I was recently pondering about constructing triangles given different attributes of it.

I am wondering whether we could construct a triangle given its Circumradius $R$ , Inradius $r$, and length of one altitude.


Yes we can! enter image description here

Let we start by drawing the circumcircle $\Gamma$ with centre at $O$ through $A$, with $OA=R$. Given some $P$ such that $AP=h_A$ (a candidate feet of the $A$-altitude), the perpendicular to $AP$ through $P$ meet $\Gamma$ at $B',C'$ and

Lemma: the intersection $I'$ between the $AB'C'$ angle bisector and the $AC'B'$ angle bisector lies on a circle. A proof through the incenter-excenter lemma can be found here.

On the other hand, since we know $R$ and $r$ we also know $OI$ by Euler's theorem, hence the incenter has to lie on the previous circle and on some circle centered at $O$ with radius $\sqrt{R^2-2Rr}$. Given $A,O,R, h_A$ and $I$, $ABC$ is straightforward to construct.

Submitted as a contribution to http://www.cut-the-knot.org/triangle/.

Alternative solution: there is a hidden gem in Poncelet's Porism configuration for a triangle.

enter image description here

It is easy to depict the family of triangles having fixed circumradius and inradius, starting from the isosceles one and exploiting Poncelet's porism. If $ABC$ is a triangle with incenter $I$, circumcenter $O$, orthocenter $H$, $H_A$ is the projection of $A$ on $BC$ and $A'$ is a point on the $OA$-ray such that $AA'=AH_A$, then both $A'$ and $H$ lie on circles.

  • $\begingroup$ where is $P$ I could not follow? $\endgroup$ – Love Everything Jul 10 '16 at 13:48
  • $\begingroup$ @LoveEverything: $P$ is the projection of $A$ on $B'C'$ in the above picture. $\endgroup$ – Jack D'Aurizio Jul 10 '16 at 14:19

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