Let $H$ be the orthocenter of triangle $\mathrm T$. The sidelines of triangle $\mathrm T_1$ pass through the midpoints of $\mathrm T$ and are perpendicular to the corresponding bisectors of $\mathrm T$. The vertices of triangle $\mathrm T_2$ bisect the bisectors of $\mathrm T$. Prove that the lines joining $H$ with the vertices of $\mathrm T_1$ are perpendicular to the sidelines of $\mathrm T_2$.
Problem
Source: Sharygin Finals 2023 9.7
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023, perpendicular lines
NO_SQUARES
02.08.2023 13:01
Not very hard by Carno, medians and bisectors formulas More details, maybe, I will add later
NO_SQUARES
16.09.2023 22:05
Also, at the some moment very good will be using this lemma: In triangle $ABC$ $A_1, C_1$ lies on sides $BC, AB$, Then $H$ lies on radial axe circles with diameters $AA_1,CC_1$.
NO_SQUARES
18.11.2023 17:44
Now I will write more detailes. Let $AL_a=l_a, BL_b=l_b, CL_c=l_c$ be a angles bisectors, $AM_a=m_a,BM_b=m_b,CM_c=m_c$ be a midpoints of sides of $T=ABC$, $I$ be its incenter. Also let $X_a,X_b,X_c$ be a vertices of $T_2$ and $P_a,P_b,P_c$ be a vertices of $T_1$. If we will prove that $HX_a^2-HX_c^2=P_bX_a^2-P_bX_c^2$, then by Carno we will done. By Carno $P_bX_a^2-P_bX_c^2=(P_bX_a^2-P_bI^2)-(P_bX_c^2-P_bI^2)=M_aX_a^2-M_aI^2 - (M_cX_c^2-M_cI^2) =^? HX_a^2-HX_c^2$. By lemma in #3 we have $HX_a^2-HX_c^2 = \frac{l_a^2-l_c^2}{4}$. By median's formula $M_aX_a^2=\frac{1}{4}(2m_a^2+2M_aL_a^2-l_a^2)$ and by Pythagorean theorem $M_aI^2=M_aK_a^2+r^2$, there $K_1 \in BC$ and $IK_a \perp BC$. Also, $l_a,l_b,l_c$ we can find by Stuart's formula. The rest is really not so interesting: of course, now we know all lengths of segments. But if someone will have questions I will write more.