Let a , b ∈ R such that E is minimum where E = a 2 + b 2 + ( a − 1 2 ) 2 + b 2 + a 2 + ( b − 5 ) 2 Let E min 2 = m + n l where m , n are co-prime integers and l is a square free integer. Then, find the value of m + n + l .
Don't Use Calculus since here it is Poisonous for you.
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I think you meant B(12,0). right???
can you explain how will p make 120 degrees .Actually , I am unable to make the figure.
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It's because in an acute angled triangle the F e r m a t ′ s P o i n t lies on the pint that makes an angle of 1 2 0 o with the lines joining the vertices of the triangle to it. For more info. you can google ,"Fermat's Point"
Could you elaborate your calculations? I seem to miss a factor somewhere
Does that point lies on the centroid of this triangle for which the value would be minimum
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No the centroid is the point that minimises the sum of the square of the distances.
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I ′ a m w a i t i n g f o r o t h e r s t o p o s t a f u l l s o l u t i o n f o r t h i s c o o l p r o b l e m . I c a n g i v e y o u h i n t t h a t a s s u m e a n r i g h t a n g l e d T r i a n g l e A B C i n X − Y p l a n e a n d l e t A ( 0 , 5 ) & B ( 1 2 , 0 ) & C ( 0 , 0 ) . N o w l e t a p o i n t P ( a , b ) T h u s w e h a v e t o c a l c u l a t e m i n i m u m v a l u e o f E = P A + P B + P C B y L o g i c a l l y t h i n k i n g w e c o n c l u d e t h a t ′ P ′ m u s t l i e i n s i d e t h e t r i a n g l e a n d m a k e e q u a l A n g l e s w i t h r e s p e c t i v e s i d e s i . e a n g l e = 1 2 0 d e g r e e N o w u s e c o s i n e l a w i n r e s p e c t i v e 3 t r i a n g l e s . A n d D o S o m e m a t h a m a t i c a l C a l c u l a t i o n a n d c o m p u t e t h e v a l u e o f E m i n 2 = 1 6 9 + 6 0 3 l = 1 6 9 m = 6 0 n = 3 ⟹ l + m + n = 2 3 2 .