Very simple!!

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Rushil

1592 posts

Rushil

#1 h Oct 14, 2005, 2:52 PM • 4 Y

Y by Adventure10, Adventure10, jhu08, Mango247

If the circumcenter and centroid of a triangle coincide, prove that it must be equilateral.

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frt

1294 posts

frt

#2 h Oct 14, 2005, 10:25 PM • 6 Y

Y by Samujjal101, quick_mafs1331, Rushil_india, Adventure10, jhu08, Mango247

One way is to use coordinate geometry...
Click to reveal hidden text
P.S.- Yeah, mine is really long

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Valentin Vornicu

7301 posts

Valentin Vornicu

#3 h Oct 14, 2005, 10:30 PM • 6 Y

Y by coolcheetah157, Adventure10, jhu08, DofL, Mango247, GottaLuvMATH

A much simpler way ...

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tanuagg13

137 posts

tanuagg13

#4 h Mar 13, 2014, 10:02 PM • 5 Y

Y by coolcheetah157, jhu08, Bgmi, Adventure10, SatisfiedMagma

Note that it is a well-known fact that the circumcenter is the point of intersection of a triangle's perpendicular bisectors and that the centroid is the point of intersection of the triangle's medians. If the centroid and circumcenter are concurrent, the medians must also be perpendicular bisectors. This implies that the triangle is isosceles when viewed from any base, as the perpendicular bisector bisects the base, so since the triangle is isosceles no matter from which side we look at it as isosceles, the triangle must be equilateral due to symmetry.

Is this rigorous enough or would this have to be tightened? I think it might be the latter...

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aayush-srivastava

137 posts

aayush-srivastava

#5 h Nov 10, 2015, 9:12 AM • 8 Y

Y by Abhinandan18, Evan-Chen123, bond98, hockey10, jhu08, Bgmi, Adventure10, Mango247

I think the simplest approach is $complex$ $numbers$ . Set $A,B,C$ as $a,b,c$ resp. also set $circumcentre$ as the $origin$ of the complex plane.
clearly $H=a+b+c$ and $G=a+b+c/3$ .(Use the fact that $HG:GC=2:1$ )... Now as $G=C$ is given we get (a+b+c=0)!!!
this implies centroid and orthocentre coincide..
the result follows by congruences.
QED

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smo

179 posts

smo

#6 h Jul 25, 2016, 1:26 PM • 3 Y

Y by jhu08, Bgmi, Adventure10

What I think is the simplest approach:
Let $G$ be the centroid and circumcenter, and $X ,Y, Z$ the midpoints of $BC , AC, AB$ respectively. Using pythagoras in $\Delta GXB$ and $\Delta GXC$ we get $GB = GC$ . Analogously we have $GC = GA = GB$ . Since medians trisect each other we also have $GZ = GX = GY$ . Thus we have all medians equal, and the triangle is equilateral.

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daniellionyang

1840 posts

daniellionyang

#7 h Jul 25, 2016, 5:13 PM • 3 Y

Y by jhu08, Bgmi, Adventure10

use barycentric coordinates.

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george_54

1585 posts

george_54

#8 h Jul 29, 2016, 6:48 PM • 6 Y

Y by Hexagrammum16, jhu08, Bgmi, abvk1718, Adventure10, Mango247

if $O$ is the circumcenter, then:
$OA = OB = OC \Leftrightarrow \frac{2}{3}{m_a} = \frac{2}{3}{m_b} = \frac{2}{3}{m_c} \Leftrightarrow$ $\boxed{a=b=c}$

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Feynmann

8 posts

Feynmann

#9 h Jan 25, 2018, 10:12 PM • 4 Y

Y by jhu08, Bgmi, Adventure10, Mango247

Recall that the area of the triangle formed by the centroid and any other two vertices of a triangle is one third of the area of the original triangle. This gives
$\frac{1}{2}R^2 \sin (2A)= \frac{1}{2}R^2 \sin(2B)=\frac{1}{2}R^2\sin(2C),$ which implies
$A=B=C$ .

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Top10AnimeBetrayal

139 posts

Top10AnimeBetrayal

#10 h Jan 25, 2018, 10:28 PM • 7 Y

Y by MaxSteinberg, programjames1, k2005, jhu08, Bgmi, Adventure10, Mango247

How do people find these posts that were posted 12 years ago.....

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ythomashu

6322 posts

ythomashu

#11 h Jan 26, 2018, 3:44 PM • 4 Y

Y by Mathphile01, jhu08, Bgmi, Adventure10

Scroll down for a long time, just like if you scroll down long enough in $\pi$ , you will eventually find 6 9s in a row.

This post has been edited 1 time. Last edited by ythomashu, Jan 26, 2018, 3:45 PM

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AlastorMoody

2125 posts

AlastorMoody

#12 h Aug 15, 2018, 5:18 PM • 6 Y

Y by MathPassionForever, Rushil_india, jhu08, Bgmi, Adventure10, Mango247

No they are preparing for RMO
So they bump in past year questions

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AlastorMoody

2125 posts

AlastorMoody

#13 h Nov 4, 2018, 7:44 AM • 3 Y

Y by jhu08, Bgmi, Adventure10

We know that, $G\equiv (1:1:1)$ and $O\equiv (\sin 2A :\sin 2B :\sin 2C)$
$\implies \sin 2A= \sin 2B = \sin 2C \implies \angle A = \angle B = \angle C \text{ (Since, A+B+C = 180)} \implies \boxed{\angle A= \angle B = \angle C =60^{\circ}}$ Hence, Equilateral

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alexheinis

10513 posts

alexheinis

#14 h Nov 4, 2018, 2:32 PM • 3 Y

Y by jhu08, Bgmi, Adventure10

Let $l$ be the perpendicular bisector of $BC$ and $A'$ the midpoint of $BC$ . Then $Z=M\in l, A'\in l$ hence $l=ZA'$ which is the median from $A$ . Hence $A\in l$ and $AB=AC$ . In the same way $BA=BC$ .

