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TrojanPoem
 one year ago
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
 one year ago
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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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the question is not so clear to me, could you draw it out showing what you mean by "heights lengths" and "heads of triangle" ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 Why did u leave me ;(

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1head of triangle means A, B, C . Heights lengths : Like drawing perpendicular lines to the line L

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1@dan815 , Go help him dan

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2is the constant same as side length ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1435578155241:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1Wait I did a mistake , the sum of height lengths squared

dan815
 one year ago
Best ResponseYou've already chosen the best response.1is there an elegant solution, not liek a calculus solution

dan815
 one year ago
Best ResponseYou've already chosen the best response.1cuz you can alwys write a calculus solution just rote

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1Lol , the solution isn't calculus related dan815, It all depends on geometry and there is no rare formulas

dan815
 one year ago
Best ResponseYou've already chosen the best response.1y=mx+b and write perpendicular equations from 3 equal points

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2then i guess areas/similar triangles works pretty well dw:1435578883770:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1@dan815, I though you'd get it easily. It's just lengths calculations

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2idk how, im still thinking.. let me grab pen paper

dan815
 one year ago
Best ResponseYou've already chosen the best response.1fomr the similar triangles we can see the hypotenuses ratios are equal i can see thats where the square relationship is popping out

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1dan so much typos u made :O

dan815
 one year ago
Best ResponseYou've already chosen the best response.1my writing gives peoples unknowns to solve

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.1it says a^2+b^2+c^2 = k

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1is this a case ? dw:1435580036617:dw

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ok so L is parallel to one side and pass the center ?( the only case allowed ?)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1hmm i seems dont understand :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1its turning into an algebra excercise on my end, am i on the wrong path then

dan815
 one year ago
Best ResponseYou've already chosen the best response.1a^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
 one year ago
Best ResponseYou've already chosen the best response.1i just wanna see what about this case ? dw:1435581045995:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.1what about that diagram?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1it'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
 one year ago
Best ResponseYou've already chosen the best response.1@ikram002p , The line (L) is not parallel to the side.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435581369777:dw

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ok we know all right triangles are congruent for some cool reason :D

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435582302688:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I have managed to show \(a+b=c\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1435582837097:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2feel free to use that if it helps, il provide the proof in the end..

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1did u use similarities properties ?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1if not i'll continue with mine lol if yes i'll take ur result

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2continue with yours..

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1\(c/b=(BH)/(HC)=\beta \\ c/a=(BG)/(GA)=\alpha \\ c=\alpha a,c=\beta b2\text{ equations i'll use also ganeshie8 result }\\ a+b=c\\ \) \((\alpha 2)^2a^2+(\beta2)^2b^2+\frac{2(\alpha2)(\beta2)}{\alpha \beta }c^2=0\) dw:1435585962332:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1The constant is 1/2 The length of the triangle side squared.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1I don't think that the three triangles are similar. Even if they are. To use similarities It must be proven.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1I have shown the similarty case, not all three are similar

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1So the solution was build on corrupted base ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1435669635846:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Here 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
 one year ago
Best ResponseYou've already chosen the best response.2Centroid 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
 one year ago
Best ResponseYou've already chosen the best response.2Sum 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
 one year ago
Best ResponseYou've already chosen the best response.1so the constant is .... from the triangle side ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1435677076734:dw