed 2,500. The same is true for tiles of other colors, meaning there are exactly 2,... | 2500 | Combinatorics | math-word-problem | olympiads | 0d4c5951-dac0-584c-a70b-b1159639b6f1 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Exercise 12. Let $A B C$ be an isosceles triangle at $A$ such that $\widehat{C B A}=61^{\circ}$. Let $E$ be the point, other than $A$, located on the circumcircle of $A B C$ such that $E B=E C$. Let $D$ be the point other than $A$ such that $D B=D C=A B$.
Determine the value in degrees of the angle $\widehat{B E D}$. | Solution to Exercise 12: We have $\widehat{B A C}=180^{\circ}-2 \widehat{C B A}=58^{\circ}$.
$D$ is the symmetric point of $A$ with respect to the line $(A B)$. We have $\widehat{B E D}=180^{\circ}-$ $\widehat{A E B}=180^{\circ}-\left(90^{\circ}-\widehat{B A E}\right)=90^{\circ}+\frac{1}{2} \widehat{B A C}=90+29=119^{... | 119 | Geometry | math-word-problem | olympiads | d388f2b0-2e60-54ba-bbfb-926566ed0cf8 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
## 56. 100 !
How many zeros does the number "100!" (i.e., the product of all integers from 1 to 100) end with? | 56. Each number ends with as many zeros as the number of times it contains 10 as a factor. But 10 is the product of 5 and 2. Therefore, the number of trailing zeros is equal to the smallest of the following two numbers: the number of factors of 2 in the prime factorization of 100! and the number of factors of 5 in this... | 24 | Number Theory | math-word-problem | olympiads | 3607fc51-51e4-58e9-b8af-343b3c607408 | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
Given the function $y=a+8\ln x$ ($x\in[\frac{1}{e},e]$), there exists a point $P$ on its graph. There also exists a point $Q$ on the graph of the function $y=-x^2-2$. Points $P$ and $Q$ are symmetric about the $x$-axis. Determine the range of values for $a$ ($\,\,$).
A: $[6-8\ln2,e^2-6]$
B: $[6-8\ln2,e^2-6]$
C: $[10+\f... | According to the problem, $a+8\ln x=x^2+2$ has a solution in $[\frac{1}{e},e]$, which means $a=x^2+2-8\ln x$ has a solution in $[\frac{1}{e},e]$.
Let $g(x)=x^2+2-8\ln x$. Then, $g'(x)=2x-\frac{8}{x}=\frac{2x^2-8}{x}=0$. When $x\in[\frac{1}{e},e]$, $x=2$.
$g(2)=6-8\ln2$, $g(\frac{1}{e})=10+\frac{1}{e^2}$, $g(e)=e^2-6$... | D | Algebra | MCQ | cn_k12 | 50c9ed50-5eae-596a-b25c-3a07353dadcb | open-r1/OpenR1-Math-220k | Dummy value | [
"dummy value"
] |
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