IMO 1984 SL 18

Origin: USA

Problem

Inside triangle ABC there are three circles k1, k2, k3 each of which is tangent to two sides of the triangle and to its incircle k. The radii of k1, k2, k3 are 1, 4, and 9. Determine the radius of k.

Solution

Suppose that circles k1(O1, r1), k2(O2, r2), and k3(O3, r3) touch the edges of the angles \angleBAC, \angleABC, and \angleACB, respectively. Denote also by O and r the center and radius of the incircle. Let P be the point of tangency of the incircle with AB and let F be the foot of the perpendicular from O1 to OP. From \triangleO1FO we obtain cot(\alpha/2) = 2\sqrtrr1/(r −r1) and analogously cot(\beta/2) = 2\sqrtrr2/(r −r2), cot(\gamma/2) = 2\sqrtrr3/(r −r3). We will now use a well-known trigonometric identity for the angles of a triangle: cot \alpha 2 + cot \beta 2 + cot \gamma 2 = cot \alpha 2 \cdot cot \beta 2 \cdot cot \gamma 2 . (This identity follows from tan(\gamma/2) = cot (\alpha/2 + \beta/2) and the formula for the cotangent of a sum.) Plugging in the obtained cotangents, we get 2\sqrtrr1 r −r1

• 2\sqrtrr2 r −r2
• 2\sqrtrr3 r −r3 = 2\sqrtrr1 r −r1 \cdot 2\sqrtrr2 r −r2 \cdot 2\sqrtrr3 r −r3 ⇒ \sqrtr1(r −r2)(r −r3) + \sqrtr2(r −r1)(r −r3)
• \sqrtr3(r −r1)(r −r2) = 4r\sqrtr1r2r3. For r1 = 1, r2 = 4, and r3 = 9 we get (r−4)(r−9)+2(r−1)(r−9)+3(r−1)(r−4) = 24r ⇒6(r−1)(r−11) = 0. Clearly, r = 11 is the only viable value for r.