(Photo Credit: Indiana University Athletics)

The No. 5 Hoosiers travel to Ann Arbor, Michigan for the Big Ten Tournament. Indiana closed the season 13-11, 6-5 in B1G play. They will play No. 12 Michigan State in the first round on Thursday, April 25 at 10:00 AM ET.

This first tournament match up is in familiar territory for the Hoosiers. Indiana ended the 2018 season at No. 5 as well in the Big Ten and played No. 12 Michigan State. They won 4-0 before falling to Minnesota in the second round 4-1.

Michigan State has struggled throughout the season and ended up going winless in the Big Ten. They only won six matches on the season with a final record 6-20. The last time these teams met was on March 30 with Indiana winning 7-0. Michigan State has one ranked player, Jack Winkler who is ranked in singles.

Indiana looks to continue to turn around its recent two match slide. The Hoosiers ended the regular season with two losses to No. 1 Ohio State (7-0) and to No. 34 Penn State (4-1). Zac Brodney was the sole victory the Hoosiers scored against the Nittany Lions in No. 3 singles.

Despite recent losses, Indiana is ranked for the second week in a row. Per the ITA Team Rankings, Indiana is now ranked No. 50 as a team from No. 49 the previous week. Antonio Cembellin is ranked No. 86 in singles and Brandon Lam, Carson Haskins are ranked at No. 57 in doubles. Indiana’s best win this season was against No. 32 Texas Tech. They also hold victories against No. 41 Northwestern and No. 46 Iowa.

They can continue to give top conference teams a battle and will play No. 4 Michigan next with a victory on Thursday. The last time these teams met was on March 31 with then No. 21 Michigan taking the victory 5-2 in Bloomington, IN.

Leave a Reply

Your email address will not be published. Required fields are marked *