A circle of radius R is inscribed in triangle ABC, tangent to side AC at point M, with AM = 2R and CM = 3R.

A circle of radius R is inscribed in triangle ABC, tangent to side AC at point M, with AM = 2R and CM = 3R. a) Prove that triangle ABC is right-angled. b) Find the distance between the centers of its inscribed and circumscribed circles, if it is known that R = 2.

Let F and K be the tangency points of the circle with sides AB and BC, respectively. AM = AF = 2R by the property of tangent line segments drawn from one point to a circle. Similarly, CM = CK = 3R. OM = OK = OF = R. O is the center of the inscribed circle. OF is perpendicular to AB, OK is perpendicular to BC, since the tangent is perpendicular to the radius drawn to the tangent point. BFOK is a square. BF = BK = OF = OK = R. Let’s check, AC ^ 2 = (2R + 3R) ^ 2 = (5R) ^ 2 = 25R ^ 2, AB ^ 2 + BC ^ 2 = (2R + R) ^ 2 + (3R + R) ^ 2 = (3R ) ^ 2 + (4R) ^ 2 = 9R ^ 2 + 16R ^ 2 = 25R ^ 2. By the theorem of the inverse Pythagorean theorem, if the square of one side of a triangle is equal to the sum of the squares of its other two sides, then such a triangle is right-angled. So triangle ABC = right-angled, with a right angle at the vertex B.
The center of the circle P circumscribed about a right-angled triangle lies in the middle of its hypotenuse. MC = 3R = 3 * 2 = 6, AC = 5R = 5 * 2 = 10. PC = AC / 2 = 10/2 = 5. MP = MC-PC = 6-5 = 1. From the triangle OMP (angle M is a straight line), by the Pythagorean theorem, we find OP. OP ^ 2 = OM ^ 2 + MP ^ 2, OP ^ 2 = 2 ^ 2 + 1 ^ 2 = 5, OP = √5.