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How many ordered triples of complex numbers ( x , y , z ) satisfy the following system of equations:
x x 2 x 3 + + + y y 2 y 3 + + + z z 2 z 3 = = = − 1 , − 1 3 + 9 5 + 3 2 i , − 4 8 i .
Details and assumptions
The imaginary unit is i , which satisfies i 2 = − 1 .
The answer is 3.
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3 solutions
how did u find the roots
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The perfect power, 1 6 i , suggested that the imaginary part would also be a power of two. To my surprise, the first one I tried worked!
Really nice solution!!!!Good work!!!After finding xyz I knda got lost putting values in the other two rqns...This considering another eqn is a really cool idea~~ :-)
I'd loved to see the solution, but actually, how can you intuitively check the roots? Is it just grabbed from thin air? I need help on this.. Thanks.
Nice; great solution. +1
Nice solutions......
Rewrite the system x + y + z = − 1 ( x + y + z ) 2 − 2 ( x y + y z + z x ) = − 1 3 + 3 2 i ( x + y + z ) ( ( x + y + z ) 2 − 3 ( x y + y z + z x ) ) + 3 x y z = 9 5 − 4 8 i Let x + y + z = a , x y + y z + z x = b , x y z = c , the system gives us a = − 1 a 2 − 2 b = − 1 3 + 3 2 i a ( a 2 − 3 b ) + 3 c = 9 5 − 4 8 i From this we get easily ( a , b , c ) = ( − 1 , 7 − 1 6 i , 2 5 ) , hence by Vieta's formulae we get that x , y , z are the roots of the polynomial p ( w ) = w 3 + w 2 + ( 7 − 1 6 i ) w − 2 5 This factors as p ( w ) = ( w − ( 1 + 2 i ) ) 2 ( w − ( − 3 − 4 i ) ) So we get three triplets ( x , y , z ) ( 1 + 2 i , 1 + 2 i , − 3 − 4 i ) ( 1 + 2 i , − 3 − 4 i , 1 + 2 i ) ( − 3 − 4 i , 1 + 2 i , 1 + 2 i ) Hence the answer is 3
How did you factorise p ? (I wrote a C program to solve cubic)
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I used computer too, I don't know any nice way to do it.
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One can use Rational Root Theorem for Gaussian Integers. It works the same way as for usual integers. Quite a bit of guess and check, but works. If you suspect a double root (which you might, because of the question asked), then taking gcd with the derivative will give it to you very fast.
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@Alexander Borisov – Instead of factorising we could show that the discriminant is 0 . After we observe that the equation doesn't have the same 3 roots, we look if there are 2 same roots. Since we aren't interested in finding the actual solutions, we could you this instead.
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@Aditya Parson – A better method would be to show that f ( x ) shares a root with f ′ ( x ) , but shares no root with f ′ ′ ( x ) .
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@Sreejato Bhattacharya – Yes, obviously. But we are supposed to avoid calculus where we can(at least on Brilliant).
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@Aditya Parson – Problems posed in Brilliant often require calculus...
@Aditya Parson – It is OK to use Calculus to solve Brilliant problems, as long as you can justify it (just like in IMO). And this is, by the way, not really Calculus. The derivative of a polynomial is an algebraic construction, that works over any field (even finite).
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And this is, by the way, not really Calculus. The derivative of a polynomial is an algebraic construction, that works over any field (even finite).
True. In fact, Algebra II classes touch upon polynomial-calculus without even realizing it: saying that a polynomial has a repeated root is another way of saying that its graph is tangent to the line y = 0 . (Or more generally, y = m x + b is tangent to a polynomial equation y = p ( x ) if and only if m x + b − p ( x ) has a repeated root.)
I've tried sharing this with my fellow math educators a number of times, but reactions have been mixed.
