Least value of complex expression
For a complex number z , find the smallest possible value of ∣ z − 3 ∣ 2 + ∣ z − 5 + 2 i ∣ 2 + ∣ z − 1 + i ∣ 2 .
Clarification: i = − 1 .
The answer is 10.000.
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5 solutions
It is good another angle of looking at the problem. Approaching the problem from the subject of Physic. I thought of from the point of Math. Different paths, same goal !! That is why I like Brilliant.
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Thanks. For a change, I was trying to take a more "visual" approach, like Ujjwal Rane would do, rather than an algebraic approach. (Ahmed's solution is very nice, though.)
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You and Ujjwal have that vision. Feel very happy about it. I lake that vision.
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@Niranjan Khanderia – Thank you. I appreciate the compliment. :)
NIce solution :) Upvoted
This is a great question.
I did it this way too. Although i got it wrong the first time because i put 2 instead of 3 in denominator while finding the center of mass! Guess that happens... ;)
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Yeah, it does to me too sometimes. I'm just glad that we get three attempts, (except on multiple-choice questions). :)
I did this way also! Putting it on a complex plane rather than using Optimization using Langrage Multiplier.
Is it always that fermats point lies on the C.O.M.
I had almost same approach but instead of triangle i thought that the complex number should lie on the centre of the circle have all these 3 coordinates on its periphery. Can u explain why did u chose triangle instead of a circle?
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By finding the center of the circle formed by the given three points you have found the circumcenter, but in general this does not correspond to the the Fermat point , which is what we are after here. For triangles with all interior angles less than 1 2 0 ∘ the Fermat point coincides with the geometric median , also known as the center of mass, (COM).
@Brian Charlesworth Sir , can you please help me with it? The question was the distance squared , then how did you call that collective distance and even I couldn't understand the reason why you opted for center of mass for minimum collective distance.
Let z = x + i y and we know, ∣ a + i b ∣ = a 2 + b 2 .
So the given expression becomes,
( x − 3 ) 2 + y 2 + ( x − 5 ) 2 + ( y + 2 ) 2 + ( x − 1 ) 2 + ( y + 1 ) 2
= 3 x 2 − 1 8 x + 3 5 + 3 y 2 + 6 y + 5
= 3 ( x − 3 ) 2 + 8 + 3 ( y + 1 ) 2 + 2
For the minimum , ( x − 3 ) 2 and ( y + 1 ) 2 both equal to 0
Therefore least value is ,
0 + 8 + 0 + 2 = 1 0
I did as you. I think this is straight forward approach.
Good!! I did it almost in the same way but instead of getting the last line I used differentiation to get the minima of the previous line.(maybeo was in a bit differentiation mood)
But anyway,I think it's also a good way & clearly much easier
Moment of inertia is minimum about center of mass
Excuse me , I didn't quite get your solution ? Do you mind elaborating on it ?
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Its like placing masses of 1 kg at the three vertices {(3,0),(5,-2),(1,-1)} given in the question , so the expression is that of moment of inertia of the system(about an axis passing through (x,y) and perpendicular to the plane). We know that it is minimum about the centre of mass which is the centroid of the triangle. Now just substitute values
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Brilliant method , Why don't you explain it a bit more a write a more elegant solution ? I'm sure everyone will appreciate it :)
( x − 3 ) 2 + y 2 + ( x − 5 ) 2 + ( y + 2 ) 2 + ( x − 1 ) 2 + ( y + 1 ) 2 . . . ( A )
= x 2 − 6 x + 9 + y 2 + x 2 − 1 0 x + 2 5 + y 2 + 4 y + 4 + x 2 − 2 x + 1 + y 2 + 2 y + 1
= 3 x 2 + 3 y 2 − 1 8 x + 6 y + 4 0
Differentiating partially w.r.t. x & y, respectively we get
6 x − 1 8 = 0 ⇒ x = 3
6 y + 6 = 0 ⇒ y = − 1
You can see that this will be minima since another differentiations of above two function is greater than zero w.r.t. both x & y.
Putting these value back into (A), we get 1 0 .
Classical approach.
Yes, that's exactly how I solved it too :)
We set up a Cartesian Coordinate system and attach masses of 1 k g on each of the three points A ( 3 , 0 ) , B ( 1 , − 1 ) and C ( 5 , − 2 ) .
Note that their centre of mass is simply G ( 3 , − 1 ) , the centroid of triangle A B C . For any point ( x , y ) in this plane, the moment of inertia of the system of masses about the axis through P ( x , y ) and parallel to the z-axis is given by
I ( x , y ) = ( ( x − 3 ) 2 + ( y − 0 ) 2 ) + ( ( x − 5 ) 2 + ( y + 2 ) 2 ) + ( ( x − 1 ) 2 + ( y + 1 ) 2 ) , which is exactly the expression whose minima is to be found out.
But we know by the parallel axes theorem, that I ( x , y ) = I ( 3 , − 1 ) + 3 ∗ ( P G ) 2 ≥ I ( 3 , − 1 ) = 1 0 , with equality iff ( x , y ) = ( 3 , − 1 ) , that is, z = 3 − i .
We need to find the point in the complex plane that is the least collective distance from the three points ( 3 , 0 ) , ( 5 , − 2 ) and ( 1 , − 1 ) . This point P is the center of mass of the triangle formed by these three points (in the complex plane), and thus has coordinates
( 3 3 + 5 + 1 , 3 0 + ( − 2 ) + ( − 1 ) ) = ( 3 , − 1 ) .
The sums of the squares of the distances between P and the three points is
[ ( 3 − 3 ) 2 + ( 0 − ( − 1 ) ) 2 ] + [ ( 5 − 3 ) 2 + ( − 2 − ( − 1 ) ) 2 ] + [ ( 1 − 3 ) 2 + ( − 1 − ( − 1 ) ) 2 ] =
1 + 5 + 4 = 1 0 .