TrojanPoem
  • TrojanPoem
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ganeshie8
  • 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
  • anonymous
@dan815 Why did u leave me ;(
TrojanPoem
  • TrojanPoem
head of triangle means A, B, C . Heights lengths : Like drawing perpendicular lines to the line L

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TrojanPoem
  • TrojanPoem
@dan815 , Go help him dan
dan815
  • dan815
|dw:1435577825081:dw|
ganeshie8
  • ganeshie8
got it, thnks
dan815
  • dan815
|dw:1435578006968:dw|
dan815
  • dan815
|dw:1435578050263:dw|
ganeshie8
  • ganeshie8
is the constant same as side length ?
ganeshie8
  • ganeshie8
|dw:1435578155241:dw|
dan815
  • dan815
oo nice
dan815
  • dan815
limiting case
perl
  • perl
*
dan815
  • dan815
|dw:1435578420056:dw|
TrojanPoem
  • TrojanPoem
Wait I did a mistake , the sum of height lengths squared
dan815
  • dan815
|dw:1435578673892:dw|
dan815
  • dan815
kk
dan815
  • dan815
is there an elegant solution, not liek a calculus solution
dan815
  • dan815
cuz you can alwys write a calculus solution just rote
dan815
  • dan815
if its true
TrojanPoem
  • TrojanPoem
Lol , the solution isn't calculus related dan815, It all depends on geometry and there is no rare formulas
dan815
  • dan815
y=mx+b and write perpendicular equations from 3 equal points
ganeshie8
  • ganeshie8
then i guess areas/similar triangles works pretty well |dw:1435578883770:dw|
TrojanPoem
  • TrojanPoem
@dan815, I though you'd get it easily. It's just lengths calculations
TrojanPoem
  • TrojanPoem
@ganeshie8 , How ?
ganeshie8
  • ganeshie8
idk how, im still thinking.. let me grab pen paper
dan815
  • 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
  • dan815
|dw:1435579184896:dw|
ikram002p
  • ikram002p
dan so much typos u made :O
dan815
  • dan815
my writing gives peoples unknowns to solve
dan815
  • dan815
people*
ikram002p
  • 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
  • dan815
|dw:1435579873776:dw|
dan815
  • dan815
it says a^2+b^2+c^2 = k
ikram002p
  • ikram002p
is this a case ? |dw:1435580036617:dw|
dan815
  • dan815
|dw:1435579995219:dw|
dan815
  • dan815
|dw:1435580085369:dw|
ikram002p
  • ikram002p
ok so L is parallel to one side and pass the center ?( the only case allowed ?)
ikram002p
  • ikram002p
hmm i seems dont understand :)
ikram002p
  • ikram002p
hello xD
dan815
  • dan815
its turning into an algebra excercise on my end, am i on the wrong path then
dan815
  • 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
  • ikram002p
i just wanna see what about this case ? |dw:1435581045995:dw|
dan815
  • dan815
|dw:1435581053518:dw|
dan815
  • dan815
|dw:1435581167187:dw|
ikram002p
  • ikram002p
ugh "sigh"
dan815
  • dan815
what about that diagram?
dan815
  • 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
  • TrojanPoem
@ikram002p , The line (L) is not parallel to the side.
TrojanPoem
  • TrojanPoem
|dw:1435581369777:dw|
ikram002p
  • ikram002p
ok we know all right triangles are congruent for some cool reason :D
ikram002p
  • ikram002p
|dw:1435582302688:dw|
ganeshie8
  • ganeshie8
I have managed to show \(a+b=c\)
ganeshie8
  • ganeshie8
|dw:1435582837097:dw|
ganeshie8
  • ganeshie8
feel free to use that if it helps, il provide the proof in the end..
ikram002p
  • ikram002p
did u use similarities properties ?
ikram002p
  • ikram002p
if not i'll continue with mine lol if yes i'll take ur result
ganeshie8
  • ganeshie8
continue with yours..
ikram002p
  • ikram002p
k :)
ikram002p
  • ikram002p
\(c/b=(BH)/(HC)=\beta \\ c/a=(BG)/(GA)=\alpha \\ c=\alpha a,c=\beta b---2\text{ equations i'll use also ganeshie8 result }\\ a+b=c\\ \) \((\alpha -2)^2a^2+(\beta-2)^2b^2+\frac{2(\alpha-2)(\beta-2)}{\alpha \beta }c^2=0\) |dw:1435585962332:dw|
TrojanPoem
  • TrojanPoem
The constant is 1/2 The length of the triangle side squared.
TrojanPoem
  • TrojanPoem
I don't think that the three triangles are similar. Even if they are. To use similarities It must be proven.
ikram002p
  • ikram002p
I have shown the similarty case, not all three are similar
TrojanPoem
  • TrojanPoem
So the solution was build on corrupted base ?
dan815
  • dan815
|dw:1435666204039:dw|
ganeshie8
  • ganeshie8
|dw:1435669635846:dw|
ikram002p
  • ikram002p
this is amazing !!!
ganeshie8
  • ganeshie8
Here is a dumb proof using coordinate geometry which of course i don't like : Let \(A = (-t,0)\) \(B=(t,0)\) \(C=(0, \sqrt{3}t)\) |dw:1435675544583:dw|
ganeshie8
  • ganeshie8
Centroid is clearly \((0, \frac{t}{\sqrt{3}})\), call this point \(G\). Then any line passing through this point is given by \[y=mx+\frac{t}{\sqrt{3}}\tag{1}\] |dw:1435676089356:dw|
ganeshie8
  • 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
  • TrojanPoem
Done ?
ganeshie8
  • ganeshie8
you don't get it ?
TrojanPoem
  • TrojanPoem
so the constant is .... from the triangle side ?
ganeshie8
  • ganeshie8
|dw:1435677076734:dw|