Complex Numbers
Area of triangle in complex plane
Grade 11
Question:
<p>The maximum area of the triangle formed by the complex coordinates \(z, z_1, z_2\) which satisfy the relations \(|z - z_1| = |z - z_2|\) and \(|z - (z_1 + z_2)/2| \leq r\), where \(r > |z_1 - z_2|\), is</p>
<p>\(\frac{1}{2}|z_1 - z_2|^2\)</p>
<p>\(\frac{1}{2}|z_1 - z_2| \cdot r\)</p>
<p>\(\frac{1}{2}|z_1 - z_2|^2 r^2\)</p>
<p>\(\frac{1}{2}|z_1 - z_2| \cdot r^2\)</p>
Step-by-Step Solution
Key Concept: The locus |z - z₁| = |z - z₂| represents the perpendicular bisector of segment z₁z₂, and combined with the disk constraint, the maximum area occurs when z is at maximum distance from the midpoint along the perpendicular bisector.
<p><strong>Step 1:</strong> The condition |z - z₁| = |z - z₂| defines the perpendicular bisector of the segment joining z₁ and z₂. Let m = (z₁ + z₂)/2 be the midpoint.</p><p><strong>Step 2:</strong> The perpendicular bisector passes through m and is perpendicular to the line z₁z₂. Let d = |z₁ - z₂|/2 be the half-distance between z₁ and z₂.</p><p><strong>Step 3:</strong> The second constraint |z - m| ≤ r restricts z to a disk of radius r centered at m. The intersection of the perpendicular bisector with this disk gives z on a line segment of length 2√(r² - d²) centered at m.</p><p><strong>Step 4:</strong> The area of triangle with vertices z₁, z₂, z equals (1/2) × base × height = (1/2) × |z₁ - z₂| × h, where h is the perpendicular distance from z to line z₁z₂.</p><p><strong>Step 5:</strong> This height h is maximized when z is at the farthest point on the perpendicular bisector within the disk, giving h_max = √(r² - d²) = √(r² - |z₁ - z₂|²/4).</p><p><strong>Step 6:</strong> Maximum area = (1/2) × |z₁ - z₂| × √(r² - |z₁ - z₂|²/4)</p><p>∴ Answer: B</p>
Correct Answer: B