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moneybird

In triangle ABC, BC = a, AC = b, AB = c, and a < (b+c)/2. Prove that angle BAC < (angle ABC + angle ACB)/2.

  • 2 years ago
  • 2 years ago

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  1. moneybird
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    |dw:1323301015086:dw|

    • 2 years ago
  2. moneybird
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    Given that \[a < \frac{1}{2}(b+c)\] Prove that \[\angle BAC < \frac{1}{2}(\angle ABC + \angle ACB)\]

    • 2 years ago
  3. dometryapple
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    WOW

    • 2 years ago
  4. asnaseer
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    my 1st thought is that you should be able to use the sine rule for this?\[\frac{a}{\sin(BAC)}=\frac{b}{\sin(ABC)}=\frac{c}{\sin(ACB)}\]

    • 2 years ago
  5. moneybird
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    ok

    • 2 years ago
  6. moneybird
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    i thought it's related to cosine rule? \[a^2 = b^2 +c^2 - 2bc (\cos \angle BAC)\]

    • 2 years ago
  7. asnaseer
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    I suppose you could use that too.

    • 2 years ago
  8. moneybird
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    \[a^2 < \frac{1}{4} (b^2 + 2bc + c^2)\]\[a < \frac{1}{2}(b+c)\] \[a^2 < \frac{1}{4} (b+c)^2\]

    • 2 years ago
  9. asnaseer
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    hmmm - needs some thought - let me think about it for a bit...

    • 2 years ago
  10. asnaseer
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    thoughts so far: BAC=180-(ABC+ACB) so what we need to prove can be adjusted to prove:\[ \angle ABC+\angle ACB>120^0\] also, we can show that:\[a=b\cos(ACB)+c\cos(ABC)\]

    • 2 years ago
  11. asnaseer
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    its past 1am here in the UK and I need to catch some sleep. I'll take a look at this again sometime tomorrow. It's definitely an interesting problem.

    • 2 years ago
  12. moneybird
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    good night

    • 2 years ago
  13. dometryapple
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    what country are you in? Im in Canada and its 8:21 pm

    • 2 years ago
  14. asnaseer
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    building on my original hunch to use the sine rule, I have been able to get far:\[\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}\]which leads to:\[b=\frac{a\sin(B)}{\sin(A)}\quad\text{and}\quad c=\frac{a\sin(C)}{\sin(A)}\]\[\therefore b+c=\frac{a(\sin(B)+\sin(C))}{\sin(A)}\]\[\therefore \frac{b+c}{2}=\frac{a(\sin(B)+\sin(C))}{2\sin(A)}\]and we are given that:\[a<\frac{b+c}{2}\]\[\therefore a<\frac{a(\sin(B)+\sin(C))}{2\sin(A)}\]\[\therefore 1<\frac{\sin(B)+\sin(C)}{2\sin(A)}\]\[\therefore\sin(A)<\frac{\sin(B)+\sin(C)}{2}\] need to work out how to get from this result to \(A<\dfrac{B+C}{2}\)

    • 2 years ago
  15. myininaya
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    since each of the angles A,B,C are between 0 and 180 degrees 0<sin(A)<=1 0<sin(B)<=1 0<sin(C)<=1 \[\sin(A)<A ; \frac{\sin(B)+\sin(C)}{2}<\frac{B+C}{2}\]

    • 2 years ago
  16. myininaya
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    I can't think if this helps or not

    • 2 years ago
  17. asnaseer
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    not sure it does, but I have been able to eliminate A from my final equation as follows. the last result I got was:\[\sin(A)<\frac{\sin(B)+\sin(C)}{2}\]but we know:\[A=180-(B+C)\]\[\therefore \sin(A)=\sin(180-(B+C))=\sin(B+C)\]so we finally get:\[\sin(B+C)<\frac{\sin(B)+\sin(C)}{2}\]so we now need to prove that this implies B+C>120

    • 2 years ago
  18. myininaya
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    you are so smart

    • 2 years ago
  19. asnaseer
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    We are ALL smart in our own ways. I strongly believe that there no such thing as a dumb person :-)

    • 2 years ago
  20. moneybird
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    |dw:1323473308275:dw| My teacher gave me some hints.

