Wernick's list problem number 69: We want to find, with straightedge and compass, the vertices of triangle $\triangle ABC$ but we're only given:

  • its incenter, $I$
  • its circumcenter, $O$
  • the midpoint of side $a$, $M_a$

What I've done:

  • draw half line $OM_a$, $r$

  • draw $s$ perpendicular to $r$ passing through $M_a$ (this line contains side $BC$)

  • draw $t$:a parallel to $r$ passing through $I$

  • $Z = t \cap s$

  • draw circle $c$ centered at $I$ passing thorugh $Z$ (this is the incircle of $\triangle ABC$)

  • $Q = c \cap t \neq Z $

  • reflect $Z$ at the point $M_a$ to find $T$.

  • draw line $QT$ ($A$ is on this line).

I don't know how to finish. I suspect we can draw the radius of the circumcircle with Steiner porisms and that would end the problem.


1 Answer 1


You can indeed find the circumcircle:

draw $IP \perp IO$ such that $P$ is in the incircle already drawn (there are two possible points for $P$ but it doesn't matter which one you pick).

let line $OP$ meet the circle centered at $P$ passing through $I$ at point $X$ ($X$ the point most distant to $O$).

The circle centered at $O$ with radius $OX$ is the circumcircle of $\triangle ABC$.

This is a result that is based on the distance $OI^2 = R^2 -2rR$.

From this, we easily get $\triangle ABC$.

  • $\begingroup$ I haven’t checked your construction, but if it is valid you have solved something an automated effort did not. poincare.matf.bg.ac.rs/%7Evesnap/animations/… $\endgroup$
    – brainjam
    Aug 19, 2020 at 22:03
  • $\begingroup$ I can't access your link. But my construction works and you can check it on geogebra quickly. Thanks for pointing it to me. I hope I'm not the first to draw it because it is not any profound new discovery $\endgroup$ Aug 19, 2020 at 22:45
  • 1
    $\begingroup$ Yes, apparently Problem 69 was solved when Wernick's list first came out. I guess the automated solver couldn't figure it out. So you can breathe a sigh of relief. Maybe you need to take the answer I gave you for Problem 82 with a grain of salt. But many people had probably tried to solve 82 before that article came out. $\endgroup$
    – brainjam
    Aug 20, 2020 at 0:40
  • $\begingroup$ I thought so. I hesitated for a while before accepting your answer but the thruth is that problem seems a lot harder and after a little research and a couple tries I don't think anyone has an easy answer for that one in particular. I will keep your answer unless somebody magically come up with the correct answer $\endgroup$ Aug 20, 2020 at 1:23

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