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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."

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  1. ganeshie8
    • one year ago
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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" ?

  2. anonymous
    • one year ago
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    @dan815 Why did u leave me ;(

  3. TrojanPoem
    • one year ago
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    head of triangle means A, B, C . Heights lengths : Like drawing perpendicular lines to the line L

  4. TrojanPoem
    • one year ago
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    @dan815 , Go help him dan

  5. dan815
    • one year ago
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    |dw:1435577825081:dw|

  6. ganeshie8
    • one year ago
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    got it, thnks

  7. dan815
    • one year ago
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    |dw:1435578006968:dw|

  8. dan815
    • one year ago
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    |dw:1435578050263:dw|

  9. ganeshie8
    • one year ago
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    is the constant same as side length ?

  10. ganeshie8
    • one year ago
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    |dw:1435578155241:dw|

  11. dan815
    • one year ago
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    oo nice

  12. dan815
    • one year ago
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    limiting case

  13. perl
    • one year ago
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    *

  14. dan815
    • one year ago
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    |dw:1435578420056:dw|

  15. TrojanPoem
    • one year ago
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    Wait I did a mistake , the sum of height lengths squared

  16. dan815
    • one year ago
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    |dw:1435578673892:dw|

  17. dan815
    • one year ago
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    kk

  18. dan815
    • one year ago
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    is there an elegant solution, not liek a calculus solution

  19. dan815
    • one year ago
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    cuz you can alwys write a calculus solution just rote

  20. dan815
    • one year ago
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    if its true

  21. TrojanPoem
    • one year ago
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    Lol , the solution isn't calculus related dan815, It all depends on geometry and there is no rare formulas

  22. dan815
    • one year ago
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    y=mx+b and write perpendicular equations from 3 equal points

  23. ganeshie8
    • one year ago
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    then i guess areas/similar triangles works pretty well |dw:1435578883770:dw|

  24. TrojanPoem
    • one year ago
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    @dan815, I though you'd get it easily. It's just lengths calculations

  25. TrojanPoem
    • one year ago
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    @ganeshie8 , How ?

  26. ganeshie8
    • one year ago
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    idk how, im still thinking.. let me grab pen paper

  27. dan815
    • one year ago
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    fomr the similar triangles we can see the hypotenuses ratios are equal i can see thats where the square relationship is popping out

  28. dan815
    • one year ago
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    |dw:1435579184896:dw|

  29. ikram002p
    • one year ago
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    dan so much typos u made :O

  30. dan815
    • one year ago
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    my writing gives peoples unknowns to solve

  31. dan815
    • one year ago
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    people*

  32. ikram002p
    • one year ago
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    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 ?

  33. dan815
    • one year ago
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    |dw:1435579873776:dw|

  34. dan815
    • one year ago
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    it says a^2+b^2+c^2 = k

  35. ikram002p
    • one year ago
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    is this a case ? |dw:1435580036617:dw|

  36. dan815
    • one year ago
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    |dw:1435579995219:dw|

  37. dan815
    • one year ago
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    |dw:1435580085369:dw|

  38. ikram002p
    • one year ago
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    ok so L is parallel to one side and pass the center ?( the only case allowed ?)

  39. ikram002p
    • one year ago
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    hmm i seems dont understand :)

  40. ikram002p
    • one year ago
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    hello xD

  41. dan815
    • one year ago
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    its turning into an algebra excercise on my end, am i on the wrong path then

  42. dan815
    • one year ago
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    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

  43. ikram002p
    • one year ago
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    i just wanna see what about this case ? |dw:1435581045995:dw|

  44. dan815
    • one year ago
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    |dw:1435581053518:dw|

  45. dan815
    • one year ago
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    |dw:1435581167187:dw|

  46. ikram002p
    • one year ago
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    ugh "sigh"

  47. dan815
    • one year ago
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    what about that diagram?

  48. dan815
    • one year ago
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    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

  49. TrojanPoem
    • one year ago
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    @ikram002p , The line (L) is not parallel to the side.

  50. TrojanPoem
    • one year ago
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    |dw:1435581369777:dw|

  51. ikram002p
    • one year ago
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    ok we know all right triangles are congruent for some cool reason :D

  52. ikram002p
    • one year ago
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    |dw:1435582302688:dw|

  53. ganeshie8
    • one year ago
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    I have managed to show \(a+b=c\)

  54. ganeshie8
    • one year ago
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    |dw:1435582837097:dw|

  55. ganeshie8
    • one year ago
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    feel free to use that if it helps, il provide the proof in the end..

  56. ikram002p
    • one year ago
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    did u use similarities properties ?

  57. ikram002p
    • one year ago
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    if not i'll continue with mine lol if yes i'll take ur result

  58. ganeshie8
    • one year ago
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    continue with yours..

  59. ikram002p
    • one year ago
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    k :)

  60. ikram002p
    • one year ago
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    \(c/b=(BH)/(HC)=\beta \\ c/a=(BG)/(GA)=\alpha \\ c=\alpha a,c=\beta b---2\text{ equations i'll use also ganeshie8 result }\\ a+b=c\\ \) \((\alpha -2)^2a^2+(\beta-2)^2b^2+\frac{2(\alpha-2)(\beta-2)}{\alpha \beta }c^2=0\) |dw:1435585962332:dw|

  61. TrojanPoem
    • one year ago
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    The constant is 1/2 The length of the triangle side squared.

  62. TrojanPoem
    • one year ago
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    I don't think that the three triangles are similar. Even if they are. To use similarities It must be proven.

  63. ikram002p
    • one year ago
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    I have shown the similarty case, not all three are similar

  64. TrojanPoem
    • one year ago
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    So the solution was build on corrupted base ?

  65. dan815
    • one year ago
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    |dw:1435666204039:dw|

  66. ganeshie8
    • one year ago
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    |dw:1435669635846:dw|

  67. ikram002p
    • one year ago
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    this is amazing !!!

  68. ganeshie8
    • one year ago
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    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|

  69. ganeshie8
    • one year ago
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    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|

  70. ganeshie8
    • one year ago
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    Sum of squares of distances from the line\((1)\) to the vertices : \[\begin{align} &\dfrac{(tm+\frac{t}{\sqrt{3}})^2+(-tm+\frac{t}{\sqrt{3}})^2+(-\sqrt{3}t+\frac{t}{\sqrt{3}})^2}{m^2+1}\\~\\ &=\dfrac{2(t^2m^2+\frac{t^2}{3}) +\frac{4t^2}{3}}{m^2+1}\\~\\ &=\dfrac{2(t^2m^2+t^2)}{m^2+1}\\~\\ &=2t^2 \end{align}\]

  71. TrojanPoem
    • one year ago
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    Done ?

  72. ganeshie8
    • one year ago
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    you don't get it ?

  73. TrojanPoem
    • one year ago
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    so the constant is .... from the triangle side ?

  74. ganeshie8
    • one year ago
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    |dw:1435677076734:dw|