## 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."

1. 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" ?

2. anonymous

@dan815 Why did u leave me ;(

3. TrojanPoem

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

4. TrojanPoem

@dan815 , Go help him dan

5. dan815

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6. ganeshie8

got it, thnks

7. dan815

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8. dan815

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9. ganeshie8

is the constant same as side length ?

10. ganeshie8

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11. dan815

oo nice

12. dan815

limiting case

13. perl

*

14. dan815

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15. TrojanPoem

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

16. dan815

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17. dan815

kk

18. dan815

is there an elegant solution, not liek a calculus solution

19. dan815

cuz you can alwys write a calculus solution just rote

20. dan815

if its true

21. TrojanPoem

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

22. dan815

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

23. ganeshie8

then i guess areas/similar triangles works pretty well |dw:1435578883770:dw|

24. TrojanPoem

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

25. TrojanPoem

@ganeshie8 , How ?

26. ganeshie8

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

27. dan815

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

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29. ikram002p

dan so much typos u made :O

30. dan815

my writing gives peoples unknowns to solve

31. dan815

people*

32. 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 ?

33. dan815

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34. dan815

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

35. ikram002p

is this a case ? |dw:1435580036617:dw|

36. dan815

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37. dan815

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38. ikram002p

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

39. ikram002p

hmm i seems dont understand :)

40. ikram002p

hello xD

41. dan815

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

42. 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

43. ikram002p

44. dan815

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45. dan815

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46. ikram002p

ugh "sigh"

47. dan815

48. 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

49. TrojanPoem

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

50. TrojanPoem

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51. ikram002p

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

52. ikram002p

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53. ganeshie8

I have managed to show $$a+b=c$$

54. ganeshie8

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55. ganeshie8

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

56. ikram002p

did u use similarities properties ?

57. ikram002p

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

58. ganeshie8

continue with yours..

59. ikram002p

k :)

60. 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|

61. TrojanPoem

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

62. TrojanPoem

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

63. ikram002p

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

64. TrojanPoem

So the solution was build on corrupted base ?

65. dan815

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66. ganeshie8

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67. ikram002p

this is amazing !!!

68. 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|

69. 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|

70. 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}

71. TrojanPoem

Done ?

72. ganeshie8

you don't get it ?

73. TrojanPoem

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

74. ganeshie8

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