The figure shows a circle with radius 1 and a right △ABC with AB=BC. Segment AB cuts the circle at D. The side lengths of △DBC form a geometric progression. If the area of △ABC can be written as \sqrt{A} - B, find A and B.
Let DB = x, then if the common ratio of the GP is c, BC = cx and CD = (c^2)x.
CD is a diameter of the circle, so (c^2)x = 2 ........................... (1)
Pythagoras says that
x^2 + (c^2)(x^2) = (c^4)(x^2),
so c^4 - c^2 - 1 = 0,
so c^2 = (1 + sqrt(5))/2, .........................................................(2)
The area of the triangle ABC = (cx)(cx)/ 2 = (c^2)(x^2)/2
= (using (1) and (2)),
\(\displaystyle \frac{1+\sqrt{5}}{2}.\frac{16}{(1+\sqrt{5})^{2}}.\frac{1}{2}=\frac{4}{1+\sqrt{5}}=\sqrt{5}-1.\)
So A = 5 and B = 1.
