[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 499
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... such an expression in the numerator:
(a-b)*(b-c)*c + (b-c)*(c-a)*a + (c-a)*(a-b)*b
Where did it come from? ...quite an unobvious monster ...
on the contrary, "monster" is quite obvious. We have three answers, hence three summands. Also recall the elementary maths: x*y/y =x (y<>0). Let's leave the denominator for now and go to the numerator:
as stated we have three options:
1) if a=b : x1=a.
2) If b=c : x1=b.
3) if c=a : x1=c.
That is, the numerator should be a*coeff1+b*coeff2+c*coeff3. For each of the options under consideration the coefficients should take the values
1) coeff1<>0, coeff2=0,coeff3=0
2) coeffeff1=0, coeff2<>0,coeffeff3=0
3) coeffeff1=0, coeff2=0,coeff3<>0
For the first variant, coeffeff2=0 and coeffeff3=0 if the multiplier (a-b) is included
for the second variant, coeffeff1=0 and coeffeff3=0 if the multiplier (b-c) is included
For the third option, coeffeff1=0 and coeffeff3=0 if multiplier (c-a) is included.
Assemble:
coeff1= (b-c)*(c-a)
coeff2= (c-a)*(a-b)
coeff3= (a-b)*(b-c)
Substitute the values and our numerator takes the form
(b-c)*(c-a)*a + (c-a)*(a-b)*b + (a-b)*(b-c)*c
Now it is time for some basic maths: x*y we already have (in any variant after zeroing there is one summand left). Now all that remains is to divide by y=coeff1+coeff2+coeff3.
Just to warn you right away: two of the three summands y equals 0, and y+0=y, so we are not violating anything by adding the coefficients and putting them in the denominator.
One last tug and we see the result:
x1=( (a-b)*(b-c)*c + (b-c)*(c-a)*a + (c-a)*(a-b)*b ) /( (a-b)*(b-c) + (b-c)*(c-a) + (c-a)*(a-b) )
OK, now it's more or less OK!
Strangely, PapaYozh got a completely different answer...
P.S. And here's another variant: x1 = ((a-b)(a-b)c + (b-c)(b-c)a + (a-c)(a-c)b ) / ( (a-b)(a-b) + (b-c)(b-c) + (a-c)(a-c) )
When a=b=x1 the right-hand side is 2*x1*(x1-x2)(x1-x2) / 2*(x1-x2)(x1-x2)
Etc.
There seems to be more than one option coming out.
P.S. I'll try to explain the logic I'm following myself. The number x1 is a common root of the original cubic equation (with roots a,b,c) and the square trinomial, which is its derivative. That's what I'm dancing around, but so far I can't get a flower of stone.
An eighth-grader is unlikely to understand it. Well, at least an 11th grader would.
Maybe that's why it doesn't work, because you're trying to look at my logic, looking for something in it that doesn't exist. And you can't find the three unknowns in two initial expressions... ...even if you can't... :) .
It is strange that PapaYozh got a completely different answer...
Another way of doing things is a different view... And who knows, it may be possible to derive one from the other...
You'd be really surprised, if you'd seen the maze (and formulas) into which I got my initial desire to get three fractions :)
Another way of doing things is a different view... And who knows, it may be possible to derive one from the other...
It does not, my solution does not allow zero in numbers a, b and c, i.e. it is incomplete.
Yours, it does.
(6-9) Write in the vertices of a regular nonagon the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and then write on each diagonal the product of the numbers at its ends. Is it possible to arrange the numbers in the vertices in such a way that all the numbers on the diagonals are different?
Well, if I understand it correctly, it's not difficult. All you have to do is eliminate one of each pair of numbers:
1*6 = 2*3
1*8 = 2*4
2*6 = 3*4
2*9 = 3*6
and number the vertices in a circle like this: 1, 6, 2, 9, 7, 5, 4, 3, 8
The diagonals in a non-pentagon are (9-3)*9/2 = 27. Have you gone through everything, ilunga?
can be counted:
works of 1: 2,9,7,5,4,3
from 6: 54,42,30,24,18,48
from 2: 14,10,8,6,16
from 9: 45, 36, 27, 72
out of 7: 28, 21, 56
from 5: 15, 40
from 4: 32
There don't seem to be any matches.