Vertices of a variable triangle are
$$(3,4)\\ (5\cos\theta,5\sin\theta) \\ (5\sin\theta,-5\cos\theta) $$
where $\theta \in \mathbb R$. Given that the orthocenter of this triangle traces a conic, evaluate its eccentricity.

I was able to find the locus after three long pages of cumbersome calculation. I found the equations of two altitudes of this variable triangle using point slope form of equation of a straight and then solved the two lines to get the orthocenter. However, the equation turned out to be of a non standard conic. I evaluated its $\Delta$ to find that it's an ellipse, but I don't know how to find the eccentricity of a general ellipse.

Moreover, there must be a more elegant way of doing this since the questions in my worksheet are to be solved within $5$ to $6$ minutes each but this took way long using my approach.

Please Help!


Note that the vertices of the triangle lie on


which is a circle of radius $5$ units and centered at origin.

Now, the circumcenter of this variable triangle is the origin, i.e,


Also, the centroid of this variable triangle is

$\text{G}\equiv\left(\dfrac{5\sin\theta + 5\cos\theta + 3}{3},\dfrac{5\sin\theta - 5\cos\theta + 4}{3}\right)$

Finally, since $\text{OG}:\text{GH}=1:2$

enter image description here

where $\text{H}$ is the orthocenter, we have

$3\left(\dfrac{5\sin\theta+5\cos\theta+3}{3}-2 \times 0 \right) = x$

$\implies x = 5\sin\theta+5\cos\theta + 3$



where $x$ and $y$ are the co-ordinates of $\text{H}$

Solving the above equations for $\sin\theta$ and $\cos\theta$, we have,



Thus, the locus is


which, on expanding gives


which is an equation of a circle.

Thus eccentricity $\boxed {e=0}$

  • 1
    $\begingroup$ Simply splendid! I had made a calculation error since I was getting an ellipse as the locus but the answer is indeed $e=0$. Bravo! $\endgroup$ – Henry Durham Dec 7 '14 at 15:37

Note that all points lie on a circle $$x^2+y^2+5^2=25$$

So the center must be $(0,0)$. Also, the centroid is $$\left( \frac{3+5\sin\theta+5\cos\theta}{3}, \frac{4-5\sin\theta+5\cos\theta}{3}\right)$$

As we know, $O,G,H$ lie on Euler's line, and: $$\frac{OG}{GH}=\frac{1}{2}$$

$$x_0=\frac{\alpha}{3}, y_0=\frac{\beta}{3} \hspace{10pt}\Rightarrow\hspace{10pt} H\text{ is }\left( 3+5\sin\theta+5\cos\theta3,4-5\sin\theta+5\cos\theta\right)$$

Squaring and adding, $$\frac{(\alpha - 3)^2}{25} + \frac{(\beta - 4)^2}{25}=2 \hspace{10pt}\Rightarrow\hspace{10pt} \text{Circle, } e = 0$$

  • 2
    $\begingroup$ "So the orthocenter must be $(0,0)$." I think it should be circumcenter instead of orthocenter. $\endgroup$ – Henry Durham Dec 7 '14 at 15:41
  • 2
    $\begingroup$ @ADG Please consider using $\LaTeX$ $\endgroup$ – Aditya Hase Dec 7 '14 at 16:19
  • 1
    $\begingroup$ @ADG, I have added TeX to your post. Please make sure I've translated everything correctly, particularly the parts which are off camera. $\endgroup$ – Alexander Gruber Dec 11 '14 at 2:19
  • $\begingroup$ @AlexanderGruber thanks for your effort and sorry for my laziness $\endgroup$ – RE60K Dec 11 '14 at 3:58

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