International Solving Contest 2012 – Solutions











8th International Solving Contest (ISC)

29th January, 2012

Solutions

Round 1:

1) Aleksandr Pankratiev, Clube Xadrez Belo Horizonte 50JT 1990-2, 1.-3.Pr.

1.Bg1? (2.S3xc4#) Kxe5 2. Qxf6 # 1. … Sc2!

1. Qxf6? (2.Sd7#) Kc5 2. Sd3 # 1. … Sc6!

1.Sd3! (5) thr./Kxe3 2. Qc5/Bg1 #

2) Don Smedley, Observer 1989-90, 1. Pr.

1.Bg8! dr. 2. f7 (1)

1. … Rc2,Rc1 2. Rxe5+ (0,5) Kxe5 3. Qxe3 #

1. … Sa4 2. Rxe5+ (0,5) Kxe5 3. Qxe3 #

1. … Sxc4 2. Rd4+ (0,5) Kxd4 3. Qxc4 #

1. … Rxc4 2. Sd6+ (0,5) Kd4 3. Qxc4 #

1. … Rd3 2. Rxd3 (0,5) Sxd3 3. Sd6 #

1. … Sd3 2. Rxd3 (0,5) Rxd3 3. Rxe5 #

1. … Sf7 2. Bxf7 (0,5) Bxg4 3. Qxg4 #

1. … Bf1 2. Qxf1 (0,5) – 3. Qh1,Qg2/Qf4 #

3) Norman MacLeod & Hans Peter Rehm, Die Schwalbe 1989, 2. HM

1. Kh3? c4! 2. Kg2 Qc5!

1. Kg2? Sc4! 2. Kh3 Sxe5!

1.Rf7! dr. 2. fxe6#

1. … Rf8 2. Kg2 Sc4 3. Kh3 (2,5) Sxe5 4. Bxh6 #

1. … exf5 2. Kh3 c4 3. Kg2 (2,5) Qc5/d4 4. Rfxf5/Re4 #

4) Oscar J. Carlsson, Mundo de Ajedrez 1975

1. Sb1 (1) (i) Rc8 (ii) 2. Sg8 (1) d4! 3. Bxd4 g1Q! (iii) 4. Bxg1 Rxc2! 5. Kg4!!
(1) (iv) Rg2+ 6. Kf3! (1) Rxg8! 7. hxg8B! Kxg1 8. Bc4 (1) (v) +-

(i) 1. h8Q? b1Q 2. Qg8 Qf1 -+; 1. Sg4? g1Q 2. Bxg1 b1Q 3. Sxb1 Kxg1 4. c3 Bxc3
=

(ii) 1. …d4 2. h8Q g1Q 3. Qa8+ +-

(iii) 3. … Rxc2 4. Bg1! Rc3+ 5. Kg4 Kxg1 6. h8Q +-

(iv) 5. h8Q? Rh2+! 6. Bxh2 =

5. Kg3? Rh2! = ;

5. Kh4? Kxg1 6. Kg5 Rg2+ 7. Kf6 Rh2 =

(v) 8….Kh2 9. Bf1 Kg1 10. Ke2 Kh2 11. Kf2 Kh1 12. Sd2 b1Q 13. Bg2+ Kh2 14.
Sf3 #

5) P. Makarenko & Aleksandr Pankratiev, The Problemist 1990,
1. HM


I) 1.Bd4 Kb4 2. Bxe3+ Kxc3 3. Bd4+ Kb4 4. Re5 Kb5 5. Rge4 c4# (2,5)

II) 1.Rg8 Ka4 2. Rxe3 Kb3 3. Re5 Kxc3 4. Rc8 Kd3 5. Rc6 c4# (2,5)

6) Friedrich Chlubna & Klaus Wenda, feenschach 1983, 1. Pr.