This post has been edited 1 time. Last edited by alexheinis, Nov 4, 2018, 2:32 PM

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dchenmathcounts

2443 posts

dchenmathcounts

#15 h Nov 4, 2018, 2:49 PM • 3 Y

Y by jhu08, Bgmi, Adventure10

Valentin Vornicu wrote:

A much simpler way ...

Is this necessarily true? This argument can be applied for altitudes, angle bisectors, and medians, but theoretically (I know this doesn't actually happen), you could have the perpendicular bisectors and the medians concur at one point while being six separate line.

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Hexagrammum16

1774 posts

Hexagrammum16

#16 h Feb 13, 2019, 7:33 PM • 4 Y

Y by jhu08, Bgmi, Adventure10, Mango247

Yes, but then they wouldn't be the medians and altitudes of the same triangle

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purplish

299 posts

purplish

#17 h Feb 13, 2019, 7:35 PM • 3 Y

Y by jhu08, Bgmi, Adventure10

dchenmathcounts wrote:

Valentin Vornicu wrote:

A much simpler way ...

Is this necessarily true? This argument can be applied for altitudes, angle bisectors, and medians, but theoretically (I know this doesn't actually happen), you could have the perpendicular bisectors and the medians concur at one point while being six separate line.

The reason this argument works in this case is because the midpoint of a side is on both its perpendicular bisector and the median to it.

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denery

180 posts

denery

#18 h Oct 27, 2020, 2:34 PM • 5 Y

Y by jhu08, Bgmi, Mango247, Mango247, Mango247

the idea or the motivation behind the problem any how is because of incircle it is the point of concurrence of angular bisectors which make angles equal and due to median which bisect sides makes sides equal and due to sas criterion we get congruent triangles on aplying cpct we get it to be equilateral

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IAmTheHazard

5000 posts

IAmTheHazard

#19 h Oct 27, 2020, 2:38 PM • 3 Y

Y by jhu08, Bgmi, Mango247

dchenmathcounts wrote:

Valentin Vornicu wrote:

A much simpler way ...

Is this necessarily true? This argument can be applied for altitudes, angle bisectors, and medians, but theoretically (I know this doesn't actually happen), you could have the perpendicular bisectors and the medians concur at one point while being six separate line.

Not in this case, since a line is defined by two points, and each of the perpendicular bisectors/medians pass through two defined points (the common point where all 6 intersect, as well as one vertex of the triangle)

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Saucepan_man02

1300 posts

Saucepan_man02

#20 h Sep 17, 2021, 5:23 AM • 3 Y

Y by jhu08, Bgmi, Sreepranad

this is from first RMO of INDIA

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Bratin_Dasgupta

598 posts

Bratin_Dasgupta

#21 h Sep 17, 2021, 6:06 AM • 1 Y

Y by Bgmi

The main idea is to note the fact that the Altitudes bisect the sides of a triangle $\iff$ it is an equilateral triangle

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OlympusHero

17020 posts

OlympusHero

#22 h Sep 18, 2021, 3:40 AM • 2 Y

Y by jhu08, Bgmi

Solution

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rmoclear

7 posts

rmoclear

#23 h Sep 18, 2021, 4:08 AM • 1 Y

Y by jhu08

simple just use coordinate geometry
its simple

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Aryamathematics

8 posts

Aryamathematics

#24 h Sep 18, 2023, 1:06 PM

Y by

This q is Ovious

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Aryamathematics

8 posts

Aryamathematics

#25 h Sep 18, 2023, 1:12 PM

Y by

If o = g then o=g=h. So AB=BC=CA.

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SomeonecoolLovesMaths

3162 posts

SomeonecoolLovesMaths

#26 h Sep 22, 2024, 3:39 PM

Y by

$[asy] import olympiad; pair B = (0,0); pair C = (10,0); pair A = (5,8.66025403784); pair D = (B+C)/2; pair E = (A+C)/2; pair F = (B+A)/2; draw(A--B--C--A); dot(A^^B^^C^^D^^E^^F); pair OG= centroid(A,B,C); draw(A--D); draw(B--E); draw(C--F); label ("$A$",A,N); label ("$B$",B,SW); label ("$C$",C,SE); label ("$OG$",OG,NW); label ("$D$",D,S); label ("$E$",E,NE); label ("$F$",F,NW); dot(OG); [/asy]$ Let the triangle be $\bigtriangleup ABC$ . $AG:GD = 2:1 = AO:GD = BO:GD$ . As $GD \perp BD$ . $\sin (\angle GBD) = \frac{1}{2}$ . Thus $\angle GBD = 30^{\circ}$ . Similarly $\angle GBF = 30^{\circ}$ . Thus $\angle ABC = 60^{\circ}$ . We had no assumptions thus every anglle is $60^{\circ}$ . Thus the triangle is equilateral.

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Jupiterballs

35 posts

Jupiterballs

#27 h Mar 7, 2025, 4:34 PM

Y by

Just use the fact that $GA = GB = GC$ , to get that the medians are equal to end it all

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