problem stringlengths 16 4.82k | reasoning stringlengths 708 112k | deepseek_solution stringlengths 0 5.46k | ground_truth_solution stringlengths 97 8.33k | correct bool 1
class | judge_reasoning stringlengths 80 22.8k |
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Let \( a, b, c \) be positive real numbers. Prove that
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{3}{1+abc},
$$
and that equality occurs if and only if \( a = b = c = 1 \). | Okay, so I need to prove this inequality: \(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{3}{1+abc}\), where \(a, b, c\) are positive real numbers, and equality holds if and only if \(a = b = c = 1\). Hmm, let me try to think through this step by step.
First, I remember that inequalities often involve ... | To prove the inequality
\[
\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{1+abc}
\]
for positive real numbers \(a, b, c\) with equality if and only if \(a = b = c = 1\), we proceed as follows:
### Key Steps and Proof:
1. **Substitution for \(abc = 1\):**
- Assume \(abc = 1\). Let \(a = \fr... | 1. Consider the given inequality:
\[
\frac{1}{a(1+b)}+ \frac{1}{b(1+c)}+ \frac{1}{c(1+a)} \geq \frac{3}{1 + abc}
\]
To simplify, we add \( \frac{3}{1 + abc} \) to both sides. The new inequality becomes:
\[
\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} + \frac{3}{1 + abc} \geq \frac{6}{1 + abc}
\]
2. Let's ... | true | The provided solution correctly demonstrates the inequality using appropriate mathematical techniques such as substitution, analysis of cases, and the AM-GM inequality. Furthermore, it establishes the equality condition accurately, matching the claims of the ground truth solution. |
Given real numbers \( a, b, c \) and a positive number \( \lambda \) such that the polynomial \( f(x) = x^3 + a x^2 + b x + c \) has three real roots \( x_1, x_2, x_3 \), and the conditions \( x_2 - x_1 = \lambda \) and \( x_3 > \frac{1}{2}(x_1 + x_2) \) are satisfied, find the maximum value of \( \frac{2 a^3 + 27 c - ... | Okay, let's try to tackle this problem step by step. The problem is about finding the maximum value of the expression (2a³ + 27c - 9ab)/λ³ given some conditions on the roots of the cubic polynomial f(x) = x³ + ax² + bx + c. The roots are real, with x₂ - x₁ = λ and x₃ > (x₁ + x₂)/2.
First, I need to recall that for a ... | Given real numbers \(a, b, c\) and a positive number \(\lambda\) such that the polynomial \(f(x) = x^3 + ax^2 + bx + c\) has three real roots \(x_1, x_2, x_3\) with the conditions \(x_2 - x_1 = \lambda\) and \(x_3 > \frac{1}{2}(x_1 + x_2)\), we need to find the maximum value of \(\frac{2a^3 + 27c - 9ab}{\lambda^3}\).
... |
We begin by analyzing the function \( f(x) = x^3 + a x^2 + b x + c \), which has three real roots \( x_1, x_2, x_3 \). We are given the following conditions:
1. \( x_2 - x_1 = \lambda \)
2. \( x_3 > \frac{1}{2} (x_1 + x_2) \)
We aim to find the maximum value of \( \frac{2a^3 + 27c - 9ab}{\lambda^3} \).
1. **Transfor... | true | Both the provided solution and the ground truth solution arrive at the same final result of \( \frac{3\sqrt{3}}{2} \). The mathematical steps taken in both solutions, including the use of Vieta's formulas, manipulation of the roots, differentiation to find maximum values, and the checks for conditions, are consistent a... |
In triangle $ABC$, $CA = CB$, and $D$ is the midpoint of $AB$. Line $EF$ passes through point $D$ such that triangles $ABC$ and $EFC$ share the same incenter. Prove that $DE \cdot DF = DA^2$. | Okay, so I have this geometry problem here. Let me try to understand it step by step. The problem says: In triangle ABC, CA equals CB, so it's an isosceles triangle with CA = CB. Point D is the midpoint of AB. There's a line EF passing through D such that triangles ABC and EFC share the same incenter. I need to prove t... | In triangle \(ABC\), \(CA = CB\), making it isosceles with \(D\) as the midpoint of \(AB\). Line \(EF\) passes through \(D\) such that triangles \(ABC\) and \(EFC\) share the same incenter. We need to prove that \(DE \cdot DF = DA^2\).
1. **Coordinate Setup**:
- Place \(D\) at the origin \((0, 0)\), \(A\) at \((-a,... | 1. **Identify Key Elements**: Consider \( \triangle ABC \) where \( CA = CB \) and \( D \) is the midpoint of \( AB \). Line \( \mathrm{EF} \) passes through \( D \) such that \( \triangle ABC \) and \( \triangle \mathrm{EFC} \) share the same incircle (inscribed circle).
