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moneybird
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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
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

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moneybird Group TitleBest ResponseYou've already chosen the best response.3
dw:1323301015086:dw
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
Given that \[a < \frac{1}{2}(b+c)\] Prove that \[\angle BAC < \frac{1}{2}(\angle ABC + \angle ACB)\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
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

moneybird Group TitleBest ResponseYou've already chosen the best response.3
i thought it's related to cosine rule? \[a^2 = b^2 +c^2  2bc (\cos \angle BAC)\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
I suppose you could use that too.
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
\[a^2 < \frac{1}{4} (b^2 + 2bc + c^2)\]\[a < \frac{1}{2}(b+c)\] \[a^2 < \frac{1}{4} (b+c)^2\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
hmmm  needs some thought  let me think about it for a bit...
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
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

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
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

moneybird Group TitleBest ResponseYou've already chosen the best response.3
good night
 2 years ago

dometryapple Group TitleBest ResponseYou've already chosen the best response.0
what country are you in? Im in Canada and its 8:21 pm
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
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

myininaya Group TitleBest ResponseYou've already chosen the best response.0
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

myininaya Group TitleBest ResponseYou've already chosen the best response.0
I can't think if this helps or not
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
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

myininaya Group TitleBest ResponseYou've already chosen the best response.0
you are so smart
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
We are ALL smart in our own ways. I strongly believe that there no such thing as a dumb person :)
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
dw:1323473308275:dw My teacher gave me some hints.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
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

moneybird Group TitleBest ResponseYou've already chosen the best response.3
this proof is awesome. It does not require any trignometric identities
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
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

moneybird Group TitleBest ResponseYou've already chosen the best response.3
Let's ask myininaya when he gets on
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
It might be worth asking JamesJ or Zarkon as well.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i will look at the hint and see if i can understand it. back later
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
ok thank you for trying
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
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

moneybird Group TitleBest ResponseYou've already chosen the best response.3
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

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
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

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
so (angle B + angle C)/2 = 90  angle A
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
angle A < 90  angle A That's what we need to prove
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
what you have wrote there now is so from this resulted that angle A < 45
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
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

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
so than angle B + angle C >= 135
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
and 135/2 > 45 allways
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
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

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
good luck bye
 2 years ago
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