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My friend gave me a question I tried my best, but I'm low on triangle concept.

Points $ O, A, B, C... $ are shown in the figure where $ OA=2AB=4BC=...$ and so on. Let $A$ be the centroid of a triangle whose orthocentre and circumcentre are $(2,4)$ and $(\frac72,\frac52)$, respectively, if an insect starts moving from the point $O=(0,0)$ along the straight line in zig zag fashions and terminates ultimately at point $P(a,b)$, then find the value of $(a+b)$.

I tried using the collinearity property of centroid, circumcentre and orhtocentre, and the distance property also, but reached nowhere. Please help.

  • $\begingroup$ Which line segments does the bug crawl on? AB, BC, CD,...? $\endgroup$ May 27, 2016 at 5:25
  • $\begingroup$ It crawls on the path shown in the figure, and the figure is also extendable. $\endgroup$ May 27, 2016 at 5:32
  • $\begingroup$ Could you use trigonometry? I don't see how this problem is possible without putting 45 degrees into play. $\endgroup$ May 27, 2016 at 5:42
  • $\begingroup$ yup, I tried using trigonometry too. $\endgroup$ May 27, 2016 at 5:44
  • $\begingroup$ Since $(a,b)=(2 r \sqrt 2, 2 r \sqrt 2 /3) $ when $A=(r,r) $, the problem is to find $r$ , given the conditions on $A.$ $\endgroup$ May 27, 2016 at 5:45

3 Answers 3


From basic trigonometry we get that that the $x$ coordinates tends to...

$$a=\lim_{N \to \infty} \sum_{n=0}^{N} \frac{d}{2^n}\cos (45)$$

Here $\frac{d}{2^n}$ denotes the magnitude of the $n+1$th hypotenuse (we start the sum at $0$ that is why it's a bit weird with $n+1$). How did I get this? You implied the distances (hypotenuses) went $d,d/2,d/4$ and I assumed this was a geometric progression. Where here $d$ is magnitude of the first hypotenuse $n=0$, aka the distance between the origin and point A. Using the assumption you will get the formula for the hypotenuse in terms of $n$. Which you can then use to find the horizontal components of each component then add them up to get $a$ as they are all positive.

Using the same method (but with vertical components alternating direction/sign), the $y$ coordinate tends to:

$$b=\lim_{N \to \infty} \sum_{n=0}^{N} \frac{d}{2^n}\sin (45) (-1)^{n}$$

Which converges to a positive number.

Note $\cos (45) = \sin (45) = \frac{1}{\sqrt{2}}$

For the first sum,

$$\sum_{n=0}^{\infty} (\frac{1}{2})^n=\frac{1}{1-(1/2)}=2$$

So as a result $a=\frac{2d}{\sqrt{2}}=\sqrt{2} d$.

For the second one,

$$\sum_{n=0}^{\infty}\frac{ (-1)^{n}}{2^n}=\sum_{n=0}^{\infty} (\frac{-1}{2})^n=\frac{1}{1-(-1/2)}=\frac{2}{3}$$

From there,

$$b=\lim_{N \to \infty} \sum_{n=0}^{N} \frac{d}{2^n}\sin (45) (-1)^{n}=\frac{2d}{3 \sqrt{2}}=\frac{\sqrt{2}d}{3}$$


Now all that is left is to find $d$....

Finding $d$ has to do with the Euler line @almagest, which passes through the orthocenter, centroid, and circumcenter. This and the fact that $A$ forms a 45-45-90 isosceles triangle with the $x$ axis so it's coordinates are $(x_0,y_0)=(x_0,x_0)$. Let's find a formula that describes the Euler line given our two points $(2,4)$ and $(3.5,2.5)$.

The slope of the line is:


So the equation of the line is:


Substituting the point $A=(x_0,x_0)$ in we get:





And thus,



Plugging $d$ back into the formulas we got for $a$ and $b$ you may get that $P=(6,2)$.

  • 1
    $\begingroup$ Check your final step, $\displaystyle\frac{4(3\sqrt{2})\sqrt{2}}{3}=8$ $\endgroup$
    – Hrhm
    May 27, 2016 at 13:41

A less complicated method that builds off of @Ahmed S. Attaalla's solution.

Because of the Euler line, we know that the centroid is 1/3 of the way between the circumcenter and the orthocenter.

Euler Line

In other words, $A=\displaystyle\frac{2*\left (\frac{7}{2},\frac{5}{2}\right)+(2,4)}{3}=\left(3,3\right)$, because $A$ is $2$ times closer to the circumcenter than the orthocenter (and on the same line).

Now, the $x$-coordinate of point P is $\displaystyle \sum_{n=0}^{\infty}\frac{3}{2^n}=6$, and the $y$-coordinate is $\displaystyle \sum_{n=0}^{\infty}\frac{3}{(-2)^n}=2.$

So the final answer is $6+2=8$.


If the two vectors are $v_1$ and $v_2$, your description is something like:

$$v_{tot} = v_1 + \frac12 v_2 + \frac14 v_1 + \frac18 v_2 \dots$$

This is equivalent to:

$$\begin{align} v_{tot} & = (v_1 + \frac12 v_2) + \frac14(v_1 + \frac12 v_2) \dots\\ &= (v_1 + \frac12 v_2)(1 + \frac14 + \frac1{16}\dots)\\ \end{align}$$

The geometric series is equal to $\frac{1}{1 - \frac14} = \frac43$. So once you figure out $v_1$ and $v_2$, your final point will be

$$\frac43(v_1 + \frac12v_2)$$

  • $\begingroup$ You assumed that $v_1=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ and that $v_2=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$. You have to find $v_1$ and $v_2$ based on the fact that coordinate A is the centroid of a triangle with a given orthocenter and circumcenter. $\endgroup$
    – Hrhm
    May 27, 2016 at 13:50
  • $\begingroup$ Yes, my $v_1$ and $v_2$ ideas were wrong. I've cut that part out. $\endgroup$ May 28, 2016 at 15:03

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