2. **Given Common Incircle**: Let the shared i... | true | The provided solution correctly follows the logical steps necessary to prove that if triangles ABC and EFC share the same incenter and D is the midpoint of AB, then DE·DF = DA². Both solutions have outlined the correct reasoning using the properties of isosceles triangles, incenters, and the Power of a Point, leading t... |
Let \( p = 2^{3009}, q = 3^{2006}, \) and \( r = 5^{1003} \). Which of the following statements is true?
(A) \( p < q < r \)
(B) \( p < r < q \)
(C) \( q < p < r \)
(D) \( r < p < q \)
(E) \( q < r < p \) | Okay, so I need to figure out which of the given options is correct by comparing the sizes of p, q, and r where p = 2^3009, q = 3^2006, and r = 5^1003. Hmm, comparing exponents with different bases and exponents... This might be tricky. Let me think about how to approach this.
First, I remember that comparing numbers ... | To determine which of the given statements is true, we need to compare the sizes of \( p = 2^{3009} \), \( q = 3^{2006} \), and \( r = 5^{1003} \).
First, we express each number with the same exponent by noting that:
- \( p = 2^{3009} = (2^3)^{1003} = 8^{1003} \)
- \( q = 3^{2006} = (3^2)^{1003} = 9^{1003} \)
- \( r =... | Given the values:
\[ p = 2^{3009}, \quad q = 3^{2006}, \quad r = 5^{1003} \]
1. Express \( p \) and \( q \) in terms of powers of the same base:
\[ p = 2^{3009} = 2^{3 \times 1003} = (2^3)^{1003} = 8^{1003} \]
\[ q = 3^{2006} = 3^{2 \times 1003} = (3^2)^{1003} = 9^{1003} \]
Note: \( r = 5^{1003} \) is already... | true | The solution provided correctly follows the steps to compare the values of the three numbers using both exponential form and logarithmic calculations, confirming that the order is indeed \( r < p < q \). Both the solution and the ground truth reached the same conclusion, producing the correct final boxed answer. |
When \( a < -1 \), the nature of the roots for the equation
$$
\left(a^{3}+1\right) x^{2}+\left(a^{2}+1\right) x-(a+1)=0
$$
is:
(A) Two negative roots.
(B) One positive root and one negative root, with the absolute value of the negative root being larger.
(C) One positive root and one negative root, with the absolute v... | Okay, so I need to figure out the nature of the roots for the quadratic equation \((a^3 + 1)x^2 + (a^2 + 1)x - (a + 1) = 0\) when \(a < -1\). The options are given as (A) two negative roots, (B) one positive and one negative root with the negative root having a larger absolute value, (C) similar to B but the negative r... | To determine the nature of the roots for the quadratic equation \((a^3 + 1)x^2 + (a^2 + 1)x - (a + 1) = 0\) when \(a < -1\), we analyze the discriminant, sum, and product of the roots.
1. **Discriminant Analysis**:
- The discriminant \(D\) is given by:
\[
D = (a^2 + 1)^2 - 4(a^3 + 1)(-(a + 1))
\]
... |
Given the quadratic equation:
\[
(a^3 + 1) x^2 + (a^2 + 1) x - (a + 1) = 0
\]
where \( a < -1 \).
1. **Identify the coefficients and analyze their signs:**
- \( a^3 + 1 \)
- \( a^2 + 1 \)
- \( -(a+1) \)
2. **Signs of the coefficients under the condition \( a < -1 \):**
- Since \( a < -1 \), we ha... | true | Both solutions correctly analyze the discriminant, the sum and product of the roots, and arrive at the conclusion that there is one positive and one negative root, confirming that the positive root has a larger absolute value. Therefore, the provided solution and the ground truth solution are aligned in the final concl... |
Suppose that $A, B, C, D$ are four points in the plane, and let $Q, R, S, T, U, V$ be the respective midpoints of $AB, AC, AD, BC, BD, CD$. If $QR = 2001$, $SU = 2002$, and $TV = 2003$, find the distance between the midpoints of $QU$ and $RV$. | Okay, let's see. So, we have four points A, B, C, D in a plane. Then there are midpoints of the segments AB, AC, AD, BC, BD, CD, named Q, R, S, T, U, V respectively. The lengths QR = 2001, SU = 2002, and TV = 2003. We need to find the distance between the midpoints of QU and RV. Hmm, okay.
First, I need to visualize ... | Given four points \(A, B, C, D\) in the plane, we consider the midpoints \(Q, R, S, T, U, V\) of the segments \(AB, AC, AD, BC, BD, CD\) respectively. We are given the distances \(QR = 2001\), \(SU = 2002\), and \(TV = 2003\), and we need to find the distance between the midpoints of \(QU\) and \(RV\).
First, we expre... | To find the distance between the midpoints of $Q U$ and $R V$, let's break down the given information and analyze each step.
