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 1 of 4: Record the incircle data

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In plain words

The right angle turns the usual tangent-length formulas into especially simple expressions.

AB=c,\quad BC=a,\quad CA=b,\quad r_1=\frac{a+b-c}{2},\quad AP=\frac{b+c-a}{2}

Detailed analysis

Because $\angle ACB=90^\circ$ , the incircle tangent lengths give $r_1=(a+b-c)/2$ . If $P$ is its tangency point with $AB$ , then $AP=(b+c-a)/2$ by equal tangent segments from $A$ .