analytic geometrygeometry
Let $p$, $q$, $r$, be the slopes of sides of a triangle, and $m$ the slope of Euler line of the triangle. Then we may say
$$m= -\frac{3+pq+pr+qr}{p+q+r+3pqr}$$
provided that the denominator isn't zero. If it's zero the euler line of the triangle is parallel to $y$ axis.
I would like hints for proving this formula.
The slope measure form 0 to 1 is basically an angle (lets call it $\alpha$), the other one is the slope, lets call it $m$. The relation is
$$ m = \tan\left( \alpha \cdot \frac{\pi}{2} \right)$$ if you are calculating with radians or
$$ m = \tan\left( \alpha \cdot 90° \right)$$ if you are calculating in degrees.
Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$, respectively.
Let $X$ be a point on $DE$ such that $\angle DAX=EAX=30^{\circ}$ ($AX$ is the angle bisector).
Observe, $$\angle DAH=\angle EAO=90^{\circ}-\angle B\implies \angle HAX=\angle OAX. $$
$$AH=2R\cos\angle A=2R\cos60^{\circ}=R=AO.$$ Hence, $\triangle AOH$ is isosceles and the angle bisector, $AX$ is perpendicular to $DE$.
Therefore, $AD=AE$ and $\angle A=60^{\circ}$ imlplies that $\triangle AED$ is equilateral.
Best Answer
You can use vectors to prove this. Let $ABC$ be the triangle, $p,q,r$ slopes of $BC,AC,AB$, respectively. Note that $\overrightarrow{BC},\overrightarrow{CA},\overrightarrow{AB}$ are respectively parallel to $(1,p),(1,q),(1,r)$. By taking into account the triangle rule $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{0}$, we may assume: \begin{align} \overrightarrow{BC}= (q-r,p(q-r)),\\ \overrightarrow{CA}= (r-p,q(r-p)),\\ \overrightarrow{AB}= (p-q,r(p-q)).\\ \end{align}
Let $T$ be the centroid. We know that $\overrightarrow{AT}= \frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC})= \frac{1}{3}(2p-q-r,pq+pr-2qr)$.
Let $H$ be the orthocenter. $\overrightarrow{AH}$ is parallel to $\overrightarrow{n_{BC}}:(-p(q-r),q-r)$ and $\overrightarrow{BH}$ is parallel to $\overrightarrow{n_{AC}}:(-q(r-p),r-p)$, so we can write: \begin{align} \overrightarrow{AH}=\alpha(-p(q-r),q-r),\\ \overrightarrow{BH}=\beta(-q(r-p),r-p). \end{align} Since $\overrightarrow{AB}= \overrightarrow{AH}-\overrightarrow{BH}$, we have: \begin{align} -\alpha\ p(q-r)+\beta\ q(r-p) = p-q,\\ \alpha(q-r)-\beta(r-p) =r(p-q). \end{align} Multiply the second equation by $q$ and add it to the first one to get: $\alpha(q-p)(q-r)=(qr+1)(p-q)$, wherefrom $\alpha=-\frac{qr+1}{q-r}$. So: $$\overrightarrow{AH}= -\frac{qr+1}{q-r}(-p(q-r),q-r)= (pqr+p, -1-qr).$$ Now $\overrightarrow{HT}= \overrightarrow{AT}- \overrightarrow{AH}=$ $$=\frac{1}{3}(2p-q-r,pq+pr-2qr)-\frac{1}{3}(3pqr+3p, -3-3qr)= \frac{1}{3}(-p-q-r-3pqr,3+pq+qr+rp).$$ The slope of the Euler line is thus equal to: $$-\frac{3+pq+qr+rp}{p+q+r+3pqr}.$$