Isolating Isosceles Inspiration

The circumradius of a triangle is $R = \cfrac{abc}{4T}$ and the inradius of a triangle is $r = \cfrac{2T}{a + b + c}$ , where $a$ , $b$ , and $c$ are the sides of a triangle and $T$ is its area.

Let the base of the acute isosceles triangle be $2x$ , its legs be $y$ , and its height be $h$ .

Then the sides of the acute isosceles triangle are $a = 2x$ , $b = y$ , $c = y$ , and $T = \cfrac{1}{2} \cdot 2x \cdot h = xh$ , so $R = \cfrac{abc}{4T} = \cfrac{2x \cdot y \cdot y}{4xh} = \cfrac{y^2}{2h}$ and $r = \cfrac{2T}{a + b + c} = \cfrac{2xh}{2x + y + y} = \cfrac{xh}{x + y}$ .

By the Pythagorean Theorem, $h = \sqrt{y^2 - x^2}$ , so $R = \cfrac{y^2}{2\sqrt{y^2 - x^2}}$ and $r = \cfrac{x\sqrt{y^2 - x^2}}{x + y}$ .

These two equations solve to $x = \cfrac{\sqrt{2}}{2R}(R \pm \sqrt{R(R-2r)})(\sqrt{R^2+rR \mp R\sqrt{R(R-2r)}}$ and $y = \sqrt{2}\sqrt{R^2+rR \mp R\sqrt{R(R-2r)}}$ for $x > 0$ and $y > 0$ , which have at most two solutions (representing at most two distinct triangles), but only one solution (or one triangle) when $\sqrt{R(R-2r)} = 0$ , which is when $R = 0$ (a degenerate triangle) and $R = 2r$ (an equilateral triangle).

Therefore, the given statement is True .