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@Peter Byers – I'd be grateful if you could explain this a bit more. I started a discussion thread about it a few months ago but nobody replied: https://brilliant.org/discussions/thread/nature-of-roots-of-a-quartic/
@Sreejato Bhattacharya – I started on that but it was quite a large amount of working (finding the roots of f'(x), which includes taking the square root of a complex number, and then substituting them all into f(x)).
Why would this method be better than the discriminant method? Does the discriminant method actually not work in some cases? (I couldn't find any discussion of it for polynomials with non-real coefficients, although the discriminant in this case does turn out to be 0).
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@Matt McNabb – I preferred not to use the discriminant method, as usually the discriminant of a cubic is pretty ugly. :)
Here's how I did it:
Note that
f
′
(
t
)
=
3
x
2
+
2
x
+
(
7
−
1
6
i
)
f
(
t
)
=
9
3
x
+
1
×
f
′
(
t
)
+
9
4
0
−
9
6
i
t
−
9
2
3
2
−
1
6
i
Any common root of
t
of
f
(
t
)
and
f
′
(
t
)
must satisfy
9
4
0
−
9
6
i
t
−
9
2
3
2
−
1
6
i
=
0
, which gives
t
=
1
+
2
i
. Plugging this value, we check that
f
(
1
+
2
i
)
=
f
′
(
1
+
2
i
)
=
0
, so
f
(
t
)
has atleast two repeated roots.
Consider
f
′
′
(
t
)
=
6
x
+
2
whose only root is
−
3
. Note that
f
(
−
3
)
=
0
, so
f
(
x
)
has exactly two repeated roots.
This is more tedious than the discriminant method, but I think we can apply this method to other cubics more easily than computing the discriminant. :)
@Matt,
I'd be grateful if you could explain this a bit more. I started a discussion thread about it a few months ago but nobody replied: https://brilliant.org/discussions/thread/nature-of-roots-of-a-quartic/
Alright, I just read that thread and the problem in question ("Matt's tangent"). I have to agree with you that the top-rated solution isn't well explained (it is nice in some ways, but personally I voted for Nhat Le's solution, which received the second largest number of votes).
But I think the deeper issue here is that tangents are traditionally taught in a highly disjointed manner: in Geometry, students are told that a tangent is a line that intersects a circle at exactly one point; in Algebra II they learn about repeated roots but make no connection with tangents; then in Calculus they see tangents defined by a formula , i.e. y = f ′ ( a ) ( x − a ) + f ( a ) . The larger concept of "tangent" is still recognizable in those cut-up pieces, but only just barely.
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in Geometry, students are told that a tangent is a line that intersects a circle at exactly one point
Incidentally, tangents to parabolas can be defined in the same way (excluding vertical lines).
The ordered triples of complex numbers ( x , y , z ) which satisfy the following system of equations are:
x = − 3 − 4 i , y = 1 + 2 i , z = 1 + 2 i
x = 1 + 2 i , y = − 3 − 4 i , z = 1 + 2 i
x = − 3 − 4 i , y = 1 + 2 i , z = − 3 − 4 i
Thus, the answer is 3
We have
x y + y z + z x = 2 ( x + y + z ) 2 − ( x 2 + y 2 + z 2 ) = 7 − 1 6 i
Also, since
x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( ( x 2 + y 2 + z 2 ) − ( x y + y z + z x ) ) = ( − 1 ) ( ( − 1 3 + 3 2 i ) − ( 7 − 1 6 i ) ) = 2 0 − 4 8 i = 9 5 − 4 8 i − 3 x y z
we have x y z = 2 5 . If we let x , y , z be roots of a cubic polynomial f , we find that
f ( k ) = k 3 + k 2 + ( 7 − 1 6 i ) k − 2 5
from Vieta's formulas. Hence, the only values of ( x , y , z ) are the permutations of the roots of this polynomial. We can quickly find the roots to be ( 1 + 2 i , 1 + 2 i , − 3 − 4 i ) , yielding 3 ! / 2 ! = 3 distinct triples.