Triangle from Angle, Inradius, and Difference of Sides

Construct $\Delta ABC,$ given angle $A=\angle BAC,$ the inradius $r,$ and the difference $c-a,$ where, as usual, $c=AB$ and $a=BC.$

Triangle from Angle, Inradius, and Difference of Sides - problem

(Note that this assumes that $AB\gt BC.)$

Analysis

Let $A',B',C'$ denote the points of tangency of the incircle with the sides of $\Delta ABC,$ as above. Let also,

$\begin{align} AB'&=AC'=q,\\ BA'&=BC'=t,\\ CA'&=CB'=s. \end{align}$

Then $c-a=(q+t)-(s+t)=q-s.$ This makes point $P$ on $AC$ at distance $q-s$ equidistant with $C$ from $B',$ making the following construction natural.

Construction

Draw angle $A$ and inscribe into it a circle of radius $r:$

Triangle from Angle, Inradius, and Difference of Sides - solution, step 1

Given that, mark point $P$ on $AC$ at distance $c-a$ from $A.$ Draw circle $C(B',P)$ centered at $B'$ through $P.$

Triangle from Angle, Inradius, and Difference of Sides - solution, step 2

Its second intersection with $AC$ is at $C.$ Drawing tangents to the incircle from $A$ and $C$ gives the remaining vertex $B.$

Acknowledgment

The construction was communicated to me by Prof. Dr. René Sperb.


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