Amber G. wrote:
Okay - here is a simple problem (as easy as 1,2,3
) I posed in a group consisting of mainly Google Engineer's Find three positive rational numbers such that their sum is 6 and the product is 6.
One answer is 1,2,3 as 1+2+3 = 1*2*3 = 6
Any other answer?
(Only fractions - rational numbers which are positives - are allowed.
(Let us see if brfites visiting this forum can find the answer(s) before the other group... computers /books etc are allowed .. googling is discouraged but I can't stop it
(Request: Please For this problem (only) _ No invoking great mathematicians
and writing about fantastic theorems ... Just the answer(s) please.. if one answer appears find a different answer - Imagine your audience just know how to add and multiply ordinary fractions! I want to see how many answers we get here in brf ...)
So this problem is:
For a given a:
Working with that last equation:
b=0.5*( (6-a) +/- sqrt( (6-a)^2 - 24/a ) )
For rational b, the discriminant has to be positive:
(6-a)^2 >= 24/a
(6-a)^2 is a parabola with a minimum at a=6; 24/a is a rectangular hyperbola with a discontinuity at a=0. We want all a where the hyperbola is below the parabola.
The hyperbola will be below the parabola for all a<0. There are also three points where the parabola and hyperbola have the same value (since this is a cubic equation). Too lazy to find the exact values of a, but roughly:
Root 1 (call it R1): a is between 0.9 and 1
R2: a is between 3 and 4
R3: a is between 7 and 8
For a<0, the hyperbola is of course below the parabola, but a can't be negative, so this case is out.
For 0<=a<R1, the hyperbola is above the parabola, hence this case is out.
For R1<=a<R2, the hyperbola is below the parabola (which is what we want).
For R2<=a<R3, the hyperbola is above the parabola, hence this case is out.
For R3<=a<Inf, the hyperbola is below the parabola, and a is also positive, but b and c will both turn out negative, hence this case is out.
So any value of a between R1 and R2 will work, except for the condition that a, b, and c all have to be rational. Not sure how to enforce this condition.
For a given a, it is easy to see that b will be one solution to the quadratic, and c will be the other solution - i.e.:
b= 0.5 * ( (6-a) - sqrt( (6-a)^2 - 24/a) )
c= 0.5 * ( (6-a) + sqrt( (6-a)^2 - 24/a) )
A further point - a=1, b=2, c=3 is a solution, but this is technically the same solution as a=2, b=1, c=3, or a=2, b=3, c=1, etc. So a further condition could be enforced, that a<=b<=c. This could modify the range of a (restrict it from the current R1<=a<=R2).