Recent content by quee

Thank you! But why are 2, 3, 6, 8, 11 included? Isn't it possible to just fold the symmetrical scheme and rewrite it with resistors 1, 4, 5, 7, 9, 10, 12?

And then use Kirchhoff's circuit laws?

I thought incorrectly that 9, 11 and 12 were out of the field made by E and F so the current wouldn't go that line. I thought about that. Thank you. But it still remains unclear to me how to get what will be the maximum value. Using symmetry?

First, I got rid of amperemeters with 0 values. These are 9. 11 and 12. Amperemeter 4 will show the maximum value of electric current as it is placed directly between E and F. But how to know its value? Will it be 18 mA? I doubt because 18 mA is not said to be the maximum value. All other...

It is okay. Thank you for your reply. If only $$AD:DB= m:n,$$ it would be completely obvious that $$\frac{DE}{AC} = \frac{n}{n+m}$$ But this is, unfortunately, a much harder problem. Or it may be an actual lack of some information in the problem.

Thank you. Actually, I considered that there may be not enough information in the problem. Now I am convinced.

Hello. There is a problem: "Through D on a side AB of the triangle ABC drawn a line parallel to AC intersecting BC in E. D is such that CD:DB = m:n. Find DE:AC" So it is easy to find out that DB:AB equals to DE:AC as DE and AC are parallel. Since DB = n, there is only one need to express AB...