The radius of the circumscribed circle of an isosceles triangle of unit base is also unity. The diameter parallel to the base of the triangle cuts off a smaller triangle. The exact lengths of the base and legs of the smaller triangle are _______

PLACEHOLDER ANSWER - the answer is two exact surds. With the circumcentre at the origin, the base of length 1 lies at distance $$\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$$ below the centre and the apex is at $$(0, 1)$$, so the height is $$1 + \frac{\sqrt{3}}{2}$$ and each leg is $$\sqrt{\frac{1}{4} + \left(1 + \frac{\sqrt{3}}{2}\right)^2} = \sqrt{2 + \sqrt{3}}$$. The diameter cuts off a similar triangle of height 1, a scale factor of $$\frac{2}{2 + \sqrt{3}} = 4 - 2\sqrt{3}$$. So the smaller triangle has base $$4 - 2\sqrt{3}$$ and legs $$(4 - 2\sqrt{3})\sqrt{2 + \sqrt{3}} = \sqrt{6} - \sqrt{2}$$.