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."

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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."

Mathematics
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the question is not so clear to me, could you draw it out showing what you mean by "heights lengths" and "heads of triangle" ?
@dan815 Why did u leave me ;(
head of triangle means A, B, C . Heights lengths : Like drawing perpendicular lines to the line L

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@dan815 , Go help him dan
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got it, thnks
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is the constant same as side length ?
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oo nice
limiting case
*
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Wait I did a mistake , the sum of height lengths squared
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kk
is there an elegant solution, not liek a calculus solution
cuz you can alwys write a calculus solution just rote
if its true
Lol , the solution isn't calculus related dan815, It all depends on geometry and there is no rare formulas
y=mx+b and write perpendicular equations from 3 equal points
then i guess areas/similar triangles works pretty well |dw:1435578883770:dw|
@dan815, I though you'd get it easily. It's just lengths calculations
@ganeshie8 , How ?
idk how, im still thinking.. let me grab pen paper
fomr the similar triangles we can see the hypotenuses ratios are equal i can see thats where the square relationship is popping out
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dan so much typos u made :O
my writing gives peoples unknowns to solve
people*
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 ?
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it says a^2+b^2+c^2 = k
is this a case ? |dw:1435580036617:dw|
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ok so L is parallel to one side and pass the center ?( the only case allowed ?)
hmm i seems dont understand :)
hello xD
its turning into an algebra excercise on my end, am i on the wrong path then
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
i just wanna see what about this case ? |dw:1435581045995:dw|
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ugh "sigh"
what about that diagram?
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
@ikram002p , The line (L) is not parallel to the side.
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ok we know all right triangles are congruent for some cool reason :D
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I have managed to show \(a+b=c\)
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feel free to use that if it helps, il provide the proof in the end..
did u use similarities properties ?
if not i'll continue with mine lol if yes i'll take ur result
continue with yours..
k :)
\(c/b=(BH)/(HC)=\beta \\ c/a=(BG)/(GA)=\alpha \\ c=\alpha a,c=\beta b---2\text{ equations i'll use also ganeshie8 result }\\ a+b=c\\ \) \((\alpha -2)^2a^2+(\beta-2)^2b^2+\frac{2(\alpha-2)(\beta-2)}{\alpha \beta }c^2=0\) |dw:1435585962332:dw|
The constant is 1/2 The length of the triangle side squared.
I don't think that the three triangles are similar. Even if they are. To use similarities It must be proven.
I have shown the similarty case, not all three are similar
So the solution was build on corrupted base ?
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this is amazing !!!
Here is a dumb proof using coordinate geometry which of course i don't like : Let \(A = (-t,0)\) \(B=(t,0)\) \(C=(0, \sqrt{3}t)\) |dw:1435675544583:dw|
Centroid is clearly \((0, \frac{t}{\sqrt{3}})\), call this point \(G\). Then any line passing through this point is given by \[y=mx+\frac{t}{\sqrt{3}}\tag{1}\] |dw:1435676089356:dw|
Sum of squares of distances from the line\((1)\) to the vertices : \[\begin{align} &\dfrac{(tm+\frac{t}{\sqrt{3}})^2+(-tm+\frac{t}{\sqrt{3}})^2+(-\sqrt{3}t+\frac{t}{\sqrt{3}})^2}{m^2+1}\\~\\ &=\dfrac{2(t^2m^2+\frac{t^2}{3}) +\frac{4t^2}{3}}{m^2+1}\\~\\ &=\dfrac{2(t^2m^2+t^2)}{m^2+1}\\~\\ &=2t^2 \end{align}\]
Done ?
you don't get it ?
so the constant is .... from the triangle side ?
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