ABC is an equilateral triangle it's center is m , straight line (L) was drawn passing through the point m and in the plane of the triangle. Prove that "The sum of heights lengths squared drawn from the heads of the triangle on the line (L) equal a constant whatever the line (L) was."

- TrojanPoem

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- chestercat

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- ganeshie8

the question is not so clear to me, could you draw it out showing what you mean by "heights lengths" and "heads of triangle" ?

- anonymous

@dan815 Why did u leave me
;(

- TrojanPoem

head of triangle means A, B, C . Heights lengths : Like drawing perpendicular lines to the line L

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## More answers

- TrojanPoem

@dan815 , Go help him dan

- dan815

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- ganeshie8

got it, thnks

- dan815

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- dan815

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- ganeshie8

is the constant same as side length ?

- ganeshie8

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- dan815

oo nice

- dan815

limiting case

- perl

*

- dan815

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- TrojanPoem

Wait I did a mistake , the sum of height lengths squared

- dan815

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- dan815

kk

- dan815

is there an elegant solution, not liek a calculus solution

- dan815

cuz you can alwys write a calculus solution just rote

- dan815

if its true

- TrojanPoem

Lol , the solution isn't calculus related dan815, It all depends on geometry and there is no rare formulas

- dan815

y=mx+b and write perpendicular equations from 3 equal points

- ganeshie8

then i guess areas/similar triangles works pretty well
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- TrojanPoem

@dan815, I though you'd get it easily. It's just lengths calculations

- TrojanPoem

@ganeshie8 , How ?

- ganeshie8

idk how, im still thinking.. let me grab pen paper

- dan815

fomr the similar triangles we can see the hypotenuses ratios are equal i can see thats where the square relationship is popping out

- dan815

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- ikram002p

dan so much typos u made :O

- dan815

my writing gives peoples unknowns to solve

- dan815

people*

- ikram002p

i didnt got this
"of heights lengths squared drawn from the heads of the triangle on the line (L) equal a constant whatever the line (L) was."
@ganeshie8 could u give an example ?

- dan815

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- dan815

it says a^2+b^2+c^2 = k

- ikram002p

is this a case ?
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- dan815

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- dan815

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- ikram002p

ok so L is parallel to one side and pass the center ?( the only case allowed ?)

- ikram002p

hmm i seems dont understand :)

- ikram002p

hello xD

- dan815

its turning into an algebra excercise on my end, am i on the wrong path then

- dan815

a^2 , b^2 and c^2 im relating them all to their hypotenuses which i know are constant and now we know the rations are same and we have different relationships for the hypotenuse as well

- ikram002p

i just wanna see what about this case ?
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- dan815

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- dan815

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- ikram002p

ugh "sigh"

- dan815

what about that diagram?

- dan815

it's all legal, just draw any line through m and show the sum of the squares of the perpendiculars from the corners of the triangle to that line through m, which is the center of the triangle are equal

- TrojanPoem

@ikram002p , The line (L) is not parallel to the side.

- TrojanPoem

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- ikram002p

ok we know all right triangles are congruent for some cool reason :D

- ikram002p

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- ganeshie8

I have managed to show \(a+b=c\)

- ganeshie8

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- ganeshie8

feel free to use that if it helps,
il provide the proof in the end..

- ikram002p

did u use similarities properties ?

- ikram002p

if not i'll continue with mine lol if yes i'll take ur result

- ganeshie8

continue with yours..

- ikram002p

k :)

- ikram002p

\(c/b=(BH)/(HC)=\beta \\
c/a=(BG)/(GA)=\alpha \\ c=\alpha a,c=\beta b---2\text{ equations i'll use also ganeshie8 result }\\ a+b=c\\ \)
\((\alpha -2)^2a^2+(\beta-2)^2b^2+\frac{2(\alpha-2)(\beta-2)}{\alpha \beta }c^2=0\)
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- TrojanPoem

The constant is 1/2 The length of the triangle side squared.

- TrojanPoem

I don't think that the three triangles are similar. Even if they are. To use similarities It must be proven.

- ikram002p

I have shown the similarty case, not all three are similar

- TrojanPoem

So the solution was build on corrupted base ?

- dan815

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- ganeshie8

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- ikram002p

this is amazing !!!

- ganeshie8

Here is a dumb proof using coordinate geometry which of course i don't like :
Let
\(A = (-t,0)\)
\(B=(t,0)\)
\(C=(0, \sqrt{3}t)\)
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- ganeshie8

Centroid is clearly \((0, \frac{t}{\sqrt{3}})\), call this point \(G\).
Then any line passing through this point is given by \[y=mx+\frac{t}{\sqrt{3}}\tag{1}\]
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- ganeshie8

Sum of squares of distances from the line\((1)\) to the vertices :
\[\begin{align}
&\dfrac{(tm+\frac{t}{\sqrt{3}})^2+(-tm+\frac{t}{\sqrt{3}})^2+(-\sqrt{3}t+\frac{t}{\sqrt{3}})^2}{m^2+1}\\~\\
&=\dfrac{2(t^2m^2+\frac{t^2}{3}) +\frac{4t^2}{3}}{m^2+1}\\~\\
&=\dfrac{2(t^2m^2+t^2)}{m^2+1}\\~\\
&=2t^2
\end{align}\]

- TrojanPoem

Done ?

- ganeshie8

you don't get it ?

- TrojanPoem

so the constant is .... from the triangle side ?

- ganeshie8

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