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source : geometry.puzzles
From : Ontadian
| Two identical isosceles triangles with angles A, (180-A)/2 and (180-A)/2
| have their vertices on the same circle.
| They have an angular displacement φ = 125 degrees to each other.
| If the overlap of the triangle areas is equal to one half of the area
| of one triangle, find angle A.
A is fixed. Point a can be moved to adjust displacement φ
B can be moved to adjust angle A.
Adjust B so that ratio = 0.5
Line OM is symmetry axis. Angle MOA = φ/2, and aA = 2.R.sin(φ/2)
We can then evaluate all angles in A :
aAO = 90░ - φ/2 from right triangle OHA, hence
aAB = 90░ - φ/2 - A/2 and
aAC = 90░ - φ/2 + A/2
This allows to calculate areas of triangles aPA, aQA = ANa and aMA
Hence area of MNPQ = (aPA) - 2*(aQA) + aMA
Finally to write that this area is 1/2 of that of triangle ABC.
Details and second case let to the reader.
Area of triangle with base c, and angles with base A and B
b/sinB = a/sinA = c/sin(180░ - A - B) = c/sin(A+B)
Hence b = c*sinB/sin(A + B)
And area = 1/2 b*c*sinA = c²/2 * sinA*sinB/sin(A+B)