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I know how to do it in normal Euclid geometry, but is it possible to do it with vector algebra?

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The translation of the claim is: Given $a,b,c$ there exists $h$ such that $\langle h-a,c-b\rangle=\langle h-b,a-c\rangle=\langle h-c,b-a\rangle=0$. Or: If $h$ is such that $\langle h-a,c-b\rangle=\langle h-b,a-c\rangle=0$, then also $\langle h-c,b-a\rangle=0$. Can you find a proof? You may assume that $h=xa+yb+zc$ with $x,y,z\in\mathbb R$ and $x+y+z=1$. –  Hagen von Eitzen Dec 10 '12 at 16:44
@HagenvonEitzen Thanks, but I don't quite follow your notation, could you please explain further? –  qed Dec 10 '12 at 17:27
Ah, maybe the use of $\langle,\rangle$ for scalar (or dot) product is unusual for you ... –  Hagen von Eitzen Dec 10 '12 at 17:59
@HagenvonEitzen, Oh, that's dot product, i see. Thanks! –  qed Dec 10 '12 at 18:48

2 Answers 2

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Let's use the figure above in our approach.

Let be $|\vec{b}|=b$, $|\vec{c}|=c$, and $\vec{b}\cdot\vec{c}= m$. The orthocenter can be calculated in two ways: $$O=A+\overrightarrow{AB}+\lambda\overrightarrow{HB}\quad (1)$$ or $$O=A+\overrightarrow{AC}+\mu\overrightarrow{JC}\quad (2)$$ But $$\overrightarrow{HB}=\vec{c}-[\frac{(\vec{b}\cdot\vec{c})}{b}]\frac{\vec{b}}{b}\quad (3)$$ and $$\overrightarrow{JC}=\vec{b}-[\frac{(\vec{b}\cdot\vec{c})}{c}]\frac{\vec{c}}{c}\quad (4)$$ If we substitute $(3)$ and $(4)$ in $(1)$ and $(2)$, we get: $$O=A+\vec{c}+\lambda[\vec{c}-(\frac{m}{b^2})\vec{b}]\quad (5)$$ and $$O=A+\vec{b}+\mu[\vec{b}-(\frac{m}{c^2})\vec{c}]\quad (6)$$ Using equations $(5)$ and $(6)$ we get: $$\vec{c}+\lambda[\vec{c}-(\frac{m}{b^2})\vec{b}]=\vec{b}+\mu[\vec{b}-(\frac{m}{c^2})\vec{c}]\Rightarrow$$


As $\vec{b}$ and $\vec{c}$ are linearly independent we can solve equation $(7)$ and we get: $$\mu=\frac{(m-b^2)}{c^2b^2-m^2}c^2$$

Now we can express $\overrightarrow{AO}$ as


If $\overrightarrow{AO}\cdot\overrightarrow{BC}=0$ then we can conclude that the three altitudes are concurrent.

So $$\overrightarrow{AO}\cdot\overrightarrow{BC}=(\vec{b}+\frac{(m-b^2)}{c^2b^2-m^2}c^2[\vec{b}-(\frac{m}{c^2})\vec{c}])\cdot(\vec{b}-\vec{c})\Rightarrow$$

$$\Rightarrow \overrightarrow{AO}\cdot\overrightarrow{BC}=b^2+\frac{(m-b^2)}{c^2b^2-m^2}c^2(b^2-\frac{m^2}{c^2})-m+\frac{(m-b^2)}{c^2b^2-m^2}c^2(-m+m)\Rightarrow$$

$$\Rightarrow \overrightarrow{AO}\cdot\overrightarrow{BC}=b^2+(m-b^2)-m+0\Rightarrow$$

$$\Rightarrow \overrightarrow{AO}\cdot\overrightarrow{BC}=0$$

Therefore the three altitudes are concurrent at point $O$.

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Denote the three vertices of the triangle by three column vectors $\mathbf{u}, \mathbf{v}, \mathbf{w} \ (\in\mathbb{R}^2)$ and the orthocenter by $\mathbf{x}$. By definition, $\mathbf{x}$ must satisfies $$ \begin{cases} (\mathbf{v}-\mathbf{w})\cdot(\mathbf{x}-\mathbf{u})=0\\ (\mathbf{w}-\mathbf{u})\cdot(\mathbf{x}-\mathbf{v})=0\\ (\mathbf{u}-\mathbf{v})\cdot(\mathbf{x}-\mathbf{w})=0 \end{cases} \ \Rightarrow \begin{cases} (\mathbf{v}-\mathbf{w})\cdot\mathbf{x}=(\mathbf{v}-\mathbf{w})\cdot\mathbf{u}\\ (\mathbf{w}-\mathbf{u})\cdot\mathbf{x}=(\mathbf{w}-\mathbf{u})\cdot\mathbf{v}\\ (\mathbf{u}-\mathbf{v})\cdot\mathbf{x}=(\mathbf{u}-\mathbf{v})\cdot\mathbf{w} \end{cases} $$ This can be rewritten in the matrix form of $A\mathbf{x}=\mathbf{b}$: $$ \underbrace{\begin{bmatrix}(\mathbf{v}-\mathbf{w})^T\\ (\mathbf{w}-\mathbf{u})^T\\ (\mathbf{u}-\mathbf{v})^T\end{bmatrix}}_{A} \ \mathbf{x} = \underbrace{\begin{bmatrix} (\mathbf{v}-\mathbf{w})\cdot\mathbf{u}\\ (\mathbf{w}-\mathbf{u})\cdot\mathbf{v}\\ (\mathbf{u}-\mathbf{v})\cdot\mathbf{w} \end{bmatrix}}_{\mathbf{b}} $$ So, the remaining question is, does there exist a unique solution for $A\mathbf{x}=\mathbf{b}$? Note that $A$ is a 3x2 matrix and $\mathbf{x}$ is a 2-vector, so the above system has three equations in two unknowns, i.e it is overdetermined. Can you show that each one of these three equations is a linear combination of the other two? Is it possible to remove one row from $A$, so that the remaining 2x2 submatrix is invertible?

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