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【Answer】105:44
【Analysis】A The grassland from being eaten clean to growing back to the original amount of grass takes 4 days. Assuming A grassland grows 1 unit of grass per day, then in 4 days it grows 4 units. The original amount of grass in A grassland is 4 units. The speed at which the cow eats grass is $(4+7) \div ... | 105:44 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | true | 7.6 | fe6ad07e9fb9c823 | 6. A uniformly growing uniform grassland is divided into two unequal parts, $\mathrm{A}$ and $\mathrm{B}$. A herd of cows first eats the grass on plot A, finishing it in exactly 7 days. Then the herd continues to eat the grass on plot B at the same speed, finishing it in 4 days, at which point plot A has just recovered... | Let $A$ and $B$ be the areas of plot A and plot B, respectively. Let $g$ be the growth rate of grass per unit area per day, and let $G$ be the initial amount of grass per unit area. Let $C$ be the rate at which the herd eats grass.
The initial amount of grass on plot A is $AG$. In 7 days, the total grass grown on plot ... |
From two inclined lines drawn from one point, the one with the larger projection is larger. Therefore, from the inequalities given in the problem, it follows that $O A \geq O B \geq O C \geq O D \geq O A$. Hence, $O A=O B=O C=O D$, which means $O$ is the center of the circle circumscribed around quadrilateral $A B C D$... | 0.5 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | true | 2 | 710648c7e0789888 | Given a quadrilateral $A B C D$ with an area of 1. From its internal point O, perpendiculars OK, OL, OM, and ON are dropped to the sides $A B, B C, C D$, and $D A$ respectively. It is known that $A K \geq K B, B L \geq L C, C M \geq M D$, and $D N \geq N A$. Find the area of the quadrilateral $K L M N$. | Final Answer: The final answer is $\boxed{1/2}$ |
Extend $C B$ and $D A$ to intersect at $F$, let $S_{\triangle A B F}=S_{3}$. Since $C E$ bisects $\angle B C D$ and $C E \perp A D$, we know that $\triangle C D F$ is an isosceles triangle, thus $S_{1} = S_{2} + S_{3}$.
Also, because $D E = 2 A E$ and $A B \parallel C D$, we know that $A F = \frac{1}{4} D F$. Therefore... | \frac{7}{8} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 3.4 | 47543e316b137c4a | Example 8 As shown in Figure 1.4.9, in trapezoid $A B C D$, $A B / / D C, C E$ is the bisector of $\angle B C D$, $C E \perp A D$ at $E, D E=2 A E, C E$ divides the trapezoid into two parts with areas $S_{1}$ and $S_{2}$, if $S_{1}=1$, find $S_{2}$. | Final Answer:
We are given that $S_1 = \text{Area}(\triangle CDE) = 1$ and $DE = 2AE$.
Since $\triangle ACE$ and $\triangle CDE$ have the same altitude from $C$ to $AD$, the ratio of their areas is equal to the ratio of their bases $AE$ and $DE$.
$\frac{S_{ACE}}{S_{CDE}} = \frac{AE}{DE} = \frac{AE}{2AE} = \frac{1}{2}$.... |
$(x-2)^2$ = perfect square.
1 < perfect square < 25
Perfect square can equal: 4, 9, or 16
Solve for $x$:
$(x-2)^2=4$
$x=4,0$
and
$(x-2)^2=9$
$x=5,-1$
and
$(x-2)^2=16$
$x=6,-2$
The sum of all integer solutions is
$4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}$ | 12 | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 1.25 | 5bae26dcf0c3215d | What is the sum of all integer solutions to $1<(x-2)^2<25$?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$ | We are asked to find the sum of all integer solutions to the inequality $1 < (x-2)^2 < 25$.
We can split this compound inequality into two separate inequalities:
\begin{enumerate}
\item $(x-2)^2 > 1$
\item $(x-2)^2 < 25$
\end{enumerate}
For the first inequality, $(x-2)^2 > 1$, we take the square root of both s... |
$$
a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2 \Rightarrow a^{2}+(2 b-1)^{2}+(3 c-2)^{2}=3,
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(1+1+1)\left[a^{2}+(2 b-1)^{2}+(3 c-2)^{2}\right] \geqslant(a+2 b+3 c-3)^{2} \\
\Rightarrow(a+2 b+3 c-3)^{2} \leqslant 9 \Rightarrow 0<a+2 b+3 c \leqslant 6 .
\end{array}
$$... | 6 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | true | 6.5 | bcc57c0c72d3fca2 | 10. Let positive real numbers $a, b, c$ satisfy $a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2$, find the minimum value of $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$. | We are given $a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2$, which can be rewritten as $a^{2}+4 b^{2}-4 b+9 c^{2}-12 c+2=0$.
Completing the square, we have $a^{2}+(2 b-1)^{2}-1+(3 c-2)^{2}-4+2=0$, so $a^{2}+(2 b-1)^{2}+(3 c-2)^{2}=3$.
Let $x=a, y=2 b-1, z=3 c-2$. Then $a=x, b=\frac{y+1}{2}, c=\frac{z+2}{3}$.
The constraint becomes... |
Each of the $\binom{9}{2} = 36$ pairs of vertices determines two equilateral triangles, for a total of 72 triangles. However, the three triangles $A_1A_4A_7$, $A_2A_5A_8$, and $A_3A_6A_9$ are each counted 3 times, resulting in an overcount of 6. Thus, there are $\boxed{66}$ distinct equilateral triangles. | 66 | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | true | 5.25 | ff388ce341b2c497 | Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$, how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$?
$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$ | We consider two types of equilateral triangles: those with exactly two vertices in $\{A_1, \dots, A_9\}$ and those with exactly three vertices in $\{A_1, \dots, A_9\}$.
Case 1: Equilateral triangles with exactly three vertices in $\{A_1, \dots, A_9\}$.
We found that there are 3 such triangles: $\{A_1, A_4, A_7\}, \{A_... |
Apply Heron's formula and the property of the angle bisector of a triangle.
## Solution
Let $B P$ and $B Q$ be the height and the angle bisector of the given triangle $A B C$. By Heron's formula,
On the other hand, $S=1 / 2$ AC$\cdot$BP.
Therefore, $B P=\frac{2 S_{\triangle A B C}}{A C}=$
$\frac{2 \cdot 336}{28}=2... | 36 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 3.75 | c8ca584a8d928f91 | [ [ Relationship,
In triangle $ABC$, the three sides are given: $AB=26$, $BC=30$, and $AC=28$. Find the part of the area of this triangle enclosed between the height and the bisector drawn from vertex $B$.
# | Let $ABC$ be a triangle with sides $AB=26$, $BC=30$, and $AC=28$. Let $BH$ be the altitude from $B$ to $AC$, and $BD$ be the angle bisector of $\angle ABC$, where $H$ and $D$ are on $AC$. We want to find the area of triangle $BHD$.
First, we find the position of $H$ on $AC$. Let $AH = x$. Then $CH = 28-x$.
By Pythagor... |
Let $M_{0}$ be the original square, and $M_{1}, M_{2}, M_{3}, \ldots$ be the polygons obtained from $M_{0}$ by sequentially cutting off corners. It is convenient to also consider the polygon $N_{k}$, whose vertices are the midpoints of the sides of $M_{k} (k=0,1,2, \ldots)$.
Let $A$ be an arbitrary vertex of the polyg... | \frac{5}{7} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 5 | e88a0f014bb545ff | Konyagin S.V.
Given a square with side length 1. From it, four corners are cut off - four triangles, each of which has two sides along the sides of the square and these sides are $1 / 3$ of their lengths. With the resulting 8-sided polygon, the same operation is performed: from each vertex, a triangle is cut off, two ... | Let $A_0 = 1$ be the area of the initial square.
In the first step, we cut off 4 triangles, each with area $\frac{1}{18}$. Total area cut off is $C_1 = 4 \times \frac{1}{18} = \frac{2}{9}$.
Area of $P_1$ is $A_1 = A_0 - C_1 = 1 - \frac{2}{9} = \frac{7}{9}$.
In the second step, we cut off 8 triangles, each with area $\f... |
Solve 1. Answer: \((2,4,4),(2,3,6)\) or \((3,3,3)\).
