Literally Too Many Variables!
a 2 p 2 + b 2 q 2 + c 2 r 2
If a p + b q + c r = 1 and p a + q b + r c = 0 , then compute the expression above.
The answer is 1.
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2 solutions
Same solution.
a 2 p 2 + b 2 q 2 + c 2 r 2 = ( a p + b q + c r ) 2 − 2 ( a b p q + a c p r + b c q r ) = ( a p + b q + c r ) 2 − 2 p q r ( a b r 1 + a c q 1 + b c p 1 ) = ( a p + b q + c r ) 2 − 2 a b c p q r ( r c + q b + p a ) = ( 1 ) 2 − 2 a b c p q r ( 0 ) = 1
good approach...+1
Did the same way ..
Relevant wiki: Algebraic Identities
Writing x = a p , y = b q a n d z = c r , we are given that x + y + z = 1 and x 1 + y 1 + z 1 = 0 . The second equation implies y z + z x + x y = 0 . Now, x 2 + y 2 + z 2 = ( x + y + z ) 2 − 2 ( y z + z x + x y ) = 1 2 − 2 × 0 = 1 .
i.e. a 2 p 2 + b 2 q 2 + c 2 r 2 = 1 .