Bruins Clinched NCAA Berth, Thanks to Pac-10 Tiebreakers

UCLA has earned the Pacific 10 Conference’s automatic bid to the NCAA basketball tournament, David Price, associate Pac-10 commissioner said Tuesday, according to the conference’s tiebreaking formula.

The Bruins already have clinched at least a tie for the conference title. They will win it outright unless they lose their final two games, Thursday night against Washington and Saturday against Washington State, and Arizona beats Stanford and Cal this week.

Even if UCLA and Arizona tie for the championship, the Bruins are in because:

--Tiebreaker No. 1 is head-to-head competition. UCLA and Arizona split, 1-1.

--Tiebreaker No. 2 is head-to-head competition against the third-place team. Either Cal or Stanford will finish third and the one that doesn’t will finish fourth, according to Price.

If Cal is third, UCLA is 2-0 in head-to-head play, Arizona 1-1, thus UCLA gets the NCAA bid. If Stanford finishes third, UCLA and Arizona are each 1-1 and they go to . . .

--Tiebreaker No. 3, which is head-to-head competition against the team finishing fourth. That would be Cal. Again, UCLA is 2-0 against Cal, Arizona 1-1, and the Bruins are the winners.