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Triangle right-angled iff a2+b2+c2=8R2
This question comes from the BMO round 1 paper. For the full discussion, see here.
By Anonymous on Thursday, January 18, 2001 - 12:11 pm:

How do you do this?

"Given a triangle of sides a,b,c, prove that it is right angled if and only if a2 + b2 + c2 = 8R2 where R is the radius of the circumcircle."

How does the condition a2 + b2 + c2 = 8R2 imply that the triangle must be right-angled?

By Kerwin Hui (Kwkh2) on Thursday, January 18, 2001 - 05:56 pm:

Use the equation a=2R sin A etc. to transform the equation

a2+b2+c2=8R2

to sin2 A+sin2 B+sin2 C=2

If A,B,C are acute, show that

sin2A+sin2B+sin2C>2

similarly for one of A,B,C is obtuse gives

sin2A+sin2B+sin2C<2

So the triangle is right-angled.

By Kerwin Hui (Kwkh2) on Friday, January 19, 2001 - 03:15 pm:

In case you don't know how to proceed, here is one approach:

tan (A+B) = - tan C <0 for A, B, C acute.

since tan A, tan B are both > 0, we have

tan A tan B >1

i.e. 2 sin A cos A sin B cos B > 2 cos2 A cos2 B

adding sin2 A cos2 B + cos2 A sin2 B to both side and conclude that

sin2 (A+B) > cos2 A + cos2 B

from which the result follows. Similarly, just reverse the inequality sign if (WLOG) C is obtuse.

Kerwin

By David Loeffler (P865) on Saturday, January 20, 2001 - 10:48 pm:

Kerwin,

There is a more direct approach to this: using trig identities one can take sin2 A + sin2 B + sin2 C = 2 and convert it into cos A cos B cos C = 0 from which the result is immediate.

David

By Anonymous on Monday, January 22, 2001 - 07:32 pm:

By the way, David

how did you deduce that if

sin2(A)+ sin2(B) + sin2(C) = 2

then it must be that cosAcosBcosC = 0

Thanks

By Kerwin Hui (Kwkh2) on Tuesday, January 23, 2001 - 02:11 pm:

Rearrange to give

sin2 C = cos2 A + cos2 B

and expanding sin2 C = sin2 (A+B) gives

cos A cos B (sin A sin B- cos A cos B)=0

from which cos A cos B cos C = 0 follows.

Kerwin