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Prove that centre of circle divides altitude in golden ratio

Source:

December 31, 2011

ratiogeometrygeometry unsolved

Problem Statement

Let

T

be an isosceles right triangle. Let

S

be the circle such that the diﬀerence in the areas of

T \cup S

and

T \cap S

is the minimal. Prove that the centre of

S

divides the altitude drawn on the hypotenuse of

T

in the golden ratio (i.e.,

\frac{(1 + \sqrt{5})}{2}

)