[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 498
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It is completely unclear where this monster for x1 comes from. Also, you have to divide it in such a way that it's not exactly zero.
No, I don't like it.
something like this:
x1 = ((a-b)*(a-c) + (b-a)*(b-c) + (c-a)*(c-b) ) / ( (b-a)*b/c + (c-b)*c/a + (a-c)*a/b)
no time...
I got it like this:
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) )
It is completely unclear where this monster for x1 comes from. Also, you have to divide it in such a way that it's not exactly zero.
No, I don't like it.
Denote the "same" number by x1, and the "other" number by x2.
1.
is reduced to a form:
2.
reduced to the form:
well and
3.
The divisor is -(A-B)^2 in either case. Yes, it is not equal to zero. And now explain the logic, RAVen_. Simple guessing is kind of insubstantial.
2 PapaYozh: x1 can be equal to zero. The solution should be suitable for any numbers.
The divisor is -(A-B)^2 in either case. Yes, it is not equal to zero. And now explain the logic, RAVen_. Simple guessing is kind of insubstantial.
2 PapaYozh: x1 can be equal to zero. The solution should be suitable for any numbers.
If the "same" numbers are zero, then the "other" can be by any.
And now explain the logic, RAVen_.
logic in getting rid of the "extra" numbers:
we have 3 options when a=b : x1= a
--- b=c : x1 = b
--- c=a : x1= c
In the numerator we use additional multipliers to zero out the "unnecessary" choices. The variant we are looking for is multiplied and divided by a non-zero multiplier.
About the guessing, you're wrong: that idea was there from the beginning. But I went the wrong way: one variant - one equation, and then we add up. The result was a constant zero in the denominator... When I realised that I had to put everything into one fraction, it took about five minutes to solve...
In your expression for the denominator
can be a division by zero (by any of the numbers a,b,c). If you stupidly multiply it (together with the numerator, of course) by abc, you get such a denominator:
If a=b=x1, then it would be (x2-x1)*x1*x2*x2 + (x1-x2)*x1*x1*x2 = x1*x2^3 - 2*x1^2*x2^2 + x1^3*x2 = x1*x2*(x2^2-2*x1*x2+x1^2) - it can be zero if at least one of x1, x2 is zero. So there's no easy way to do it.
By the way, here's RAVen_'s solution seems to be correct. But I still want to see the logic of the solution.
P.S. RAVen_, I see. Still don't like it, sorry. You need a clear mathematical logic of the solution from the beginning. Of course, the immediately written out formula in the Olympiad problem is formally a solution. But it's... ...like it fell out of the sky...
I'll try to do it myself.
P.S. RAVen_, I see. Still don't like it, sorry. You need a clear mathematical logic of the solution from the beginning. Of course, immediately written out the formula in the Olympiad problem is formally the solution. But it is so...
What is not to like about the given logic? No more detailed "logic" was used in the solution. Cutting off variants in the formula by zeroing them out (in the absence of condition and switches) is not a new method. That's what it's based on.
But it's so... it's like it fell out of the sky...
So parse the formula in terms of the logic I described... and you will see that what I have said is enough for a fairly down-to-earth solution :)
No offence, please. Your final formula is very similar to the correct one. Score!
But just imagine: you are an 8th grader, and you are asked to explain how you arrived at the solution. And you give this explanation:
логика в избавлении от "лишних" чисел:
we have 3 options when a=b : x1= a
--- b=c : x1 = b
--- c=a : x1= c
In the numerator we zero out the "unnecessary" choices using additional multipliers. The variant we are looking for is multiplied and divided by a non-zero multiplier.
Do you think other eighth graders will understand you? Especially this expression in the numerator:
(a-b)*(b-c)*c + (b-c)*(c-a)*a + (c-a)*(a-b)*b
Where did it come from? So I'm trying to find a solution that consistently explains where this totally non-obvious monster in the numerator came from - without all the "getting rid of extra" and "zeroing out unnecessary choices".
P.S. I'll try to explain the logic I follow 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 it's not coming out like a stone flower.
An eighth-grader is unlikely to understand it. Let at least an 11th grader understand it.