Question

If three successive terms of a G.P. with common ratio $r \, (r>1)$ are the lengths of the sides of a triangle and $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r$, then $3\lfloor r \rfloor + \lfloor -r \rfloor$ is equal to:

Answer 1

Correct Answer: 1

Let's denote the three successive terms of the geometric progression (G.P.) as \( a, ar, ar^2 \). According to the triangle inequality, the sum of the lengths of any two sides must be greater than the third side. Thus, we have:

• \( a + ar > ar^2 \) which simplifies to \( 1 + r > r^2 \) (Dividing through by \( a \)).
• \( a + ar^2 > ar \) which simplifies to \( 1 + r^2 > r \) (Dividing through by \( a \)).
• \( ar + ar^2 > a \) which simplifies to \( r + r^2 > 1 \) (Dividing through by \( a \)).

Analyzing inequality \( 1 + r > r^2 \), we rearrange to \( r^2 - r - 1 < 0 \). The roots of \( r^2 - r - 1 = 0 \) are \( r = \frac{1 \pm \sqrt{5}}{2} \). Since \( r > 1 \), we have \( 1 < r < \frac{1+\sqrt{5}}{2} \), approximately \( 1 < r < 1.618 \).
The inequality \( 1 + r^2 > r \) is trivially satisfied for \( r > 1 \), as \( r^2 \) increases faster than \( r \). Similarly, \( r + r^2 > 1 \) is trivially satisfied for \( r > 1 \).
Therefore, the only non-trivial inequality is \( r^2 - r - 1 < 0 \), confirming:

• \( 1 < r < 1.618 \).

Now, let's determine \( 3\lfloor r \rfloor + \lfloor -r \rfloor \). Since \( 1 < r < 1.618 \), it follows \( \lfloor r \rfloor = 1 \) and \( -r \) is roughly between \(-1.618\) and \(-1\), so \( \lfloor -r \rfloor = -2 \).
Thus, we need:

• \( 3\lfloor r \rfloor + \lfloor -r \rfloor = 3(1) + (-2) = 3 - 2 = 1 \).

Therefore, \( 3\lfloor r \rfloor + \lfloor -r \rfloor = 1 \). This value matches the specified range of [1,1].

Answer 2

Let the three successive terms of the G.P. be \( a, ar, ar^2 \), where \( r \(>\) 1 \). Since these terms form the sides of a triangle, they must satisfy the triangle inequality:
\[ a + ar \(>\) ar^2, \quad ar + ar^2 \(>\) a, \quad a + ar^2 \(>\) ar \]

Checking the Triangle Inequality Conditions
1. \( a + ar \(>\) ar^2 \):
\[ a(1 + r) \(>\) ar^2 \implies 1 + r \(>\) r^2 \implies r^2 - r - 1 \(<\) 0 \]
Solving the quadratic inequality:
\[ r = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
Since \( r \(>\) 1 \), we have:
\[ 1 \(<\) r \(<\) \frac{1 + \sqrt{5}}{2} \approx 1.618 \]

2. \( ar + ar^2 \(>\) a \):
\[ ar(1 + r) \(>\) a \implies r(1 + r) \(>\) 1 \]
This condition is always satisfied for \( r \(>\) 1 \).

3. \( a + ar^2 \(>\) ar \):
\[ a(1 + r^2) \(>\) ar \implies 1 + r^2 \(>\) r \implies r^2 - r + 1 \(>\) 0 \]
This condition is always true for all values of \( r \).

Finding the Value of \( \lfloor r \rfloor \)
Since \( 1 \(<\) r \(<\) \frac{1 + \sqrt{5}}{2} \approx 1.618 \), the greatest integer less than or equal to \( r \) is:
\[ \lfloor r \rfloor = 1 \]

Calculating \( 3\lfloor r \rfloor + \lfloor -r \rfloor \)
\[ 3\lfloor r \rfloor + \lfloor -r \rfloor = 3 \times 1 + \lfloor -r \rfloor \]
Since \( \lfloor -r \rfloor \) is the greatest integer less than or equal to \( -r \), and \( -1.618 \(<\) -r \(<\) -1 \), we have:
\[ \lfloor -r \rfloor = -2 \]
Thus:
\[ 3\lfloor r \rfloor + \lfloor -r \rfloor = 3 \times 1 + (-2) = 1 \]

Conclusion: \( 3\lfloor r \rfloor + \lfloor -r \rfloor = 1 \).