Let ABC be a right-angled isosceles triangle where AB = BC = a. Assume that C is its centroid and I is its incenter. Find, in terms of a, the distance between C and I.

Answer : $CI= \frac{{a \cdot (3\sqrt{2}-4})}{12}$

How to find it ?

  • $\begingroup$ You began by designating the point C to be one of the vertices of triangle ABC, so you are not free to label the centroid as C. There are plenty more letters of the alphabet. Use them. $\endgroup$ – David H Dec 31 '13 at 10:21

If Coordinate Geometry is allowed, WLOG we can assume $\displaystyle B(0,0),A(a,0), C(0,a)$

So, the equation of $AC$ will be $\displaystyle\frac xa+\frac ya=1\iff x+y-a=0$

So, the centroid will be $\displaystyle C\left(\frac a3,\frac a3\right)$

Now, if $I(p,q)$ then we have the perpendicular distance of $I$ from $AB,BC,CA$ will be same

$\displaystyle\implies |p|=|q|=\frac{|p+q-a|}{\sqrt{1^2+2^2}} $

If $\displaystyle p,q>0,p=q=\frac{2p-a}{\sqrt2}\implies (2-\sqrt2)p=a\iff p=\frac a{2+\sqrt2}$

Now, $IC=|\sqrt{(p-a)^2+(q-a)^2}|=\sqrt2|p-a|$

Can you take it from here?

  • $\begingroup$ I think so ! Thank you ! +1 $\endgroup$ – Ewin Dec 31 '13 at 17:00
  • $\begingroup$ @Ewin, my pleasure. $\endgroup$ – lab bhattacharjee Dec 31 '13 at 18:04

I'm not seeing how to get the answer you posted, but here's my try.

Take the triangle to have vertices $(\pm \frac{\sqrt{2}a}{2}, 0), (0, \frac{\sqrt{2}a}{2})$.

The centroid and the incenter will both lie on the $y$-axis by symmetry.

The centroid is the average of the vertices: $(0, \frac{ \sqrt{2} a}{6})$.

By Heron's formula, the radius of the incircle is $$r = \frac{2K}{P} = a \left(1-\frac{\sqrt{2}}{2}\right),$$

where $K = a^2/2$ is the area of the triangle and $P = (2+\sqrt{2})a$ is the perimeter.

So the distance between these is

$$\frac{3 - \sqrt{2}}{3}a.$$


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