Centroids of triangles inscribed/circumscribed to 2 circles.

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Centroids of triangles inscribed/circumscribed to 2 circles.

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Source: Moldova 2008 IMO-BMO First TST Problem 3

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337 posts

#1 h Mar 3, 2008, 4:04 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $\Gamma(I,r)$ and $\Gamma(O,R)$ denote the incircle and circumcircle, respectively, of a triangle $ABC$ . Consider all the triangels $A_iB_iC_i$ which are simultaneously inscribed in $\Gamma(O,R)$ and circumscribed to $\Gamma(I,r)$ . Prove that the centroids of these triangles are concyclic.

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Amir.S

#2 h Mar 4, 2008, 7:20 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

the nine-point circles of $A_iB_iC_i$ are tangent to incircle with a constant Radius hence the locus of $H$ (homothecy point of Circumcircle and ninpoint circle) is a cirlce hence the locus of centroids is a cricle.

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Inequalities Master

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Inequalities Master

#3 h Mar 29, 2008, 2:02 PM • 1 Y

Y by Adventure10

Can you explain with more details your solution,please?I am weak at geometry and I do not know many things.Thanks

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Amir.S

#4 h Mar 29, 2008, 10:28 PM • 2 Y

Y by Adventure10 and 1 other user

Inequalities Master wrote:

Can you explain with more details your solution,please?I am weak at geometry and I do not know many things.Thanks

cause the nine point circle is circum circle of median triangle its radius is half of circumradius, hence it's constant ,also nine point circle is tangent to incircle so the locus of its center $N$ is a circle.also we have $\frac{ON}{OG}=\frac{3}{2}$ ,hence the locus of G is a circle.I think i didn't explained it enough in details,if you didn't get it.say me to explain more in details

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Inequalities Master

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Inequalities Master

#5 h Apr 10, 2008, 12:57 PM • 2 Y

Y by Adventure10, Mango247

Sorry,but I think I need more details.Sorry for the late answer.

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#6 h Apr 11, 2008, 5:11 PM • 2 Y

Y by Adventure10, Mango247

Here's my solution from the TST:

Apply a symmetry to the triangle wrt to $OI$ . Clearly, the new triangle, if different from the original will also be circumscribed/inscribed to $\Gamma(I,r)$ and $\Gamma(O,R)$ respectively. Hence for the problem to be true, we need to show that all centroids lie on a circle of center lying on $OI$ . Assume we have found this center $M$ . Moreover assume it lies on $(OI)$ . Let $IM/IO = \lambda$ . For $MG$ to be a constant value (not depending on the chosen triangle), we are looking to express $MG$ as a function only of $R$ and $r$ (not depending on other elements of the triangle). Since $R$ and $r$ remain constant, the solution will be done. In fact we are looking for a constant $\lambda$ so that $MG = f(R,r)$ .

So here we go:
From Stewart Relation $GM^2\cdot OI = GI^2\cdot MO + GO^2\cdot IM - IM\cdot MO\cdot IO$ , or $GM^2 = (1 - \lambda)GI^2 + \lambda GO^2 - OI^2\lambda(1 - \lambda)$ .
Well, $OI^2\lambda(1 - \lambda) = g(R,r)$ , hence we need only look at $E = (1 - \lambda)GI^2 + \lambda GO^2$ . Use Leibniz to obtain
$3E = (1 - \lambda)\left [IA^2 + IB^2 + IC^2 - \frac13(a^2 + b^2 + c^2)\right] + \lambda\left[3R^2 - \frac13(a^2 + b^2 + c^2)\right]$ .

Hence we are looking for a $\lambda$ so that $E' = 3(1 - \lambda)(IA^2 + IB^2 + IC^2) - (a^2 + b^2 + c^2)$ to equal $g(R,r)$ . Let $3(1 - \lambda) = \alpha$ . Note that $IA^2 + IB^2 + IC^2 = 3r^2 + (p - a)^2 + (p - b)^2 + (p - c)^2$ , so
$E' = 3\alpha r^2 + \alpha[(p - a)^2 + (p - b)^2 + (p - c)^2] - (a^2 + b^2 + c^2)$ . Finally, we must find an $\alpha$ so that $F = \alpha[(p - a)^2 + (p - b)^2 + (p - c)^2] - (a^2 + b^2 + c^2) = h(R,r)$ .
Using $a + b + c = 2p$ and $ab + bc + ca = r^2 + p^2 + 4Rr$ , we get
$F = - \alpha p^2 + (\alpha - 1)(\sum a^2) = - \alpha p^2 + (\alpha - 1)[4p^2 - 2(r^2 + p^2 + 4Rr)]$ . Further $F = Q(R,r) + (\alpha - 2)p^2$ .
So taking $\alpha = 2$ , works. That means, $\lambda = \frac13$ . Hence, all centroids lie on a circle with center $M\in(OI)$ , so that $IO = 3IM$ . Moreover, tracking back the 'ignored' functions of $R$ and $r$ , we easily get $GM^2 = \frac {(R - 2r)^2}9 = \frac49\left(\frac R2 - r\right)^2$ .
Actually, as I now realize, using $\omega O = 3\omega G$ , the above is a proof for $I\omega = \frac R2 - r$ , from which follows Feuerbach's Theorem: the Nine-Point Circle is tangent to the incircle (because radius of 9-point circle is $\frac R2$ ).

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