Let \(x \leqslant y \leqslant z\) be natural numbers, and \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\). Then \(x1000^{10}\), so the number \(2^{100}\) has at least 31 digits. On the other hand,
\[
\frac{1024^{10}}{1000^{10}}<\left(\frac{1025}{1000}\right)^{10}=\left(\... | (2,4,4),(2,3,6),(3,3,3) | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 2 | 5c5ca9d90e4ec4e6 | 3-ча 1. The sum of the reciprocals of three positive integers is equal to 1. What are these numbers? Find all solutions. | Let the three positive integers be x, y, and z.
We are given the equation:
$$ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 $$
Without loss of generality, assume $x \le y \le z$.
Since x, y, z are positive integers, we must have $x > 1$, otherwise if $x=1$, then $\frac{1}{1} + \frac{1}{y} + \frac{1}{z} = 1$, which implie... |
3.11 Let's construct the following table:
| Apple Variety | Cost, rub | Quantity, tons | Price, rub/kg |
| :---: | :---: | :---: | :---: |
| First | 228 | $x$ | $\frac{228}{x}$ |
| Second | 180 | $x+5$ | $\frac{180}{x+5}$ |
According to the problem, we have the equation
$\left(\frac{228}{x}-0.9\right)(2 x+5)=408$,
... | 85 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | General algebraic systems | true | 3 | f20bc40e5e1ede97 | 3.11 Apples of the first grade for a total of 228 rubles and apples of the second grade for a total of 180 rubles were delivered to the store. During unloading, the delivered apples were accidentally mixed. It was found that if all the apples are now sold at one price - 90 kopecks lower than the price per kilogram of f... | Let $m_1$ be the mass of first-grade apples in kilograms, and $m_2$ be the mass of second-grade apples in kilograms.
Let $P_1$ be the price per kilogram of first-grade apples in rubles, and $P_2$ be the price per kilogram of second-grade apples in rubles.
We are given that the total cost of first-grade apples is 228 r... |
Solution. The rounding error of the measurement can be considered as a random variable $X$, which is uniformly distributed in the interval between two adjacent integer divisions. The density of the uniform distribution $f(x)=1 /(b-a)$, where $(b-a)$ is the length of the interval in which the possible values of $X$ are ... | 0.6 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Probability theory and stochastic processes | true | 2 | d0e773c9bc2f9a5b | 308. The scale division value of the amperemeter is 0.1 A. The readings of the amperemeter are rounded to the nearest whole division. Find the probability that an error exceeding $0.02 \mathrm{~A}$ will be made during the reading. | Let the scale division value of the amperemeter be $d = 0.1 \mathrm{~A}$.
The readings of the amperemeter are rounded to the nearest whole division, which means the readings are rounded to the nearest multiple of $0.1 \mathrm{~A}$.
Let $I$ be the true value of the current.
Let $R$ be the reading of the amperemeter.
Wh... |
The increase in value was
$$
149.70-29.90=119.80 \text { reais }
$$
which corresponds to:
$$
\frac{119.80}{29.90} \times 100 \% = 400.66 \%
$$
The correct option is (d). | 400 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematics education | true | 2.4 | 34a69a3a5da3136b | In 1972, filling the tank of a small car cost $R \$ 29.90$, and in 1992, it cost $\$ 149.70$ to fill the same tank. Which of the following values best approximates the percentage increase in the price of gasoline over this 20-year period?
(a) $20 \%$
(b) $125 \%$
(d) $300 \%$
(d) $400 \%$
(e) $500 \%$ | The price of filling the tank in 1972 was $P_{1972} = \$ 29.90$.
The price of filling the tank in 1992 was $P_{1992} = \$ 149.70$.
The increase in price is $P_{1992} - P_{1972} = 149.70 - 29.90 = \$ 119.80$.
The percentage increase in the price of gasoline is given by:
Percentage increase $= \frac{\text{Increase in pri... |
The zero constant function is a solution, but we exclude this case. By setting $x=0$, we get $f(0)=0$, and by setting $y=-1$, we get $f(-1)=-1$. By setting $x=-1$ and $y=-\frac{1}{2}$, we get $f\left(-\frac{1}{2}\right)=-\frac{1}{2}$. Finally, by setting $x=-\frac{1}{2}$ and $y=1$, we get $f(1)=1$. Then, by setting $x=... | f(x)=x | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Difference and functional equations | true | 5.75 | 604c3458242cf84a | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying:
$$
f(x(1+y))=f(x)(1+f(y))
$$ | Final Answer: The solutions are $\boxed{f(x) = 0, f(x) = x}$ |
The answer is $\dfrac{\text{number of students who preferred spaghetti}}{\text{number of students who preferred manicotti}}$
So,
$\frac{250}{100}$
Simplify,
$\frac{5}{2}$
The answer is $\boxed{\textbf{(E)}\ \dfrac{5}{2}}$ | \textbf{(E)}\\frac{5}{2} | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematics education | true | 3.2 | da4a7ae24e4bf1c0 | $650$ students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?
$\mathrm{(A)} \frac{2}... | Rationale:
Let $S$ be the number of students who preferred spaghetti and $M$ be the number of students who preferred manicotti. We are asked to find the ratio $\frac{S}{M}$. We need to read the number of students who preferred each type of pasta from the bar graph (which is not provided here). However, we can test each... |
Regardless of the number of years, the Gregorian calendar offers 97 days every 400 years, which is $97 \cdot 24=2328$ hours, compared to 5 hours, 48 minutes, and 46 seconds 400 times over. It is reasonable to count the minutes as $\frac{4}{5}$ of an hour, thus in 400 years, $400 \cdot\left(5+\frac{4}{5}\right)=$ 2320 h... | 3323 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Numerical analysis | true | 5.25 | 4c6ae5502cd56a0f | According to the Gregorian calendar, 97 leap days must be inserted over 400 consecutive years. After how many years will there be a 1-day difference between the Gregorian calendar and the "precise" calendar, if the length of a year is 365 days 5 hours 48 minutes 46 seconds? | The length of a Gregorian year is, on average, $365 + \frac{97}{400} = \frac{146097}{400} = 365.2425$ days.
The length of the precise year is given as 365 days 5 hours 48 minutes 46 seconds.
Convert 5 hours 48 minutes 46 seconds to days:
5 hours = $5/24$ days
48 minutes = $48/60$ hours = $48/(60 \times 24)$ days = $48/... |
Suppose the pair $ (k,m) $ satisfies the conditions of the problem.
If the inequality $ m\geq k $ holds, then
$$(m+3)^2=m^2+6m+9>m^2+5m\geq m^2+5k>m^2,$$
and since $ m^2+5k $ is a square of an integer, it follows that one of the equalities $ m^2+5k=(m+1)^2 $ or $ m^2+5k=(m+2)^2 $ must hold.
If $ m^2+5k=(m+1)^2=m^2+2... | (1,2),(9,22),(8,9) | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 5 | 2a9114fe3d1f7eb1 | LVIII OM - I - Problem 2
Determine all pairs of positive integers $ k $, $ m $, for which each of the numbers $ {k^2+4m} $, $ {m^2+5k} $ is a perfect square. | We are given that $k^2+4m = a^2$ and $m^2+5k = b^2$ for some positive integers $a, b$.
From the first equation, $4m = a^2-k^2$. Since $m>0$, $a>k$. Let $a = k+x$ for some positive integer $x$. Then $4m = (k+x)^2-k^2 = 2kx+x^2$, so $m = \frac{x(2k+x)}{4}$.
From the second equation, $5k = b^2-m^2$. Since $k>0$, $b>m$. Le... |
Answer: 999.
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$ also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the s... | 999 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 6 | ccc3ed17d9261dd9 | 5. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Final Answer: The final answer is $\boxed{999}$ |
Answer: $320^{\circ}$
Solution: We first apply sum-to-product and product-to-sum:
$$
\begin{array}{c}
\frac{\sin 4 x+\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x} \\
2 \sin (2.5 x) \cos (1.5 x) \sin (2 x)=\sin (4 x) \sin (3 x)
\end{array}
$$
Factoring out $\sin (2 x)=0$,
$$
\sin (2.5 x) \cos (1.5 x)=\cos (2 x) \sin (3 x... | 320 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | General algebraic systems | true | 10 | 9cca75520f6d37fe | 20. [10] There exist several solutions to the equation
$$
1+\frac{\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x},
$$
where $x$ is expressed in degrees and $0^{\circ}<x<180^{\circ}$. Find the sum of all such solutions. | We are asked to find the sum of all solutions to the equation $1+\frac{\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x}$ for $0^{\circ}<x<180^{\circ}$.
