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## anonymous 5 years ago In triangle ABC, BC = a, AC = b, AB = c, and a < (b+c)/2. Prove that angle BAC < (angle ABC + angle ACB)/2.

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1. anonymous

|dw:1323301015086:dw|

2. anonymous

Given that $a < \frac{1}{2}(b+c)$ Prove that $\angle BAC < \frac{1}{2}(\angle ABC + \angle ACB)$

3. anonymous

WOW

4. asnaseer

my 1st thought is that you should be able to use the sine rule for this?$\frac{a}{\sin(BAC)}=\frac{b}{\sin(ABC)}=\frac{c}{\sin(ACB)}$

5. anonymous

ok

6. anonymous

i thought it's related to cosine rule? $a^2 = b^2 +c^2 - 2bc (\cos \angle BAC)$

7. asnaseer

I suppose you could use that too.

8. anonymous

$a^2 < \frac{1}{4} (b^2 + 2bc + c^2)$$a < \frac{1}{2}(b+c)$ $a^2 < \frac{1}{4} (b+c)^2$

9. asnaseer

hmmm - needs some thought - let me think about it for a bit...

10. asnaseer

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:$a=b\cos(ACB)+c\cos(ABC)$

11. asnaseer

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.

12. anonymous

good night

13. anonymous

what country are you in? Im in Canada and its 8:21 pm

14. asnaseer

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}$$

15. myininaya

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 $\sin(A)<A ; \frac{\sin(B)+\sin(C)}{2}<\frac{B+C}{2}$

16. myininaya

I can't think if this helps or not

17. asnaseer

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

18. myininaya

you are so smart

19. asnaseer

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

20. anonymous

|dw:1323473308275:dw| My teacher gave me some hints.

21. asnaseer

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}

22. anonymous

this proof is awesome. It does not require any trignometric identities

23. asnaseer

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.

24. anonymous

Let's ask myininaya when he gets on

25. asnaseer

It might be worth asking JamesJ or Zarkon as well.

26. anonymous

i will look at the hint and see if i can understand it. back later

27. anonymous

ok thank you for trying

28. jhonyy9

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)

29. anonymous

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.

30. jhonyy9

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

31. jhonyy9

so (angle B + angle C)/2 = 90 - angle A

32. anonymous

angle A < 90 - angle A That's what we need to prove

33. jhonyy9

what you have wrote there now is so from this resulted that angle A < 45

34. anonymous

A/2 = 90 -(angle B + angle C)/2 angle A = 180 - (angle B + angle C) (angle B + angle C) = 180 - angle A

35. jhonyy9

so than angle B + angle C >= 135

36. jhonyy9

and 135/2 > 45 allways

37. anonymous

myininaya says $\angle BAC < \frac{1}{2} (\angle ABC + \angle ACB)$ $\frac{3}{2} \angle BAC < \frac{1}{2}(180)$ $\angle BAC < 60$

38. jhonyy9

good luck bye

39. anonymous

thanks

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