# Given a triangle's circumcenter, incenter, and midpoint of one side, construct its vertices

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.

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 $$\triangle ABC$$.

• 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/… Aug 19, 2020 at 22:03
• 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 Aug 19, 2020 at 22:45
• 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. Aug 20, 2020 at 0:40
• 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 Aug 20, 2020 at 1:23