We rewrite the equation as $\cos \frac{9x}{2} (\cos \frac{5x}{2} - \cos \frac{3x}{2}) = 0$.
Case 1: $\cos \frac{9x}{2} = 0$.
$\frac{9x}{2} = 90^{\circ} + k \cdot 180^{\circ}... |
A2. In the graph, there is a rational function with a pole at $x=-1$, a double zero at $x=2$, and an initial value of 2, that is, the function $f(x)=\frac{(x-2)^{2}}{2 x+2}$. | f(x)=\frac{(x-2)^{2}}{2x+2} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | true | 2 | bcac71a51263e3e9 | A2. The graph of the function on the image is:
(A) $f(x)=\log _{3}(x+1)-1$
(B) $f(x)=\frac{(x-2)^{2}}{2 x+2}$
(C) $f(x)=2^{x+1}+3$
(D) $f(x)=\frac{2}{3} x-1$
(E) None of the above. | Rationale:
We analyze the given graph and compare it with the properties of the functions in the options.
The graph has a vertical asymptote at $x = -1$ and as $x \to -1^+$ the function tends to $-\infty$. The function is increasing.
Let's check each option:
(A) $f(x)=\log _{3}(x+1)-1$
Domain: $x+1 > 0 \Rightarrow x >... |
Isosceles triangles $E O F, G O H$ and $E F G$ are similar.
## Otвет
$3: 1$.
## Answer
$3: 1$. | 3:1 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 3 | d0c0cd67f88b221b | When triangle $E F G$ is rotated by an angle $\arccos _{1 / 3}$ around point $O$, which lies on side $E G$, vertex $F$ moves to vertex $E$, and vertex $G$ moves to point $H$, which lies on side $F G$. Find the ratio in which point $O$ divides side $E G$. | Let $O$ be the origin. Let $\vec{OE} = \mathbf{e}$ and $\vec{OG} = \mathbf{g}$. Since $O$ lies on side $EG$, $\mathbf{e}$ and $\mathbf{g}$ are collinear and in opposite directions. Let $OE = x$ and $OG = y$. Then $\mathbf{e} = -x \mathbf{u}$ and $\mathbf{g} = y \mathbf{u}$ for some unit vector $\mathbf{u}$.
Let $\vec{O... |
There are at most $5$ games played.
If team $B$ won the first two games, team $A$ would need to win the next three games. So the only possible order of wins is $BBAAA$.
If team $A$ won the first game, and team $B$ won the second game, the possible order of wins are: $ABBAA, ABABA,$ and $ABAAX$, where $X$ denotes that... | \frac{1}{5} | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 3 | 3ace2f546c0e4578 | Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first gam... | Let $E$ be the event that team B wins the second game.
Let $F$ be the event that team A wins the series.
Let $G$ be the event that team B wins the first game.
We want to find $P(G | E \cap F) = \frac{P(G \cap E \cap F)}{P(E \cap F)}$.
We need to find the sequences of games where team B wins the second game and team A ... |
WLOG, let $CD=1$.
Looking at square GHDC, we see that $\angle DHC=45$, which implies that $DC=CH=1$ and $DH=\sqrt{2}$
Taking each cross-section one at a time, we look at square DHFB. We obviously know that CHB is a $30$ degree angle, giving $BH=\frac{2\sqrt{3}}{3}$, and $BC=\frac{\sqrt{3}}{3}$.
Looking at square ABCD,... | \frac{\sqrt{6}}{4} | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 4 | 9f060aab2f7d7203 | In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$. Find the cosine of $\angle BHD$.
$\text {(A)} \frac{\sqrt{3}}{6} \qquad \text {(B)} \frac{\sqrt{2}}{6} \qquad \text {(C)} \frac{\sqrt{6}}{3} \qquad \text{(D)}\frac{\sqrt{6}}{4}\qquad \text{(E)}\frac{\sqrt{6}-\sqrt{2}}{4... | Let $H$ be the origin. Let $\vec{HD} = (0,0,z)$, $\vec{HG} = (x,0,x)$. Then $\cos(\angle DHG) = \frac{\vec{HD} \cdot \vec{HG}}{||\vec{HD}|| \cdot ||\vec{HG}||} = \frac{x z}{\sqrt{z^2} \sqrt{2x^2}} = \frac{1}{\sqrt{2}}$, so $\angle DHG = 45^\circ$.
Let $\vec{HE} = (0,y,0)$. Then $\vec{HF} = \vec{HG} + \vec{HE} = (x,y,x)... |
4. $\frac{2+\sqrt{2}}{4}$.
Assume the edge lengths of the quadrilateral pyramid $P-ABCD$ are all 2. Taking the center $O$ of the square $ABCD$ as the origin, and the directions of $\overrightarrow{DA}$, $\overrightarrow{DC}$, and $\overrightarrow{OP}$ as the positive directions of the $x$-axis, $y$-axis, and $z$-axis,... | \frac{2+\sqrt{2}}{4} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Linear and multilinear algebra; matrix theory | true | 4 | 8af90ab84c548e84 | 4. Given a regular quadrilateral pyramid $P-A B C D$ with all edges of equal length. Taking $A B C D$ as one face, construct a cube $A B C D-E F G H$ on the other side of the pyramid. Then, the cosine value of the angle formed by the skew lines $P A$ and $C F$ is | Let $A = (0, 0, 0)$, $B = (a, 0, 0)$, $C = (a, a, 0)$, $D = (0, a, 0)$.
The center of the square $ABCD$ is $O = (\frac{a}{2}, \frac{a}{2}, 0)$. The height of the pyramid is $PO = \frac{\sqrt{2}a}{2}$.
So $P = (\frac{a}{2}, \frac{a}{2}, \frac{\sqrt{2}a}{2})$.
The cube $ABCD-EFGH$ is constructed on the other side of the ... |
# Solution
Notice that 345 and $5 y^{2}$ are divisible by 5, so $3 x^{2}$ must also be divisible by 5. Therefore, $\quad x=5 t, t \in Z$. Similarly, $y=3 n, n \in Z$. After simplification, the equation becomes $5 t^{2}+3 n^{2}=23$. Therefore, $t^{2} \leq \frac{23}{5}$, $n^{2} \leq \frac{23}{3}$ or $|t| \leq 2,|n| \leq... | 13 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 3 | ca34d2811a28a389 | 1. Find the maximum value of the expression $x+y$, where $x, y-$ are integer solutions of the equation $3 x^{2}+5 y^{2}=345$ | We are given the equation $3 x^{2}+5 y^{2}=345$ and we need to find the maximum value of $x+y$ where $x, y$ are integer solutions.
First, we can simplify the equation by dividing by the greatest common divisor if possible.
We can see that $345 = 3 \times 115 = 3 \times 5 \times 23$.
Divide the equation by 3: $x^{2} + ... |
There are (n+2)(n+1)/2 matches, so the total score is (n+2)(n+1)/2. Let the other players score k each. Then 8 + nk = (n+2)(n+1)/2, so n 2 - (2k-3) - 14 = 0. We know this equation has one root which is a positive integer. The product of the roots is -14, so the possibilities for the roots are: 1, -14; 2, -7; 7, -2; 14,... | 7or14 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | true | 6 | 84c0ad44e41a1698 | 8th CanMO 1976 Problem 3 n+2 students played a tournament. Each pair played each other once. A player scored 1 for a win, 1/2 for a draw and nil for a loss. Two students scored a total of 8 and the other players all had equal total scores. Find n. | Let $n+2$ be the number of students. Two students scored a total of 8, and the other $n$ students scored equal scores $x$. The sum of scores is $8+nx$. The total number of games played is $\binom{n+2}{2} = \frac{(n+2)(n+1)}{2}$.
Thus, $8+nx = \frac{(n+2)(n+1)}{2}$.
$16+2nx = (n+2)(n+1) = n^2+3n+2$.
$2nx = n^2+3n-14$.
