<p>A line is divided at random in three parts, what is the chance that they form the sides of a possible triangle?</p>
Step-by-Step Solution
Key Concept: For three lengths to form a triangle, each side must be less than the sum of the other two. When a line of length L is divided into three parts at two random points, the condition simplifies to: each part must be less than L/2.
<p><strong>Step 1: Set up the problem</strong> Let the line have length 1. Two random points divide it into three parts with lengths x, y, and z where x + y + z = 1 and x, y, z > 0.</p><p><strong>Step 2: Apply triangle inequalities</strong> For a valid triangle, we need:<br>• x + y > z ⟹ x + y > 1 - x - y ⟹ x + y > 1/2<br>• y + z > x ⟹ y + z > 1 - y - z ⟹ y + z > 1/2<br>• z + x > y ⟹ z + x > 1 - z - x ⟹ z + x > 1/2</p><p><strong>Step 3: Geometric probability approach</strong> Let the two division points be at distances a and b from the start (0 < a < b < 1). Then x = a, y = b - a, z = 1 - b. The sample space is a triangle with area 1/2.</p><p><strong>Step 4: Find favorable region</strong> The constraints become:<br>• a + (b-a) > 1/2 ⟹ b > 1/2<br>• (b-a) + (1-b) > 1/2 ⟹ a < 1/2<br>• (1-b) + a > 1/2 ⟹ a + 1 - b > 1/2 ⟹ b < a + 1/2</p><p><strong>Step 5: Calculate probability</strong> The favorable region (intersection of b > 1/2, a < 1/2, and b < a + 1/2) forms a triangle with vertices at (0, 1/2), (1/2, 1/2), and (1/2, 1). Its area = 1/8.</p><p><strong>Step 6: Final answer</strong> Probability = (1/8)/(1/2) = 1/4</p><p>∴ <strong>Answer: 1/4</strong></p>
Correct Answer: 1