MathLabs

Problem 4

A semicircular arc γ\gamma has diameter ABAB. Let CC be a point of the arc other than A,BA,B, and let DD be the foot of the perpendicular from CC to ABAB. Three circles γ1,γ2,γ3\gamma_1,\gamma_2,\gamma_3 are tangent to line ABAB; γ1\gamma_1 is inscribed in triangle ABCABC, while γ2\gamma_2 and γ3\gamma_3 are tangent to CDCD and to γ\gamma, on opposite sides of CDCD. Prove that the three circles have a second common tangent.
Step 2 of 4: Compute the two auxiliary radii
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In plain words

The semicircle supplies a common right-triangle relation, so the two radii are controlled by the legs a,ba,b.

r2=a−a2c,r3=b−b2c,r2+r3=a+b−c=2r1r_2=a-\frac{a^2}{c},\qquad r_3=b-\frac{b^2}{c},\qquad r_2+r_3=a+b-c=2r_1
Detailed analysis

Let X,YX,Y be the contact points of γ2,γ3\gamma_2,\gamma_3 with ABAB. A right-triangle calculation using the semicircle gives r2=a−a2/cr_2=a-a^2/c; the symmetric calculation gives r3=b−b2/cr_3=b-b^2/c. Since c2=a2+b2c^2=a^2+b^2, their sum simplifies to a+b−c=2r1a+b-c=2r_1.