$... |
(1) $\left[0, \frac{\sqrt{6}}{6}\right]$ | [0,\frac{\sqrt{6}}{6}] | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | true | 3 | 2e3cd9da5d2f3982 | (1) The range of the function $f(x)=\sqrt{\frac{x+1}{x^{2}+4 x+7}}$ is $\qquad$ | The domain of the function $f(x)=\sqrt{\frac{x+1}{x^{2}+4 x+7}}$ is determined by the condition $\frac{x+1}{x^{2}+4 x+7} \ge 0$.
Since $x^2+4x+7 = (x+2)^2+3 > 0$ for all real $x$, we need $x+1 \ge 0$, which means $x \ge -1$.
Let $g(x) = \frac{x+1}{x^2+4x+7}$. We want to find the range of $f(x) = \sqrt{g(x)}$ for $x \ge... |
[Solution] The sum of the squares of six odd numbers must be even, so we exclude $(D)$.
Let these six odd numbers be $2 k_{1}+1, 2 k_{2}+1, \cdots 2 k_{6}+1$. Note that
$$
\begin{aligned}
& \left(2 k_{1}+1\right)^{2}+\left(2 k_{2}+1\right)^{2}+\cdots+\left(2 k_{6}+1\right)^{2} \\
= & 4\left(k_{1}^{2}+k_{2}^{2}+\cdots+k... | C | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 2.67 | 22098cad61a4eab0 | $30 \cdot 35$ Six odd numbers, their sum is 42, their sum of squares can only be
(A) 280.
(B) 368.
(C) 382.
(D) 423.
(China Jiangxi Nanchang Junior High School Mathematics Competition, 1990) | Let the six odd numbers be $x_1, x_2, x_3, x_4, x_5, x_6$. We are given that $\sum_{i=1}^6 x_i = 42$. Since the sum of six odd numbers is even, this is possible.
Let $x_i = 7 + \delta_i$, where $\sum_{i=1}^6 \delta_i = 0$. Since $x_i$ are odd, $\delta_i = x_i - 7$ must be even integers. Let $\delta_i = 2k_i$, where $k_... |
Let $S$ be the area of the base of the pyramid, $S_1$ and $S_2$ be the areas of the largest and smallest sections, respectively. The largest and smallest sections are polygons similar to the polygon of the base of the pyramid with coefficients $\frac{99}{100}$ and $\frac{1}{100}$, respectively. Therefore,
$$
S_1 = \le... | 9801 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 2 | 4e50a84d2fe4fa46 | [ [ Cross-sectional area $]$
The lateral edge of the pyramid is divided into 100 equal parts, and planes parallel to the base are drawn through the points of division. Find the ratio of the areas of the largest and smallest of the resulting cross-sections.
# | Let H be the height of the pyramid, and A be the area of the base.
Let the lateral edge of the pyramid be divided into 100 equal parts by 99 division points.
Planes parallel to the base are drawn through these 99 division points. This results in 99 cross-sections.
Let the heights of the smaller pyramids formed by these... |
First, let us list the statement "If a quadrilateral is a square, then it is a rectangle" as a statement of the form "If $p$, then $q$". In this case, $p$ is "a quadrilateral is a square", and $q$ is "it is a rectangle".
The converse is then: "If $q$, then $p$". Plugging in, we get "If a quadrilateral is a rectangle, t... | D | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | true | 1.4 | 89e808a74469eb7d | Given the true statement: If a quadrilateral is a square, then it is a rectangle.
It follows that, of the converse and the inverse of this true statement is:
$\textbf{(A)}\ \text{only the converse is true} \qquad \\ \textbf{(B)}\ \text{only the inverse is true }\qquad \\ \textbf{(C)}\ \text{both are true} \qquad \\ \... | Let the given statement be "If a quadrilateral is a square, then it is a rectangle."
Let P be the statement "A quadrilateral is a square."
Let Q be the statement "It is a rectangle."
The given statement is in the form "If P, then Q", which is $P \implies Q$. We are given that this statement is true.
The converse of th... |
Let $CH$ be the height of the isosceles triangle $ABC$ dropped to the lateral side $AB$, with $AH=2$, $BH=1$. Then $AC=AB=3$.
By the Pythagorean theorem from the right triangles $ACH$ and $BCH$, we find that
$$
CH^2=AC^2-AH^2=9-4=5, BC^2=BH^2+CH^2=1+5=6 \text{. }
$$
Therefore, $BC=\sqrt{6}$.
## Answer
$\sqrt{6}$. | \sqrt{6} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 2 | c268418b704d8f8d | [Pythagorean Theorem (direct and inverse).]
The height of an isosceles triangle, dropped to the lateral side, divides it into segments of 2 and 1, counting from the vertex of the triangle. Find the base of the triangle. | Let the isosceles triangle be ABC, with AB = AC. Let CH be the height dropped from vertex C to the lateral side AB.
The height CH divides the side AB into segments AH and HB. We are given that these segments are of lengths 2 and 1, counting from the vertex of the triangle. Since CH is dropped to the lateral side AB, th... |
$1347$ | 1347 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | true | 6 | d283b9d7a2bbac85 | 7. Let $[a]$ denote the greatest integer not exceeding $a$, for example: $[8]=8, [3.6]=3$.
Some natural numbers can be expressed in the form $[x]+[2 x]+[3 x]$, such as 6 and 3:
$$
\begin{array}{c}
6=\left[\frac{5}{4}\right]+\left[2 \times \frac{5}{4}\right]+\left[3 \times \frac{5}{4}\right], \\
3=[0.8]+[2 \times 0.8]+[... | We want to find the number of natural numbers $n$ in the range $1 \le n \le 2020$ that can be expressed in the form $n = [x] + [2x] + [3x]$ for some real number $x$.
Let $f(x) = [x] + [2x] + [3x]$. We analyzed the values of $f(x)$ for $x \in [k, k+1)$ for integer $k \ge 0$.
For $x \in [k, k+1)$, $f(x)$ takes values $6k... |
3. When $n=3$, the conclusion is 3.
When $n>4$, construct a regular $n$-sided polygon and slightly adjust its shape to meet the conditions described in the problem. In this $n$-sided polygon, it can be directly calculated that for any vertex, the lines connecting it to its two adjacent vertices are the closest.
When $... | 3 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | true | 4.2 | 4f35a9b05dd20b43 | 3. Given $n(n>2)$ points on a plane, satisfying that no three points are collinear. Draw a line through any two points, and mark the point among the remaining points that is closest to this line (it is known that in any case, the point satisfying this condition is unique). Question: For each given $n$, what is the maxi... | Final Answer: For each given $n$, we want to find the maximum number of points that can be marked.
We consider small values of $n$.
For $n=3$, let the points be $A, B, C$.
For line $AB$, the closest point is $C$.
For line $AC$, the closest point is $B$.
For line $BC$, the closest point is $A$.
All 3 points $A, B, C$ ar... |
Let's inscribe a hexagon $K L M N O P$ into the triangle $A B C$ in the prescribed manner.
Given that both figures as a whole are symmetric with respect to three axes of symmetry, the centro... | 30\mathrm{~}^{2} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 3 | d2a78d72e8bce47b | For an equilateral triangle $A B C$, a regular hexagon $K L M N O P$ is inscribed such that points $K, M, O$ lie sequentially at the midpoints of sides $A B, B C$, and $A C$. Calculate the area of the hexagon $K L M N O P$ if the area of triangle $A B C$ is $60 \mathrm{~cm}^{2}$. (K. Pazourek) | Let $Area(ABC)$ denote the area of triangle $A B C$. We are given $Area(ABC) = 60 \mathrm{~cm}^{2}$.
Let $s$ be the side length of equilateral triangle $A B C$. Then $Area(ABC) = \frac{\sqrt{3}}{4} s^2 = 60$.
Since $K, M, O$ are midpoints of sides $A B, B C, C A$, triangle $KMO$ is equilateral with side length $\frac{s... |
Let pile $A$ have $A$ rocks, and so on.
The total weight of $A$ and $C$ can be expressed as $44(A + C)$.