    • 2 years ago
  21. asnaseer
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    I have thought of another way to reason about the problem in order to come up with the desired solution. Lets start off by looking at an equilateral triangle where \(a=b=c\) and \(\angle A=\angle B=\angle C=60^0\): |dw:1323541342145:dw| \[ \begin{align} \text{projection of AB onto BC will have length=}\frac{c}{2}\\ \text{projection of AC onto BC will have length=}\frac{b}{2}\\ \text{and we can see that }a=\frac{b+c}{2}\\ \end{align} \] Now, one way for \(a<\dfrac{b+c}{2}\) is to move point B towards point C along line BC by a small amount. This would reduce the projected length of AB onto BC. Moving point B towards point C would also increase \(\angle B\) by a small amount - lets call this increase \(\delta\). It would also decrease \(\angle A\) by this same amount. Thus we would end up with:\[ \begin{align} a&<\frac{b+c}{2}\\ \angle A&=60^0-\delta\\ \angle B&=60^0+\delta\\ \angle C&=60^0\\ \therefore \angle B + \angle C&=120^0+\delta\\ \therefore \frac{\angle B+\angle C}{2}&=60^0+\frac{\delta}{2}\\ \therefore \angle A=60^0-\delta&<\frac{\angle B+\angle C}{2}\\ \end{align} \]

    • 2 years ago
  22. moneybird
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    this proof is awesome. It does not require any trignometric identities

    • 2 years ago
  23. asnaseer
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    I am not sure if it counts as a /true/ mathematical proof. It is just a description of the reasoning I went through to try and prove it to myself.

    • 2 years ago
  24. moneybird
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    Let's ask myininaya when he gets on

    • 2 years ago
  25. asnaseer
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    It might be worth asking JamesJ or Zarkon as well.

    • 2 years ago
  26. satellite73
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    i will look at the hint and see if i can understand it. back later

    • 2 years ago
  27. moneybird
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    ok thank you for trying

    • 2 years ago
  28. jhonyy9
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    moneybird how do you think if you begin from : the sum of angles in one triangle is 180 degree so than angle A + angle B + angle C = 180 so than angle A = 180 -(angle B + angle C)

    • 2 years ago
  29. moneybird
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    Yea so you know that if you can prove either angle A is less than 60 or angle B + angle C is 120, then you are done.

    • 2 years ago
  30. jhonyy9
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    try angle A=180-(angle B + angle C) so divide both sides by 2 and will get angle A/2 = 90 -(angle B + angle C)/2

    • 2 years ago
  31. jhonyy9
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    so (angle B + angle C)/2 = 90 - angle A

    • 2 years ago
  32. moneybird
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    angle A < 90 - angle A That's what we need to prove

    • 2 years ago
  33. jhonyy9
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    what you have wrote there now is so from this resulted that angle A < 45

    • 2 years ago
  34. moneybird
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    A/2 = 90 -(angle B + angle C)/2 angle A = 180 - (angle B + angle C) (angle B + angle C) = 180 - angle A

    • 2 years ago
  35. jhonyy9
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    so than angle B + angle C >= 135

    • 2 years ago
  36. jhonyy9
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    and 135/2 > 45 allways

    • 2 years ago
  37. moneybird
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    myininaya says \[\angle BAC < \frac{1}{2} (\angle ABC + \angle ACB)\] \[\frac{3}{2} \angle BAC < \frac{1}{2}(180)\] \[\angle BAC < 60\]

    • 2 years ago
  38. jhonyy9
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    good luck bye

    • 2 years ago
  39. moneybird
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    thanks

    • 2 years ago
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