Centroid Distance Identity in Triangle

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Centroid Distance Identity in Triangle

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zeta1

#1 h Apr 16, 2025, 12:28 PM • 1 Y

Y by cubres

Let M be any point inside triangle ABC, and let G be the centroid of triangle ABC. Prove that:

$|MA|^2 + |MB|^2 + |MC|^2 = |GA|^2 + |GB|^2 + |GC|^2 + 3|MG|^2$

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#2 h Apr 16, 2025, 12:59 PM • 2 Y

Y by zeta1, cubres

Let's translate in the complex plane such that $G=0$ . Then $\sum |z-a_k|^2=\sum (|z|^2+|a_k|^2-2{\rm Re}(a_k\overline{z}))=3|z|^2 +\sum |a_k|^2-2{\rm Re} ((\sum a_k)\overline{z})=3|z|^2 +\sum |a_k|^2$ .
It doesn't matter if $M$ is inside the triangle or not.

This post has been edited 1 time. Last edited by alexheinis, Apr 16, 2025, 1:01 PM

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zeta1

#3 h Apr 16, 2025, 1:18 PM • 1 Y

Y by cubres

Thank you for the solution. I haven’t learned the complex plane yet, so I was wondering if there’s a more geometric way to prove it — maybe using triangle properties or classical geometry???

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10588 posts

#4 h Apr 16, 2025, 2:33 PM • 2 Y

Y by zeta1, cubres

Let's avoid complex notation but again we may assume that $G=(0,0)$ . Then $AM^2+BM^2+CM^2=(x-a_1)^2+(y-a_2)^2+(x-b_1)^2+(y-b_2)^2+(x-c_1)^2+(y-c_2)^2=$
$3(x^2+y^2)-2(a_1+b_1+c_1)x-2(a_2+b_2+c_2)y+OA^2+OB^2+OC^2=3OM^2+OA^2+OB^2+OC^2$ .

This post has been edited 1 time. Last edited by alexheinis, Apr 16, 2025, 2:34 PM

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zeta1

#5 h Apr 16, 2025, 3:55 PM • 1 Y

Y by cubres

Thank you! This is exactly the kind of solution I was looking for

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DottedCaculator

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DottedCaculator

#6 h Apr 16, 2025, 4:03 PM • 3 Y

Y by zeta1, mathfan2020, compoly2010

Follows from the parallel axis theorem.

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#7 h Apr 17, 2025, 1:16 AM

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DottedCaculator wrote:

Follows from the parallel axis theorem.

:flushed:

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