Poland's Jerzy Janowicz, one of the most improved players on the ATP Tour in 2012, has been confirmed for the Heineken Open in Auckland in January.
Janowicz has climbed almost 200 places to No 26 in the world rankings during the past 12 months, highlighted by a fairytale run to the final of last month's Paris Masters, where he beat five top 20-ranked players including US Open champion Andy Murray before falling at the last hurdle to Heineken Open champion David Ferrer.
"Janowicz has had one of those seasons that most players only dream about,'' tournament director Karl Budge said.
''Players outside the top 50 simply don't make the finals of masters level events. You only need to look at how long it took Ferrer to win one to see how treasured they are.
''Maybe the Heineken Open will be Janowicz's chance for a rematch with our defending champion."
Janowicz made his Grand Slam debut as a qualifier at Wimbledon this year and went through to the third round.
But Paris was where he made history. With a ranking of 69, Janowicz had to firstly qualify for the Masters tournament.
Then he went on his giant-killing run - beating three top 20 players and two top 10 players (Murray and Janko Tipsarevic) to reach his first tour title.
If Janowicz can continue the form he showed in Paris and claim the Heineken Open title, he will be the first Pole to win an ATP title since 1982.
World No 5 Ferrer is the top seed for the Heineken Open and he will be attempting to win his third consecutive title.
- © Fairfax NZ News
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