Shevchenko’s Adelaide run ends with loss to Humbert

Alexander Shevchenko was eliminated in the quarterfinals in Adelaide / Photo: Tennis.com.au

Kazakhstan’s second-ranked tennis player Alexander Shevchenko (World No. 104) was eliminated in the quarterfinal against France’s Ugo Humbert (36) at the ATP-250 Adelaide International 2026 tournament with 0-6, 3-6.

Before today, Jan. 15, head-to-head score of Shevchenko with Humbert was 0-2. However, every one of those matches was fiercely contested. For example, at the start of Wimbledon 2024, athletes played five sets: 6-1, 4-6, 7-6 (7-2), 6-7 (3-7), 6-1.

The match in Adelaide began poorly for Shevchenko. Humbert won nine consecutive games. The Kazakhstani struggled on court. After losing the first set 0-6 and falling behind 0-3 in the second, Shevchenko managed to hold his service. However, he then lost his opponent’s service in the following game, making it 1-4.

As a result, Humbert’s lone break in the second set proved sufficient for victory, 6–0, 6–3. The match lasted just 57 minutes. Humbert won 58 points, while Shevchenko claimed 27. The Frenchman’s semifinal opponent is Spain’s Alejandro Davidovich Fokina (15), seeded first.

The Adelaide International, held since 2020, is part of the hard-court tournaments leading up to the Australian Open, the first Grand Slam of the calendar year. The women’s WTA event, with $1.2 million in prize money, is classified as a WTA 500 tournament, while the men’s ATP event, with $770,000, is an ATP 250 tournament.