I can't find the solution to this problem.

One is given the circumscribed circle and the inscribed circle of a triangle, but not the triangle itself. One has to find this triangle, using only a ruler and a compas.

Any help would be welcome!

Note that it does not work to start from an arbitrary point on the circumscribed circle and to draw the 2 tangents to the inscribed circle that contain this point. Indeed, if B and C denote the other intersections of these 2 tangents with the circumscribed circle, the line BC is in general not tangent to the inscribed circle.


Once the radiuses $R$ and $r$ are given, the distance between the centers $d$ satisfies $d^2 = R(R-2r)$. This ensures the closing condition ( $BC$ tangent to the inner circle). When $d$ is as such then you can start from any point on the circle.

More generally, there is a condition for $d$ when considering polygons with $n$ sides inscribed and circumscried $d= d_n = f_n(R,r)$. ( Poncelet theorem).

  • $\begingroup$ Wikipedia's "Poncelet's Closure Theorem" article includes a nice animated diagram. $\endgroup$ – Blue Jan 11 '15 at 14:46
  • $\begingroup$ This solves the problem, thanks! $\endgroup$ – arkani Jan 11 '15 at 17:37
  • 1
    $\begingroup$ @arkani: No worries! Oh, if you want to determine the triangle up to isometry you need one more element, perhaps $p$, the semi-perimeter. Now the lenghts $p-a$, $p-b$, $p-c$ will be the roots of a third degree equation which must have all the roots real and positive. This equation has coefficients some expressions in $R$, $r$ and $p$. From here one gets conditions on $R$,$r$ and $p$ for the existence of a triangle. $\endgroup$ – Orest Bucicovschi Jan 11 '15 at 17:46
  • $\begingroup$ @Blue: Yes, nice! Also the links there have some nice animations, for pentagons and hexagons. $\endgroup$ – Orest Bucicovschi Jan 11 '15 at 17:48

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