Two equilateral triangles are inscribed in a circle radius r. Show that the area common to both triangles is at least r2(√3)/2.
Solution
Let the circle have center O. Suppose the triangles are ABC and PQR and that PQ and AB meet at X, PR and AB meet at Y, PQ and AC meet at Z. If we reflect in the line OX, then PXY becomes AXZ, so PX = AX. Similarly, reflecting in OY, we find that PY = BY. So the triangle PXY has perimeter AX + XY + YB = AB = r√3. If we rotate through 120o twice, we see that the part of PQR which lies outside ABC is three triangles congruent to PXY. So the area common to both triangles is area ABC - 3 area PXY.
But of all triangles with a given perimeter, the equilateral triangle has largest area (if we keep X and Y fixed and vary P subject to PX + XY + YP constant, then we maximise the area by taking P as far as possible from the line XY and hence taking PX = PY. Similarly for the other pairs of sides). So area PXY ≤ 1/9 area ABC. Hence the common area >= 2/3 area ABC = (2/3) r2 (3√3)/4 = r2 (√3)/2.
© John Scholes
jscholes@kalva.demon.co.uk
3 Oct 2002