To get the total weight of $B$ and $C$, we add the weight of $B$ and subtract the weight of $A$: $44(A + C) + 50B - 40A = 4A + 44C + 50B$
Therefore, the mean of $B$ and $C$ is $\frac{4A + 44C + 50B}{B + C}$, which i... | 59 | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | true | 3.4 | 2280828ca5f9d7b3 | $A$, $B$, $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the gr... | Let $n_A, n_B, n_C$ be the number of rocks in piles $A, B, C$ respectively, and let $\mu_A, \mu_B, \mu_{AB}, \mu_{AC}$ be the mean weights given. We are given $\mu_A = 40$, $\mu_B = 50$, $\mu_{AB} = 43$, $\mu_{AC} = 44$. Let $\mu_{BC}$ be the mean weight of rocks in combined piles $B$ and $C$. We want to find the great... |
Reference: 1995 HG5 . . $37^{100} \ldots$ 157-digit number, $37^{15} \ldots n$-digit ....
$$
\begin{array}{l}
P^{50}=y, 10<P \leq 99,10^{68} \leq y<10^{69} \\
P=y^{\frac{1}{50}} ; 10^{68 \div 50}<P<10^{69 \div 50} \\
1.34<\log P<1.38 \\
\log 22=\log 2+\log 11=1.3424 ; \log 24=3 \log 2+\log 3=1.3801 \\
\log 22<\log P<\l... | 23 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Field theory and polynomials | true | 3 | 07fbf42f4e0b4aef | I2.1 If the $50^{\text {th }}$ power of a two-digit number $P$ is a 69-digit number, find the value of $P$.
(Given that $\log 2=0.3010, \log 3=0.4771, \log 11=1.0414$.)
I2.2 The roots of the equation $x^{2}+a x-P+7=0$ are $\alpha$ and $\beta$, whereas the roots of the equation $x^{2}+b x-r=0$ are $-\alpha$ and $-\beta... | I2.1 Let $P$ be a two-digit number. The $50^{\text {th }}$ power of $P$, $P^{50}$, is a 69-digit number.
This means that $10^{68} \le P^{50} < 10^{69}$.
Taking logarithm base 10 on all sides, we get
$\log_{10}(10^{68}) \le \log_{10}(P^{50}) < \log_{10}(10^{69})$
$68 \le 50 \log_{10} P < 69$
Divide by 50:
$\frac{68}{50}... |
Let $O$ be the center of the circle with $ABCDE$ on it.
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$.
$\angle ECA$ is, therefore, $5y$ by way... | 58 | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | General algebraic systems | true | 4.2 | 67f8fe346b09317f | Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$. | Let $\widehat{AB} = \widehat{BC} = \widehat{CD} = \widehat{DE} = x$ and $\widehat{EF} = \widehat{FG} = \widehat{GH} = \widehat{HI} = \widehat{IA} = y$.
We are given that $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$, so $\angle ABD - \angle AHG = 12^\circ$.
$\angle ABD = \frac{1}{2} \widehat{AD} = \frac{1}{2} (\wideh... |
To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \boxed{\textbf{(D)}\ 225}.$ | 225 | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | true | 4 | 92778e98834aa268 | A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | Let $k$ be the number of distinct values in the list.
Let one of these values be the unique mode, which occurs exactly $10$ times.
The remaining $k-1$ distinct values must each occur at most $9$ times, since the mode is unique.
To minimize $k$, we want to maximize the frequencies of these $k-1$ distinct values.
So we a... |
The given information is equivalent to the first two dice being 20 and 19 and there being at least two 20's among the last three dice. Thus, we need to find the probability that given at least two of the last three dice are 20's, all three are. Since there is only one way to get all three 20's and $3 \cdot 19=57$ ways ... | \frac{1}{58} | KbsdJames/Omni-MATH/HMMT_2 | {'domain': ['Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations'], 'difficulty': 4.0} | Combinatorics | true | 2 | a4457562dde4bba7 | Your math friend Steven rolls five fair icosahedral dice (each of which is labelled $1,2, \ldots, 20$ on its sides). He conceals the results but tells you that at least half of the rolls are 20. Assuming that Steven is truthful, what is the probability that all three remaining concealed dice show $20 ?$ | Let $N$ be the number of rolls that are 20 in five rolls of fair icosahedral dice.
We are given that at least half of the rolls are 20, which means $N \ge 3$. We want to find the probability that all three remaining concealed dice show 20. We interpret this as the probability that all five rolls are 20, given that at l... |
13. A.
Notice that all two-digit palindromic numbers $11, 22, \cdots$, 99 are multiples of 11.
Since $N$ is the sum of two-digit palindromic numbers, $N$ is also a multiple of 11.
Among the multiples of 11, the smallest three-digit number is 110, and $110=77+22+11$, so $N=110$.
Therefore, the sum of the digits of $N... | 2 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 1 | 2cb6a700624bc334 | 13. If a number reads the same from left to right as it does from right to left, it is called a "palindromic number." For example, 12321 is a palindromic number. If $N$ is the smallest three-digit number that satisfies the following conditions: $N$ is equal to the sum of three different two-digit palindromic numbers an... | The two-digit palindromic numbers are 11, 22, 33, 44, 55, 66, 77, 88, 99.
We want to find the smallest three-digit number $N$ that is the sum of three different two-digit palindromic numbers and $N$ is not a palindromic number.
We want to minimize $N$, so we choose the smallest two-digit palindromic numbers.
Let's cons... |
Solution 1
Since $A B$ and $E D$ are parallel, quadrilateral $A B D E$ is a trapezoid.
We know that $A B=30 \mathrm{~cm}$.
Since $A B C F$ is a rectangle, then $F C=A B=30 \mathrm{~cm}$.
Suppose that $D C=x \mathrm{~cm}$.
Then $E D=F C-F E-D C=(30 \mathrm{~cm})-(5 \mathrm{~cm})-(x \mathrm{~cm})=(25-x) \mathrm{cm}$... | 17\mathrm{~} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Game theory, economics, social and behavioral sciences | true | 2.4 | c54660382eb88b03 | In the diagram, $A B C F$ is a rectangle with $A B=30 \mathrm{~cm}$ and $A F=14 \mathrm{~cm}$. Points $E$ and $D$ are on $F C$ so that $F E=5 \mathrm{~cm}$ and the area of quadrilateral $A B D E$ is $266 \mathrm{~cm}^{2}$. The length of $D C$ is
(A) $17 \mathrm{~cm}$
(B) $16 \mathrm{~cm}$
(C) $19 \mathrm{~cm}$
(D) $20 ... | Let $A B C F$ be a rectangle with $A B=30 \mathrm{~cm}$ and $A F=14 \mathrm{~cm}$.
The area of rectangle $A B C F$ is $A B \times A F = 30 \times 14 = 420 \mathrm{~cm}^{2}$.
We are given that $F E=5 \mathrm{~cm}$ and points $E$ and $D$ are on $F C$. Let $D C = x$.
Since $F C = A B = 30 \mathrm{~cm}$, we have $F D = F C... |
Since there are two possible prizes that Jamie can win and each is equally likely, then the probability that Jamie wins $\$ 30$ is $\frac{1}{2}$ and the probability that Jamie wins $\$ 40$ is $\frac{1}{2}$. If Jamie wins $\$ 30$, then for the total value of the prizes to $\$ 50$, Ben must win $\$ 20$. The probability t... | \frac{1}{3} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Probability theory and stochastic processes | true | 1 | b87ec707839457ba | Ben participates in a prize draw. He receives one prize that is equally likely to be worth $\$ 5, \$ 10$ or $\$ 20$. Jamie participates in a different prize draw. She receives one prize that is equally likely to be worth $\$ 30$ or $\$ 40$. What is the probability that the total value of their prizes is exactly $\$ 50$... | Let $B$ be the value of Ben's prize and $J$ be the value of Jamie's prize.
Ben's prize can be $\$ 5, \$ 10$ or $\$ 20$, each with probability $\frac{1}{3}$.
Jamie's prize can be $\$ 30$ or $\$ 40$, each with probability $\frac{1}{2}$.
We want to find the probability that $B+J = 50$.
We consider the possible values of $... |
6.11 Let's use formula (6.10), valid under the condition $|x|<1$:
a) $2 x+1+\frac{x^{2}}{1+x}=\frac{13}{6} \Rightarrow x_{1}=\frac{1}{2}, x_{2}=-\frac{7}{9}$;
b) $\frac{1}{x}+\frac{x}{1-x}=\frac{7}{2} \Rightarrow x_{1}=\frac{1}{3}, x_{2}=\frac{2}{3}$.
