# Whyatt Medal winner: Geoff Foster

### 26 Jan. 2012 | by Peter Wong

The Australian Chess Federation has announced the winner of the Whyatt Medal, presented once every few years for outstanding achievements in the chess problem field. The well-deserved recipient is Geoff Foster. His compositions – a diverse range of orthodox and fairy problems – have garnered plenty of awards from around the world. One highlight is a two-mover that won the Brian Harley Award, the first time in over thirty years that an Australian has gained this major prize. Geoff is also the expert editor of the ‘Problem Potpourri’ column in Australasian Chess. And recently he took over the reins of The Problemist Supplement, a periodical that accompanies the world’s leading chess problem journal, The Problemist. Well done, Geoff!

Geoff Foster

The Problemist 2003, 2nd Prize

Mate in 2

Here are two of Geoff’s prize-winning problems. The two-mover demonstrates cyclic refutation, a theme that’s very difficult to achieve when it involves the maximum amount of changed play. Tries (all waiting moves): 1.Qa8? 1…exd5 2.Qxd5, 1…d3 2.Qa4, but 1…exf4! 1.Qc8? 1…exf4 2.Qxe6, 1…exd5 2.Qxf5, but 1…d3! 1.Qh4? 1…d3 2.Re3, 1…exf4 2.Qxf4, but 1…exd5! In these try phases, the three thematic pawn moves rotate their functions as the refutation and as defences that induce changed mates. After the flight-giving key, 1.Sb5! (waiting), the three pawn moves yield further new mates, to bring about the maximum nine different mating moves in total: 1…exd5 2.Sxd6, 1…d3 2.Sc3, 1…exf4 2.Qxd4, and 1…Kxd5 2.Qa8.

Geoff Foster

Die Schwalbe 1996, 4th Prize

Series-helpstalemate in 21, 2 solutions

In the series-mover, Black plays 21 moves consecutively to reach a position where White can give stalemate. To incorporate more than one solution in a problem of this length is rare, and the two precise sequences are pleasingly varied. 1. Rh1 2.g1=S 3.Bg2 4.Sf3 5.Rhg1 6.h1=B 7.h2 8.Bh3 9.Rg2 10.Kg1 11.Rgf2 12.B1g2 13.h1=R 14.Kh2 15.Rfg1 16.Bf1 17.Bhg2 18.Kh3 19.Rh2 20.Bh1 21.g2 Rxg5. 1.h1=R 2.Rh2 3.Rgh1 4.g1=S 5.Rg2 6.h2 7.Sh3 8.Rhg1 9.h1=B 10.Rh2 11.Bhg2 12.Rgh1 13.Kg1 14.Rf2 15.Bf1 16.Rhg2 17.Kh2 18.Sg1 19.Kh3 20.Rgh2 21.g2 Kxg5. The two parts see Black promoting to the same triplet of R, B, and S, but in a shifted order. Black’s final configurations also display a change of the pinned piece on f3, which in turn forces a different white stalemating move.