Problem 4

A semicircular arc

\gamma

has diameter

AB

. Let

C

be a point of the arc other than

A,B

, and let

D

be the foot of the perpendicular from

C

AB

. Three circles

\gamma_1,\gamma_2,\gamma_3

are tangent to line

AB

;

\gamma_1

is inscribed in triangle

ABC

, while

\gamma_2

and

\gamma_3

are tangent to

CD

and to

\gamma

, on opposite sides of

CD

. Prove that the three circles have a second common tangent.

Step 4 of 4: Reflect the known tangent

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In plain words

A line through the midpoint of two circle centers exchanges the two circles under reflection and fixes the middle circle, producing the second tangent automatically.

O_1\in O_2O_3,\qquad O_1P=r_1,\qquad O_2X=r_2,\qquad O_3Y=r_3,\qquad O_1\text{ is the midpoint of }O_2O_3

Detailed analysis

All three centers lie on perpendiculars to $AB$ through their base contact points. Since $P$ is the midpoint of $XY$ and $r_2+r_3=2r_1$ , the center $O_1$ is the midpoint of $O_2O_3$ . Reflect the common tangent $AB$ in the line $O_2O_3$ : a reflection preserves tangency, so the reflected line is a second common tangent to all three circles.