Hi everyone.....
I have a problem that say:
if ( 18*cos(a)*cos(b)*cos(c)=0) then (a=b=c=60 degree)
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I solved it
18*cos(a)*cos(b)*cos(c)=
1  8*cos A cos B cos C =1  4*cos A * [cos (BC) + cos (B+C)]=[sin (BC)]^2 + [cos (BC)]^2  4*cos (A) cos (B  C) + 4*[cos A]^2 =[sin (BC)]^2 +[cos (BC)  2*cos A]^2 >=0********************************
but they said that when 18*cos A * cos B * cos C=0
that mean a=b=c=60 deg
?????
how can i handle this problem.?
Tanks Grandad
sorry about my mistake ....
I mean
if
18*cos(a)*cos(b)*cos(c)=0
that mean a=b=c=60 degree
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I know that cos (60)=0.5 ==>18*cos 60*cos 60* cos 60=0
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but i mean how can i improve that ??
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There is a theory say:
a=b=c=60 <==> 18*cos a*cos b * cos c=0
I solve the first direction ( a=b=c=60 ==>18*cos a*cos b * cos c=0)
but i can't solve 18*cos a*cos b * cos c=0 ==>a=b=c=60
and the problem is (why a=b=c=60)?
THX again....
but the theory say that
18*cos a*cos b * cos c=0 ==>a=b=c
but a+b+c=180 deg==>a=b=c=60 deg
I solve it:
18*cos(a)*cos(b)*cos(c)= 0
==>1  8*cos A cos B cos C =0==>1  4*cos A * [cos (BC) + cos (B+C)]=0==>[sin (BC)]^2 + [cos (BC)]^2  4*cos (A) cos (B  C) + 4*[cos A]^2 =0==>[sin (BC)]^2 +[cos (BC)  2*cos A]^2 =0
==> [cos (BC)  2*cos A]^2=[sin (BC)]^2
==>I have 2 solution
either (first solution) cos (BC)  2*cos A=sin (BC)
or (second solution) cos (BC)  2*cos A=sin (BC)
let's talk about first solution:
cos (BC)  2*cos A=sin (BC)
==>cos (BC)+sin(bc)=2cos(a)
?????
but i didn't know what i'll do after that
so i can't reach that a=b=c ?
let's talk about second solution:
cos (BC)  2*cos A=sin (BC)
==>cos (BC)sin(bc)=2cos(a)
I i didn't know what i'll do after that too?
Hello blue_blue.
Ah! I see. It's a pity you mention this sooner!I take it, then, that A, B and C are the angles of a triangle.
In that case:
This can be considered as a quadratic in , which has real roots only if
i.e.
But, of course, for all .
which is only possible, for and , if .
In the same way, we may eliminate and prove .
Therefore the only solution is
Grandad
blue_blue wrote: "if ( 18*cos(a)*cos(b)*cos(c)=0) then (a=b=c=60 degree)"
suggests that, 1  8(cos 60)(cos 60)(cos 60) = 1  8(1/2)(1/2)(1/2) = 0.
How about THIS:
i) 1  8(cos 20)(cos 40)(cos 80) = 1  8(1/8) = 0?
ii) 1  8(cos (pi/7) cos (2pi/7) cos (3pi/7)) = 1  8(1/8) = 0?
beautiful isn't it?
Now i know WHY A = B = C = 60 degrees is the solution because of the restriction A + B + C = 180 degrees! THANKS

Found this nice thread on Google, so I will add one short solution using an advanced formula using Power of a Point:
If O is the circumcentre, and H is the orthocentre, then it is well known that .
Then in this case, OH must be 0, and the only possible case for this is an equilateral triangle, so we are done.
Best wishes,
PowerofPoint