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#1
Mar1105, 07:25 PM

P: 6

In triangle ABC if sin (A/2) sin (B/2) sin (C/2) = 1/8
prove that the triangle is equilateral plz show steps 


#2
Mar1105, 11:11 PM

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Perhaps you can use the sine rule (?)



#3
Mar1205, 04:59 AM

P: 6

Sure .. I tried but had no success .. If you find an answer plz post your steps



#4
Mar1205, 05:51 AM

P: 137

Trig identities
one method suggested:
(an absurd reasonning) if it is an equilateral triangle then: A = B = C = pi/3 rad implies > A/2 = B/2 = C/2 = pi/6 rad implies > sin(A/2) = sin(B/2) = sin(C/2) = 1/2 implies > sin(A/2)sin(B/2)sin(C/2) = 1/2*1/2*1/2 = 1/8 thus it is indeed an equilateral triangle if i come with another one i will post it :) hope it will help 


#5
Mar1205, 08:33 AM

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P: 1,421

Try expand the equation
[tex]\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}} = \frac{1}{8}[/tex] to [tex]4\sin{\frac{C}{2}}^{2}  4\sin{\frac{C}{2}}\cos{\frac{A  B}{2}} + 1 = 0[/tex] Then to [tex](2\sin{\frac{C}{2}}  \cos{\frac{A  B}{2}})^{2} + (\sin{\frac{A  B}{2}})^{2} = 0[/tex] Now you have something like [itex]A^{2} + B^{2} = 0[/itex] so [tex]\left\{ \begin{array}{c} A = B \\ \sin{\frac{C}{2}} = \frac{1}{2}\cos{\frac{A  B}{2}} \end{array}\right[/tex] So you will have A = B = C = 60 degrees, which implies the triangle ABC is equilateral. Hope it help. Viet Dao, 


#6
Mar1205, 11:34 AM

P: 6

For AI thank you but this won't do



#7
Mar1205, 11:39 AM

P: 6

For VietDao29 If A^2 + B^2 = 0 then we're stuck because no two +ve numbers addto zero ...right ? Then it should be A^2 =  B^2
How did you expand 1st step How did you get last step plz go in more details 


#8
Mar1205, 12:04 PM

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Both A and B are real.So their square is larger or equal to zero.In order for the sum of the squares to be 0,each if the squares must be 0.
Daniel. 


#9
Mar1205, 02:33 PM

P: 6

Well that's a good point. How did I miss that :) Now for the first step plz how did we expand Sin (A/2) Sin (B/2) Sin (C/2) to next step
ie. How to start .... the rest is ok 


#10
Mar1205, 03:35 PM

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Use this IDENTITY:
[tex] \sin x\sin y\equiv \frac{1}{2}[\cos(xy)\cos(x+y)] [/tex] The result is immediate. Daniel. 


#11
Mar1205, 03:40 PM

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Try starting by eliminating a variable. Since you know that A, B, and C are all in the same triangle, you have:
A+B+C=180 C=180AB See where that gets you. 


#12
Mar1205, 05:20 PM

P: 6

Thank you all ... I can do it now following your steps
The rule supplied by Dextercioby did not look familiar (but it's correct I checked) ..well memory is not what it used to be :) isn't that a bit complicated though ... I thought the answer should be more straight forward .. any way thank you all again 


#13
Mar1205, 05:25 PM

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Daniel. 


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