euclidean-geometrygeometrytrianglesvectors
Given an acute-angled triangle $\Delta ABC$ having it's Orthocentre at $H$ and Circumcentre at $O$. Prove that $\vec{HA} + \vec{HB} + \vec{HC} = 2\vec{HO}$
I realise that $\vec{HO} = \vec{BO} + \vec{HB} = \vec{AO} + \vec{HA} =\vec{CO} + \vec{HC}$ which leads to $3\vec{HO} = (\vec{HA} + \vec{HB} + \vec{HC}) + (\vec{AO} + \vec{BO} + \vec{CO})$
How can I prove that $(\vec{AO} + \vec{BO} + \vec{CO}) = \vec{HO}$ in order to solve the problem?
Thank you for answering!
Let $P$ and $Q$ be, respectively, the orthocenter and circumcenter of $\triangle ABC$. Let $R$ be the intersection of the perpendicular bisector of $\overline{AB}$ with the altitude from $B$; and let $S$ be the intersection of the perpendicular bisector of ${AC}$ with the altitude from $C$.
Introduce $B^\prime = \overleftrightarrow{AB}\cap\overleftrightarrow{QS}$ and $C^\prime = \overleftrightarrow{AC}\cap\overleftrightarrow{QR}$. Then $\triangle ABC^\prime$ and $\triangle AB^\prime C$ are equilateral.
Observe that $R$ is orthocenter, circumcenter, and incenter for $\triangle ABC^\prime$; likewise, $S$ for $\triangle AB^\prime C$. Therefore, $R$ and $S$ lie on the bisector of $\angle A$. A little angle-chasing shows that $\triangle PRS$ and $\triangle QRS$ are equiangular, hence equilateral, so that $\square PSQR$ is a rhombus, and the result follows.
Barycentric coordinates provide a simple method. Let we assume that our triangle is $ABC$ and the side lengths are $a,b,c$. $a,b,c$ can be easily computed from the Pythagorean theorem, and the following barycentric coordinates $$ I=[a,b,c],\qquad O=[a^2(b^2+c^2-a^2),b^2(a^2+c^2-b^2),c^2(a^2+b^2-c^2)]$$ $$ H=\left[\frac{1}{b^2+c^2-a^2},\frac{1}{a^2+c^2-b^2},\frac{1}{a^2+b^2-c^2}\right] $$ that we may summarize as $P=[p_a,p_b,p_c]$, give the vector identity $$ P = \frac{p_a A+p_b B+p_c C}{p_a+p_b+p_c}$$ from which it is straightforward to compute the cartesian coordinates of $P$ from the coordinates of $A,B,C$. It is interesting to point out that Euler's theorem gives a further shortcut, since $3G=A+B+C=2O+H$ allows us to compute the coordinates of $H$ from the coordinates of $O$ (or the opposite) in a very simple way.
Here, as usual, $G,I,O,H$ stand for the centroid, incenter, circumcenter and orthocenter of a triangle. The derivation of their barycentric coordinates is straightforward from the computation of their trilinear coordinates, i.e. from the computation of their distances from the triangle sides in terms of $a,b,c$.
About your second question: up to reflections we may assume that $A$ lies on the positive $x$ axis, $B$ lies on the positive $y$ axis and $C$ lies on the positive $z$ axis. We are claiming that the projection of $O$ on the $ABC$-plane $\pi$ is given by the orthocenter of $ABC$. Let we denote such projection with $P$ and consider the plane $\pi_A$ through $O,A,P$. By minimality of $OP$, $\pi_A$ has to be orthogonal to the $BC$ line: otherwise, it would be possible to move a bit the $\pi_A$ plane and decrease the distance between $O$ and $\pi\cap\pi_A$. It follows that $\pi\cap\pi_A$ is orthogonal to $BC$, and by repeating the same argument we get that $P$ is the orthocenter of $ABC$.
Best Answer
If $G$ is the centroid of the triangle, that relation follows from $$ \vec{AO} + \vec{BO} + \vec{CO}= \vec{AG} + \vec{BG} + \vec{CG}+3\vec{GO} =3\vec{GO}=\vec{HO} $$ where the last one is a well-known equality, arising in the proof for the Euler line, see e.g. https://en.wikipedia.org/wiki/Euler_line#A_vector_proof .