Answer: a) $x_{1}=\frac{1}{2}, x_{2}=-\frac{7}{9} ;$ b) $x_{1}=\f... | )x_{1}=\frac{1}{2},x_{2}=-\frac{7}{9};b)x_{1}=\frac{1}{3},x_{2}=\frac{2}{3} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Sequences, series, summability | true | 6 | 70030b526bd06158 | 6.11 Solve the equations:
a) $2 x+1+x^{2}-x^{3}+x^{4}-x^{5}+\ldots=\frac{13}{6}$, where $|x|<1$;
b) $\frac{1}{x}+x+x^{2}+\ldots+x^{n}+\ldots=\frac{7}{2}$, where $|x|<1$. | 6.11 Solve the equations:
a) $2 x+1+x^{2}-x^{3}+x^{4}-x^{5}+\ldots=\frac{13}{6}$, where $|x|<1$;
We can rearrange the terms as:
$1 + (2x + x^{2}-x^{3}+x^{4}-x^{5}+\ldots)=\frac{13}{6}$
Consider the series $S = 2x + x^{2}-x^{3}+x^{4}-x^{5}+\ldots = 2x + (x^{2}-x^{3}+x^{4}-x^{5}+\ldots)$.
The series in the parenthesis ... |
The sum is always 300. Pairs are (43, 257), (100, 200), (98, 202), (125, 175) | 300 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | true | 1.4 | 0da8ae73f72e3733 | ## Task 4
Pair the following numbers so that they always add up to the same sum.
$\begin{array}{llllllll}43 & 202 & 100 & 175 & 98 & 257 & 125 & 200\end{array}$ | To solve this task, we need to pair the given numbers so that the sum of each pair is the same.
First, let's calculate the sum of all the given numbers:
$43 + 202 + 100 + 175 + 98 + 257 + 125 + 200$
To make the addition easier, let's rearrange and group the numbers:
$(43 + 98) + (100 + 200) + (125 + 175) + (202 + 257... |
Solution
a) 4. We need to choose 3 points out of a total of 4. In situations like this, when we choose any triangle, for example, triangle $A B C$, all permutations of points $A, B$, and $C$ still form triangle $A B C$. Therefore, when using the fundamental principle of counting, we must divide by the number of permut... | 4,0,12 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | true | 2.5 | 07c207fd33282d24 | Juquinha marks points on a circumference and draws triangles by connecting 3 of these points. The lengths of the arcs between 2 consecutive points are equal.
a) By marking 4 points on the circumference, how many triangles can he draw?
b) By marking 5 points on the circumference, how many equilateral triangles can he ... | Final Answer:
a) By marking 4 points on the circumference, the number of triangles he can draw is the number of ways to choose 3 points out of 4, which is $\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$.
b) By marking 5 points on the circumference, we want to find the number of equilateral triangl... |
Solution. Let $F$ be the midpoint of segment $AB$.
Since triangle $ABC$ is half of a square, $CF = AF$, and triangles $ACF$ and $BCF$ are isosceles right triangles. $\angle CAP = \angle PCG$,... | 90 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 4.25 | 81aa2b100e4230cd | In the isosceles right triangle $ABC$, the right angle is at $C$. Let $P$ be any point on the side $BC$, and let $G$ be the orthogonal projection of point $C$ onto $AP$. Let $H$ be the point on segment $AP$ such that $AH = CG$. At what angle is the segment $GH$ seen from the midpoint of $AB$? | Let $M$ be the midpoint of $AB$. We want to find the angle $\angle GMH$. We calculated $\vec{MG} \cdot \vec{MH} = 0$ for any point $P$ on $BC$. This means that the vectors $\vec{MG}$ and $\vec{MH}$ are orthogonal, so the angle $\angle GMH = 90^\circ$, unless one of the vectors is zero.
We found that $\vec{MG} = 0$ and ... |
3. We rewrite the equation in the form $37=n^{3}-27^{m}$ and factor the difference of cubes:
$$
37=\left(n-3^{m}\right)\left(n^{2}+n \cdot 3^{m}+9^{m}\right) .
$$
The number 37 is a prime number, and on the right side of the equation, there is a product of two integers, with the second factor being greater than 1. Th... | =1,n=4 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 3 | ae2fd1d2abb14500 | 3. Determine all pairs of positive integers $m$ and $n$ for which $37+27^{m}=n^{3}$.
The written part of the school round in category B takes place
## on Thursday, January 21, 2010
so that it starts in the morning and the participants have 4 hours of pure time to solve the problems. For each problem, participants ca... | We are given the equation $37+27^{m}=n^{3}$, where $m$ and $n$ are positive integers.
We rewrite the equation as $n^{3}-27^{m}=37$, or $n^{3}-(3^{m})^{3}=37$.
Let $x = 3^{m}$. Then the equation becomes $n^{3}-x^{3}=37$.
Factoring the difference of cubes, we have $(n-x)(n^{2}+nx+x^{2})=37$.
Since $m$ and $n$ are positiv... |
$13$ | 13 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 3 | 83658df674e1a7ab | 3. Given an integer $k$ satisfying $1000<k<2020$, and such that the system of linear equations in two variables $\left\{\begin{array}{l}7 x-5 y=11 \\ 5 x+7 y=k\end{array}\right.$ has integer solutions, then the number of possible values for $k$ is $\qquad$. | We are given the system of linear equations
$$\left\{\begin{array}{l}7 x-5 y=11 \\ 5 x+7 y=k\end{array}\right.$$
We solved for $x$ and $y$ in terms of $k$:
$$x = \frac{77 + 5k}{74}$$
$$y = \frac{7k - 55}{74}$$
For integer solutions $(x, y)$, we need $77 + 5k$ to be divisible by 74 and $7k - 55$ to be divisible by 74.
W... |
4. The answer is $\mathbf{( C )}$. Substituting $a=b=c=674$ (integers that indeed satisfy $a+b+c=$ 2022) we get $3 p(674)=p(674)$, which means $p(674)=0$. Substituting then $a=b=0$ and $c=2022$ we obtain
$$
2 p(0)+p(2022)=p(674)=0 \Rightarrow p(2022)=-2 p(0)=5392 \text {. }
$$ | 5392 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Commutative algebra | true | 4 | 1319b26beaf63d6d | 4. The polynomial $p(x)$ has the following property: for every triplet of integers $a, b, c$ such that $a+b+c=2022$ we have that $p(a)+p(b)+p(c)=p(674)$. It is also known that $p(0)=-2696$. What is the value of $p(2022)$?
(A) -2696
(B) 674
(C) 5392
(D) 8088
(E) It is not possible to determine it with the given data. | Let $p(x) = Ax+B$.
For any integers $a, b, c$ such that $a+b+c=2022$, we have $p(a)+p(b)+p(c) = (Aa+B)+(Ab+B)+(Ac+B) = A(a+b+c)+3B = 2022A+3B$.
We are given that $p(a)+p(b)+p(c) = p(674)$.
$p(674) = A(674)+B$.
Thus, $2022A+3B = 674A+B$.
$2022A - 674A + 3B - B = 0$.
$1348A + 2B = 0$.
$2B = -1348A$.
$B = -674A$.
So $p(x)... |
20. 7
20. Rewrite the given equation as $(a+b)^{2}+16(a-b)^{2}=16$. Hence we may let $a+b=4 \cos x$ and $a-b=\sin x$. Note that
$$
\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}=\sqrt{(4 a-2 b)^{2}-3(4 a-2 b)+9}=\sqrt{\left(4 a-2 b-\frac{3}{2}\right)^{2}+\frac{27}{4}} .
$$
Since $4 a-2 b=(a+b)+3(a-b)=4 \cos x+3 \sin x$, wh... | 7 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Algebraic geometry | true | 7 | 94f0c9001cc9bf03 | 20. Let $a$ and $b$ be real numbers such that $17\left(a^{2}+b^{2}\right)-30 a b-16=0$. Find the maximum value of $\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$.
(2 marks)
設 $a$ 、 $b$ 為實數 , 使得 $17\left(a^{2}+b^{2}\right)-30 a b-16=0$ 。 求 $\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$ 的最大值。 | Let $E = \sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$. We rewrite the expression inside the square root as
$16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9 = (4a-2b)^2 - 3(4a-2b) + 9$.
