# Math Contest #25 Results and Solution

in #puzzle4 years ago

## Solution

The problem of this contest was to find a and x so that these vectors would be orthogonal:
(a, 5a, 2a/x) and (3a, 4a, x)
being orthogonal means that the scalar product gives 0:
`a*3a + 5a*4a + 2a/x *x = 0`
`3a² + 20a² + 2a = 0`
`(23a + 2)a = 0`

`a = 0` or `23a+2 = 0``a = -2/23`
Is a=0 really a solution?
After all it gives the 0-vector which should have no direction and therefor shouldn't be orthogonal to anything?
It depends on how you define orthogonality.
I didn't define it, so I guess I have to count a=0 as a solution.

One important thing is still missing: the values of x:
x can be anything, because the solution does not depend on it, but it mustn't be 0 because 2a/0 is not defined!
So x != 0.

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### List of participants with their entries:

Namesolution foundcomment
@jeffjagoea = 0You forgot to define x and you didn't find the second solution :(
@lalloa = 0 or a = -2/23, x != 0

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## Winner draw:

Not necessary since exactly 2 successful participant:
Congratulations @jeffjagoe and @lallo , you won 1 SBI each!
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### This contest will continue as a 2-week contest as it's getting harder to find a unique problem and I have more stress lately.

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