Complex Numbers
Geometric Applications of Complex Numbers
Grade 11
Question:
<p>Let <i>z₁, z₂, z₃</i> be complex numbers denoting the vertices of an equilateral triangle ABC having circumradius equal to unity. If <i>P</i> denotes any arbitrary point on its circumcircle, then the value of <i>(1/2)((PA)² + (PB)² + (PC)²)</i> equals:</p>
Step-by-Step Solution
Key Concept: For an equilateral triangle inscribed in a circle, the sum of squared distances from any point on the circle to the three vertices is constant, found using angle addition formulas.
<p><strong>Analysis:</strong> Let the circumcircle have center at the origin and radius <i>R = 1</i>. The vertices of an equilateral triangle inscribed in this circle are at angles 0°, 120°, 240°:</p><p><i>z₁ = 1</i>, <i>z₂ = e^(i·2π/3)</i>, <i>z₃ = e^(i·4π/3)</i></p><p>Let <i>P = e^(iθ)</i> be an arbitrary point on the circumcircle.</p><p>The distances are:</p><p><i>|PA|² = |e^(iθ) - 1|² = 2 - 2cos(θ)</i></p><p><i>|PB|² = |e^(iθ) - e^(i·2π/3)|² = 2 - 2cos(θ - 2π/3)</i></p><p><i>|PC|² = |e^(iθ) - e^(i·4π/3)|² = 2 - 2cos(θ - 4π/3)</i></p><p>Sum: <i>|PA|² + |PB|² + |PC|² = 6 - 2[cos(θ) + cos(θ - 2π/3) + cos(θ - 4π/3)] = 6 - 0 = 6</i></p><p>(The sum of cosines at equally spaced angles is 0.)</p><p>Therefore, <i>(1/2)(|PA|² + |PB|² + |PC|²) = 6/2 = <b>3</b></i>.</p>
Correct Answer: P