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shubhamsrg Group Title

now the question which follows up is : If all the sides and area of a triangle were rational numbers then show that the triangle is got by `pasting' two right-angled triangles having the same property

  • one year ago
  • one year ago

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  1. sauravshakya Group Title
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    same property means?

    • one year ago
  2. shubhamsrg Group Title
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    both triangles should have rational sides.

    • one year ago
  3. shubhamsrg Group Title
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    |dw:1345129143913:dw| if ABC has rational sides, we have to show AMB and MBC also has rational sides.

    • one year ago
  4. mukushla Group Title
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    AC.MB is rational because area is rational and AC is rational so MB is rational ...now u just proveAM and MC are rational

    • one year ago
  5. mukushla Group Title
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    lol...not that easy

    • one year ago
  6. sauravshakya Group Title
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    I think it will be enough to show just one side Am or MC rational

    • one year ago
  7. mukushla Group Title
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    yeah..

    • one year ago
  8. sauravshakya Group Title
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    beause if one is rational than the other must be rational

    • one year ago
  9. shubhamsrg Group Title
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    hmm..

    • one year ago
  10. shubhamsrg Group Title
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    if we say tanA = BC/AC = rational also tanA = MB/AM is it good enough to say here that AM will be rational ?

    • one year ago
  11. shubhamsrg Group Title
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    sorry,,tanA=BC/Ab ,,but still rational..

    • one year ago
  12. mukushla Group Title
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    ABC is right triangle !!? work on \[\sqrt{AB^2-MB^2}+\sqrt{BC^2-MB^2}=AM+MC\]

    • one year ago
  13. mukushla Group Title
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    i thought its not right

    • one year ago
  14. shubhamsrg Group Title
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    hmm,,well,,how will sum of sides help ?

    • one year ago
  15. mukushla Group Title
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    sorry that will not help..

    • one year ago
  16. sauravshakya Group Title
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    AM^2 and MC^2 will be rational

    • one year ago
  17. sauravshakya Group Title
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    right?

    • one year ago
  18. mukushla Group Title
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    right

    • one year ago
  19. shubhamsrg Group Title
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    yep..thats where even i was wondering

    • one year ago
  20. shubhamsrg Group Title
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    this ques is a subpart of my previous ques,,do both have any relation ?

    • one year ago
  21. shubhamsrg Group Title
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    how ?

    • one year ago
  22. mukushla Group Title
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    i mean ABC is a right triangle?

    • one year ago
  23. sauravshakya Group Title
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    Dont think so

    • one year ago
  24. shubhamsrg Group Title
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    aha...thats what i had assumed!! so its not a right triangle!! hmm..

    • one year ago
  25. mukushla Group Title
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    sauravshakya mentiond a good point \[(AM+MC)^2=AC^2=AM^2+2AM.MC+MC^2\]so \(AM.MC\) is rational

    • one year ago
  26. mukushla Group Title
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    now our problem is like this if \(a+b\) and \(ab\) are rational prove that both \(a\) and \(b\) are rational

    • one year ago
  27. sauravshakya Group Title
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    YEP

    • one year ago
  28. shubhamsrg Group Title
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    well why were AM^2 and MC^2 rational ?

    • one year ago
  29. sauravshakya Group Title
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    Take right triangle AMB and triangle BMC

    • one year ago
  30. shubhamsrg Group Title
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    ohh leave it got it..

    • one year ago
  31. shubhamsrg Group Title
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    if ab and and a+b are rational,,its not necessary both are rational,, eg. -sqrt(2) +1 and sqrt(2) + 1

    • one year ago
  32. shubhamsrg Group Title
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    has to be an other way around..

    • one year ago
  33. mukushla Group Title
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    yeah i was thinkin of that

    • one year ago
  34. shubhamsrg Group Title
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    hmm..

    • one year ago
  35. mukushla Group Title
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    guys i gotta go...but i love to know what is the answer

    • one year ago
  36. shubhamsrg Group Title
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    |dw:1345131550151:dw| here,,we have AC/sin(a+b) = BC/sinx now sin(a+b) = sina cosb + cosa sinb = AM/AB . BM/BC + BM/AB . MC/BC = (AM+BM)/AB + (BM +MC)/BC also, sin x =BM/AB on substituing, AC . BM/AB = BC . ( (AM+BM)/AB + (BM +MC)/BC ) AC.BM = (AM+BM)BC + (MB +MC)AB AC BM = AM BC + BM BC + MB AB + MC AB I DONT KNOW IF IT HELPS OR NOT!! lol.. :P

    • one year ago
  37. experimentX Group Title
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    hasn't this been solved yet?

    • one year ago
  38. shubhamsrg Group Title
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    its a diff ques..

    • one year ago
  39. experimentX Group Title
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    i lost track of it ... it's fuzzy ... well what's the Q?

    • one year ago
  40. sauravshakya Group Title
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    |dw:1345132117021:dw|if ABC has rational sides, we have to show AMB and MBC also has rational sides.

    • one year ago
  41. sauravshakya Group Title
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    @experimentX thats the question

    • one year ago
  42. experimentX Group Title
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    |dw:1345138702333:dw| a^2 + b^2 = c^2 b^2 = rational AM + MC = rational AM^2 + MC^2 = rational. AM*MC = rational.

    • one year ago
  43. experimentX Group Title
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    Let me finish it man!! AM + MC = rational <-- either both are irrational or rational) AM*MC = rational <--- either both are irrational or rational Let's assume both to be rational, AM ( rational - AM) = rational AM (rational) - AM^2 = rational irrational - rational = rational (contradiction) hence both must be rational

    • one year ago
  44. experimentX Group Title
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    typo:- Let's assume both to be irrational,

    • one year ago
  45. mukushla Group Title
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    here is my proof|dw:1345140385316:dw|

    • one year ago
  46. mukushla Group Title
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    \(x+y=c\) is rational Area=\(ch/2\)=rational so \(h\) is rational since \(c\) is rational we have\[x^2+h^2=a^2\]\[y^2+h^2=b^2\]\[(x-y)(x+y)=(x-y).c=a^2-b^2\]so we obtain\[x-y=\frac{a^2-b^2}{c}\]\[x+y=c\]it gives\[x=\frac{a^2+c^2-b^2}{2c}\]\[y=\frac{b^2+c^2-a^2}{2c}\]\(x\) and \(y\) both are rational

    • one year ago
  47. experimentX Group Title
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    lol ... your method is more obvious!!

    • one year ago
  48. shubhamsrg Group Title
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    aha..i see and understand both methods! thanks again @mukushla and @experimentX

    • one year ago
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