Triangle from Side, Inradius, and Altitude

Construct $\Delta ABC,$ given side $BC=a,$ the altitude $h_a,$ and the inradius $r.$

Triangle from Side, Inradius, and Altitude - problem

Construction

Let $s, t,u$ be the lengths of the tangents from the vertices of $\Delta ABC$ to the incircle $(I),$ as in the diagram:

Triangle from Side, Inradius, and Altitude - solution

We can construct the triangle if we know the length $s.$ Indeed, drawing a tangent to $(I)$ of length $s$ ($EF$ in the diagram) we then add circle $C(I,F)$ centered at $I$ and through $F$ and find $A$ at its intersection with line $pp$ parallel to $BC$ at distance $h_a.$

To determine $s,$ we express the area $[\Delta ABC]$ of triangle $ABC$ in two ways:

$[\Delta ABC]=\frac{1}{2}a\cdot h_a$ and
$[\Delta ABC]=\frac{1}{2}[(s+t)r + (t+u)r +(u+s)r]=(s+t+u)r = (s+a)r,$

from which $s=\displaystyle\frac{a h_a}{2r}-a.$

The ratio $\displaystyle\frac{a h_a}{2r}$ is easily constructed using Thales' theorem:

Triangle from Side, Inradius, and Altitude - lemma

Acknowledgment

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


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