Monday, March 13, 2017

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UCLA Secures No. 3 Seed in NCAA Tournament

The Bruins (29-4) will face Kent State (22-13) on Friday evening in Sacramento, Calif. 

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LOS ANGELES – The UCLA men's basketball team has earned a No. 3 seed in the 68-team NCAA Tournament field, as announced late Sunday afternoon.

The Bruins (29-4) will face No. 14-seed Kent State (22-13) in a first-round meeting in Sacramento on Friday, March 17. The winner of the matchup between UCLA and Kent State will face either No. 6-seed Cincinnati or No. 11-seeds Kansas State or Wake Forest in a second-round contest. UCLA will play the final game Friday evening at the Golden 1 Center in Sacramento (game time is approximately 7:00 p.m. PT).

UCLA has advanced to the NCAA Tournament for the 48th time in school history and will be making its 10th appearance in the national tournament in the past 13 seasons. This marks UCLA's highest seed in the NCAA Tournament since securing the No. 1 seed in the West Regional in 2008.

UCLA enters this season's NCAA Tournament having won 10 of its last 11 games. The Bruins tied the school record for regular-season victories (28), a feat only once before accomplished by UCLA's 2007-08 squad.

In addition, the Bruins defeated every opponent they faced on their schedule, having avenged December and January losses to Oregon, Arizona and USC with victories over those Pac-12 schools in February.

This year's tournament selection marks the third year in four seasons under head coach Steve Alford in which the Bruins will compete in the NCAA Tournament. UCLA advanced to the Sweet 16 (regional semifinal) under Alford in 2014 and 2015.

Through games played Saturday, March 11, the Bruins continued to lead the nation in scoring offense (90.4 ppg), field goal percentage (51.9%), assists per game (21.5), total assists (709) and assist-turnover ratio (1.86).

Zo, T.J. and Bryce: Reaction to bracket announcement link

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