Triangular Stage in a Circular Hall: Two Equilateral Candidates
A triangular stage is being built inside a circular exhibition hall of radius 5 m centred at the origin of a coordinate plane. Two anchor points are fixed at A = 1 + 2i and B = 3 + 5i, in metres. The third anchor C must make triangle ABC equilateral. Either orientation is allowed in principle, but every anchor must lie strictly inside the hall, so |z| < 5. A junior designer rotates the vector B − A by +60° and by −60° and obtains two candidates. The first candidate has modulus about 5.27 m; the second has modulus about 4.93 m. The site engineer says the first is obviously wrong because it is outside, while the designer says both rotations are mathematically valid and the choice is a design constraint. The hall manager also requires the chosen anchor to be at least 0.5 m from the wall. The stage area and final anchor coordinates must be reported.
1. Find both possible coordinates for C using complex rotations, and decide which one is admissible under the hall constraint |z| < 5. Show the rotation calculation.
Read the worked answer
Let d = B − A = (3 − 1) + (5 − 2)i = 2 + 3i. An equilateral triangle on AB has third vertex C = A + d e^(±iπ/3). Since e^(iπ/3) = 1/2 + i√3/2 and e^(−iπ/3) = 1/2 − i√3/2, compute:
d e^(iπ/3) = (2 + 3i)(1/2 + i√3/2) = (1 − 3√3/2) + i(√3 + 3/2).
Adding A gives C+ = (2 − 3√3/2) + i(7/2 + √3) ≈ −0.598 + 5.232i, with |C+| ≈ 5.27. Similarly,
d e^(−iπ/3) = (1 + 3√3/2) + i(3/2 − √3),
so C− = (2 + 3√3/2) + i(7/2 − √3) ≈ 4.598 + 1.768i, with |C−| ≈ 4.93. Both points are mathematically valid equilateral third vertices. The hall constraint |z| < 5 rejects C+ and admits C− only.
2. Find the area of the stage. Does the orientation change the area?
Read the worked answer
The side length is |AB| = |d| = sqrt(2^2 + 3^2) = sqrt(13). For an equilateral triangle with side s, area = (√3/4)s^2. Therefore
Area = (√3/4)(13) = 13√3/4 ≈ 5.63 m².
Both orientations produce an equilateral triangle with the same side AB, so both have exactly the same area. The choice of C+ or C− affects position, not area.
3. Find the clearance of the admissible anchor from the wall. Does it satisfy the 0.5 m rule? What should be reported?
Read the worked answer
For C−, |C−| ≈ 4.927 m, so the distance to the circular wall is 5 − 4.927 = 0.073 m. This is less than the required 0.5 m. C− is inside the hall but violates the wall-clearance rule. C+ is already outside the hall, so it is not an option. The correct report is: the only inside candidate is C−, but the 0.5 m clearance condition fails, so the placement must be redesigned or the anchor points moved.