Diagonals AC and BD of rectangle ABCD intersect at point O. Radii of circles inscribed in triangles AOB and BOC

Diagonals AC and BD of rectangle ABCD intersect at point O. Radii of circles inscribed in triangles AOB and BOC are equal to 1 and r, respectively. What values can r take?

a) Let:

AB = a;
BC = b;
AC = c;
∠BAC = α;
r1 = 1, r2 = r are the radii of the triangles AOB and COB.
Then:

a = c * cosα;
b = c * sinα.
b) Triangle semi-perimeter AOB and COB:

p1 = (a + c) / 2 = (c * cosα + c) / 2 = c / 2 * (1 + cosα);
p2 = (b + c) / 2 = (c * sinα + c) / 2 = c / 2 * (1 + sinα).
c) The areas of triangles AOB and COB are equal:

p1r1 = p2r2;
c / 2 * (1 + cosα) r1 = c / 2 * (1 + sinα) r2;
(1 + cosα) r1 = (1 + sinα) r2;
k (α) = r2 / r1 = (1 + cosα) / (1 + sinα);
The angle α varies within (0 °; 90 °):

k1 = k (0 °) = (1 + cos0 °) / (1 + sin0 °) = 2;
k2 = k (90 °) = (1 + cos90 °) / (1 + sin90 °) = 1/2.
1/2 <k <2;
1/2 <r2 / r1 <2;
1/2 <r2 <2;
r2 ∈ (1/2; 2);
r ∈ (1/2; 2).
Answer:

1) two solutions: (-3; -2), (-2; -1);
2) r ∈ (1/2; 2).