View topic - Centroids of triangles inscribed/circumscribed to 2 circles. • Art of Problem Solving

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freemind

Posts: 337
Location: MIT or Moldova
Moldova, Republic of

Posted: Mar 03, 2008, 9:04 am • # 1

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

Posts: 777
Location: Islamic Republic of Iran
Iran, Islamic Republic of

Posted: Mar 04, 2008, 12:20 pm • # 2

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 Posts: 595	Posted: Mar 29, 2008, 7:02 am • # 3 Can you explain with more details your solution,please?I am weak at geometry and I do not know many things.Thanks

Amir.S

Posts: 777
Location: Islamic Republic of Iran
Iran, Islamic Republic of

Posted: Mar 29, 2008, 3:28 pm • # 4

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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Say: Tell me if it is from Allah; then you disbelieve in it, who is in greater error than he who is in a prolonged opposition?

Inequalities Master Posts: 595	Posted: Apr 10, 2008, 5:57 am • # 5 Sorry,but I think I need more details.Sorry for the late answer.

freemind

Posts: 337
Location: MIT or Moldova
Moldova, Republic of

Posted: Apr 11, 2008, 10:11 am • # 6

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
$3E = (1 - \lambda)\left [IA^2 + IB^2 + IC^2 - \frac13(a^2 + b^2 + c^2)\right] + \lambda\left[3R^2 - \frac13(a^2 + b^2 + c^2)\...$ .

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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Centroids of triangles inscribed/circumscribed to 2 circles.

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