If 2 s = a + b + c and ( s − a ) 2 + ( s − b ) 2 + ( s − c ) 2 + s 2 = a x + b x + c x for a real x , find x 2 .
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( s − a ) 2 + ( s − b ) 2 + ( s − c ) 2 + s 2 = s 2 − 2 s a + a 2 + s 2 − 2 s b + b 2 + s 2 − 2 s c + c 2 + s 2 4 s 2 + a 2 + b 2 + c 2 − 2 s ( a + b + c ) = ( 2 s ) 2 + a 2 + b 2 + c 2 − 2 s ( 2 s ) = a 2 + b 2 + c 2 = a x + b x + c x x = 2 x 2 = 4
I was however deceived to see 2 s = a + b + c and assume pure geometric solution about semi-perimeter and triangle, because the expression is also close to Heron's formula. Good problem, good solution though
It is clear that the degree of expression is 2 hence x cannot be any other number except 2 .
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X = ( s − a ) 2 + ( s − b ) 2 + ( s − c ) 2 + s 2 = s 2 − 2 a s + a 2 + s 2 − 2 b s + b 2 + s 2 − 2 c s + c 2 + s 2 = 4 s 2 − 2 s ( a + b + c ) + a 2 + b 2 + c 2 = 4 s 2 − 4 s 2 + a 2 + b 2 + c 2 = a 2 + b 2 + c 2
Therefore, x 2 = 4 .