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

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1my 1st thought is that you should be able to use the sine rule for this?\[\frac{a}{\sin(BAC)}=\frac{b}{\sin(ABC)}=\frac{c}{\sin(ACB)}\]

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

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

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

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1thoughts 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
 3 years ago
Best ResponseYou've already chosen the best response.1its 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.

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1building 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
 3 years ago
Best ResponseYou've already chosen the best response.0since 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 \[\sin(A)<A ; \frac{\sin(B)+\sin(C)}{2}<\frac{B+C}{2}\]

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1not 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

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

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1I 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:\[ \begin{align} a&<\frac{b+c}{2}\\ \angle A&=60^0\delta\\ \angle B&=60^0+\delta\\ \angle C&=60^0\\ \therefore \angle B + \angle C&=120^0+\delta\\ \therefore \frac{\angle B+\angle C}{2}&=60^0+\frac{\delta}{2}\\ \therefore \angle A=60^0\delta&<\frac{\angle B+\angle C}{2}\\ \end{align} \]

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1I 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.

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

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

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

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

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0moneybird 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)

moneybird
 3 years ago
Best ResponseYou've already chosen the best response.3Yea 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
 3 years ago
Best ResponseYou've already chosen the best response.0try 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
 3 years ago
Best ResponseYou've already chosen the best response.0so (angle B + angle C)/2 = 90  angle A

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

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

moneybird
 3 years ago
Best ResponseYou've already chosen the best response.3A/2 = 90 (angle B + angle C)/2 angle A = 180  (angle B + angle C) (angle B + angle C) = 180  angle A

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

moneybird
 3 years ago
Best ResponseYou've already chosen the best response.3myininaya says \[\angle BAC < \frac{1}{2} (\angle ABC + \angle ACB)\] \[\frac{3}{2} \angle BAC < \frac{1}{2}(180)\] \[\angle BAC < 60\]
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