11.4. In a semicircle with a radius of 18 cm, a semicircle with a radius of 9 cm is constructed on one of the halves of the diameter, and a circle is inscribed, touching the larger semicircle from the inside, the smaller semicircle from the outside, and the second half of the diameter. Find the radius of this circle.

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# Answer: 8 cm.

Solution. Let O, O_1, O_2 be the centers of the large semicircle, the small semicircle, and the inscribed circle, respectively, and let P, Q, R be the points of tangency of the inscribed circle with the diameter of the large semicircle, the small semicircle, and the large semicircle, respectively. Then the points O_1, Q, O_2 lie on the same line, and the points O, O_2, R lie on the same line. Let the radius of the inscribed circle be x. From the Pythagorean theorem for triangle OO_2P, we have OP = \sqrt{(18-x)^2 - x^2} = \sqrt{324 - 36x}. From the Pythagorean theorem for triangle O_1O_2P, we have O_1P = \sqrt{(9 + x)^2 - x^2} = \sqrt{81 + 18x}. Therefore, O_1P = OP + 9 = \sqrt{324 - 36x} + 9 = \sqrt{81 + 18x}. Dividing by 3, we get \sqrt{36 - 4x} + 3 = \sqrt{9 + 2x}. From the last equation, x = 8.

Criteria. There is no clear justification for the calculation of OP and O_1P: deduct 2 points. The equation for x is correctly set up, but not solved: 3 points.