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Algebra Level 4

Given that a 2 + b 2 + c 2 + d 2 − a b − b c − c d − d + 2 5 = 0 a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac{2}{5}=0 , find the value of a + b + c + d a+b+c+d .

Source: Mathematical Circles.


The answer is 2.

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Department 8
Oct 11, 2015

Well it took me a lot of time but I knew it is something related to square:-

a 2 + b 2 + c 2 + d 2 − a b − b c − c d − d + 2 5 = 0 a 2 + b 2 4 + 3 b 2 4 + ( 3 4 ) ( 4 c 2 9 ) + 2 c 2 3 + ( 9 d 2 16 ) ( 2 3 ) + 5 d 2 8 − 2 a b 2 − ( 4 b c 3 ) ( 3 4 ) − ( 2 c 3 ) ( 3 d 4 ) ( 2 ) − ( 8 d 5 ) ( 5 8 ) + ( 16 25 ) ( 5 8 ) = 0 ( a 2 + b 2 4 − 2 a b 2 ) + ( 3 b 2 4 + ( 4 c 2 9 ) ( 3 4 ) − ( 4 b c 3 ) ( 3 4 ) ) + ( 2 c 2 3 + ( 9 d 2 16 ) ( 2 3 ) − ( 2 c 3 ) ( 3 d 4 ) ( 2 ) ) + ( 5 d 2 8 + ( 16 25 ) ( 5 8 ) − ( 8 d 5 ) ( 5 5 ) ) = 0 ( a − b 2 ) 2 + ( 3 4 ) ( b − 2 c 3 ) 2 + ( 2 3 ) ( c − 3 d 4 ) 2 + ( 5 8 ) ( d − 4 5 ) 2 = 0 \large{{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }-ab-bc-cd-d+\frac { 2 }{ 5 } =0\\ { a }^{ 2 }+\frac { { b }^{ 2 } }{ 4 } +\frac { 3{ b }^{ 2 } }{ 4 } +\left( \frac { 3 }{ 4 } \right) \left( \frac { { 4c }^{ 2 } }{ 9 } \right) +\frac { 2{ c }^{ 2 } }{ 3 } +\left( \frac { 9{ d }^{ 2 } }{ 16 } \right) \left( \frac { 2 }{ 3 } \right) +\frac { 5{ d }^{ 2 } }{ 8 } -\frac { 2ab }{ 2 } -\left( \frac { 4bc }{ 3 } \right) \left( \frac { 3 }{ 4 } \right) -\left( \frac { 2c }{ 3 } \right) \left( \frac { 3d }{ 4 } \right) \left( 2 \right) -\left( \frac { 8d }{ 5 } \right) \left( \frac { 5 }{ 8 } \right) +\left( \frac { 16 }{ 25 } \right) \left( \frac { 5 }{ 8 } \right) =0\\ \left( { a }^{ 2 }+\frac { { b }^{ 2 } }{ 4 } -\frac { 2ab }{ 2 } \right) +\left( \frac { 3{ b }^{ 2 } }{ 4 } +\left( \frac { 4{ c }^{ 2 } }{ 9 } \right) \left( \frac { 3 }{ 4 } \right) -\left( \frac { 4bc }{ 3 } \right) \left( \frac { 3 }{ 4 } \right) \right) +\left( \frac { 2{ c }^{ 2 } }{ 3 } +\left( \frac { 9{ d }^{ 2 } }{ 16 } \right) \left( \frac { 2 }{ 3 } \right) -\left( \frac { 2c }{ 3 } \right) \left( \frac { 3d }{ 4 } \right) \left( 2 \right) \right) +\left( \frac { 5{ d }^{ 2 } }{ 8 } +\left( \frac { 16 }{ 25 } \right) \left( \frac { 5 }{ 8 } \right) -\left( \frac { 8d }{ 5 } \right) \left( \frac { 5 }{ 5 } \right) \right) =0\\ { \left( a-\frac { b }{ 2 } \right) }^{ 2 }+\left( \frac { 3 }{ 4 } \right) { \left( b-\frac { 2c }{ 3 } \right) }^{ 2 }+\left( \frac { 2 }{ 3 } \right) { \left( c-\frac { 3d }{ 4 } \right) }^{ 2 }+\left( \frac { 5 }{ 8 } \right) { \left( d-\frac { 4 }{ 5 } \right) }^{ 2 }=0}

this gives:-

d = 4 5 , c = 3 5 , b = 2 5 , a = 1 5 \large{d=\frac { 4 }{ 5 } ,c=\frac { 3 }{ 5 } ,b=\frac { 2 }{ 5 } ,a=\frac { 1 }{ 5 } }

4 Helpful 0 Interesting 0 Brilliant 0 Confused

I did it the same way. Just complete the squares, a straightforward if a bit tedious task.

More generally, you can consider the analogous equations for n n variables when the constant term is n 2 n + 2 \frac{n}{2n+2} ... in our case we have n = 4 n=4 . These computations come up in graph theory.

Otto Bretscher - 5 years, 8 months ago

Well i yried bashing on it and it just got solved in a mere 2 min......not a good way to approach problems..uet fascinating enough.... :)

Mohit Gupta - 5 years, 8 months ago

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Explain bashing (I don't know what you mean)

Department 8 - 5 years, 8 months ago

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