Complex Numbers
Geometry in complex plane
Grade 11
Question:
<p>One vertex of the triangle of maximum area that can be inscribed in the curve \(|z - (1+i)| = \sqrt{10}\) is \(-2 + 2i\), then the remaining vertices are</p>
<p>\(\left(\dfrac{5 - \sqrt{3}}{2}\right) + i\left(\dfrac{1 + 3\sqrt{3}}{2}\right)\)</p>
<p>\(\left(\dfrac{5 - \sqrt{3}}{2}\right) + i\left(\dfrac{1 - 3\sqrt{3}}{2}\right)\)</p>
<p>\(\left(\dfrac{5 + \sqrt{3}}{2}\right) + i\left(\dfrac{1 - 3\sqrt{3}}{2}\right)\)</p>
<p>\(\left(\dfrac{5 + \sqrt{3}}{2}\right) i\left(\dfrac{1 + 3\sqrt{3}}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: For a triangle inscribed in a circle, maximum area occurs when the triangle is equilateral. Given one vertex, find the other two by rotating it 120° about the circle's center.
<p><strong>Step 1:</strong> Identify the circle parameters. Center: C = 1+i, Radius: r = √10</p><p><strong>Step 2:</strong> Verify the given vertex A = -2+2i lies on the circle: |(-2+2i)-(1+i)| = |-3+i| = √(9+1) = √10 ✓</p><p><strong>Step 3:</strong> For maximum area, the inscribed triangle must be equilateral. The other two vertices B and C are obtained by rotating A about the center C by ±120°.</p><p><strong>Step 4:</strong> Express A relative to center: A - C = (-2+2i)-(1+i) = -3+i</p><p><strong>Step 5:</strong> Rotate by 120° counterclockwise using e^(i·2π/3) = -1/2 + i√3/2:<br>(-3+i)·(-1/2 + i√3/2) = 3/2 - 3i√3/2 - i/2 - √3/2 = (3-√3)/2 - i(3√3+1)/2</p><p><strong>Step 6:</strong> Rotate by 120° clockwise (or -120°) using e^(-i·2π/3) = -1/2 - i√3/2:<br>(-3+i)·(-1/2 - i√3/2) = 3/2 + 3i√3/2 - i/2 + √3/2 = (3+√3)/2 + i(3√3-1)/2</p><p><strong>Step 7:</strong> Add center back to get vertices:<br>B = (1+i) + [(3-√3)/2 + i(3√3-1)/2] = (5-√3)/2 + i(5+3√3)/2<br>C = (1+i) + [(3+√3)/2 + i(3√3-1)/2] = (5+√3)/2 + i(5+3√3)/2</p><p>∴ Answer: A</p>
Correct Answer: A