Here's the question you clicked on:

## moneybird Group Title 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

• This Question is Closed
1. moneybird Group Title

|dw:1323301015086:dw|

2. moneybird Group Title

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

3. dometryapple Group Title

WOW

4. asnaseer Group Title

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. moneybird Group Title

ok

6. moneybird Group Title

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

7. asnaseer Group Title

I suppose you could use that too.

8. moneybird Group Title

$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 Group Title

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

10. asnaseer Group Title

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 Group Title

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. moneybird Group Title

good night

13. dometryapple Group Title

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

14. asnaseer Group Title

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 Group Title

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 Group Title

I can't think if this helps or not

17. asnaseer Group Title

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 Group Title

you are so smart

19. asnaseer Group Title

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

20. moneybird Group Title

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

21. asnaseer Group Title

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. moneybird Group Title

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

23. asnaseer Group Title

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. moneybird Group Title

Let's ask myininaya when he gets on

25. asnaseer Group Title

It might be worth asking JamesJ or Zarkon as well.

26. satellite73 Group Title

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

27. moneybird Group Title

ok thank you for trying

28. jhonyy9 Group Title

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. moneybird Group Title

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 Group Title

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 Group Title

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

32. moneybird Group Title

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

33. jhonyy9 Group Title

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

34. moneybird Group Title

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

35. jhonyy9 Group Title

so than angle B + angle C >= 135

36. jhonyy9 Group Title

and 135/2 > 45 allways

37. moneybird Group Title

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 Group Title

good luck bye

39. moneybird Group Title

thanks