Results after Problem 5

The list of the contest leaders after the fifth problem:

Points for Problem No 5:

Correct solutions (ordered according to the arrival time; best solutions in bold):

1. Szabó Attila (Hungary)

2. Nikita Sopenko (Russia)

3. Ilie Popanu (Moldova)

4. Papimeri Dumitru (Moldova)

5. Dinis Cheian (Moldova)

6. Cristian Zanoci (Moldova)

7. Ng Fei Chong (Malaysia)

8. Lorenzo Comoglio (Italy)

9. Jakub Šafin (Slovak)

10. Lars Dehlwes (Germany)

11. Petar Tadic (Montenegro)

12. Ion Toloaca (Moldova)

13. Teoh Yee Seng (Malaysia)

14. Selver Pepić (Bosnia-Herzegovina)

(The list is ordered according to the arrival time)

Also, there are two incorrect solutions and two solutions, which are based on correct principles, but contain some mistakes.

Overall number of registered participants: 247 from 42 countries.

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Since the number of correct solutions was relatively low, the following hints were given by the end of the third week:

– study the formula sheet (section V-3) to get an idea of how many eigenfrequencies you need to find; if in difficulties counting the number of oscillators, take it equal to the number of degrees of freedom (how many independent loop currents are needed for a superposition to represent arbitrary current distribution of the circuit);

– make use of the strong inequalities at as early stage as possible, to simplify the mathematical task;

– study the section VIII of the formula sheet; particularly useful are pts. 11, 5, 3, 2 (though, depending on your approach, you may not need all of these formulae).

– in order to find the natural frequencies, you can write down the system of differential equations, and based on that system, write down the characteristic equation, the solutions of which are the natural frequencies. However, note that you can avoid writing down the differential equations. Instead, you can make use of the concept of current and voltage resonances. So, for a voltage resonance, there will be a non-zero voltage amplitude U between two nodes A and B, even if there is no current flowing into the node A (and from the node B). Indeed, if the impedance between the nodes A and B is , we can write |U| = |I||Z|, hence a non-zero U is compatible with I=0 if . Therefore, natural frequencies can be found as the solutions of the equation .