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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- anonymous

|dw:1323301015086:dw|

- anonymous

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

- anonymous

WOW

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## More answers

- asnaseer

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

- anonymous

ok

- anonymous

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

- asnaseer

I suppose you could use that too.

- anonymous

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

- asnaseer

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

- 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:\[a=b\cos(ACB)+c\cos(ABC)\]

- 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.

- anonymous

good night

- anonymous

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

- 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 \(A<\dfrac{B+C}{2}\)

- myininaya

since each of the angles A,B,C are between 0 and 180 degrees
0

- myininaya

I can't think if this helps or not

- 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

- myininaya

you are so smart

- asnaseer

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

- anonymous

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

- 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:\[
\begin{align}
a&<\frac{b+c}{2}\\
\angle A&=60^0-\delta\\
\angle B&=60^0+\delta\\
\angle C&=60^0\\
\therefore \angle B + \angle C&=120^0+\delta\\
\therefore \frac{\angle B+\angle C}{2}&=60^0+\frac{\delta}{2}\\
\therefore \angle A=60^0-\delta&<\frac{\angle B+\angle C}{2}\\
\end{align}
\]

- anonymous

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

- 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.

- anonymous

Let's ask myininaya when he gets on

- asnaseer

It might be worth asking JamesJ or Zarkon as well.

- anonymous

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

- anonymous

ok thank you for trying

- 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)

- 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.

- 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

- jhonyy9

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

- anonymous

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

- jhonyy9

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

- anonymous

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

- jhonyy9

so than angle B + angle C >= 135

- jhonyy9

and 135/2 > 45 allways

- anonymous

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

- jhonyy9

good luck bye

- anonymous

thanks

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