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Is x + y > z?

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Is x + y > z?

by sanju09 » Tue Feb 10, 2009 2:11 am
In a triangle with angles of degree measures x, y, and z, which are opposite sides A, B, and C, respectively, is x + y > z?

(1) x = y

(2) C^2 <A^2 + B^2

IMO B and OA :?:
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Re: Is x + y > z?

by billzhao » Wed Feb 11, 2009 2:06 am
sanju09 wrote:In a triangle with angles of degree measures x, y, and z, which are opposite sides A, B, and C, respectively, is x + y > z?

(1) x = y

(2) C^2 <A^2 + B^2

IMO B and OA :?:
From (1), we cannot tell whether x+y>z
eg. x=10, y=10, z=160, which satisfies: x=y, but x+y<z
x=80, y=80, z=20, which satisfies: x=y, but x+y>z

from (2), we have A^2+B^2>C^..............(m), and according to law of cosines: C^2=A^2+B^2-2*a*b*cos(z).................(n), from (m) and (n), we have A^2+B^2>A^2+B^2-2*a*b*cos(z) => 2*a*b*cos(z) >0. Since a>0 and b>0, we have cos(z)>0, which is z<90 degree. Since x+y+z=180 and z<90, x+y>90>z. So (2) is sufficient.

My answner is (B)
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Re: Is x + y > z?

by sanju09 » Wed Feb 11, 2009 4:14 am
billzhao wrote:
sanju09 wrote:In a triangle with angles of degree measures x, y, and z, which are opposite sides A, B, and C, respectively, is x + y > z?

(1) x = y

(2) C^2 <A^2 + B^2

IMO B and OA :?:
From (1), we cannot tell whether x+y>z
eg. x=10, y=10, z=160, which satisfies: x=y, but x+y<z
x=80, y=80, z=20, which satisfies: x=y, but x+y>z

from (2), we have A^2+B^2>C^..............(m), and according to law of cosines: C^2=A^2+B^2-2*a*b*cos(z).................(n), from (m) and (n), we have A^2+B^2>A^2+B^2-2*a*b*cos(z) => 2*a*b*cos(z) >0. Since a>0 and b>0, we have cos(z)>0, which is z<90 degree. Since x+y+z=180 and z<90, x+y>90>z. So (2) is sufficient.

My answner is (B)
Good! But use of Cosine Law could be your personal approach, can you come up with some "Globally Accepted" way to crack it?
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by moorthy76 » Wed Feb 11, 2009 4:31 am
IMO B...

I applied pythagorus theorem...
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by sanju09 » Wed Feb 11, 2009 4:46 am
moorthy76 wrote:IMO B...

I applied pythagorus theorem...
:wink: Howdy! But where and how did you apply that?
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by moorthy76 » Wed Feb 11, 2009 5:04 am
I thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. ...
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by sanju09 » Wed Feb 11, 2009 5:11 am
moorthy76 wrote:I thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. ...
B-) It is still an incomplete elucidation!

You thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. .. which angle? or then what? :!:
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by x2suresh » Wed Feb 11, 2009 7:45 am
sanju09 wrote:
moorthy76 wrote:I thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. ...
B-) It is still an incomplete elucidation!

You thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. .. which angle? or then what? :!:


Here is my theory



IF C^2= A^2+B^2 is true then z=90

Assume A and B lengths are fixed.

If you decrease the angle z.. then opposite side lenght (C) will also decrease.

If you increase the angle z.. then opposite side lenght (C) will also increase.

So C^2< A^2+B^2 --> means z<90.

x+y+z=180 and z<90 --> x+y>90>z

B sufficient.
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