Kei Nishikori and Grigor Dimitrov will be going head-to-head for the fourth time in their careers when they battle for the Brisbane International title on Sunday night. Nishikori has won all three of their previous encounters, all on hard courts. After suffering straight-set losses to the Japanese star in 2013 (Shanghai) and 2014 (Miami), Dimitrov snagged a set for the first time last summer in Toronto but ended up falling 6-3, 3-6, 6-2.
“We played a pretty close match last year,” the Bulgarian reflected. “I felt like I had my chance and my momentum…. I’m not playing anything next week, so I’m going to get out there and leave it all on the court.”
Despite dealing with a difficult draw, Dimitrov has not been required to expend too much energy so far in Brisbane. The world No. 17 has surrendered just a single set in victories over Steve Johnson, Nicolas Mahut, Dominic Thiem, and No. 1 seed Milos Raonic. After scraping past Thiem 6-3, 4-6, 6-3 in the quarterfinals, Dimitrov upset Raonic 7-6(7), 6-2 on Saturday.
Both players finished in relatively strong form to end the 2016 campaign. Dimitrov reached the Toronto quarters, the Cincinnati semis, the U.S. Open last 16, the Chengdu semis, the Beijing final, and the Stockholm semis. Nishikori made a run to the U.S. Open semis, finished runner-up in Basel, and advanced through round-robin competition at the World Tour Finals. The world No. 5 earned a first-round bye this week before defeating Jared Donaldson, Jordan Thompson, and Stan Wawrinka. Like his opponent, Nishikori has dropped only one set–his first of the tournament against Donaldson.
Dimitrov is just 4-5 lifetime in ATP finals and is 0-4 in his last four appearances dating back to the fall of 2014. His first-ever title match came at this same event back in 2013, when he lost to Andy Murray 7-6(0), 6-4. Facing another extremely tough opponent, Dimitrov may see an outstanding week come to an end in less than perfect fashion.
Pick: Nishikori in 3
[polldaddy poll=9629473]
Leave a Reply