This question is an investigation starting here, with a straightedge and compass construction of $ABC$ given $(R,r,h_A)$. The key lemma is the following one:

enter image description here

Let $\Gamma$ be a circle with centre $O$, radius $R$ and let $A\in\Gamma$. Let $H_A$ be any point inside $\Gamma$ such that $AH_A=h_A$ and let $B',C'$ be the intersections between $\Gamma$ and the perpendicular to $AH_A$ through $H_A$. Claim: the incenter $I'$ of $AB'C'$ lies on a circle.

I was able to prove the above claim by trigonometry, but I am not satisfied by my solution and I strongly believe that more elegant ones exists. There are many interesting facts in the above configuration:

  • $bc=AB'\cdot AC' = 2 h_A R$ is constant;
  • $OI'^2=R^2-2Rr$ is fixed too, by Euler's theorem;
  • $I'$ lies on the angle bisector of $\widehat{H_A A O}$ since the circumcenter $O$
    and the orthocenter $H$ are isogonal conjugates;
  • By Carnot's theorem, the sum of the distances of $O$ from the sides of $ABC$ equals $R+r$;
  • $\frac{1}{r}=\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C} = \frac{1}{h_A}\left(1+\frac{b+c}{a}\right);$
  • the circumradius of the nine point circle equals $\frac{R}{2}$ and by Feuerbach's theorem
    the circumcircle of the medial/orthic triangle is tangent to the incircle.

Update: solved with an elementary proof in my own answer.

An equivalent formulation comes from considering the Poncelet's porism configuration.
The set of triangles with given inradius and circumradius is given by the case $n=3$:

enter image description here

In this configuration, if we take $A'$ on the $OA$ ray in such a way that $AA'=AH_A$,
then both $A'$ and the orthocenter $H$ lie on fixed circles.

Are you able to provide a simple proof of the last claim?

Update: $H\in\text{circle}$ is solved with an elementary proof in my own answer.
We just need to find an elementary proof of $A'\in\text{circle}$.

I give an extra picture for clarity: clearly, the purple circles we are interested in are fixed by the two isosceles triangles having the given inradius and circumradius.

enter image description here

up vote 1 down vote accepted

Let we start from the second question, second part.

If we consider the centre of the nine point circle, $N=\frac{O+H}{2}$, Feuerbach's theorem gives $$ IN = \frac{R}{2}-r, $$ hence $N$ travels on a circle centered at $I$ with a constant radius.
By homothety, $H$ travels on a circle, too, having radius $R-2r$. Summarizing:

The orthocenter $H$ of a triangle always lies on a circle having radius $R-2r$ centered at
the symmetric of the circumcenter $O$ with respect to the incenter $I$.

In a similar way, assuming that $A'$ actually travels on a circle, the centre of such a circle has to be the symmetric of the incenter $I$ with respect to the circumcenter $O$ (by simply considering the isosceles triangles in Poncelet's configuration). Let we call such a point $J$. About the first part, we just have to prove that $$ A'J^2 = (R+h_A)^2+(R^2-2Rr)+2(R+h_A)\sqrt{R^2-2Rr} \cos(\widehat{AOI})$$ does not really depend on $h_A$. That is the unpleasant part I cracked through trigonometry:
it boils down to expressing $\frac{b+c}{a}$ in terms of $R,r$ and $h_A$, or expressing $\frac{a}{a+b+c}$ in terms of
$R,r$ and $h_A$, where: $$ \frac{a}{a+b+c}=\frac{ar}{r(a+b+c)}=\frac{a r}{2\Delta} = \frac{ar}{ah_A}=\frac{r}{h_A}.$$


About the first question, we may consider $I'_A$, the $A$-excenter of the $AB'C'$ triangle. Then $IB' I'_A C'$ is a cyclic quadrilateral with circumcircle $\Gamma_A$, and the claim follows from $$\text{pow}_A(\Gamma_A) = AI'\cdot AI'_A = f(h_A, R). $$ The incenter-excenter lemma gives that the centre of $\Gamma_A$ is the midpoint $N$ of the minor $B'C'$ arc, hence: $$\begin{eqnarray*}\text{pow}_A(\Gamma_A) = AN^2 - R_A^2 = AN^2 - MB'^2 &=& 2R^2\left(\cos A+\cos(B-C)\right)\\ &=& 4R^2 \sin B \sin C\\&=& bc = 2Rh_A.\end{eqnarray*}$$

We may avoid trigonometry at all by just noticing that $\Gamma_A$ is symmetric with respect to the $A$-angle bisector $l_A$, hence $\text{pow}_A(\Gamma_A)=bc=2R h_A$ readily follows by considering the symmetric of $B',C'$ with respect to $l_A$, that still lie on $\Gamma_A$.

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