Restore a triangle by one of its vertices, the circumcenter and the Lemoine's point. (The Lemoine's point is the intersection point of the reflections of the medians in the correspondent angle bisectors)
Problem
Source: Sharygin geometry olympiad 2016, grade 10, Final Round, Problem 7.
Tags: construction, symmedian point, Circumcenter, geometry, circumcircle, geometric transformation, reflection
anantmudgal09
05.08.2016 16:42
Consider the lemma.
Lemma In triangle $ABC$ with Lemoine's point $K$, the symmedian $AK$ meets the circumcircle again at point $L$ and the perpendicular bisector of $BC$ at point $T$. Then $(A,L;K,T)$ is fixed.
Proof Since $K$ is projectively defined (the other points as well), consider a projective transformation sending $(ABC)$ to another circle and $K$ to the center of this circle. Then, the newly formed triangle $A'B'C'$ is equilateral. There this ratio is fixed. Hence the result follows.
Now, the given facts indicate that we know the circumcircle, and the points $A,K,L$. Finding $T$ is enough to locate $B,C$. This follows from the lemma.
nikolapavlovic
01.10.2016 18:59
This problem was quite fun! Let the one vertex be $A$. As we are given $O$ and $A$ we can construct $(O,OA)=\Omega$ i.e the circumcircle of triangle $\triangle ABC$.Let $AL_e\cap \Omega=\{A,P\}$.We have the circumcenter so construct the tangents from $A,P$ which cut in $Q$ specifically in the center of the $A$-Appolonius circle.Now its well-known that the isodynamic points lie on the line $O,L_e$ and that their pedal triangles are equilateral.And so $(Q,QA)\cap OL_e =\{A_p,A_p'\}$Pick one ,say the first one, and invert thru it with center with radius $AA_p$.As the its pedal triangle is equilateral the image of $\triangle ABC$ is also so after inverting we get the following: Construct the equilateral triangle given one of its vertexes $A$ and its circumcircle $\Omega_{1}$ The above is done immediately by trisecting $\Omega_{1}$ on three equal arcs 2 of which endpoint is $A$. So now we have constructed $\triangle AB_1C_1$ .$\Omega \cap \{A_pB_1,A_pC_1 \}=\{B,C\}$ ,where the order is $A_p-B_1-B$ so we are done.
FlyingCucumber
01.10.2016 19:07
What does restoring a triangle mean?
nikolapavlovic
01.10.2016 19:09
You are given $L_e,O,A$ ,construct the triangle $\triangle ABC$ from them.