1. **Identify the midpoints**:
- $Q$, $R$, $S$, $T$, $U$, $V$ are midpoints of segments $AB$, $AC$, $AD$, $BC$, $BD$, $CD$ respectively.
2. **Recall Midpoint theorem**:
- The Midpoint T... | true | The solution accurately applies the properties of midpoints and parallelograms to determine the distance between the midpoints of segments in a geometric configuration, leading to the correct conclusion that the distance is 2001. |
Given that \(1 \leq x, y, z \leq 6\), how many cases are there in which the product of natural numbers \(x, y, z\) is divisible by 10? | Okay, let's see. The problem is asking how many ordered triples (x, y, z) there are, where each of x, y, z is a natural number between 1 and 6 inclusive, such that their product x*y*z is divisible by 10. Hmm, so I need to count all possible triples where the product is a multiple of 10.
First, let me recall that a nu... | To determine the number of ordered triples \((x, y, z)\) where \(1 \leq x, y, z \leq 6\) and their product is divisible by 10, we use the principle of inclusion-exclusion. A product is divisible by 10 if it contains at least one factor of 2 and one factor of 5.
1. **Total number of triples**: \(6 \times 6 \times 6 = ... | Given the constraints \(1 \leq x, y, z \leq 6\), we are to find the number of natural number combinations \((x, y, z)\) such that their product can be divided exactly by 10.
To begin, we observe:
1. The total number of combinations of \(x, y, z\) is \(6^3\):
\[
6^3 = 216
\]
2. To be divisible by 10, the pro... | true | The provided solution correctly uses the principle of inclusion-exclusion to count the number of ordered triples
(x, y, z) such that 1 ≤ x, y, z ≤ 6 and their product is divisible by 10. It accurately counts the total number of combinations and the valid counts while also considering overlaps of cases where factors of... |
Find all prime numbers \( p \) such that for any prime number \( q < p \), if \( p = kq + r \) with \( 0 \leq r < q \), then there does not exist an integer \( a > 1 \) such that \( a^2 \) divides \( r \). | Okay, let's try to tackle this problem. So, we need to find all prime numbers \( p \) such that for any prime number \( q < p \), when we divide \( p \) by \( q \) and get a remainder \( r \) (so \( p = kq + r \) with \( 0 \leq r < q \)), there's no integer \( a > 1 \) where \( a^2 \) divides \( r \). Hmm, that's a bit... | To find all prime numbers \( p \) such that for any prime number \( q < p \), if \( p = kq + r \) with \( 0 \leq r < q \), then there does not exist an integer \( a > 1 \) such that \( a^2 \) divides \( r \), we need to ensure that the remainder \( r \) when \( p \) is divided by any smaller prime \( q \) is square-fre... |
**Step 1:** Identify the prime numbers \( p \) that meet the given conditions:
For every prime \( q < p \), \( p = k q + r \) where \( 0 \leq r < q \). It must hold that no integer \( a > 1 \) exists such that \( a^2 \mid r \).
**Step 2:** Check small prime numbers:
- For \( p = 2 \):
- \( p = 2 \) and there are... | true | The solution provided correctly identifies all prime numbers that satisfy the condition laid out in the problem. Both the solution and ground truth lead to the same result: the prime numbers 2, 3, 5, 7, and 13 meet the criteria, and the evaluation process is meticulously followed to affirm neutrality in cases evaluated... |
"Find all functions \\( f: \\mathbb{R} \\rightarrow \\mathbb{R} \\) that satisfy for all \\( x, y \\(...TRUNCATED) | "Okay, so I need to find all functions \\( f: \\mathbb{R} \\rightarrow \\mathbb{R} \\) that satisfy (...TRUNCATED) | "To find all functions \\( f: \\mathbb{R} \\rightarrow \\mathbb{R} \\) that satisfy the functional e(...TRUNCATED) | "\nNous voulons trouver toutes les fonctions \\( f: \\mathbb{R} \\rightarrow \\mathbb{R} \\) qui sat(...TRUNCATED) | true | "Both the given solution and the ground truth solution arrive at the same conclusion: the only funct(...TRUNCATED) |
"Points $M$ and $N$ are taken on the sides $AB$ and $BC$ respectively of triangle $ABC$. It turns ou(...TRUNCATED) | "Okay, so I need to prove that triangle ABC is isosceles given some conditions about the perimeters (...TRUNCATED) | "Given triangle \\(ABC\\) with points \\(M\\) on \\(AB\\) and \\(N\\) on \\(BC\\), we know the perim(...TRUNCATED) | "\n1. Denote the perimeter with the letter $P$. According to the conditions of the problem, we know (...TRUNCATED) | true | "The provided solution correctly derives that triangles AMC and CAN are congruent based on equal sid(...TRUNCATED) |
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