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Inequality in triangle

by harry654
Tags: inequality, triangle
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harry654
#1
Nov26-10, 03:12 PM
P: 58
Prove that in any triangle ABC with a sharp angle at the peak C apply inequality:(a^2+b^2)cos(α-β)<=2ab
Determine when equality occurs.

I tried to solve this problem... I proved that (a^2+b^2+c^2)^2/3 >= (4S(ABC))^2,
S(ABC) - area
but I dont know prove that (a^2+b^2)cos(α-β)<=2ab :(
thanks for your help
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tiny-tim
#2
Nov26-10, 05:09 PM
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hi harry654! welcome to pf!

(try using the X2 icon just above the Reply box )

by "sharp angle", i assume you mean an acute angle, less than 90?

hint: where in that triangle can you find (or construct) an angle α-β ?
harry654
#3
Nov27-10, 03:16 AM
P: 58
by "sharp angle", i assume you mean an acute angle, less than 90? yes :)
hint: where in that triangle can you find (or construct) an angle α-β ?
oh. I know that apply π-(α+β)=γ. But i cant see in that triangle an angle α-β. I have tried to solve this problem already 3 days, but I always proved other inequality.

tiny-tim
#4
Nov27-10, 03:28 AM
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Inequality in triangle

Quote Quote by harry654 View Post
But i cant see in that triangle an angle α-β.
then make one!

you know that α-β is in the answer, so you know there must be an α-β somewhere
where could you draw an extra line to make an angle α-β ?
harry654
#5
Nov27-10, 03:59 AM
P: 58
I can draw paralell straight line with BC. Then i draw paralell straight line with AC. And then I depict triangle ABC in axial symmetry according AB. And at the peak B(under) I have an angle α-β. Is it OK?
tiny-tim
#6
Nov27-10, 04:08 AM
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sorry, i don't understand

can you supply some extra letters (D, E, ), to make it clearer?
harry654
#7
Nov27-10, 04:19 AM
P: 58
OK.
We have triangle ABC.
I draw paralell line p with BC and paralell line l with AC. p intersects l in the point D.
Then angle DBA is equivalent with α. I depict triangle ABC in axial symmetry according AB. this triangle mark AEB.Then angle EBA is equivalent with β. And angle DBE is α-β
tiny-tim
#8
Nov27-10, 04:34 AM
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oh i see!

yes, but a lot simpler would be to draw just one line

draw CD equal to CA with D on BC then triangle DBC has two sides the same as ABC, and angle DCB is α-β

(btw, "depict triangle ABC in axial symmetry according AB" is weird write "reflect triangle ABC in AB" )
harry654
#9
Nov27-10, 04:54 AM
P: 58
(btw, "depict triangle ABC in axial symmetry according AB" is weird write "reflect triangle ABC in AB" )[/QUOTE]

sorry my english is not enough good I know :(

draw CD equal to CA with D on BC ... When I draw D on BC so B,C,D are on one line... and it isnt triangle DBC so I am confused... sorry :(
tiny-tim
#10
Nov27-10, 04:56 AM
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sorry, i meant with D on AB
harry654
#11
Nov27-10, 05:13 AM
P: 58
oh yes.
Now, Should I compare areas of triangles ABC and DBC, shouldnt I?
tiny-tim
#12
Nov27-10, 05:35 AM
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no!!

(why are you so keen on areas anyway? you'll hardly ever need them, and certainly not here )

forget about triangle ABC now

just use triangle DBC
harry654
#13
Nov27-10, 05:55 AM
P: 58
Ok... I use cosinus law and I have equality. But I think I must something compare because I need prove inequality
tiny-tim
#14
Nov27-10, 06:03 AM
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what equality?

(btw, we say "sine" and "cosine" )
harry654
#15
Nov27-10, 06:14 AM
P: 58
Ok.Maybe dont understand me. When I use cosine law I get a^2+b^2-2abcos(α-β)=h^2
when h is DB. But I need (a^2+b^2)cos(α-β)<=2ab I now see that equality occurs when triangle is isosceles. But I try from a^2+b^2-2abcos(α-β)=h^2 get (a^2+b^2)cos(α-β)<=2ab.
tiny-tim
#16
Nov27-10, 06:47 AM
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Quote Quote by harry654 View Post
When I use cosine law I get a^2+b^2-2abcos(α-β)=h^2
when h is DB.
(try using the X2 icon just above the Reply box )

hint: eliminate ab
harry654
#17
Nov27-10, 07:28 AM
P: 58
hint: eliminate ab

I found that I am stupid...
when I eliminate ab cosine law will not apply or no?
harry654
#18
Nov27-10, 08:43 AM
P: 58
apply (a+b)cos(α-β)>= a+b-h

and then I dont know aaaaa :(


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