Let X be the set consisting of twenty positive integers n, n + 2,...,n + 38. The smallest value of n
for which any three numbers a, b, c ∈ X, not necessarily distinct, form the sides of an acute-angled
triangle is

To solve this problem, we need to find the smallest value of \( n \) such that any three numbers \( a \), \( b \), \( c \) in the set \( X \) can form the sides of an acute-angled triangle. The set \( X \) consists of twenty positive integers: \( n, n+2, n+4, \ldots, n+38 \).

### Key points to consider:
1. For three sides \( a \), \( b \), and \( c \) to form a triangle, the following must hold:
   - \( a + b > c \)
   - \( b + c > a \)
   - \( c + a > b \)

2. For the triangle to be acute-angled, the square of the longest side should be less than the sum of the squares of the other two sides:
   - \( c^2 < a^2 + b^2 \) (assuming \( c \) is the longest side).

### Finding the smallest \( n \):
The set \( X = \{ n, n+2, n+4, \dots, n+38 \} \) contains 20 consecutive even numbers.

Let's consider the largest three numbers in the set, \( a = n+34 \), \( b = n+36 \), and \( c = n+38 \). For \( n \) to be the smallest possible, we must ensure the triangle formed by these three numbers is acute-angled.

We must check the condition for an acute triangle:
\[
(n+38)^2 < (n+34)^2 + (n+36)^2
\]
Expanding and simplifying:
\[
n^2 + 76n + 1444 < n^2 + 68n + 1156 + n^2 + 72n + 1296
\]
Simplifying further:
\[
n^2 + 76n + 1444 < 2n^2 + 140n + 2452
\]
Subtract \( n^2 + 76n + 1444 \) from both sides:
\[
0 < n^2 + 64n + 1008
\]
This inequality is true for all positive \( n \), so we need to check if this holds when \( n \) is small enough to still satisfy the acute triangle condition.

### Testing specific values of \( n \):

To ensure that \( X \) contains all positive numbers:
- \( n + 38 \) must be positive, so \( n \geq 1 \).

Next, let's consider the smallest \( n \) such that the smallest elements in \( X \) are also large enough to form an acute-angled triangle.

**Try \( n = 14 \):**

Set \( X = \{ 14, 16, \dots, 52 \} \). The largest three numbers are 48, 50, and 52.

Check:
\[
52^2 < 48^2 + 50^2 
\]
\[
2704 < 2304 + 2500 = 4804
\]
This inequality is true, so the triangle formed is acute-angled.

**Thus, the smallest \( n \) that satisfies the conditions is \( \boxed{14} \).**

Would you like more details, or do you have any questions?

Here are five related questions:
1. What conditions must be met for three sides to form an obtuse-angled triangle?
2. How would the problem change if the numbers were odd instead of even?
3. What is the general method to determine if a set of numbers can form an acute-angled triangle?
4. How would the solution change if the set consisted of only 10 numbers?
5. Can you explain how the triangle inequality theorem applies in different scenarios?

**Tip:** When solving geometric problems involving triangles, always check the type of triangle (acute, right, or obtuse) based on the given conditions.

Let X be the set consisting of twenty positive integers n, n + 2,...,n + 38. The smallest value of n for which any three numbers a, b, c ∈ X, not necessarily distinct, form the sides of an acute-angled triangle is

Solve the problem of finding the smallest value of n for which any three numbers in the set of 20 positive integers can form the sides of an acute-angled triangle. This problem involves the Triangle Inequality Theorem and acute triangle conditions, suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation