Directions: Consider the following: In a triangle ABC, a, b and c are the lengths of the sides and p, q and r are the lengths of its medians.
Which one of the following is correct?
2(p + q + r) < 3(a + b + c)
Understanding Triangle Side and Median Lengths
This problem asks us to identify the correct mathematical relationship between the lengths of the sides (\(a, b, c\)) and the lengths of the medians (\(p, q, r\)) of a triangle, referred to as ABC. We need to analyze the comparison between the sum of medians and the perimeter of the triangle.
Inequalities Governing Triangle Medians
For any triangle ABC, let \(a, b, c\) represent the lengths of the sides opposite to vertices A, B, and C, respectively. Let \(p, q, r\) be the lengths of the medians drawn from vertices A, B, and C to the midpoints of the opposite sides, respectively.
A well-known inequality in geometry relates the sum of the lengths of the medians to the perimeter of the triangle. This inequality states that the sum of the medians is always less than the perimeter:
\(p + q + r < a + b + c\)
This inequality is a crucial starting point for evaluating the given options.
Derivation of the Relationship \(2(p + q + r) < 3(a + b + c)\)
We can use the fundamental inequality \(p + q + r < a + b + c\) to find the correct relationship among the choices:
- Start with the base inequality:
\(p + q + r < a + b + c\)
- Multiply both sides of this inequality by 2. Since 2 is a positive number, the direction of the inequality remains unchanged:
\(2 \times (p + q + r) < 2 \times (a + b + c)\)
\(2(p + q + r) < 2(a + b + c)\)
- Now, consider the expression \(3(a + b + c)\)\(. Because \)a, b, c\( are the lengths of the sides of a triangle, they must be positive values (\)a > 0, b > 0, c > 0\(). Consequently, the perimeter \)a + b + c\( is also positive. This means that \)2(a + b + c)\( is always less than \)3(a + b + c)$:
$\(2(a + b + c) < 3(a + b + c)\)
- By combining the results from step 2 and step 3, we can establish a chain of inequalities:
\(2(p + q + r) < 2(a + b + c) \quad \text{and} \quad 2(a + b + c) < 3(a + b + c)\)
This implies:
\(2(p + q + r) < 2(a + b + c) < 3(a + b + c)\)
- From this chain, we can directly conclude that:
\(2(p + q + r) < 3(a + b + c)\)
Evaluating the Given Options
Let's compare our derived inequality with each of the options provided:
- Option 1: \(2(p + q + r) = (a + b + c)\)\( - This is incorrect. Our analysis shows \)\(2(p + q + r)\)\( is strictly greater than \)\(a + b + c\), not equal to it.
- Option 2: \(2(p + q + r) > 3(a + b + c)\)\( - This is incorrect. Our derivation established that \)\(2(p + q + r)\)\( is strictly less than \)\(3(a + b + c)\).
- Option 3: \(2(p + q + r) < 3(a + b + c)\) - This is correct. It perfectly matches the inequality we derived from the fundamental property that the sum of medians is less than the perimeter.
- Option 4: \(11(p + q + r) > 10(a + b + c)\)\( - This inequality is equivalent to \)\(p + q + r > \frac{10}{11}(a + b + c)\). While this condition might hold true for many triangles, it is not universally guaranteed for all triangles, especially for highly 'squeezed' or degenerate triangles. Option 3 represents a relationship that holds true for any triangle.
Summary
The analysis confirms that the correct relationship between the sum of the medians (\(p + q + r\)) and the perimeter (\(a + b + c\)) of a triangle ABC is \(2(p + q + r) < 3(a + b + c)\).
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