Given that $ABC$ is a triangle where $AB < AC$. On the half-lines $BA$ and $CA$ we take points $F$ and $E$ respectively such that $BF = CE = BC$. Let $M,N$ and $H$ be the mid-points of the segments $BF,CE$ and $BC$ respectively and $K$ and $O$ be the circumcenters of the triangles $ABC$ and $MNH$ respectively. We assume that $OK$ cuts $BE$ and $HN$ at the points $A_1$ and $B_1$ respectively and that $C_1$ is the point of intersection of $HN$ and $FE$. If the parallel line from $A_1$ to $OC_1$ cuts the line $FE$ at $D$ and the perpendicular from $A_1$ to the line $DB_1$ cuts $FE$ at the point $M_1$, prove that $E$ is the orthocenter of the triangle $A_1OM_1$.
Problem
Source: Balkan MO BMO Shortlist 2016 G3
Tags: geometry, circumcircle, orthocenter
Pathological
06.07.2019 01:17
First of all, it's easy to obtain that $O$ is the incenter of $\triangle ABC.$ We need only prove two things: 1) $OA_1 \perp EM_1$ 2) $OE \perp A_1M_1$ Lemma 1. $OA_1 \perp EM_1$. Proof. Note that it suffices to prove $OK \perp EF$, which by Carnot is equivalent to $$OE^2 - OF^2 = KE^2 - KF^2 \Leftrightarrow OB^2 - OC^2 = KE^2 - KF^2.$$Now, note that $KE^2 = KC^2 + (a-\frac{b}{2})^2 - (\frac{b}{2})^2 = KC^2 + a(a-b)$ and $KF^2 = KB^2 + a(a-c)$ similarly. Therefore, we have that $$KE^2 - KF^2 = a(c-b)$$because $KB = KC$. Finally, $OB^2 - OC^2 = (s-b)^2 - (s-c)^2 = a(c-b)$, where $a, b, c$ are the side lengths of the triangle and $s$ is the semi-perimeter of $\triangle ABC.$ We are done. $\blacksquare$ Lemma 2. $OE \perp A_1M_1.$ Proof. Since $A_1M_1 \perp DB_1$, we just need to show that $DB_1 || OE.$ Let $X = A_1B_1 \cap EF.$ We wish to show that $$\frac{XE}{XD} = \frac{XO}{XB_1}.$$From the definition of $D$, we know that $$\frac{XD}{XC_1} = \frac{XA_1}{XO}.$$Hence, if we multiply these equations, we simply wish to show that $$\frac{XE}{XC_1} = \frac{XA_1}{XB_1},$$however this is obvious since $\triangle XA_1E \sim \triangle XB_1C_1$ from $A_1E || B_1C_1.$ $\blacksquare$ From the two previous lemmas, we are done. $\square$
Mahdi_Mashayekhi
21.02.2022 16:29
It's really easy though it was a little hard to draw for me... Claim1 : $O$ is incenter of $ABC$. proof : $BM = BH$ and $O$ lies on perpendicular bisector of $MH$ so $O$ lies on angle bisector of $\angle B$. with same approach for $C$ we can prove $O$ is incenter. Claim2 : $EM_1 \perp A_1O$. Proof : we will prove $KF^2 - KE^2 = OF^2 - OE^2$. Let $OX$ and $OY$ be perpendicular to $AB$ and $AC$. $OF^2 = r^2 + FX^2$ , $OE^2 = r^2 + EY^2$ so $OF^2 - OE^2 = (FX+EY)(FX-EY) = a(b-c)$. $KF^2 = (R^2 - c^2/4) + (a - c/2)^2$ , $KE^2 = (R^2 - b^2/4) + (a - b/2)^2$ so $KF^2 - KE^2 = -ac + ab = a(b-c)$. Claim3 : $M_1A_1 \perp EO$. Proof : we know $M_1A_1 \perp B_1D$ so we need to prove $EO || B_1D$. Let $OK$ meet $EF$ at $Q$. we need to prove $QE/QD = QO/QB_1$.we know $A_1D || OC_1$ so $QD/QC_1 = QA_1/QO$ and we also know $A_1E || HN$ so $QE/QC_1 = QA_1/QB_1$ so with only a division we will have $QE/QD = QO/QB_1$ as wanted so our claim is true. Claims 2,3 mean that $E$ is orthocenter of $M_1OA_1$. we're Done.