Geometrical inequality involving the centroid and circumcent

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Geometrical inequality involving the centroid and circumcent

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Source: Balkan MO 1996, Problem 1

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Valentin Vornicu

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Valentin Vornicu

#1 h Apr 24, 2006, 11:41 PM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the circumcenter and $G$ be the centroid of a triangle $ABC$ . If $R$ and $r$ are the circumcenter and incenter of the triangle, respectively,
prove that $OG \leq \sqrt{ R ( R - 2r ) } .$
Greece

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#2 h Apr 25, 2006, 12:11 AM • 3 Y

Y by Adventure10, Shinichi-123, Mango247

$OG^2 = R^2 - \frac{1}{9}(a^2+b^2+c^2)$

$R(R-2r) = R^2-2Rr = R^2 - \frac{2R}{s}\cdot \frac{abc}{4R} =R^2 - \frac{abc}{a+b+c}$ since $rs = \frac{abc}{4R}$ is the area of the triangle.

So it reduces to prove:

$(a^2+b^2+c^2)(a+b+c) \geq 9abc$ which is true by AM-GM.

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#3 h Apr 25, 2006, 5:31 PM • 1 Y

Y by Adventure10

Euler`s relation in a triangle gives us $OI^2=R(R-2r)$ then all we have to do is prove that $OG \leq OI$ .And this seems like a well-known inequality(it looks too beautifull).

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#4 h May 3, 2006, 4:27 PM • 2 Y

Y by Adventure10, Mango247

Yes, it is a (fairly) well-known fact that $I$ lies in the circle diameter GH,
and with a diagram showing the Euler line and this circle,
$OG \leq OI$ becomes immediate

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#5 h May 3, 2006, 5:00 PM • 1 Y

Y by Adventure10

well we do use that
$OG^2=R^2- \frac{1}{9}(a^2+b^2+c^2)$
But
$R(R-2r)=R^2-2Rr=R^2-2\frac{2S}{p} \cdot \frac{abc}{4S}=R^2- \frac{abc}{a+b+c}$
So we need to demonstrate that
$\frac{1}{9}(a^2+b^2+c^2) \geq \frac{abc}{a+b+c}$
...

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#6 h May 3, 2006, 5:04 PM • 1 Y

Y by Adventure10

...

fuzzylogic wrote:

$OG^2 = R^2 - \frac{1}{9}(a^2+b^2+c^2)$

$R(R-2r) = R^2-2Rr = R^2 - \frac{2R}{s}\cdot \frac{abc}{4R} =R^2 - \frac{abc}{a+b+c}$ since $rs = \frac{abc}{4R}$ is the area of the triangle.

So it reduces to prove:

$(a^2+b^2+c^2)(a+b+c) \geq 9abc$ which is true by AM-GM.

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#7 h Aug 17, 2006, 7:15 AM • 2 Y

Y by Adventure10, Mango247

tim1234133 wrote:

Yes, it is a (fairly) well-known fact that $I$ lies in the circle diameter GH,
and with a diagram showing the Euler line and this circle,
$OG \leq OI$ becomes immediate

What is the meaning of " $I$ lies in the circle diameter GH"
Could you explain more clearly,thank you.

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#8 h Aug 17, 2006, 10:01 AM • 2 Y

Y by Adventure10, Mango247

On the Euler line,

O (the circumcentre) lies at one end.
H (the orthocentre) lies at the other.
G (the centroid) lies one third of the way along, nearer O.
N (the nine-point centre) lies half-way between O and H.

Imagine the point halfway between G and H (and so two-thirds of the way from O to H). If a circle is drawn, with a centre at this point, through G and H, then GH is its diameter.

It is known that the incentre (I) lies within this circle. (See here for example).
Clearly, I lies on the diameter GH if and only if the triangle is isoceles. As I tends towards the nine-point centre, the triangle tends towards being equilateral. For more on the this circle, and the points that lie within it, see here

Moving back to the actual question, it is clear, since OGH is a diamater of the circle, G is the point on the circle that is closest to O. (Draw a picture of this circle and the Euler line- it should look like a lollipop).
Therefore $OI \geq OG$

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#9 h Aug 17, 2006, 10:19 PM • 1 Y

Y by Adventure10

Remark. It is well-known the $\triangle ABC$ - inequality $p^{2}\ge 3\sum (p-b)(p-c)$ $\Longleftrightarrow$ $p^{2}\ge 3r(4R+r)$ . The proposed inequality $OG\le OI$ is equivalently with the inequality $p^{2}\ge 3\sum (p-b)(p-c)+r(R-2r)$ $\Longleftrightarrow$ $p^{2}\ge 13Rr+r^{2}$ what is stronger than the first.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=53811

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#10 h Jan 24, 2020, 8:13 AM • 1 Y

Y by Adventure10

...........

This post has been edited 1 time. Last edited by Feridimo, Jan 24, 2020, 9:09 AM

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#11 h Jan 24, 2020, 8:15 AM • 1 Y

Y by Adventure10

Feridimo wrote:

https://artofproblemsolving.com/community/c1049667h1991756_distance_between_circumcenter_and_center
$OG\leq R\leq \sqrt{R(R-2r)} \implies 2r\leq R$

are you sure?

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