Chess Problem Art 
by Paz Einat
אומנות בעיות השחמט
מאת פז עינת

Annual Meeting 2011

מפגש הפרובלמאים השנתי - 2011


תחרות בפתרון בעיות שחמט

Chess problem solving tourney 

Director: Ofer Comay

Participants had 3 hours for solving 8 problems

Results were as follows:

1st Place: Mordechai Chubnik       40 points (maximum)
2nd Place: Yedael Stepak              39 points
3rd Place: Omer Friedland            38 points
4th Place: Haim Temes                18.3 points
5th Place Dmitrij Baibikov            14 points

Here are the Problems the participant received for solving.
To see the solutions drag the cursor below the "solution"
Joe Bunting
To Alain White 1945

#2                           ( 6 + 5 )
Solution:
1.Qc6 ! (2.Qxc5 #)
1...Kc4 2.Rxb4 #
1...Ke5 2.Qd6 #
1...Ke3 2.Qe4 #
1...Se4 2.Qxe4 #
1...B~  2.Bf2 #
Ofer Comay
2009

#3                           ( 9 + 8 )

Solution:
1.Sf3! (2.Bf2 #)
1...Rd6+ 2.Kf4!
1...Rc6+ 2.Ke4!
1...Ra6+ 2.Kd3!
1...Rxb7+ 2.Ke4!
1...Rb5+ 2.Kd3!
1...Rb4+ 2.Kd2!
Milan Vukcevic
3rd Prize Olympic Tourney, Thesaloniki 1984

#4                         ( 10 + 7 )

Solution:
1.Sg4! (2.Sc3#)
1...Rxg4 2.Kb1! Qb7+ 3.Rb4+
1...Bxg4 2.Ka1! Qf6+ 3.Re4+
Ofer Comay
Mat Plus 2008

H#2   b) wBc8-->a4  ( 5 + 6 )

Solution:
a) 1.Kf5 Rxd7 2.Se7+ Rd5#
b) 1.Ke6 Bxc6 2.d5+ Bd7#
Petko Petkov
2nd Prize Stella Polaris 1969

S#3                     ( 10 + 12 )

Solution:
1.Bf7! 2.e8=Q+ Qxd8 3.Qe5+
1...Qxd8 2.e8=S+ Qxe8 3.Sxe4+
1...Sc3 2.e8=B+ Qxd8 3.Rxd6+
1...Kxf7 2.e8=R+ Kg6 3.Qg5+
Jean Michel Trillon
Themes 64 1981

#5 Circe                  ( 6 + 2 )

In "Circe" when a piece is captured it returns to its birth place. If the birth square is occupied the piece is removed from the board.

Solution:
Set play
1...Ke4 2.Ba3 Kf4 3.Bc1+ Ke4 4.Kd2 Kf4 5.Kde#

Sol:
1.Kd2! Ke4 2.Bf8 Kf4 3.Bh6+ Ke4 4.Bf4! Kxf4 (Bc1) 5.Kd3#
Uri Avner & Yuri Berezhnoy
2nd Prize Phenix 1993

S#3                      ( 10 + 14 )

Solution:
1.Ke6!  2.Sb6+ Bxb6 3.Qd5+
1...Bg2,f1 2.Rc5+ Bxc5 3.Qd5+
1...Bxf5+ 2.Kd6+ Be6 3.Se5+
1...Rxg6+ 2.Kd7+ Re6 3.Rc5+
D. Petrov
1932

win                          ( 4 + 3 )


Solution:
1.Bc1 Rf2+ 2.Kg3 Rf1 3.Bf4+ Rxf4
4.Bd3 ! Rd4/Rb4 5.Sb5+/Sd5+ wins