Bulgaria’s Dimitrov wins in Australian Open first round, Pironkova out
Grigor Dimitrov won his opening match at the Australian Open, the first tennis Grand Slam event of the year, in straight sets on January 17, while Bulgaria’s top ranked female player Tsvetana Pironkova was ousted in the first round for the second year running.
Dimitrov, seeded 15th and one week removed from winning his fifth career title in Brisbane, where he defeated three top-10 players to lift the trophy, was handed an easy draw to face Australian wild card Christopher O’Connell, ranked 231st in the world.
O’Connell held his own against Dimitrov in the first set, but never threatened an upset (he had no breakpoints during the match) and was quickly overpowered in the tie-break. Dimitrov upped his game in the next two sets, winning them to record a final score line of 7-6 (7-2), 6-3, 6-3.
He faces another low-ranked opponent next, Hyeon Chung of South Korea, who is currently 105th in the ATP rankings, but faces a potential third-round tie against 18th seed Richard Gasquet. Dimitrov has only one win in six career meetings against the Frenchman, which came in their last match at the 2016 Shanghai Masters.
Should the Bulgarian prevail, he would likely face world number two and defending champion Novak Djokovic, who is bidding for a record seventh Australian Open title, in the fourth round.
Pironkova had a much tougher draw in third-seed Agnieszka Radwanska of Poland. After going down 6-1 in the first set, Pironkova fought back to take the second frame 6-4, but was unable to keep the momentum going and lost the final set 6-1.
(Photo: Brad Touesnard/flickr.com)
