A Pre-RMO question! -4

Algebra Level 1

If 2 s = a + b + c 2s=a+b+c and ( s a ) 2 + ( s b ) 2 + ( s c ) 2 + s 2 = a x + b x + c x (s-a)^2+(s-b)^2+(s-c)^2+s^2=a^x+b^x+c^x for a real x x , find x 2 x^2 .


The answer is 4.

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2 solutions

Chew-Seong Cheong
May 31, 2020

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 \begin{aligned} X & = (s-a)^2 + (s-b)^2 + (s-c)^2 + s^2 \\ & = s^2 -2as + a^2 + s^2 -2bs + b^2 + s^2 - 2cs + c^2 + s^2 \\ & = 4s^2 - 2s(a+b+c) + a^2 + b^2 + c^2 \\ & = 4s^2 - 4s^2 + a^2+b^2+c^2 \\ & = a^2+b^2+c^2 \end{aligned}

Therefore, x 2 = 4 x^2 = \boxed 4 .

Bro he askefx^2

vivek teja Sapavath - 8 months, 1 week ago

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Thanks. I have changed it. I don't why people likes to play game.

Chew-Seong Cheong - 8 months, 1 week ago
Zakir Husain
May 31, 2020

( 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 (s-a)^2+(s-b)^2+(s-c)^2+s^2=\blue{s^2}\green{-2sa}\red{+a^2}\blue{+s^2}\green{-2sb}\red{+b^2}\blue{+s^2}\green{-2sc}\red{+c^2}\blue{+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 4s^2+a^2+b^2+c^2-2s(a+b+c)=\cancel{(2s)^2}+a^2+b^2+c^2\cancel{-2s(2s)}=a^2+b^2+c^2=a^x+b^x+c^x x = 2 x=2 x 2 = 4 \boxed{x^2=4}

I was however deceived to see 2 s = a + b + c 2s = 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

Mahdi Raza - 1 year ago

It is clear that the degree of expression is 2 2 hence x x cannot be any other number except 2 2 .

Vilakshan Gupta - 1 year ago

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