A circle is inscribed in a right-angled triangle, the radius of which is 1 cm, and the point of contact divides the hypotenuse

A circle is inscribed in a right-angled triangle, the radius of which is 1 cm, and the point of contact divides the hypotenuse in a ratio of 2: 3. Find the legs of the triangle and the radius of the circumscribed circle.

By the property of tangents to the circle drawn from one point AK = AE, KB = BM, CM = CE.

CM and CE are equal to the inscribed circle radius CM = CE = 1 cm.

By condition, BK / AK = 2/3. Let us denote the ratio of the parts of the hypotenuse X, then BK = 2 * X, AK = 3 * X, and the hypotenuse AB = 5 * X.

Then the leg BC = BM + CM = BK + 1 = 2 * X + 1.

AC leg = AE + CE = 3 * X + 1.

By the Pythagorean theorem (5 * X) ^ 2 = (2 * X + 1) ^ 2 + (3 * X + 1) ^ 2.

25 * X ^ 2 = 4 * X ^ 2 + 4 * X + 1 + 9 * X ^ 2 + 6 * X + 1.

12 * X ^ 2 – 10 * X – 2 = 0.

6 * X ^ 2 – 5 * X – 1 = 0.

Let’s solve the quadratic equation.

D = b ^ 2 – 4 * a * c = (-5) ^ 2 – 4 * 6 * (-1) = 25 + 24 = 49.

X1 = (5 – √49) / (2 * 6) = (5 – 7) / 12 = -2/12 = -1/6. (Not suitable, since there cannot be X <0).

X2 = (5 + √49) / (2 * 6) = (5 + 7) / 12 = 12/12 = 1.

X = 1, then:

BC = 2 * 1 + 1 = 3 cm.

AC = 3 * 1 + 1 = 4 cm.

AB = 5 * 1 = 5 cm.

The radius of the circumscribed circle about a right-angled triangle is half the hypotenuse.

R = AB / 2 = 5/2 = 2.5 cm.

Answer: R = 2.5 cm, AC = 4 cm, BC = 3 cm.