Chaminade to open with UCLA in Maui Invitational

Updated: August 3, 2006, 10:54 AM ET
Associated Press

LAHAINA, Hawaii -- NCAA runner-up UCLA will play host Chaminade in the opening round of the 2006 EA Sports Maui Invitational.

The matchups for the eight-team tournament to be held Nov. 20-22 were announced Thursday. The winner of the UCLA-Chaminade game will advance to a semifinal against the winner of the Kentucky-DePaul first-round game.

The other bracket has Oklahoma, under first-year coach Jeff Capel, against Memphis, and Purdue playing Georgia Tech.

Memphis, which was ranked No. 4; No. 7 UCLA, which lost to Florida in the NCAA title game; and No. 24 Oklahoma were all ranked in The Associated Press' final poll of 2005-06.

It will be tough for the tournament to match the excitement of last year, when No. 3 Connecticut beat No. 8 Gonzaga 65-63 on a last-second shot by Denham Brown. In the semifinals, Gonzaga beat No. 12 Michigan State 109-106 in triple overtime.

"We've carried the momentum of an unforgettable 2005 event right into this year," tournament chairman Dave Gavitt said. "This tournament promises to set the stage for the 2006-07 college basketball season with play from an assembly of decorated teams, including a pair in UCLA and Kentucky that collectively boasts 18 NCAA titles."

Kentucky, which won the Maui Invitational in 1993, is the only former champion in the 2006 field.

For the second consecutive year, ESPN will televise all 12 games of the tournament on one its networks.

The 2007 Maui Invitational field includes Duke, Arizona State, Illinois, LSU, Marquette, Oklahoma State, Princeton and Chaminade.


Copyright 2006 by The Associated Press