Let $x = 4a-2b$. Then $E = \sqrt{x^2 - 3x + 9}$.
We want to find the maximum value of $E$ subject to $17\left(a^{2}+b^{2}\right)-30 a b-16=0$.
We need... |
Solution. Let
$$
\begin{aligned}
A & =\{\text { a white ball was added to the urn }\}, \\
C & =\{\text { a white ball was drawn }\}, \\
D & =\{\text { a white ball remains }\} .
\end{aligned}
$$
Note that $A=C D$. Then
$$
\begin{aligned}
\mathrm{P}(D \mid C)=\frac{\mathrm{P}(D C)}{\mathrm{P}(C)}=\frac{\mathrm{P}(A)}... | \frac{2}{3} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Probability theory and stochastic processes | true | 2 | 16f94773a3a8c5c4 | 14. A ball was added to an urn containing one white ball - either white or black (with equal probabilities of selection). After this, one ball was randomly drawn from the urn. It turned out to be white. What is the conditional probability that the remaining ball in the urn is also white? | Let $W$ be the event that a white ball is added, and $B$ be the event that a black ball is added. We are given $P(W) = P(B) = 1/2$.
Let $D$ be the event that a white ball is drawn from the urn.
We want to find the conditional probability that the remaining ball in the urn is also white, given that a white ball was draw... |
【Analysis】For a proper fraction, when the difference between the numerator and the denominator is constant, the larger the denominator, the larger the fraction.
【Solution】Solution: $\frac{2007}{2009}<\frac{2008}{2010}, \frac{2008}{2010}<\frac{2009}{2011}, \frac{2009}{2011}<\frac{2010}{2012}$,
Therefore: $\frac{2007}{20... | C | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | true | 3 | 361d65db5df96f8d | 5. (3 points) $a=\frac{2007 \times 2008}{2009 \times 2010}, b=\frac{2008 \times 2009}{2010 \times 2011}, c=\frac{2009 \times 2010}{2011 \times 2012}$, then we have ( )
A. $a>b>c$
B. $a>c>b$
C. $a<b<c$
D. $a<c<b$ | We are given $a=\frac{2007 \times 2008}{2009 \times 2010}, b=\frac{2008 \times 2009}{2010 \times 2011}, c=\frac{2009 \times 2010}{2011 \times 2012}$.
Let $n=2007$. Then $a=\frac{n(n+1)}{(n+2)(n+3)}, b=\frac{(n+1)(n+2)}{(n+3)(n+4)}, c=\frac{(n+2)(n+3)}{(n+4)(n+5)}$.
To compare $a$ and $b$, we consider the ratio $\frac{a... |
Problem 10. The correct answer is (D).
The possible triples $(a, b, c)$ are: $\quad(1,2,4), \quad(1,2,6), \quad(1,2,12), \quad(1,3,6), \quad(1,3,12), \quad(1,4,12), \quad(1,6,12)$, $(2,4,12), \quad(2,6,12), \quad(3,6,12)$. In total, there are 10. | 10 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | true | 3.25 | fc82239ac0387506 | 10. How many triples of distinct natural numbers $(a, b, c)$ are there such that the number $a$ is a divisor of $b$, the number $b$ is a divisor of $c$, and the number $c$ is a divisor of 12?
(A) 6
$\begin{array}{ll}\text { (B) } 7 & \text { (C) } 8\end{array}$
(D) 10
(E) 9 | We want to find the number of triples of distinct natural numbers $(a, b, c)$ such that $a|b$, $b|c$, and $c|12$.
The divisors of 12 are $\{1, 2, 3, 4, 6, 12\}$. Thus, $c \in \{1, 2, 3, 4, 6, 12\}$.
Since $b|c$ and $b \neq c$, $b$ must be a proper divisor of $c$.
Since $a|b$ and $a \neq b$, $a$ must be a proper divisor... |
7. $\sqrt{2007}$ Hint: $f(x)=x^{4}\left(x^{2}-2 \sqrt{2006} x-1\right)+x\left(x^{2}-2 \sqrt{2007} x+1\right)+x-\sqrt{2006}-\sqrt{2007}$ $+\sqrt{2007}, f(\sqrt{2006}+\sqrt{2007})=\sqrt{2007}$. | \sqrt{2007} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Field theory and polynomials | true | 5.4 | a9d469aee7388520 | $$
\begin{array}{l}
\text { 7. If } f(x)=x^{6}-2 \sqrt{2006} x^{5}-x^{4}+x^{3}-2 \sqrt{2007} x^{2}+2 x-\sqrt{2006} \text {, then } f(\sqrt{2006}+ \\
\sqrt{2007})= \\
\end{array}
$$ | Let $a = \sqrt{2006}$ and $b = \sqrt{2007}$. Then $f(x) = x^6 - 2ax^5 - x^4 + x^3 - 2bx^2 + 2x - a$. We want to find $f(a+b)$.
Let $x = a+b$. Then $x-a = b$ and $x-b = a$.
From $x-a = b$, we have $(x-a)^2 = b^2 = 2007$. So $x^2 - 2ax + a^2 = 2007$. $x^2 - 2ax + 2006 = 2007$. $x^2 - 2ax = 1$.
From $x-b = a$, we have $(x... |
(16) The correct answer is $(A)$.
Triangle $A^{\prime} B^{\prime} C^{\prime}$ is equilateral for evident reasons of symmetry (or, if you prefer, by observing the congruence of triangles $A^{\prime} B^{\prime} A, B^{\prime} C^{\prime} B, C^{\prime} A^{\prime} C$).
Let $D$ be the orthogonal projection of $A^{\prime}$ o... | 3\sqrt{4+\sqrt{3}} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Difference and functional equations | true | 3.4 | d7633ed40d738e57 | 16. On the sides of the equilateral triangle $A B C$, which have length 1, three squares are constructed, as shown in the figure. What is the perimeter of the triangle $A^{\prime} B^{\prime} C^{\prime}$?
(A) $3 \sqrt{4+\sqrt{3}}$
(D) $\frac{3}{2} \sqrt{3}$
(E) $\frac{3}{2}+\frac{3}{4} \sqrt{3}$
(C) $3 \sqrt{1+2 \sqrt{3... | Final Answer: The final answer is $\boxed{(A)}$ |
$$
V_{K L M D}=\frac{D K}{D A} \cdot \frac{D L}{D B} \cdot \frac{D M}{D C} \cdot V_{A B C D}=\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{2}{5} \cdot 1=\frac{4}{45}
$$
Therefore,
$$
V_{A B C K L M}=V_{A B C D}-V_{K L M D}=1-\frac{4}{45}=\frac{41}{45}
$$
## Answer
$\frac{41}{45}$ | \frac{41}{45} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 3 | e5df6d985c2919a1 | The volume of the pyramid $ABCD$ is 1. Points $K, L$, and $M$ are taken on the edges $AD, BD, CD$ respectively, such that $2AK = KD, BL = 2LD$, and $2CM = 3MD$. Find the volume of the polyhedron $ABCKLM$.
# | Let $V_{ABCD}$ be the volume of the pyramid $ABCD$. We are given $V_{ABCD} = 1$.
We are given points $K, L, M$ on edges $AD, BD, CD$ such that $AK = \frac{1}{3} AD, KD = \frac{2}{3} AD$, $BL = \frac{2}{3} BD, LD = \frac{1}{3} BD$, $CM = \frac{3}{5} CD, MD = \frac{2}{5} CD$.
The volume of the pyramid $KLMD$ is given by ... |
Answer. In eight ways.
Solution. From part a), it follows that the coloring of all points with integer coordinates is uniquely determined by the coloring of the points corresponding to the numbers $0,1,2,3,4,5$, and 6. The point $0=14-2 \cdot 7$ must be colored the same as 14, i.e., red. Similarly, the point $1=71-10 ... | 8 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | true | 6.67 | 4f38e3866a17a7f2 | 5. On the number line, points with integer coordinates are painted red and blue according to the following rules: a) points whose coordinate difference is 7 must be painted the same color; b) points with coordinates 20 and 14 should be painted red, and points with coordinates 71 and 143 - blue. In how many ways can all... | The color of a point with integer coordinate $x$ depends only on $x \pmod 7$. Let $C(x)$ denote the color of the point with coordinate $x$. We are given that $C(x) = C(y)$ if $x \equiv y \pmod 7$. We need to determine the colors $C(0), C(1), C(2), C(3), C(4), C(5), C(6)$.