1. Ra4? (2.Ba2) Rd8!

1. Ra3? (2.Ba2) e5!

1. Be5? Bg6!

1.Kf1! dr. 2.d3 dr.,Rxf8 3.Sd4+ (1) Bxd4#

1…. … 2…. dr.,Bg6,e5 3.Se3+ (1) Bxe3#

1…. Rxf8 2.Se3+ Ke6 3.Sf5+ (1) Be3#

1…. Bg6 2.Sd4+ Kxf6 3.Sde6+ (1) Bd4#

1…. e5 2.Se1+ e4 3.Rxa5+ (1) Bc5#

7) Liew Chee Meng, The Problemist 1985-I, 3. Pr.

1. Re4? Rd5! 1. Se5? Rd4!

1.Sb8! (5) dr. 2. Sd7 #

1. … Be5, Rd5 2. R(x)d5 #

1. … Re4, Bd3+ 2. S(x)d3 #

1. … Rd4, Be4 2. Q(x)d4 #

1. … Rg7 2. Rd5 #

1. … Re5 2. Qf8 #

8) Carel Sammelius, Schweizerische Arbeiterzeitung 1980-1, 1. Pr.

1.Bg3! dr. 2. Qa3+ (1) Rd3/Bd3 3. Qe7/Qc1 #

1. … Rd1 2. Qb3+ (1) Rd3/Bd3 3. Qe6/Qxb6 #

1. … Bc4 2. Qc3+ (1) Rd3/Bd3 3. Qe1/Qd2 #

1. … Qh2,Qh7,Qg7,Qf8 2. Rxf3+ (1) Kxf3 3. Qf4 #

1. … Qxf4 2. Bxf4+ (1) Kd3 3. Rd2,Qc4 #

9) Dieter Kutzborski & Stefan Eisert, Sächsische Zeitung
1980, 2. Pr.


1. Kb2! (dr. 2. Bd7+ Kd5 3. Sf6+ Kxc4 4. Bb5 #)

1. … bxa3+! 2. Kb3 Rxh3+ 3. Rf3! Rh5! 4. Sf6 Bh3 5. Rf5 (5) Rxf5/Bxf5 6. Bd7/Bd5
#

1. … d3 2. Bd7+ Kd5 3. Rf4 (5#)

10) Siegfried Hornecker, Original

1. Rh1+ (1) (i) Kg2 2. Ba7! (1) (ii) Kxg3 3. Bxd4 Kg2 4. Rh4!! (1) (iii) e2+

5. Ke1 b2 (iv) 6. Rg4+ Kh3 7. Bxb2 (1) axb2 (v) 8. Rb4 Kg2 9. Rg4+ (1) Kh3 10.
Rb4 =

(i) 1. Se2+? fxe2+ 2. Kxe2 b2! 3. Rh3 b1Q 4. Rg3+ Kh2 5. Rxe3+ Qxb8 -+

(ii) 2. Be5? d3! -+

(iii) 4. Rh8? e2+ 5. Ke1 (5. Kd2 e1Q+ 6. Kxe1 f2+ 7. Bxf2 b2 -+) f2+ 6. Bxf2
b2 7. Kxe2 b1Q 8. Rg8+ Kh3 9. Rh8+ Kg4 10. Rg8+ … Kh7 -+

(iv) 5… f2+ 6. Bxf2 b2 7. Kxe2 b1Q 8. Rg4+ Kh3 9. Rh4+ =

(v) 7. … Kxg4 8. Ba3 =

11) Christer Jonsson, The Problemist 1991

I) 1. Ke4 Bg3 2. Rh3 Bb8 3. Rd3 Rf4 #

II) 1. e6 Be7 2. Sg3 Bc5 3. Se4 f4 #

III) 1. Sf3 Bg5 2. Sd4 f3 3. e6 Bf4 #

12) Bertil Gedda, Stella Polaris 1972, 1. Pr.

1.Qa7!

1…. g3 2.Bf8 g2 3.Se7+ Kd6 4.Seg6+ Kc6 5.Se5+ (2,5) Sxe5 #

1…. gxh3 2.Ra3 bxa3 3.b4 cxb4 4.Kb3 Kb5 5.Sd4+ (2,5) Sxd4#

Problems

 



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