I was going through physics competition problems on http://fykos.org/ when I stumbled across this one:

... The Square The Joker

The electric circuit in the image is made from the infinite number of wire-frame-squares. The next square is exactly sqrt(2) smaller then the previous. The resistance of used wire is R per one side of the largest square. What is electrical resistance between most left and most right point?

I have tried to solve it but just can't seem to get it. They've attached their solution to the same page which can be found here although it is in Czech and I can't understand it nor the mathematics behind what they're doing.


closed as off-topic by Kyle Kanos, user36790, Daniel Griscom, Norbert Schuch, ACuriousMind Feb 1 '16 at 16:41

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  • $\begingroup$ Have you used the property of the circuit that it is self-similar? $\endgroup$ – CuriousOne Feb 1 '16 at 1:52
  • $\begingroup$ @CuriousOne I'm not aware of the term self-similar, are you referring to the fact that the circuit repeats itself but with smaller dimensions? $\endgroup$ – Craig Feb 1 '16 at 2:00
  • 1
    $\begingroup$ Yes. That's the major clue to the puzzle. You only have to calculate the value of a single layer of the onion and then you can use that value, together with a scaling constant, to insert into the inside of the same onion. That will give you an equation that allows you to calculate the resistance of the whole. $\endgroup$ – CuriousOne Feb 1 '16 at 2:02
  • $\begingroup$ @CuriousOne I'm having a lot of trouble executing the whole calculating the value of a single layer and then adding the rest of the layers in with a scaling constant since I'm not entirely sure how these resistors are connected. I'm currently trying to solve it by using the fact that since it is a symmetric circuit no current will flow through the center points of this circuit so we can consider this an open circuit and disconnect all of these points. This is where I'm stuck, I understand that in just the top or bottom I have $2R_{eq}$ since it's in parallel but I can't see how to reduce it. $\endgroup$ – Craig Feb 2 '16 at 5:53
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    $\begingroup$ Hmmm... after looking at it, it seems that the thing actually completely unravels because of the symmetry. You don't even have to solve for the self-similarity, the inner part just falls out. Yes, write down the currents and voltages that you know because of the symmetry, eliminate all the connections between points on the same potential and simply solve for the resistance of the remaining circuit, which is merely series and parallel circuits. $\endgroup$ – CuriousOne Feb 2 '16 at 6:46

Google translate is a wonderful thing:

Cut and paste the solution (in parts) into Google translate. It will not be a perfect translation but it should be good enough to provide an explanation as to how to solve the puzzle.

For example here is the first paragraph:

20th edition , the role of VI . 3 ... čtverákčtverec ( 4 points , average 2.73 ; dealt with 15 students ) Fig. First wire mesh the unknown resistance. The circuit of Figure 1, formed by the merger of infinitely many wire squares , each following a √ 2 times smaller. The wire from which the circuit is made ​​, with a length equal to the biggest part Siho square has a resistance R . Determine the resistance of the circuit between the extremes Points left and right. Marek role devised Pechal:


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