From rule b), we have:
$C(20) = R$. Since $20 \... |
# Answer: 8.5 times.
## Solution:
The car was on the road for 16 minutes longer than usual, due to spending 8 minutes catching up to Dima and 8 minutes returning home. The car caught up with Dima at 8:28, and during the 68 minutes from 7:20 to 8:28, Dima ran the same distance that the car traveled in 8 minutes, i.e.,... | 8.5 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | true | 8.5 | 9302a67fc7240fd4 | 2. Usually, Dima leaves home at $8:10$ AM, gets into Uncle Vanya's car, who delivers him to school by a certain time. But on Thursday, Dima left home at 7:20 and ran in the opposite direction. Uncle Vanya waited for him and at $8:20$ drove after him, caught up with Dima, turned around, and delivered him to school 26 mi... | Let $v_D$ be Dima's running speed and $v_V$ be Uncle Vanya's car speed.
Usually, Dima leaves home at $8:10$ AM. Let $T_{usual}$ be the usual travel time from home to school by car. Then the usual arrival time at school is $8:10 \text{ AM} + T_{usual}$.
On Thursday, Dima left home at $7:20$ AM and ran in the opposite d... |
Solution. We apply formula (1); based on the given parametric equations of the cardioid and the formula for the differential of the arc of a plane curve (Chapter 5, §), we transform the line integral in formula (1) into an ordinary integral with the variable $t$.
$$
\begin{gathered}
\dot{x}=-2 a \sin t+2 a \sin 2 t, \... | 16a | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Differential geometry | true | 3.75 | 2afb6d56bee6fe8c | 884. Find the length of the cardioid $x=2 a \cos t-a \cos 2 t, y=$ $-2 a \sin t-a \sin 2 t$. | To find the length of the cardioid $x=2 a \cos t-a \cos 2 t, y=-2 a \sin t-a \sin 2 t$, we use the formula for the length of a parametric curve:
$L = \int_{0}^{2\pi} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$
First, we calculate the derivatives of $x$ and $y$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(2 a... |
Answer: 29 The main idea is the following lemma:
Lemma. For any non-negative integer $n$ and prime $p, \sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv 1(\bmod p)$.
Proof. Note that $a^{b}$ depends only on the value of $a(\bmod p)$ and the value of $b(\bmod p-1)$. Since $p$ and $p-1$ are relatively prime, the Chinese Remainder Th... | 29 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 6 | d2136761d0326573 | 29. [18] Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by 101 . | We want to compute the remainder when $\sum_{k=1}^{30303} k^{k}$ is divided by 101.
We have shown that $\sum_{k=1}^{30303} k^{k} \equiv \sum_{k=1}^{30300} k^{k} + \sum_{k=30301}^{30303} k^{k} \pmod{101}$.
We have $\sum_{k=1}^{30300} k^{k} \equiv 98 \pmod{101}$ and $\sum_{k=30301}^{30303} k^{k} \equiv 32 \pmod{101}$.
Th... |
In equilibrium, the algebraic sum of static moments is $0\left(\sum P p=0\right)$. Let $x$ be the distance from the support point to the end of the rod where the $20 \mathrm{~kg}$ weight hangs, then
$$
\begin{gathered}
20 x+(x-1) 30+(x-2) 40+(x-3) 50+(x-4) 60=0 \\
20 x+30 x-30+40 x-80+50 x-150+60 x-240=0 \\
200 x=500 ... | 2\frac{1}{2} | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mechanics of particles and systems | true | 1.8 | 5c909ddd972b74cc | A four-meter-long rod has a $20 \mathrm{~kg}$ weight attached to one end, and $30,40,50 \mathrm{~kg}$ weights are attached at one, two, and three meters from this end, respectively. At the other end of the rod, a $60 \mathrm{~kg}$ weight is attached. Where should the rod be supported so that it remains in balance? The ... | Let $x=0$ be the position of the end of the rod where the $20 \mathrm{~kg}$ weight is attached.
The weights are attached at the following positions:
$w_1 = 20 \mathrm{~kg}$ at $x_1 = 0 \mathrm{~m}$
$w_2 = 30 \mathrm{~kg}$ at $x_2 = 1 \mathrm{~m}$
$w_3 = 40 \mathrm{~kg}$ at $x_3 = 2 \mathrm{~m}$
$w_4 = 50 \mathrm{~kg}$ ... |
17. Ans: 16
Let $A B=y \mathrm{~cm}$. Since $\triangle D C F \sim \triangle G B F$, we have
$$
\frac{D C}{D F}=\frac{G B}{G F} \Longrightarrow \frac{y}{24}=\frac{G B}{x} \Longrightarrow G B=\frac{x y}{24} .
$$
Since $\triangle D C E \sim \triangle G A E$, we have
$$
\frac{D C}{D E}=\frac{G A}{G E} \Longrightarrow \fra... | 16 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | true | 3 | e088e8c13a64aa7e | 17. In the diagram below, $A B C D$ is a square. The points $A, B$ and $G$ are collinear. The line segments $A C$ and $D G$ intersect at $E$, and the line segments $D G$ and $B C$ intersect at $F$. Suppose that $D E=15 \mathrm{~cm}, E F=9 \mathrm{~cm}$, and $F G=x \mathrm{~cm}$. Find the value of $x$. | Let $A B C D$ be a square. Points $A, B, G$ are collinear, so $G$ is on the line $A B$. Line segments $A C$ and $D G$ intersect at $E$. Line segments $D G$ and $B C$ intersect at $F$. We are given $D E=15$, $E F=9$, $F G=x$. We want to find $x$.
Since $B C \parallel A D$, we have similar triangles $\triangle E C F \sim... |
The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant.
It now suffices to illustrate an example for which $N = 1$: ... | 1 | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Field theory and polynomials | true | 1.67 | fee6bcf53ae0f073 | Suppose that $P(z), Q(z)$, and $R(z)$ are polynomials with real coefficients, having degrees $2$, $3$, and $6$, respectively, and constant terms $1$, $2$, and $3$, respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$. What is the minimum possible value of ... | We are given that $P(z), Q(z), R(z)$ are polynomials with real coefficients, having degrees $2$, $3$, and $6$, respectively, and constant terms $1$, $2$, and $3$, respectively. We want to find the minimum number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$.
Let $F(z) = P(z) \cdot Q(z... |
(a) $|x-5|=2$ means that the distance of $x$ from 5 is 2. Therefore, the roots are 3 and 7.
(b) $|x+3|=1$ means that the distance of $x$ from -3 is 1. Therefore, the roots are -4 and -2.
... | 3,7,-4,-2,-\frac{2}{3},\frac{16}{3},1.5 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | true | 1 | 6c33eb4308ff6d5f | Solve the given equations geometrically, that is, interpreting the absolute value $|a-b|$ as the distance between $a$ and $b$.
(a) $|x-5|=2$
(c) $|3 x-7|=9$
(b) $|x+3|=1$
(d) $|x+2|=|x-5|$ | We are asked to solve the given equations geometrically by interpreting $|a-b|$ as the distance between $a$ and $b$.
(a) $|x-5|=2$
This equation states that the distance between $x$ and $5$ is $2$. On the number line, we locate the point $5$. We need to find the points that are at a distance of $2$ from $5$. These poi... |
Here there are many possibilities.
- $A_{2}=\left\lfloor\frac{100}{2}\right\rfloor=50$ numbers less than 100 divisible by 2.
- $A_{3}=\left\lfloor\frac{100}{3}\right\rfloor=33$ numbers less than 100 divisible by 3
- $A_{5}=\left\lfloor\frac{100}{5}\right\rfloor=20$ numbers less than 100 divisible by 5
- $A_{2,3}=\... | 74 | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | true | 1.33 | 5c81816f0547fb9f | How many numbers less than 100 are divisible by 2, 3, or 5? | Let $A$ be the set of numbers less than 100 divisible by 2, $B$ be the set of numbers less than 100 divisible by 3, and $C$ be the set of numbers less than 100 divisible by 5. We want to find $|A \cup B \cup C|$.
We use the Principle of Inclusion-Exclusion:
$|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C... |
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