Medvedev sent packing by Dimitrov at Indian Wells

Norrie through to last eight after beating American Paul

Daniil Medvedev was knocked out in the fourth round of Indian Wells by Bulgarian Grigor Dimitrov as the world No 2’s near-flawless start turned nightmarish in a 4-6 6-4 6-3 defeat as British No 2 Cameron Norrie beat American Tommy Paul in three sets to reach the quarter-finals.

Medvedev broke Dimitrovs serve to open the match, easing through a near-perfect first set in which he won all but four of his first-serve points and fired off three aces with just five unforced errors.

The Russian was up two breaks in the second set and seemingly on track for an easy victory. But Dimitrov came roaring back, winning five games in a row to wrest control of the game.

Medvedev told reporters he struggled to control the ball, particularly on the first serve. “I don’t remember myself losing three service games, even four service games ever, I guess, on hard courts,” he said.

World No 26 Norrie won 6-4 4-6 6-2 to claim his 44th win on the ATP Tour this year.