a , b , c and d are non-zero complex numbers that satisfy
a + b + c + d = a 3 + b 3 + c 3 + d 3 = 0
The sum of all possible values of ( a + b ) ( a + c ) ( a + d ) can be expressed as p + q i where p and q are real numbers and i is the imaginary unit that satisfies i 2 = − 1 . What is p + q ?
This problem is inspired by a problem that appeared in ITT - 1 9 9 4 .
This problem is from the set "Olympiads and Contests Around the World -1". You can see the rest of the problems here .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
But good hypothesis and proof nice
Another way to do it is show that the polynomial P ( x ) = ( x − a ) ( x − b ) ( x − c ) ( x − d ) is a polynomial in x 2 by using vieta's / newton's sums
we can also prove that one of (a+b),(a+c) and(a+d) is zero,by putting a=1,b=(omega),c=(omega)^2 an d=(omega)^3, where omega is cube root of unity,as they are found to satisfy the equation a + b + c + d =0
Log in to reply
Your claim, as Akshay has said, is not true. Even if it were true, that is not a 'proof'. Just because a specific set ( a , b , c , d ) satisfies a certain property doesn't mean that all other sets will do that too. Notice that the problem asks for the sum of all possible values of the product. So, 0 may not be the only answer. You have to prove beyond doubt that no matter what a , b , c and d are, the product will always equal zero.
Log in to reply
@Mursalin Habib yes really... happy following you... plzz follow me too .. and the claim he stated is.. not right as i said...
@Pranav Kilambi how actually as per ur assumptions a +b+c+d = 1+ omega + omegâ^2+ omegâ^3 But omegâ^2+ omega +1=0 and omrgâ^3=1 thus eqn becomes... =0+1 =1 which actually not equal to zero...
I will be using a lot of sigma summations in this solution, just a heads up.
Interestingly, ∑ c y c a = ∑ c y c a 3 = 0 leads to a lot of other zeros.
It turns out if you expand the desired product you get: a 3 + a 2 ( b + c + d ) + ∑ c y c a b c = ∑ c y c a b c , which is a symmetric sum.
Now consider two constructed products:
( ∑ c y c a ) ( ∑ c y c a 2 ) = ∑ c y c a 3 + ∑ c y c a 2 b
⇒ ∑ c y c a 2 b = 0
( ∑ c y c a ) 3 = ∑ c y c a 3 + 3 ∑ c y c a 2 b + 6 ∑ c y c a b c
⇒ ∑ c y c a b c = − 2 ∑ c y c a 2 b = 0 , and that's our answer.
According to me its bit obivous that two of things a b c d must be equal to 1 and other two equals to -1 thus... let us say that a =b=1 and c=d=-1 to get the equality that a+b+c+d=â^3+b^3+c^3+d^3=0 which satisfies my asumption so directly puting values.. (a+b)(a+c)(a+d)= (1+1)(1+-1)(1+-1) =0 thus p+qi=0+0i thus the answer is 0 ... Thats it
Problem Loading...
Note Loading...
Set Loading...
We claim that at least one of ( a + b ) , ( a + c ) and ( a + d ) is equal to zero. So, ( a + b ) ( a + c ) ( a + d ) is always equal to zero.
Now we're going to prove this.
Notice that a 3 + b 3 = − ( c 3 + d 3 )
⇒ ( a + b ) ( a 2 − a b + b 2 ) = − ( c + d ) ( c 2 − c d + d 2 )
As − ( c + d ) = a + b , this equation can be rewritten as:
( a + b ) ( a 2 − a b + b 2 ) = ( a + b ) ( c 2 − c d + d 2 )
Now if a + b = 0 , we're done. If it's not equal to zero, we can divide both sides of the equation by ( a + b ) to get:
a 2 − a b + b 2 = c 2 − c d + d 2 ⋯ ( 1 )
⇒ ( a + b ) 2 − 3 a b = ( c + d ) 2 − 3 c d .
Since a + b = − ( c + d ) , ( a + b ) 2 = ( c + d ) 2 and that means a b = c d .
Go back to ( 1 ) and re-write it as:
a 2 + b 2 = c 2 + d 2
⇒ ( a 2 − c 2 ) + ( b 2 − d 2 ) = 0
⇒ ( a + c ) ( a − c ) + ( b + d ) ( b − d ) = 0
Substitute b + d with − ( a + c ) :
( a + c ) ( a − c ) − ( a + c ) ( b − d ) = 0
⇒ ( a + c ) ( a − c − b + d ) = 0
Since − c − b = a + d , we can rewrite the whole thing as:
2 ( a + c ) ( a + d ) = 0
This means at least one of ( a + c ) and ( a + d ) is equal to zero.
So, ( a + b ) ( a + c ) ( a + d ) = 0 and that is the only possible value of that.
That means the sum of all possible values of the given expression is equal to 0 + 0 i .
And p + q = 0 .
I honestly have no idea what ITT is. I was reading a seminar lecture and came across a problem similar to this.